INTEGRATION
BY
ADRIAAN CORNELIS ZAANEN
PROFESSOR OF MATHEMATICS
AT THE UNIVERSITY OF LEIDEN, NETHERLANDS
COMPLETELY REVISED EDITION OF
AN INTRODUCTION TO THE THEORY OF INTEGRATION
c
1 ,--
--
-
1967
NORTH-HOLLAND PUBLISHING COMPANY - Al\ISTERDAM

@ NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM - 1967
A II rights reserved. No part of this book may be reproduced in any form by print,
Photoprint, microfilm or any other means without written permission
from the publisher
First edition 1958
Second printing 1961
Third printing 1 965
Second edition 1967
Library of Congress CatalDg Card Number 67-21975
PUBLISHERS:
NORTH-HOLLAND PUBIJISHING CO. - AMSTERDAM
SOLE DISTRIBUTORS FOR THE U.S.A. AND CANADA:
INTER SCIENCE PUBLISHERS, a division of
JOHN WILEY & SONS, INC. - NEW YORK
PRINTED IN THE NETHERLANDS

CONTENTS
PREFACE . . . . . . . . . . . . . . . . . . . . . . .. IX
CHAPTER 1. POINT SETS, ZORN'S LEMMA, AND METRIC SPACES 1
1. Point Sets. . .. .............. 1
2. Partial Ordering, the Axiom of Choice and Zorn's
Lemma. . . . . . . . . . . . . . . .
3. Metric Space, Baire's Category Theorem. .
CHAPTER 2. MEASURE. . . . . . . . . . . . . . . . . . .
4. Semi-rings and a-rings of Point Sets . . .
5. Measure on a Semi-ring. . .. ....
6. Exterior Measure. . . . . . . . .. .....
7. Measurable Sets . . . . . . . . . . . . . . . . .
8. Null Sets, and Sequences of Measurable Sets. . . . .
9. Approximation Theorems, and the Sum Measure of
two lVleasures . . . . . . . . . . . . . . . . . .
10. Examples, in particular Lebesgue Measure and Stiel-
tjes-Lebesgue Measure . . . .. .......
11. Metric Space of Measurable Sets . . . .. ...
CHAPTER 3. DANIELL INTEGRAL . . . . . . . . . . . .
12. Introduction of the Daniell Integral . . . . . .
13. The Daniell Integral for Non-negative Functions.
14. The Daniell Integral for Real Functions. . . . . . .
15. Null Functions and Null Sets . . . . . . .
16. The Metric Space of all Summable Functions . . . .
7
14
24
24
28
35
38
46
50
58
67
73
73
79
95
102
107

VI
CONTENTS
CHAPTER 4. STIELTJES-LEBESGUE INTEGRAL. . . . · . · · ·
17. The Induced Measure in X, and the Representation
Theorem . . . . . . . . . . . . . . . .
18. Further Properties of the Integral . . . . . . . . -.
19. The Integral for Complex Functions . . . . . . . .
20. The Metric Space of all Complex Summable Functions
21. Riemann Integral and Lebesgue Integral, and the
Riesz Representation Theorem. . .. .....
22. The Exterior Measure of Ordinate Sets . . . . .
CHAPTER 5. FUBINI'S THEOREM . . . . . .
23. Fubini's Theorem . . . . . . . . .
24. Partial Integration. . . . . . . . . .
25. The Gamma Function and Fractional Integration
CHAPTER 6. BANACH SPACE AND HILBERT SPACE; Lp SPACES. .
26. N ormed Linear Spaces and Banach Spaces. . . . . .
27. The Hahn-Banach Extension Theorem . . . . . . .
28. The Problem of a Measure, defined for all Bounded
Subsets of the Real Line . . . .. ......
29 . Hilbert Space . . . . . . . . . . . . . . . . . .
30. Banach Function Spaces; Lp Spaces ......
31. Bochner Integral. . . . . . . . . . . . .
CHAPTER 7. THE RADON-NIKODYM THEOREM
32. The Radon-Nikodym Theorem.
33. The Radon-Nikodym Theorem for
Case . . . . . . . . . . . . . .
34. Partial Order in the Measurable Functions, and the
Finite Subset Property . . . . . . . . . .
35. Localizable Measure . . . . . .. ......
111
111
125
134
136
139
150
154
154
171
175
184
184
190
194
199
205
217
. . . . . . 230
. . . . . . 230
the Non-a-finite
. . . . . . 248
CHAPTER 8. DIFFERENTIATION. . . . . . . . . . . . . . .
36. Differentiation in Abstract Sets . . . . . . . . . .
37. Differentiation in Finite-dimensional Number Space .
255
262
267
267
276

CONTENTS
VII
CHAPTER 9. NEW VARIABLES IN A LEBESGUE INTEGRAL.
38. A Simple Example in R 1 . . . . . . . . .
39. Linear Transformations in Rk . . . . . .
40. Change of Variables in a Lebesgue Integral .
284
284
287
290
CHAPTER 10. MEASURES AND FUNCTIONS OF THE REAL LINE. . 301
41. Measures and Functions on the Real Line . ... 301
42. Functions of Finite Variation . . . . . . . . . 308
CHAPTER 11. SIGNED MEASURES AND COMPLEX MEASURES. 318
43. Signed Measures. . . . . . . . . . . . . . . 318
44. Complex Measures . . . . . . . . . . . . . . 326
45. Absolutely Continuous Signed and Complex Measures. 328
CHAPTER 12. CONJUGATE SPACES AND WEAK SEQUENTIAL
CONVERGENCE. . . . . . . . . . . . . . . . . . . . . 342
46. Borel Sets in Euclidean Space . . . . . . . . . 342
47. Function Spaces. . . . . . . . . . . . . . . 344
48. Decomposition of a Real Bounded Linear Functional 346
49. The Conjugate Space of C K . . . . . . . . . . . . 350
50. The Conjugate Space of Lp . . . .. ..... 355
51. The Banach-Steinhaus Theorem . . . . . . 369
52. Weak Sequential Convergence in L p . ... 371
53. The Second Conjugate Space and Reflexivity 378
CHAPTER 13. FOURIER TRANSFORMATION . . . . . . . . . . 387
54. Unitary Transformations and Projections in Hilbert
Space. . . . . . . . . . . . . . . . . . . . . . 387
55. Unitary Transformations in a Hilbert Function Space 394
56. The Fourier Transformation. . . . . . . . . . . . 397
CHAPTER 14. ERGODIC THEORY . . . . . . . . . . . .
57. Measurable Transformations. . . .
58. Measure Preserving Transformations .
59. Poincare's Recurrence Theorem . . . . .
60. Mean Ergodic Theorem. . . . . . . . . . . . . .
408
408
409
412
413

VIII
CONTENTS
61. Individual Ergodic Theorem.
62. Ergodic Transformations · · · ·
CHAPTER 15. N ORl\IED KOTHE SPACES. . . . . . . . . . . .
63. Function Seminorms and Function Norms; N ormed
Kothe Spaces and Banach Function Spaces . . . . .
64. Norm Completeness . . . . . . . . . . ·
65. The Fatou Property . . . . . . . .. ....
66. The Lorentz Seminorm . . . . . . · .
67. Saturated Function Seminorms . . . . . . . . . .
68. Associate Function Seminorms.. ...
69. The Associate Space of a Normed Kothe Space. . . .
70. Decomposition of a Bounded Linear Functional into
an Integral and a Singular Functional . . . . . . .
71. The Second Associate Function Norm. . . . . . . .
72. The Subspace of all Functions of Absolutely Continuous
Norm. . . . .. ... . . .
73. Reflexivity . . . . . . . . . . . . .
SOLUTIONS. . . . . . . . . . . . . . . . . . . . . . . .
BIBLIOGRAPHICAL REFERENCES. . . . .
INDEX . . . . . . . . . . .
416
431
441
441
443
446
450
453
457
459
464
469
475
483
486
589
597

PREFACE
The present book is a revised and extended edition of my earlier
book "An Introduction to the Theory of Integration", which appeared
first in 1958. Although anyone who wishes to compare the two editions
will find major or minor changes everywhere, one thing has remained
the same, and that is the attempt to produce an advanced textbook
on integration theory, which makes the student familiar not only with
the measure theoretic 'approach and the linear functional approach to
the theory, but also with the fact that the integral of a non-negative
function has something to do with the measure (in a certain specified
sense) of the ordinate set of the function. In fact, one of the things I
want to make clear is that in the linear functional approach, where
the introduction of the notion of "measure" in the underlying point
set is postponed as long as possible, measure theory in disguise is used
from the very beginning. Some further remarks on this point are in
order here. Within the framework of each of the two or three main
approaches there are many smaller variations possible in what will
finally be the definition of an integral, so that, in the end, there are
almost as many definitions as there are textbooks. It may, therefore,
be asked if different mathematicians, speaking about integrals, really
men the same thing. The problem arises already in the simplest
possible case, the Lebesgue integral for functions on the real line. If
we choose the measure theoretic approach, we may either first define
Lebesgue measure for subsets of the real line and then extend the
corresponding step function integral by one of the usual methods, or
we may immediately define the step function integral only for finite
linear combinations of characteristic functions of half open intervals
and then extend the integral. There is another possibility; in the
linear functional approach we can extend the classical Cauchy integral

x
PREFACE
/
for continuous functions possessing a compact carrier. Not only does the
question arise whether the measure induced in the real line by the
extended integral itself is again Lebesgue measure, but it may also
be asked whether the extended integral is the same in all cases. This
question makes sense even in the case of the first extension procedure
referred to above. As long as we do not possess satisfactory answers
to these questions, it is also impossible, for example, to understand
fully what the significance of the Radon-Nikodym theorem is (cf.
Exercises 32.1-32.4 on a converse of the Radon-Nikodym theorem).
For these reasons mainly I believe that the Theorems 7, 8, 9 in sec. 17,
where such satisfactory answers are indeed supplied, are among the
most important ones in the book.
As in the former edition, the book consists of a main part (Ch. 1-8)
and a number of additions and applications (Ch. 9-15). The Stone
version of the Daniell integral is introduced in Ch. 3, but from Ch. 4 on
it is assumed (with only one exception in sec. 22) that the integrals are
of the Stieltjes-Lebesgue-Radon type, i.e., if X is the abstract point
set over which the integration is performed, the integral may be
thought of as being generated by the step functions with respect to
the measure induced in X by the integral itself.
It has not been my aim to obtain the greatest possible generality in
each separate theorem; the attempt has been made, however, to
adhere to a small and fixed set of initial hypotheses (supplemented by
a a-finiteness condition where this seemed required), and to cover all
vital points of the theory in a manner sufficiently elementary for a
first reading and sufficiently general for later purposes (such as, for
instance, being able to comprehend the contents of one of the steadily
growing number of advanced textbooks on probability theory or
harmonic analysis). It should be observed, at this point, that the
exercises in almost all sections form an integral part of the whole.
The number of exercises is considerably larger than before. There are
448 exercises now. And even though the student cannot be expected
to pay attention to everyone of them, he should at least make a serious
effort to work through a good number if he wishes to master the subject.
In each section the exercises are divided into small sets with subtitles
indicating the subjects involved; I believe that this now more widely
accepted custom facilitates a quick survey. All solutions are put to-

PREFACE
XI
gether at the end of the book. Some of the exercises may be of interest
also for the expert; I shall list a few which are probably not so well-
known, and for which either the problem itself or its solution has an
elegant, curious or unexpected aspect. They are Ex. 9.6 (sufficient
conditions for #e == #Ie); Ex. 17.5-17.7 (a necessary and sufficient
condition for a Daniell integral to be. a Stieltjes-Lebesgue integral);
Ex. 17.12 (Loomis' summability condition for an abstract Riemann
integral); Ex. 21.23 (Luxemburg's observation that every non-
negative linear functional on the space of all real continuous functions
on Rk is an integral); Ex. 30.12 (an inclusion theorem about Banach
function spaces); Ex. 32.1-32.4 (converse of the Radon-Nikodym
theorem); Ex. 32.18 (an extension of Kakutani's theorem on product
measures); Ex. 36.3 (a curious counterexample on differentiability);
Ex. 37.7-37.9 (S. Kurepa's theorem on the difference set of two sets);
Ex. 42.14 (the theorem that if y == g(x) is continuous and strictly
increasing with inverse x == G(y), then g' (x) == 0 holds for almost
every x if and only if G'(y) == 0 holds for almost every y); Ex. 56.11
(J. W. Green's theorem about convex sets in R 2 ) and Ex. '56.14 (the
observation that a certain natural extension of Green's theorem to R3
does not hold).
The following topics, which in this edition are either new or receiving
a more extensive treatment than before, are sufficiently important
to be listed below.
(i) The motivation behind Caratheodory's measurability condition
in sec. 7 is improved; it is shown also that if the exterior measure #*
is derived in the usual way from a measure # on a semi-ring of subsets
of a point set X, then the a-field of all #-measurable sets is the largest
ring of subsets of X on which #* behaves as a measure.
Attention is paid to contracted measures and finitely additive
measures and to the corresponding integrals; the notion of a measure
possessing atoms is introduced.
(ii) In sec. 21 and sec. 23 are included lists of examples in Lebesgue
integration; several classical integrals (for functions of one or two
variables) are explicitly computed. The discussion of the properties of
the Gamma function in sec. 25 is considerably extended.
(iii) More attention is paid to product measures and product integrals
for an infinite number of factors, and also to Bochner integrals.

XII
PREFACE
(iv) The discussion of the Radon-Nikodym theorem is more com-
plete. The equivalence of the integral version and the measure version
of the theorem is proved; the case of a non-a-finite measure is considered,
and it is proved that the theorem holds in exactly the same form as in the
a-finite case if and only if the measure is localizable in the sense of Segal.
(v) There are new sections on functions of finite variation and on
reflexive spaces respectively.
(vi) The individual ergodic theorem is proved not only for measure
preserving transformations, but more generally for positive strong
contractions in the space L I .
(vii) The last chapter of the book, on normed Kothe spaces, is new.
Normed Kothe spaces are normed linear spaces, the elements of which
are measurable functions. The Lp spaces are very special examples.
Several characteristic features of the theory of such spaces are dis-
cussed. Many of the results on integration and normed linear spaces,
which were proved in the earlier chapters, are applied here, together
with various properties derived from the order structure of the set
of real measurable functions, in building up the theory.
Not included are:
(a) The theory of measure and integration on Boolean algebras
("Somentheorie" in German); the "elements" on which the measures
and functions are defined will be, therefore, subsets of one or several
fixed point sets.
(b) The theory of directed limits; convergence will always be
sequential convergence.
(c) Many details regarding differentiation theory; the discussion
on this subject involves only the most essential parts, and the use of
the Vitali covering theorem is altogether avoided.
(d) The theory of integration in locally compact spaces; for an excel-
lent discussion I refer either to the book by L. Nachbin on the Haar
integral or to the book by E. Hewitt and K. Stromberg on real and
abstract analysis (cf. the references at the end of the present book.)
A glance at the bibliographical references will show that the theory
of measure and integration is a popular subject with the authors of
textbooks. Although it seems an impossible task to include a complete
list of textbooks on this particular subject, it may be of interest to add
the following items to the list:

PREFACE
XIII
B. SZ.-NAGY, Introduction to real functions and orthogonal ex-
pansions, . Budapest (1964), translated from the 1961 Hungarian
edition.
H. BAUER, Wahrscheinlichkeitstheorie und Grundziige der Masz-
theorie, Berlin (1964).
R. E. EDWARDS, Functional analysis, New York (1965).
J. F. C. KINGMAN and S. J. TAYLOR, Introduction to measure and
probability, Cambridge (1966).
R. G. BARTLE, The elements of integration, New York (1966).
N. DINCULEANU, Vector measures, Berlin (1966).
Of all the textbooks which have been published since 1958, A. E.
Taylor's book is the nearest in spirit to the present book.
I would like to thank Prof. W. Roelcke for some valuable comments
on the subject of Fubini's theorem and Dr. B. Ramachandran for his
critical reading of the text of the 1958 edition, which resulted in a
number of improvements in the present presentation, in particular
in the chapter on differentiation. Finally, I am appreciative of the
staff at the North-Holland Publishing Company for their assistance
and cooperation.
LEIDEN JUNE 1967
A.C.ZAANEN


CHAPTER 1
POINT SETS, ZORN'S LEMMA AND METRIC SPACES
The first chapter has an introductory character; it is, in fact, devoted to
several subjects of a general nature, some knowledge of which is indispensable
nowadays to every student of mathematics. Section 1 contains the simplest ele-
ments of point set theory (no topology included); in section 2 it is shown that
the axiom of choice and Zorn's lemma are equivalent, and in section 3 the
notion of a metric space is introduced. The discussion in this section culminates
in Baire's category theorem, asserting essentially that a complete metric space
is never the union of an at most countable number of nowhere dense sets.
1. Point Sets
Let X be a non-empty set; the elements x, y, ... of X will be called
the points of X. We shall consider the collection of subsets A, B, ...
of X. If the point x is an element of the subset A, we write xEA.
Given the subsets A and B, we write A cB or, equivalently, B'::)A,
whenever A is a subset of B, i.e. whenever xEA implies XEB. In parti-
cular, A c A for every A. By definition, A ==B whenever A c Band
Be A hold simultaneously. The empty set, that is, the set containing
no points at all, is denoted by 0 (so 0 c A for every A eX). Given the
sets A and B, the set of all points belonging to A, but not to B, is
called the difference of A and B, and denoted by A-B. By A-B-C
we mean (A-B)-C. The set Ae==X-A is called the comPlement of
A (with respect to X).
If A ex is a collection of subsets of X (where the index ex runs through
an index set which is not necessarily countable), then the set of all
points belonging to at least one Aex is called the union of the sets Aex,
and this union is denoted by U ex A ex. The set of all points belonging to
all A ex simultaneously is called the intersection of the sets A ex, and de-
noted by nex Aex. If the number of Aex is finite or countable, i.e. if we

2
POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,  1
have a finite or countable sequence of sets An (n== 1 , 2, . . .), we shall
usually write  An or Al +A2+' .. instead of U An, and II An or
AIA2. . . instead of n An.
If AB==0, that is, if the sets A and B have no common points, then
A and B are called disjoint.
Given an infinite sequence of sets An, the set of all points x satis-
fying xEA n for an infinity of values of n is called the upper limit of
An, and denoted by lim sup An. The set of all points x for which there
exists an index no(x), depending therefore on x, such that xEA n for all
nno(x), is called the lower limit of A n, and denoted by lim inf An.
Obviously lim inf An clim sup An. If lim inf An==lim sup An, the se-
quence A n is said to be convergent; in this case the set lim inf A n==
lim sup An is called the limit of An, and denoted by lim An.
THEOREM 1.
00 00
(a) lim sup An= II  An,
k=l n=k
00 00
(b) lim inf An==  II An.
k=l n=k
PROOF. (a) xElim sup AnxE;;=k An for all kXEII=l ;;=k An
(the notation  means that the statements on the left and the right
of it are equivalent).
(b) xElim inf AnxErr;;=ko An for some ko (depending on x) <:>
XE=l II;;=k An.
If, for a sequence An of sets, we have AncAn+I for all n, the
sequence is called ascending or non-decreasing; if An+I cAn for all n,
the sequence is called descending or n01t-increasing. Ascending and de-
scending sequences are said to be monotone".
THEOREM 2.
(a) Any monotone (infinite) sequence is convergent.
(b) If A n is ascending, then lim A n==  An.
(c) If An is descending, then lim An== II An.
PROOF. (a) We have to show that lim sup An clim inf An. Assume,
for this purpose, that xElim sup An, so xEA n for an infinity of values
of n. Hence, if An is ascending, xEAno for some no, and then xEA n for

Ch. 1,  1 ]
POINT SETS
3
all n?::-no, so xElim inf An. If, on the other hand, An is descending and
xEA k , then xEA n for all nk. Hence, since xEA k holds for infinitely
many k, we have xEA n for all n, so certainly xElim inf An.
(b) If An is ascending, then
00 00 00
lim An==lim inf An==  II An==  Ak.
k=l n=k k=l
(c) If An is descending, then
00 00 00
im An==lim sup An== II  An== II Ak.
k=l n=k k=l
For complementary sets we have the following simple properties:
xe==0, 0 e ==X, (Ae)e==A, and AcB implies Ae=>Be. Furthermore
A -B==ABe.
THEOREM 3.
(a) (U AlX)e==n A and (n AlX)e==u A.
(b) (lim sup An)e==lim inf A and (lim inf An)e==lim sup A.
PROOF. (a) XE(U AlX)e<=>XEA lX holds for no ex<=>xEA for all ex<=>
XEn A.
XE(n AlX)e<=>XEA for at least one ex<=>XEU A.
(b) xE(lim sup An)e<=>XEA n holds for only a finite number of values
of n<=>xEA for all n?::-no(x) <=>xElim inf A.
xE(lim inf An)e<=>XEA for infinitely many n<=>xElim sup A.
The set A -B is sometimes called the complement of B with respect
to A. Hence, if all sets AlX and An in the last theorem are subsets of
some set A c X, then the theorem remains true if all complements are
taken with respect to A (the set A takes over the role of X).
Given the non-empty point sets X and Y (not necessarily common
subsets of another point set given in advance), the collection of ordered
pairs (x, y), with XEX and YEY, is called a function (or a maPPing) on
X into Y whenever for each XOEX there is exactly one ordered pair in
the collection having Xo as its first element. If for each Yo E Y there is
at least one ordered pair in the collection having Yo as its second ele-
ment, then the function is said to be on X onto Y. A function is usu-

4
POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,  1
ally denoted by a letter, e.g. by the letter f, in the following way:
Given XEX, the point YE Y which occurs in the (uniquely determined)
pair (x, y) of the function is denoted by y==f(x), and we say then that
the function f assumes the "value" f(x) at the point XEX. The point
f(x) E Y is called the image of the point XEX. Observe that, regardless
of whether the function is onto Y or properly into Y, the same point
YEY may be the image of several different points in X. If A eX, then
the set of all f(x) for xEA is called the image of A, and denoted by
f(A). The set X is called the domain of the function f, and the set of
all YEY for which there exists at least one XEX satisfying f(x)==y is
called the range of f. Hence, the range of f is the image f(X) of X. The
range of f is the whole of Y if and only if the function f is onto Y.
If Y is the real line R I (that is, the set of all real numbers), the
function f on X into Y is said to be a real function. Similarly, we have
a comPlex function whenever Y is the complex plane. In the classical
case, from which the above definition of a function is derived by gener-
alization, the set Y is the real line, and the set X is either also the real
line or an interval on the real line.
If, for each y in the range f(X) of f, there exists only one XEX such
that f(x) =y, the correspondence between the domain X and the range
f(X) of the function f is one-one, and the inverse function -If, with f(X)
as its domain and X as its range, is defined by -If(y)==x for all y=f(x).
ExamPles: f(x)=x 3 is one-one on RI onto R I ; g(x)==X 3 _X 2 is also on
RI onto Rl, but not one-one; h(x) ==sin x is on RI properly into R I , the
range of h is the interval -1 y 1, and the correspondence between
domain and range is not one-one; k(x)==x/(I+lxl) is on Rl properly
into RI, the range of k is the interval -1 <y< 1, and the correspond-
ence between domain and range is one-one.
Finally, we mention the notion of a choice function. Given a col-
lection of non-empty point sets F rx, subsets of the point set Z, we con-
sider each F rx as one point of the set X =={F rx}, and we denote by Y the
union of the sets F rx, so Y == U rx F rx e Z. The function f on X into Y is
now called a choice function whenever f(F rx) EF rx for each cx. The function
f selects, therefore, one point from F rx for each cx, and the range of f
consists of the thus selected points.

Ch. 1, 9 1 ]
POINT SETS
5
We return to the situation that all point sets considered are subsets
of the fixed point set X. Given A eX, the real function XA(X) which is
equal to one at all points xEA, and to zero at all points xEAe, is called
the characteristic functon of A. Evidently, if A== An, and all An are
disjoint, then XA(X)== X.An(x), If the sequence An is monotone, then
the sequence of the characteristic functions X.An(x) is at every poij1t x
also monotone, non-decreasing if An is ascending and non-increasing
if A n is descending.
THEOREM 4. If lim sup An==P and lim inf An==Q, then
Xp(X) ==lim sup X.A n (x),
XQ(X) ==lim inf X.An(x),
The sequence An converges to the set A if and only if
XA(X) ==lim X.An(X)'
PROOF. Xp(x)==I<=>xEP==1im sup An<=>XEAn for infinitely many
n<=>X.An(x)== 1 for infinitely many n<=>lim sup X.An(x)== 1. Since Xp(x), as
well as lim sup XAn(x), can only assume the values zero and one, the
equality Xp(x)==lim sup X.An(x) holds therefore at all XEX.
Similarly, it will be sufficient to prove that XQ(x) == 1 if and only if
liminfXAn(x)==I. This follows from: XQ(x)==I<=>xEQ==liminfA n <=>
xEA n for nno(x)<=>XAn(X)== 1 tor nno(x)<=>lim inf X.An(x)== 1.
Let now A==limAn, so A=P=Q. Then
lim sup X.An(x) == Xp(x) == XA(X) == XQ(x) ==lim inf X.A n (x),
so XA(X) ==lim X.A n (x).
If, conversely, XA(x)==lim X.An(X) , then XA(X)==XP(x)==XQ(x), so A=
P==Q, and this shows that A ==lim An.
Exercises
POINT SETS
1.1) Show that A(B+C)=AB+AC. This property is similar to the
distributive law a(b+c)==ab+ac for real numbers.
1.2) Show that A+BC==(A+B)(A+C). This distributive law for
sets has no parallel for real numbers.

6 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1, 9 1
1.3) Show that (U A ex) -B== U (A ex-B) and (n A ex) -B==n (A a-B).
1.4) ShowthatB-(U Aa)==n (B-Aex) andB-(n Aa)==U (B-A a ).
1.5) Show that
A-B==A-AB==(A+B)-B,
A(B-C)==AB-AC,
(A-B)-C==A-(B+C),
A -A 1 -. . . -An==A -(AI +. . . +An),
A -(B-C)==(A -B)+AC,
(A -B)(C -D)==AC -(B+D).
1.6) Determine lim sup An and lim inf An, if X is the real line, and
{ {X:OXI}
An==
{x: lx2} for n odd.
(If P is some property, then {x: P} is the set of all x having the proper-
ty P.)
1.7) Let X be the real line, and
for n even,
{ {x: O<x< I-n- 1 }
An==
{x: n-lx< I}
for n even,
for n odd.
Show that the sequence An is convergent, but not monotone.
FUNCTIONS
1.8) Let f be a function on X into ¥. Show that, for any subsets A ex
of X, the images in ¥ satisfy f(U A ex) == U f(A a) and f(n A a) C n f(A a).
Show that the last inclusion may be proper even for two sets Al and A 2 .
1.9) Let f be a function on X into ¥. For any subset Be }T, the
inverse image -If(B) is defined to be the set of all XEX such that f(x) EB
(note that the inverse function of f does not necessarily exist, even
though the notation -If is used). Show that, for any subsets Bex of ¥,
the inverse images in X satisfy -If(U Ba)==U -If(Ba) and -If(n Ba)==
n -If(Ba). Show also that for complements we have -If(Be)=={-lf(B)}e;
more generally, -If(BI-B2)==-lf(Bl)--lf(B2)'

Ch. 1, 9 2J
AXIOM OF CHOICE AND ZORN'S LEMMA
7
2. Partial Ordering, the Axiom of Choice and Zorn's Lemma
We consider the collection of all subsets A, B, . .. of the infinite
point set X, and we recall that A cB denotes that A is included in B.
The inclusion relation' c has the following properties: .
(a) A cB and Be C implies A c C (the relation is transitive),
(b) A cA for all A (the relation is reflexive),
(c) A cB and BcA implies A==B (the relation is anti-symmetric).
On account of these properties it is said that the collection @f subsets
A, B, . .. of X is partially ordered by the inclusion relation c. Note
that not every pair of sets A, B is comparable with respect to inclusion,
that is, there exist pairs A, B such that neither A cB nor BcA. We
extend the notion of a partial ordering to more general situations by
the following definition.
DEFINITION. Let V be a set with elements a, b, . . ., and let --< be a re-
lation which exists between some, but not necessarily all, pairs a, b. The
set V is said to be partially ordered by the relation --< whenever --< is
transitive, reflexive and anti-symmetric, i.e., whenever
(a) a--<b and b--<c imPlies a--<c,
(b) a--<a for all aE V,
(c) a--<b and b--<a imPlies a==b.
Sometimes the term partial ordering is already used if only (a) is
satisfied or (a) and (b) are satisfied; we shall assume here that ( a), (b)
and (c) hold. Any pair of elements a, b, for which neither a--<b nor b--<a
is satisfied, is called incomparable, and any pair a, b satisfying one at
least of the relations a--<b and b--<a, is said to be comparable. If each
pair of elements is comparable, the ordering is said to be a linear order-
ing (or simPle ordering). Any partial ordering in V induces automatic-
ally a partial ordering in any subset C of V. If the induced partial
ordering in the subset C is a linear ordering, then C is called a chain
in V.
Examples. (1) V is the set of all real numbers, and a--<b means that
ab. The ordering is linear.
(2) V is the set of all points x== (Xl, X2) in real two-dimensional

8
POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1, 9 2
space, and x.-<y means that Y1-X10 and Y2-X20. The ordering is
not linear.
(3) V is the set of points along a river and its tributaries. We shall
allow that the river has a delta, provided the delta begins downstream
of the point where the last tributary joins the main river. If P and Q
are two points, then P.-<Q means that either Q==P or Q is downstream'
of P.
The chain C in the partially ordered set V is called a maximal chain
whenever there does not exist any chain C' in V such that C is a
proper subset of C'. In the Examples (1), (2) and (3) maximal as well
as non-maximal chains are easily visible.
If D is a subset of V, and the element aEV satisfies d.-<a for all
dED, then a is called an upper bound of D. If, in the Example (2), D
consists of all (Xl, X2) such that x+xI, and a==(a1, a2) where a11,
a2 1, then a is an upper bound of D. If, however, V itself is restricted
to all (Xl, X2) such that x+x 1 (the partial ordering as in Example
(2)), there exist many subsets, even subsets containing only two points,
having no upper bound. The same is true in Example (3) if the river
possesses a delta.
If Dc V has an upper bound a with the property that a.-<a' for any
other possibly existing upper bound a' of D, then a is said to be a least
upper bound of D. In this case a is uniquely determined, for if a and a'
are least upper bounds of D, then a.-<a' and a' .-<a, so a==a'. If in Ex-
ample (2) D is again the set of all (Xl, X2) such that x+xI, then a==
( 1, 1) is the least upper bound of D. If in the example of the subsets of
X, partially ordered by inclusion, D is a collection of sets Aex, then
U A ex is the least upper bound of D.
The element mE V is called a maximal element if m.-<a implies m==a
(observe that this is not the same as requiring that a.-<m for all aEV).
If, again, V is restricted to all (Xl, X2) such that x + x 1, then all
points satisfying X10, X20, x+x== 1 are maximal.
THEOREM 1. The following four statements are equivalent:
(a) The axiom of choice: Given a non-empty collection of non-empty
subsets F ex of the set Z, there exists a choice function f on this collection,
i.e., f(F ex) EF ex for each F ex.

Ch. 1, 9 2J
AXIOM OF CHOICE AND ZORN'S LEMMA
9
(b) The Hausdorff-Kuratowski princiPle: Every partially ordered set
V contains a maximal chain.
(c) Zorn's lemma (first version): If V is any partially ordered set, and
each chain in V has an upper bound, then V contains a maximal element.
(d) Zorn's lemma (second version): If V is any partially ordered set,
and each chain in V has a least upper bound, then V contains a maximal
element.
PROOF. We first prove (b)=>(c)=>(d)=>(b), and then (d)=>(a)=>(d).
(b)=>(c). We assume that each chain in the partially ordered set V
has an upper bound. Then V contains a maximal chain C by (b), so
this chain C has an upper bound m. Hence, a--<.m for all aEC. If m is
not a maximal element of V, there exists an element b=l=-m such that
m--<.b. Then b is not contained in C, for bEC would imply b--<.m, so
b=m (since also m--<.b). It follows that the union of C and the element
b is still a chain. This contradicts the maximal character of C, so m is
a maximal element in V.
(c)=>(d). Each chain in V has a least upper bound by the hypothe-
sis of (d), so the chain has an upper bound. Then V contains a maximal
element by (c).
(d)=>(b). The chains in V form a set W, which is partially ordered
by inclusion. If {C} is a chain in W with elements C (each C is therefore
a chain in V), then the union UCE{C} C is a chain in V, so this union
is an element of W which is obviously the least upper bound of {C}.
Hence, by (d), W contains a maximal element, i.e. V contains a maxi-
mal chain.
(d)=>(a). Let the non-empty collection of non-empty subsets Frx of
the set Z be given. There exist functions g(F rx), perhaps not defined
for all F rx, but such that g(F rx) E F rx for all F rx for which g is defined.
Take for example one single F rx o ' let Yo be an arbitrary element of F rxo'
and define g only for F rx=F rxo by g(F rxo) ==yo. The set V of all such
functions g is therefore not empty, and V is partially ordered by in-
clusion; i.e., gl --<.g2 whenever g2 (F rx) == gl (F rx) for all F rx for which gl (F rx)
is defined. In other words, denoting the domain of such a function g by
D(g), the notation gl--<.g2 means that D(gl) cD(g2), and gl==g2 on D(gl)'
Let C be a chain in V. Then, if F rx is an arbitrary element of U gEC D (g),
there is at least one gEC for which g(F rx) is defined, and if gl and g2
are any two elements of C such that gl(F rx) and g2(F rx) are both de-

10 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,9 2
fined, then either gl«g2 or g2«gl, so gl(F ex)==g2(F ex). Hence, all gEC
for which g(F ex) is defined, yield the same value for g(F ex). We define
go(F ex) to be this common value. Then goE V, and go is defined on
U gEC D(g), with go»g for all gEC. This shows that go is the least upper
bound of C. Hence, on account of (d), V contains a maximal element
f. Then f is defined for all F ex, for otherwise, if f is not defined for F ex o '
we can choose an element YoEF ex o and define f(F ex o ) ==yo, thereby ex-
tending the domain of f properly, which contradicts the maximal
character of f.
(a)=:>(d). We first prove the following lemma.
Let V be non-empty and partially ordered, and let each chain in V have
a least upper bound. Furthermore, let a'f(a) be a function on V into
itself such that a«a' for all aEV. Then each subset of the form V o ==
{a: a »ao} contains an invariant element, i.e. an element a". such that
a;==a". (we recall that if P is some property, then {a: P} is the set of all a
having the property P).
The proof of this lemma is divided into three parts.
(1) As stated above, let Vo=={a: a»ao}. The subset Gc V o is, for
the purposes of the present proof, said to be closed if
(ex) aoEG,
(fJ) aEG implies a' EG,
(y) given M c G such that the least upper bound s of M exists, then
sEG.
Note, for better understanding of (fJ) and (y), that a E V 0 implies
a' E V 0, and that any upper bound of any subset of V 0, if it exists, is in
V o . The set V 0 itself is closed therefore, and the intersection of an arbi-
trary collection of closed sets is closed. Observe that such an inter-
section is never empty since ao is one of its elements. Denoting the
intersection of all closed sets by Go, it will be sufficient to prove that
Go is a chain. In fact, in that case the least upper bound a". of Go exists
by the hypothesis of the lemma, and a". E Go by (y), so a; E Go by (fJ).
But a; »a"., and a". is the least upper bound of Go; hence a; a"..
(2) We consider therefore the set Go; note that aEG o implies a' EGo,
and any existing least upper bound of any subset of Go is an element of
Go. All elements mentioned in what follows will be elements of Go. The
element n is called normal whenever an implies a' «n (so ao is normal,

Ch. 1,  2J
AXIOM OF CHOICE AND ZORN'S LEMMA
1 1
since there exists no element aao). Let n be normal, and let Bn==
{b: b--<.n or b>-n'}. Then Bn is closed, since
(ex) aoEBn on account of ao--<.n,
(fJ) bEBn implies either bn (so b'--<.n), or bn (so b'==n', which
implies b'>-n'), or b>-n' (so b'>-b>-n'), so b' EBn in all cases,
(y) given Me Bn such that the least upper bound s of M exists (so
.sEGO), we have either m--<.n for all mEM (so s--<.n), or m1>-n' for some
1n1EM (so s>-m1>-n'), hence sEBn in either case.
It follows that Bn==G o (since Bn c Go, and Go is the intersection of
all closed sets), and this shows that any normal element n is compa-
rable with all elements of Go (since for any aEG o we have now either
.a--<.n or a>-n'>-n).
(3) We finally prove that the set N of all normal elements n is
,closed.
( ex) ao is normal.
(fJ) Given nEN, we have to prove that n' EN, in other words, we
have to prove that an' implies a'--<.n'. If an', then a--<.n since aEG o
Bn. It follows that either an (so a' --<.n--<.n'), or a==n (so a' ==n').
Hence a' --<.n' in either case.
(y) Given MeN such that the least upper bound s of M exists (so
.sEG o ), we have to prove that s is normal, i.e. that as implies a' --<.s.
The element a is comparable with all normal elements, so in particular
with all mEM. If a';>m would hold for all mEM, then also a>-s in
contradiction to as; hence am1, for some m1EM. Since m1 is
normal, this implies a' --<.m1 --<.s, and the proof is complete.
It follows that N is closed, i.e. N==G o . This shows that each pair
of elements of Go is comparable, so Go is a chain. This completes the
proof of the lemma.
We finally show how (a)=>(d) follows from this lemma. Let, there-
fore, each chain in V have a least upper bound, and assume that V
contains no maximal element. Then all sets F a{b: bf;;;a} are non-
empty, so that by the axiom of choice there exists a choice function
defined on the collection of all Fa. In other words, there exists a
function a'f(a) such that aa' for all aEV. On account of the lemma,
however, there is at least one element a". such that a; ==a".,. a contra-
diction.

12 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,  2
This proof of (a) =>(d) is due to E. WITT (1950, [lJ); it avoids the
use made in older proofs of Zermelo's wellordering theorem (cf. Exer-
cise 2.5).
We shall assume in this book that the axiom of choice, and hence
the Hausdorff-Kuratowski principle and Zorn's lemma, are true. Some
mathematicians feel doubt about the validity of the axiom of choice,
but it has been proved that if mathematics is consistent without the
axiom of choice, then it remains consistent if this axiom is added. A
few of the sometimes rather surprising consequences of Zorn's lemma
will be discussed later (in sec. 10, Theorem 2, in Exercises 26.1 and
26.2, and in sec. 28).
Exercises
EXAMPLES
2. 1 ) We recall the conditions for partial ordering:
(a) a--<.b and b--<.c implies a«c,
(b) a --<.a for all a,
(c) a--<.b and b--<.a implies a=b.
Note that (c) may be formulated equivalently as: a=l=-b implies that
a--<.b and b--<.a do not hold simultaneously. Show that if in the set of
all real numbers a--<.b means that a<b, then conditions (a) and (c) are
satisfied but (b) is not. Show also that if in the set of all triangles
a, b, . .. in the plane a--<.b means that a and b are congruent (in the
usual elementary geometric sense), then conditions (a) and (b) are
satisfied but (c) is not. The same holds if in the set of all integers a--<.b
means that a == b (mod n), where n is some integer 2.
2.2) Show that the set of all natural numbers is partially ordered
if a--<.b means that b is a divisor of a. Show that m== 1 is the only
maximal element. Show also that if we consider the same partial order-
ing in the set of all natural numbers 2, then each prime number is
maximal.
2.3) Let a be a decomposition of the non-empty point set X into
disjoint subsets: X ==U Aa with all Aa disjoint. Show that the col-
lection of all such decompositions is partially ordered if a--<.b means

Ch. 1,  2J
AXIOM OF CHOICE AND ZORN'S LEMMA
13
that b is a refinement of a, i.e. if b is the decomposition X == U B p , then
each Bp is a subset of some Aa. Determine the maximal element.
THE WELLORDERING THEOREM
2.4) The partially ordered set V is called wellordered whenever each
non-empty subset We V contains a "smallest" element Wo (i.e., wo.-<w
for all WE W). Show that if V is wellordered, then V is linearly ordered.
It is sufficient to observe that each subset, consisting of two elements,
contains a smallest element.
2.5) Show that each non-empty set 5 can be wellordered (Zermelo's
theorem, 1904, [IJ), in the following way. Note first that all subsets
consisting of one element are trivially wellordered, so that the col-
lection W of wellordered subsets V c 5 is not empty. Introduce a partial
ordering in W as follows: V 1 .-< V 2 means that
(ex) VIe V 2,
((3) the wellordering in VIis identical with the wellordering in-
duced in VI by the wellordering in V 2,
(Y) VI is an initial segment of V 2 , Le., aEV1, bEV 2 and b.-<a implies
bE VI.
Then each chain in W has a least upper bOUlld, so W contains a
maximal element V o by Zorn's lemma, and it is not difficult to derive
now that Vo=S.
THE CONTINUUM HYPOTHESIS
2.6) If V is wellordered with respect to .-< and a is an element of
V, then the set of all bE V satisfying b -<.a is called an initial segment
of V. The continuum hypothesis states that the set R1 of all real
numbers can be wellordered in such a manner that all initial segments
with respect to this wellordering are at most countable. Note that
there may exist many other wellorderings of Rl not having this proper-
ty. Assume that the continuum hypothesis holds, let .-< be a well-
ordering of R1 with the stated property, and consider the subset E1 of
the real plane R2 consisting of all points (x, y) ER 2 such that x.-<y.
Show that E1 intersects each horizontal line in the plane in an at most
countable number of points, whereas the intersection of E1 and any

14 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,93
vertical line is the entire vertical line with an at most countable number
of points left out. Note that the complementary set E 2 of E 1 has simi-
lar properties, with the roles of horizontal and vertical lines inter-
changed.
3. Metric Space, Baire's Category Theorem
If X is a non-empty point set, and if there exists a real function
d(x, y), defined on the set of all pairs XEX, YEX, such that
(1) d(x, y) ==0 if and only if x==y,
(2) d(x, y)d(z, x)+d(z, y) for x, y, ZEX, then X is called a metric
space, and d(x, y) is called the distance between the points x and y.
Choosing y==x in (2), we obtain Od(z, x)+d(z, x), so
(3) d(z, x)O for x, ZEX.
Next, choosing z==y in (2), we obtain d(x, y)d(y, x), and similarly,
the choice z==x in d(y, x)d(z, y)+d(z, x) yields d(y, x)d(x, y). Hence
(4) d(x, y) ==d(y, x) for x, YEX.
It follows therefore from (1), (2) and the derived properties (3), (4)
that d(x, y) has all properties usually required from a distance function.
The inequality (2) is called the triangle inequality.
ExamPle. The k-dimensional real number space of all points x==
(Xl, . . ., Xk) with Xl, . . ., Xk real and
d(x, y) =={(X1-Y1)2+ . . . + (Xk-Yk)2}!
is a metric space. Note that the same point set is also a metric space
with respect to the distance functions
d 1 (x, y) == I X 1-Y11 + . . · + !xk-Ykl
and
d 2 (x, y) == max IXi-Yil.
lik
Given the subset A of X, the number
15(A) == sup d(x, y)
XEA. ,YEA.
is called the diameter of A. If 15(A) <00, then A is said to be bounded.

Ch. 1, 9 3J
METRIC SPACE, BAIRE'S CATEGORY THEOREM
15
Given the subsets A and B of X, the number
d(A, B) == inf d(x, y)
XEA.,YEB
is sometimes called the distance between A and B, although this is not
a very appropriate name (cf. Exercise 3.1). In particular, if A eX and
XEX, then
d(x, A) ==inf d(x, y)
YEA.
is the distance between the point x and the set A.
Given the point XEX and the number r>O, the set of all points
YEX satisfying d(x, y)<r is called the (spherical) neighbourhood of x
of radius r.
If A eX, xEA, and there exists a neighbourhood of x which is en-
tirely included in A, then x is called an interior point of A. If OeX,
and all points of 0 are interior points of 0, then 0 is said to be an open
set. Evidently, X itself and the empty set 0 are examples of open sets.
Furthermore, any spherical neighbourhood of any point is open (by
the triangle inequality). If FeX, and the complement Fe==X-F is
open, then F is called a closed set. Evidently X and 0 are closed. Also,
if XEX and rO, the set {y: d(x, y)r} is closed; in particular any set
consisting of one point is closed.
THEOREM 1. (a) The union of any collection of open sets is open, and
the intersection of a finite number of open sets is open.
(b) The intersection of any collection of closed sets is closed, and the
union of a finite number of closed sets is closed.
PROOF. (a) Let O==U Ocx, where all Ocx are open. If XEO, then XEO cx (}
for some cx===cxo. Hence, there exists a neighbourhood of x which is con-
tained in Ocx o ' so in O==U Ocx. This shows that any point XEO is an
interior point of 0, i.e. 0 is open.
Let
k
0== II On,
n=l
where 01,..., Ok are open.
If XEO, then XEO n for n== 1, . . " k. Hence, there exist neighbourhoods
N n={Y: d(x, y) <r n } of x such that N neOn for n== 1, . . " k. Setting
r==min{r1, "', rk} and N=={y: d(x, y)<r}, we have then NeNneO n

16 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,  3
for n==l, "', k, so NeO. This shows that any point XEO is an in-
terior point of 0, Le. 0 is open. Observe that the success of the proof
is due to the property that the minimum of a finite number of positive
numbers is positive.
(b) Let F == n F ex, where all F ex are closed. Then Fe== (n F ex)e== U F
is open by (a), since all F are open. This shows that F is closed.
Let
k
F==  Fn,
n=l
where F 1, . . " F k are closed.
Then
k k
Fe==( Fn)e==II F is open by (a), since all F are open.
1 1
This shows that F is closed.
The union of a countable number of closed sets is not necessarily
closed, as the following example shows: If X ==R1 (the real line with
the usual distance), then the set A of all rational points is the union of
a countable number of closed sets (each point is a closed set), but A is
not closed (it is easy to see that A e==x -A is not open). Hence, the
intersection of a countable number of open sets is not necessarily open.
Given A eX, the set of all points XEX such that each neighbour-
hood of x contains at least one point of A, is called the closure of A,
and denoted by A . Evidently A eA, and A eB implies A e B .
THEOREM 2. (a) For each set A, the set 1 is closed.
(b) If F is closed, then F ==F.
Hence, for each A the closure of 1 equals A.
PROOF. (a) We have to prove that (.A)e is open. Let, for this purpose,
xE(A)e. Then there exists a neighbourhood N x of x containing no point
of A. It follows that N x contains no point of A either, for if YENxn A ,
there exists a neighbourhood Ny eN x such that Ny (and hence also
N x) contains a point of A. It follows that x is an interior point of (1) e,
and this shows that (A)e is open.
(b) Let F be closed, i.e. Fe is open. Since any point xEFe is an in-
terior point of Fe, there exists a neighbourhood of x which contains no

Ch. 1, 9 3J
METRIC SPACE, BAIRE'S CATEGORY THEOREM
17
point of F. But then x is not a point of F, i.e., xE(F)e. This shows that
Fec(F)e, so FcF. Since we have FcE' as vvell, it follows that F==F.
THEOREM 3. ..!fBcAB and A+B==A+B for any pair of sets A eX,
BcX.
PROOF. In view of ABc A andABcB wehave AB cA and AB c B ,
- --
so AB cAB.
In view of A eA +B and B cA +Bwe have A' +:8 c A +B . Conversely,
A+BcA+ B , and A+B is closed. Hence, A+BcA+ B ==A+B.
The point XOEX is called a point of accumulation of the set A cX
whenever each neighbourhood of Xo contains at least one point xEA
which differs from Xo (and then each neighbourhood of Xo contains
at least a countably infinite number of different points of A). Note that
a point of accumulation Xo of A is not necessarily a point of A. It is
true in that case, however, that xoEA. Conversely, if xOEA, but xoEA
fails to hold, then each neighbourhood of Xo contains a point of A
which cannot be Xo itself, so Xo is now a point of accumulation of A.
Hence, if the set of all points of accumulation of A is denoted by A +,
we have A==A+A+.
The point sequence xnEX (n==l, 2, ...) is said to be a fundamental
sequence or Cauchy sequence whenever lim d(xm, xn)==O as m, n--x:>o.
The point Xo is called the limit of the sequence Xn whenever lim d(xn, xo)
==0 as noo, and this is denoted by xo==lim Xn. Any point sequence
can have at most one limit (if Xo and-yo are limits of the sequence X n ,
then d(xo, yo)d(xo, xn)+d(yo, X n ), so d(xo, yo)=O, i.e. xo=yo). Any
point sequence possessing a limit is said to be convergent. Every con-
vergent sequence is a fundamental sequence; the converse is not gener-
ally true (take for X the set of all rational points on the real line with
the usual definition for the distance).
Whenever in the metric space X every fundamental sequence is con-
vergent, the space X is said to be comPlete.
THEOREM 4. The metric space X is comPlete if and only if for each
sequence F 1  F 2  . .. of non-empty closed sets, whose diameters tend to
zero, the intersection n;;= 1 F n consists of one single point.
PROOF. Let X be complete, and F 1  F 2:::> . . ., where all F n are non-

18 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1,9 3
empty and closed, and (F n) O. Selecting a point X n in each F n, we
have d(xm, x n ) O as m, noo on account of (F n) O, so xo===lim X n
exists (since X is complete). Since XiEFicFn for i?;::-n, we have XOEFn
==Fn for all n, so xoEIT=l Fn. Evidently, since (Fn)O, Xo is the
only point of rrr Fn.
Let now, conversely, for each descending sequence of non-empty
closed sets F n such that (F n) O, the intersection IT r F n consist of
one point only. Let X n be a fundamental sequence. Then, for En==
{xn, Xn+1, . . .}, we have (En) O, so also (F n) O for F n==En. Further-
more, the sequence F n is descending (since En is descending). Hence,
by hypothesis, rr r F n consists of one point Xo, so d(xo, x n ) O on ac-
count of d(xo, xn)(F n). This shows that xo==lim Xn. In other words,
X is complete.
We continue with some definitions.
The set A c X is called dense if A ==X, and A is called a boundary
set if its complement A e is dense, i.e. if A e ==X (the set of all rational
points and the set of all irrational points on the real line are both dense
as well as boundary sets). The set A cX is called nowhere dense if its
closure A is a boundary set (the sets just mentioned are boundary sets,
but not nowhere dense). In twodimensional Euclidean space R 2 the set
of all points X==(X1, X2), satisfying xi+x== 1, is nowhere dense. Obvi-
ously, every subset of a nowhere dense set is nowhere dense. The set
A c X is said to be of the first category (or meager) if A is the union of
an at most countable number of nowhere dense sets. Every subset of X
which is not of the first category is said to be of the second category. In
R 2 the union of all sets xi+x==r2, where r is rational and non-negative,
is therefore of the first category; this set is dense as well as a boundary
set.
THEOREM 5. (a) If A is nowhere dense, then A is a boundary set.
(b) If A is closed and a boundary set, then A is nowhere dense.
(c) The set A is a boundary set if and only if A contains no (open)
sphere.
PROOF. (a) If A is nowhere dense, then X-A is dense, so X-A is
dense since X-A  X - A. It follows that A is a boundary set.
(b) Follows immediately from the definitions.

Ch. 1, 9 3J
METRIC SPACE, BAIRE'S CATEGORY THEOREM
19
(c) The set A contains an open sphere if and only if X -A *X (ob-
serve that X -A is closed), that is, if and only if A is not a boundary set.
THEOREM 6 (BAIRE'S CATEGORY THEOREM). Let X be a comPlete
metric space. Then the following three equivalent statements hold.
(a) Every set of the first category is a boundary set.
(b) If A ==  A n, where all A n are closed boundary sets, then A is a
boundary set.
(c) If B== II Bn, where all Bn are open and dense, then B is dense.
PROOF. (b) follows from (a) in view of the fact that any closed
boundary set is nowhere dense, and (a) follows from (b) by observing
that if A ==  A n, where all A n are nowhere dense, then A * ==  An is a
boundary set by (b), so A is a boundary set as well on account of
A cA*. Finally, (b) and (c) are equivalent by complementation.
We shall prove now that (c) holds. Let, therefore, the sets Bn (n==
1, 2, . . .) be open and dense, and let U be an open sphere in X. We
have to show that U contains at least one point of II Bn. Since B1 is
dense, U contains a point of B1, and hence, since both U and B1 are
open, U contains a closed sphere U 1 consisting entirely of points of B 1,
and such that the open sphere U1 has positive radius. We may assume
that 015( U 1)< 1 (it can happen that 15( [] 1)==0; cf. Exercise 3.3). By
the same argument the open sphere U 1 contains a closed sphere U 2
consisting entirely of points of B2, and we may assume that 015( U 2)
< t. Continuing, we obtain a sequence of closed spheres U n (n== 1, 2,
...) such that U U 1 U 2..., 015( U n)<n-1, and U n consists en-
tirely of points of Bn. Then, by Theorem 4, the intersection II U n con-
sists of one point Xo (here, we use the hypothesis that X is complete).
Evidently, Xo E II Bn (since U n c Bn) and XOE U. This is the desired re-
suit.
The theorem was established for sets in Euclidean space by R. BAIRE
(1899, [lJ) and, almost simultaneously, by W. F. OSGOOD (1900, [IJ).
For sets in a complete metric space the theorem is due to F. HAUS-
DORFF (1914, [lJ, p. 327-328).
THEOREM 7. A ny comPlete metric space X is of the second category, in
other words, if X ==  An, then An contains an open sPhere for at least one
value of n.

20 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1, 9 3
PROOF. Evidently, X is no boundary set, so (by Baire's theorem) X
is not of the first category, i.e. X is of the second category.
Note that the completeness of the space is essential. If X is the
metric space of all rational points on the real line with the usual dis-
tance, then X is the countable union of its points, and since each point
is a nowhere dense subset, X is of the first category.
The metric space X is said to be separable if there exists a countable
subset of X which is dense.
Evidently, if X is a metric space with distance function d(x, y) and
Q is a subset of X, then Q is a metric space with respect to the same
distance function d(P, q), restricted now to points p, qEQ. It is not im-
mediately evident, however, that separability of X implies separability
of Q, since the points of the countable dense subset of X (which exists
whenever X is separable) may not be points of Q. In order to prove
separability of Q it will be necessary, therefore, to replace the dense
subset of X by another countable set which is dense in Q.
THEOREM 8. A ny subset of a separable metric space is a separable
metric space with respect to the same distance function.
PROOF. Let Q be a subset of the separable metric space X with
distance d(x, y). If {Xl, X2, . . .} is dense in X, we define for m, n==
1, 2, . . . the spheres Smn by Smn=={x: d(x, x m ) < 1/2n}. For each value of
n, each point qEQ is contained in one at least of the spheres SIn, S2n,
. . . . Now, if Smn contains points of Q, we select one of these points
and call it qmn. Hence, given qEQ and the index n, there exists at
least one qmn satisfying d(q, qmn) < 1 In, and this shows that the set
of all points qmn is dense in Q.
We finally recall, for the case that X is k-dimensional Euclidean
space Rk, two well-known theorems.
BOLZANO- WEIERSTRASS THEOREM: A ny bounded set in Rk, containing
an infinity of different points, has at least one point of accumulation.
HEINE-BoREL-LEBESGUE COVERING THEOREM: If FcR k is bounded
and closed, and FeU 0 a, where all 0 ex are open, then there exists a finite
number of the sets Oex such that F is already contained in (covered by) the
u1tion of that finite number.

Ch. 1, 9 3J
METRIC SPACE, BAIRE'S CATEGORY THEOREM
21
Exercises
DISTANCE BETWEEN SETS
3.1) Show that the definition of the distance d(A, B) between two
subsets A and B of the metric space X does not nlake a metric space
out of the collection of all subsets of X.
CLOSURE OF A UNION
3.2) Show that Al +. . . +A p==A1 + . . . +Ap for a finite number p
of sets. Show also that U Aa C UA lX, where the inclusion may be proper
even when the number of AlX is countable.
METRIC SPACES WITH INTEGERS AS POINTS
I
3.3) The set of all integers is a complete metric space if the distance
is defined by d(m, n)==lm-nl. If m is a point in the space, show that
the closure of the open sphere, having centre m and radius 1, is a proper
subset of the closed sphere having the same centre and radius.
3.4) Show that the set of all natural numbers is a metric space if the
distance is defined by d(m, n)= Im- 1 -n- 1 /, and show that this metric
space is not complete.
BOUNDED METRIC SPACE
3.5) Let X be a metric space with distance d(x, y), and let
d ) d(x, y)
1(X, Y = 1 +d(x, y) ·
Show that X, with d1(x, y) as a distance function, becomes a metric
space Xl. The space Xl is bounded (more precisely, 15(X1)I). Show
that the space Xl is complete if and only if X is complete.
COUNTEREXAMPLES
3.6) Show that there exists a complete metric space X with a de-
scending sequence of non-empty closed sets F n such that the inter-
section rrr F n is empty. Compare this to the statement in Theorem 4.

22 POINT SETS, ZORN'S LEMMA, AND METRIC SPACES [Ch. 1, 9 3
3.7) Show that there exists a complete metric space X with a de-
scending sequence of non-empty closed sets F n such that the sequence
of diameters (Fn) is bounded and the intersection nl Fn is empty.
Compare this to the statement in Theorem 4.
DIFFERENT METRICS IN FINITE-DIMENSIONAL SPACE
3.8) Show that the space of all real k-tuples X=== (Xl, . . . , Xk) is com-
plete with respect to any of the distances
k
d(x, y) == [ (Xi-Yi)2J!,
1
k
d 1 (x, y) ==  IXi-YiL
1
or
d2(X, y) == max IXi-Yil.
lik
Show also that the notions of open set, closure of a set and point of
accumulation are the same for all these distances. For k==2, draw
pictures of the "unit circle" around the origin for each of these distances.
CLOSED SETS
3.9) We recall that the set of all points of accumulation of the subset
A of the metric space X is denoted by A +. Show that A is closed if and
only if A+cA.
BAIRE'S CATEGORY THEOREM
3.10) Show that, on the real line R1 with the usual distance, the set
of all rational points is not the intersection of a countable number of
open sets.
3.11) In the plane R2 with the usual distance we assign to each real
number X the open circle C x of radius rx>O and centre (x, r x ). Let
A =  C x for all rational x, and B == U C x for all irrational X. Show that,
irrespective of how the radii r x are selected, the sets A and B are not
disjoint.
3.12) Let I(x) be a continuous and realvalued function on the inter-
val I ={x: OxI}, let 11 be any primitive function of I (i.e. d/1fdx==1
at all XEI), 12 any primitive function of 11, and so on. Assume that for
each xEI there is a natural number k (possibly a different k for differ-
ent x) such that Ik(x) ==0. Show that I(x) ==0 at all xEI.

Ch. 1,  3J
METRIC SPACE, BAIRE'S CATEGORY THEOREM
23
3.13) A real infinitely differentiable function f(x) on an open inter-
val I cR1 is called analytic at the point aEI whenever the Taylor series
of f about a converges to f in some neighbourhood of a, and f is called
analytic in I whenever f is analytic at each point of I. Show that if f
is infinitely differentiable on I and its Taylor series, about each aEI,
has a positive radius of convergence, then f is analytic on some sub-
interval of I. Show, in fact, that the set of points of non-analyticity is
nowhere dense.

CHAPTER 2
MEASURE
This chapter contains the principal points of the theory of measure, where
measure is to be understood as the notion which generalizes the length of a
line segment, the area of a plane surface or the content of a volume in space,
as well as the total amount of mass contained within a certain volume. One of
the essential features of measure is the property of additivity: the measure of
a finite union of disjoint sets is the sum of the measures of the separate sets.
Except in some groups of exercises, we shall restrict ourselves to measures having
the stronger property of countable additivity, that is to say, the measure of any
countable union of disjoint sets is the sum of the measures of the separate sets.
4. Semi-rings and a-rings of Point Sets
If we start out to endow certain point sets in the xy-plane with an
area, the usual way to begin is to define the area of a rectangle, with
sides parallel to the x-axis and y-axis respectively, as the product of
its length and width. The next step is then to extend the definition to
finite unions of disjoint rectangles of this type, and from there on a
limit procedure yields the area of triangles, polygons, circles, and so on.
As long as the approach along these lines is on a more or less intuitive
level, no distinction is made between closed anQopen rectangles (i.e.,
between rectangles of the types {(x, y): axb, cyd} and {(x, y):
a<x<b, c<y<d}), but anyone with some mathematical feeling will
guess that this is probably assuming too much. More explicitly, by de-
fining the area of closed and open rectangles with the same vertices to
have the same value, we assume something which in the course of a
strict mathematical development we should be able to prove. It might
seem at first as if the proper way out is to restrict ourselves in the
beginning to either closed or open rectangles, but if we do so the bounda-
ries still continue to cause trouble. If A and B are closed rectangles,

Ch. 2, 9 4J
SEMI-RINGS AND a-RINGS OF POINT SETS
25
and B is included in A, then we surely wish the area of A -B to be the
difference of the areas of A and B, but there is no easy method to check
this since A -B is not a finite union of disjoint closed rectangles. A
similar difficulty arises if A and B are open rectangles. The difficulties
disappear, however, if we consider half open rectangles {(x, y): a<xb,
c<yd}, and it is easy to verify that the collection T of all such left
half open rectangles has the properties that ABET whenever A ET and
BET, and if A and B belong to T with B included in A, then A - B is
a finite union of disjoint sets of T. Any collection of sets possessing
these properties is an example of what is called a semi-ring of sets, and
this notion will be the starting point in our more abstract discussion of
measure theory.
Let X be a non-empty point set.
DEFINITION. The collection T of subsets of X is called a semi-ring
whenever
(a) 0ET,
(b) if AET and BET, then ABET,
(c) if AET and BET, where BcA, then A-B= C n , where  C n
is a finite or countable union, all CnET, and all C n are mutually disfoint.
We proceed with another definition. Any set O==r An, where all
An belong to the semi-ring T, will be called a a-set (with respect to T).
Condition (c) in the definition of a semi-ring T may be read, therefore,
as follows: I f A E T, BET, and B c A, then A - B is a a-set with disfoint
terms.
THEOREM 1. (1) Countable unions and finite intersections of a-sets are
a-sets.
(2) If A ET, and AnET for n= 1, . . " p, then A -A1-A2-' . . -A p
is a a-set with disjoint terms.
(3) Any a-set can be written as a a-set with disfoint terms.
PROOF. (1) The assertion concerning countable unions is evident.
As regards a finite intersection, note that (Am)f1( Bn) = m,nAmBn,
and AmBnET whenever AmET and BnET.
(2) A-A1==A-AA1== C n , where all C n are disjoint and belong

26
MEASURE
[Ch. 2, 9 4
to r by condition (c). Then
A-A1-A2==( Cn)-A2== (Cn-A2)== (Cn-CnA2)==  C nj ,
n j
where all CnjEr, and C nj C nk ==0 for ii=-k. It follows now, since the sets
C n are disjoint, that all C nj are disjoint. Hence A-A1-A2== Bk,
where all BkEr, and the sets Bk are disjoint. The general result is de-
rived by induction.
(3) If O== Anisaa-set, thenO==A1+(A2-A1)+(A3-A1-A2)+
. . " and by (2) each of the disjoint sets on the right is a a-set with
disj oint terms.
In most of the examples which will follow, and in particular in the
example of the rectangles cited in the beginning of this section, the
semi-ring r has the property that if AEr and BEr, then A-B=
A -AB is a a-set consisting of a finite number of disjoint sets of r.
Let us consider, in this case, the collection A of all finite unions of sets
of r (i.e., the collection of all a-sets with only a finite number of non-
empty terms). Evidently, any finite union of sets of A is again a set of
A, and it is also true that if A EA and BEA, then A-BEA. Indeed, in
view of the additional condition on r, an inspection of the proofs of
(2) and (3) in the last theorem shows first that A may be written as a
finite union l An of disjoint sets of r. Then
p p p
A-B==( An)-B== (An-B)= (An-AnB),
1 1 1
and since each of the disjoint sets on the right is now, by (2) in the last
theorem, a finite union of disjoint sets of r, the same holds for A -B,
i.e., A-BEA. The collection A is an example of what is called a ring
of sets. We present the formal definition.
DEFINITION. The non-empty collection A of subsets of the non-empty
point set X is called a ring whenever
(a) any finite union of sets of A is again a set of A,
(b) if AEA and BEA, then A-BEA.
If, instead of (a), the ring A satisfies the stronger condition:
(a') any finite or countable union of sets of A is again a set of A,
then A is called a a-ring of sets.

Ch. 2, 9 4J
SEMI-RINGS AND a-RINGS OF POINT SETS
27
THEOREM 2. If A is a ring, then
(1) 0EA,
(2) if An EA (n== 1, ..., P), then IIi An EA ,
(3) A is a semi-ring.
(4) If A is a a-ring, and AnEA (n== 1,2, .. .), then II AnEA,
lim sup AnEA and lim inf AnEA.
PROOF. (1) A contains at least one set A, hence 0==A-AEA
by (b).
(2) If AnEA (n== 1, "', P), then
p
A== AnEA
1
by (a), so A-AnEA for all n by (b). Hence, since
p p
A-II An== (A-An),
1 1
we ha ve
p p
II An==A- (A-An)EA.
1 1
(3) Follows from (1), (2) and condition (b) in the definition of a ring.
(4) If A is a a-ring, and AnEA for n== 1,2, ..., then the proof that
II An EA is similar to the proof in part (2), replacing i by . It
follows then that
00 00
lim sup An== II  An EA
k=l n=k
and
00 00
lim inf An==  II An EA .
k=l n=k
Any ring or a-ring A of subsets of X such that X itself is a set of A
is sometimes called a field or a-field respectively. Obviously, if A is
a field, and A EA, then A e==X -A EA.
The following diagram shows the implications between the intro-
duced concepts:
7' a-rIng
a-field ring  semI-rIng.
 field 7'

28
MEASURE
[Ch. 2,  5
In what follows we shall have to deal mainly with semi-rings and a-
fields.
The notion of a semi-ring, as it is defined here, is a slightly general-
ized version, due to N. G. DE BRUIJN, of a notion due to J. VON
NEUMANN ([lJ, p. 85); rings (fields) and a-rings (a-fields) have
appeared in measure theoretic investigations from the beginning on.
Exercises
THE SMALLEST a-FIELD INCLUDING A GIVEN COLLECTION OF SUBSETS
4.1) Show that if To is an arbitrary non-empty collection of subsets
of X, then there exists a smallest a-field Ao::> To (i.e., any other a-field
AI::> To satisfies AI::> Ao). Show, similarly, that there exist a smallest
field including To, a smallest a-ring including To, and a smallest ring
including To.
A a-FIELD IN A METRIC SPACE
4.2) Show that in any metric space the collection of all sets of the first
category is a a-ring, and the collection of all sets A, such that either
A or its complement A e is of the first category, is a a-field.
THE COLLECTION OF ALL a-SETS
4.3) Show that if A and B are a-sets (with respect to the semi-ring T)
such that B contains only a finite number of non-empty terms, then
A-B is a a-set. Show also that if B contains an infinite number of
non-empty terms, then- A-B is not necessarily a a-set. Hence, the
collection of all a-sets is not necessarily a a-ring.
5. Measure on a Semi-ring
In the following, we shall frequently encounter functions assuming
real values, the values +00 and -00 not excluded. We add, therefore,
the symbols +00 and -00 to the set R 1 of all real numbers, and in this

Ch. 2, 9 5J
MEASURE ON A SEMI-RING
29
extended real number system we define
-oo<a< +00 for any finite real a,
a+(:l::oo)==(:l::oo)+a==:l::oo for a=l=- + oo,
(+00) + (-00) == (-00) + (+00) == (:1::00) - (:1::00) ==0,
a(:l::oo)==(:l::oo)a==:l::oo for a>O,
a (:I:: 00) == (:I:: 00) a == + 00 for a<O,
0(:1:: 00 ) == (:1::00)0==0,
a/(:l::oo) ==0,
a/O== +00.
It may be useful to make some remarks. In the extended real number
system the associative law does not hold without restrictions; we have
for example {1+(+00)}+(-00)==0, but 1+{(+00)+(-00)}==1. In
view of this fact we shall agree that a1 +a2+a3 means strictly (a1 +a2)
+a3, and generally, by induction, Sn==  ak is defined by Sn==Sn-1 +a n .
The notation r an==a will mean that a==lim  ak exists as noo,
where it is not excluded that a==:l::oo. It is also worth while to ob-
serve that a+b==c is not always equivalent to a==c-b (e.g., 1 +00==00,
but 1 =1=-00-00). The two equalities are equivalent, however, if b is
finite. There is, fortunately, an important special case where we are
not hampered by the fixed ordering of the terms in a finite or infinite
sum; if Oanoo for n==l, 2, ..., then a==r an==lim ak always
exists, and the sum a of the series r an is unaltered if the terms are
reordered arbitrarily. Note that a==+oo if either an==+oo for at least
one value of n, or the series r an of finite non-negative numbers an
diverges in the classical sense. The foregoing statements about a== r an
remain true if some terms an are negative, provided the finite or infinite
series formed by these negative terms has a sum which is not -00.
Finally, we recall one well-known property of double series which will
be used several times in the next sections. If Oamnoo for m, n==
1, 2, . . " then
00 00 00 00 p q
s==  ( a mn )==  (  a mn )== lim   a mn ,
m=l n=l n=l m=l p,q-+oo m=l n=l
where S may be +00.

30
MEASURE
[Ch. 2,  5
DEFINITION. Let T be a semi-ring of subsets of X, and assume that to
every set A E T corresponds a real number #(A). This set function # is
called a measure on T, if
(a) #(0)==0, and O#(A)oo for every A ET,
(b) if A ET, AnET (n=== 1,2, . . .) and A c r An, then #(A)
r #(An),
(c) if A ET, AnET (n== 1, . . " P) where AI, . . " A p are mutually dis-
joint, and A l An, then #(A)l #(An).
THEOREM 1. If # is a measure on the semi-ring T, then
(1) # is monotone, i.e., if A ET, BET, and A cB, then #(A)#(B),
(2) # is a-additive (also called countably additive or comPletely ad-
ditive), i.e., if A ET, AnET (n== 1,2, . . .) with all An mutually disjol:nt,
and A==r An, then #(A)==r #(An).
Conversely, if the set function #(A) on T is monotone and a-additive,
and if in addition #(0) ==0, then # is a measure on T.
PROOF. The proof of (1) is derived from the defining condition (b)
by setting A 1 ==B and An==0 for n2; (2) follows by observing that
00
#(A) #(An)
1
by (b),
and
p
#(A) #(An)
1
for all p by (c).
Conversely, if #(0) ==0 and # is monotone as well as a-additive on T,
property (a) follows immediately from the monotony. If the hypotheses
of (c) are satisfied, then
p
A- An==A-A1-" .-A p
1
is a a-set  Bk with disjoint terms by sec. 4, Theorem 1. Hence
p p p
#(A) ===#( An+  B k )===  #(An) +  #(Bk)  #(An),
111
so (c) is satisfied. It follows now immediately that if the a-set r An
with disjoint terms is included in A ET, then r #(An)#(A).

Ch. 2, 9 5J
MEASURE ON A SEMI-RING
31
Let now the hypotheses of (b) be satisfied, i.e. A c r An, where
A ET and all AnET. It is no restriction of the generality to assume that
A== An, since we may replace each An by AAnET, and #(A)
 #(AAn) implies #(A) #(An). Then, once more by sec. 4, Theo-
rem 1,
A== An==A1 +(A2-A1)+(A3-A1-A2)+.. .
==A 1 +  B2i+  B 3j + . . . ,
i j
where all sets in the last expression are disjoint, and belong to T.
Hence, using now that  BncBETfor disjoint BnETimplies  #(Bn)
#(B), we obtain
#(A)==#(A1)+  #(B 2i ) +  #(B3j) + . ..
i j
<#(A 1 )+#(A 2 )+#(A 3 )+. . . .
We present some examples.
ExamPle 1. Let X be k-dimensional real number space Rk, con-
sisting of all real points x== (Xl, . . " Xk). Given two points a== (aI, . . .,
ak) and b== (b 1 , . . " b k ) such that ai< b i for i== 1, . . " k, the left half
open interval (aI, b 1 ; . . . ; ak, b k ], i.e., the set of all points X such that
ai<xibi for i== 1, . . " k, is called a cell.
THEOREM 2. The collection T, consisting of all cells and the empty set
0, is a semi-ri1g.
PROOF. Evidently 0ET, and A ET, BET implies ABET. Let now
A ET, BET and B cA. If B==0, then A -B==A ET; and if B==A, then
A -B==0ET; if B is a cell properly contained in A, then, by extending
the faces of B sufficiently far, the cell A is divided by these extended
faces into a finite number of disjoint cells, one of which is exactly B.
In either case A -B is, therefore, the empty set or a a-set consisting of
a finite number of disjoint terms.
THEOREM 3. If #(0) ==0, and #(A) == II= 1 (bi-ai) for the cell A ==
(aI, b 1 ; . . . ; ak, b k ], then # is a measure on the semi-ring T.
PROOF. Evidently #(0)==0, O#(A)<oo for all A ET, and # is mo-
notone on T. It remains to show that # is a-additive on T. Let, for this

32
MEASURE
[Ch. 2, 9 5
purpose, A==r An, whereAET, AnET(n==l, 2, ...) and the sets An
are disjoint. It is no restriction of the generality to assume that neither
A nor any of the sets An is the empty set. Considering, for a moment,
only AI, . . . , A p, and extending all faces of these cells, the cell A is
divided into disjoint cells B1, . . " Bq such that every Ai (i== 1, . . " P)
is the union of some of these Bj, and any Bj included in Ai is included
in none of the cells A k for k =l=-i. Hence, it follows easily from the defi-
nition of # that
q q p
#(A)==#( Bj)== #(Bj) #(A i ).
111
This holds for all p, so #(A)r #(An).
Assume now that #(A) > r #(A n). Then there exists a number cx>O
such that
00
#(A»  #(An)+2cx.
1
(1)
We choose the closed interval F c A such that #(A) <#(F e) +cx, where
Fe is the cell having the same vertices as F, and we also choose open
intervals On and cells Bn, satisfying AncOncBn and #(Bn)<#(An)
+cx/2n for n== 1,2, ... . Then FcA==r An c r On, so that in view
of the Heine-Borel-Lebesgue theorem F is already covered by the union
of a finite number of the open sets On. It follows that the cell Fe is
covered by the union of a finite number of the cells Bn, so (by a similar
argument as above) #(Fe)' #(Bn), where ' denotes summation
over that finite number. Then #(F e)r #(Bn) < r #(A n) +cx, so
#(A)<#(Fe)+cx<r #(An)+2cx, which contradicts (1). Hence #(A)==
r #(An).
There exist some variants of this example. If we restrict ourselves to
all cells with rational vertices, we obtain a semi-ring Tr, and the same
definition for # furnishes a measure on Tr. Similarly, if we consider only
the cells with integer vertices (that is, the coordinates of the vertices
are integers), we obtain a semi-ring T i and a measure on T i .
ExamPle 2. X is an arbitrary non-empty point set; T consists of 0
and all sets containing only one point; #(0)==0 and #(A)== 1 if A con-
tains one point.

Ch. 2, 9 5J
MEASURE ON A SEMI-RING
33
ExamPle 3. X is an infinite point set; r consists of 0, X itself, all
finite subsets of X and their complements; #(O) ==0, #(A) ==n if A con-
tains n points, and #(A)==oo if A is the complement of a finite set.
Note that r is a field, but not a a-field.
ExamPle 4. X is an arbitrary non-empty point set; r consists of all
subsets of X (the empty set included); #(O)==0 and #(A) ==00 for A *0.
Note that r is a a-field.
ExamPle 5. X is an uncountable point set; r consists of all subsets
of X; #(A) ==0 for finite or countable A, and #(A) =00 for uncountable
A.
ExamPle 6. X ==R2 (two-dimensional space of all real points x==
(Xl, X2)) ; r consists of ° and all "horizontal" one-dimensional cells A ==
{(Xl, X2): a<x1b, X2=C}; #(O)=0 and #(A)==b-a.
ExamPle 7. X ==R 2 ; r consists of O, all horizontal cells A =={(X1, X2) :
a<x1b, X2=C}, all vertical cells B=={(X1, X2): X1==a, c<x2d}, and all
sets C consisting of one point; #(O)==0, #(A)=b-a, #(B)=d-c and
#(C) ==0. In order to see that the collection r is a semi-ring, it is suf-
ficient to observe that a horizontal cell A with one point removed is
either a horizontal open interval or the union of such an interval and
another horizontal cell. Since any horizontal open interval is the union
of a countable number of horizontal cells, it follows that the set con-
sisting of A minus one point is a a-set.
ExamPle 8. X ==R1, and g(x) is a real increasing function, defined
and right continuous on R1, i.e., g(Xl)g(X2) for Xl <X2, andg(x+h) ig(x)
as hi 0; r is the semi-ring consisting of ° and all cells; #(O) ==0 and
#(A)==g(b)-g(a) for A=(a, bJ. The proof that # is a measure on r is
exactly the same as in Example 1. Note that the right continuity of
g(x) is used where we choose the closed interval FcA such that #(A)<
/l(Fe) + ex, and also where we choose the open intervals On and the cells
Bn such that AncOncBn and #(Bn)<#(An)+exf2n for n==l, 2, ... .
Note also that # is identically zero if g(x) is constant on Rl, and that
for g(x) ==x we obtain the one-dimensional case of Example 1.

34
MEASURE
[Ch. 2,  5
The thus defined measure # on T is said to be generated by the
function g(x), and we shall prove now, conversely, that any measure #
on T, satisfying #(A)<oo for all A ET, is generated by an increasing
right continuous function g(x) on R1. For this purpose, given the
measure # on T such that #(A) < 00 for all A E T, let
o for x==o
,
g(x) == #(0, xJ for x>O,
-#(x, OJ for x<O.
Then g(X2) -g(X1) ==#(X1, X2J0 for Xl <X2, so g(x) is increasing and
generates the measure #. Furthermore, g(x) is right continuous. In-
deed, if xn!XO and a is an arbitrary finite number satisfying a<xo, let
An==(a, xnJ
for n==O, 1, 2, . . .
and
Dn==(X n +1, x n J==An- A n+1
for n == 1, 2, . . . .
Then A1==Ao+r Dn, so #(A1)==#(Ao)+r #(Dn) by the a-addi-
tivity of #, and since #(A 1 )<00, this implies that
00
 #(Dk)!O
k=n
as
noo.
I t follows, on account of (xo, xnJ == r=n Dk, that
00
g(x n ) -g(xo) ==#(xo, xnJ ==  #(D k )! 0
k=n
as
noo,
i.e., g(x) is right continuous.
Exercises
FURTHER EXAMPLES OF MEASURES
5.1 ) We consider the following generalization of Example 7. Let
X ==R2, and let To be the collection of all (one-dimensional) cells on the
xl-axis. Any set A cR 2 , which can be obtained from some cell AOETO
by rotating the xl-axis around any of its points through an angle
qJ(OqJ<n), will be called a general (one-dimensional) cell. Show that

Ch. 2,  6J
EXTERIOR MEASURE
35
the collection r, consisting of 0, all general cells and all sets consisting
of one point, is a semi-ring. Show also that if #(0)==0, #(A)==b-a for
any general cell A obtained by rotation from A O =={(X1, 0) : a<x1b},
and #(C)==O if C consists of one point, then # is a measure on r.
5.2) We consider the following generalization of Example 8. Let
X ==R2, and let the real function g(X1, X2) be defined on R2, such that
L1g(a, b; c, dJ ==g(b, d) -g(a, d) -g(b, c) +g(a, c)O for ab, cd,
and
L1g(a, b+h; c, d+kJ !L1g(a, b; c, dJ as h! 0, k! o.
Let, finally, r be the semi-ring consisting of 0 and all cells in R2, and
let #(0) ==0, #(A) ==L1g(a, b; c, dJ for A == (a, b; c, dJ. Show that # is a
measure on r. Consider the case that g(X1, x2)==f1(x1)f2(x2) where f1
and f2 are increasing and right continuous on R1. Formulate and prove
the generalization to the space R3.
5.3) Show that if A c X, and A is not empty, then there exists a
measure # on the semi-ring of all subsets of X such that #(A)==#(X)== 1.
6. Exterior Measure
Given the measure # on the semi-ring r of subsets of .X and given
the set 5 c X, the set 5 is said to be sequentially covered by r if there
exists a a-set r An such that 5cr An. Note that if r is a field,
then any 5 c X is sequentially covered by r since in this case X E r.
If in Example 2 of the preceding section (where r consists of 0 and all
sets containing one point, #(0)==0 and #(A)== 1 for all other A Er) the
entire set X is uncountable, then any uncountable subset of X fails to
be sequentially covered. If in Example 6, the example of the horizontal
one-dimensional cells in R2, the set 5 c R2 contains points (Xl, X2) for
an uncountable number of values of X2, then 5 fails to be sequentially
covered. In Example 1, the example of the "usual" measure in Rk, the
set X ==Rk itself, and therefore every subset 5 c Rk, is sequentially
covered.
DEFINITION. Given the set 5 eX, sequentially covered by r, the ex-
terior measure #*(5) of 5 is defined by #*(5)==inf  #(An) for all possi-

36
MEASURE
[Ch. 2,  6
ble sequential coverings  A n  5 by sets An E T. If 5 fails to be se-
quentially covered by T, the exterior measure #*(5) of 5 is defined by
#*(5)==00. We shall say that #* is generated by (X, T, #).
THEOREM 1. (1) #*(0)==0, and 0#*(5)00 for every SeX.
(2) #* is a-subadditive, i.e., #*(r 5n)r #*(5 n ).
(3) #* is monotone, i.e., if 5 e T, then #*(5)#*(T).
(4) If A ET, then #*(A)==#(A).
PROOF. (1) #*(5) is either +00 or the greatest lower bound of a set
of non-negative numbers, so 0#*(5)00. If 5==0, then 5e An with
A n =0 for all n; hence #*(S) #(An)=O, and this shows that #*(0)==0.
(2) Let  #*(Sn) <00 (otherwise there is nothing to prove), and let
8>0. Then, for n= 1,2, . . ., there exist AnjET (j== 1,2, . . .) such that
Sne Ani
i
and
 #(Anj) <#*(5 n ) +8/2 n .
i
Hence,
 5ne Ani
ni
and
 #(Anj)< #*(Sn)+8.
ni
This shows that
#*( Sn)< #*(Sn)+8
holds for all 8>0;
it follows that
#*( Sn) #*(5 n ).
(3) Let 5 e T, and let T be sequentially covered (otherwise there is
nothing to prove). Then any sequential covering of T is also a se-
quential covering of 5, so #*(S)#*(T).
(4) If A ET, and L An is any sequential covering of A by sets AnET,
then #(A) #(An) by property (b) in the definition of a measure, so
#(A)#*(A). On the other hand #*(A)#(A), since A+0+0+. .. is a
sequential covering of A. Hence #*(A)==#(A).
The next theorem will be of importance for the theory of the Stiel-
tjes-Lebesgue integral (cf. sec. 17, Lemma ex). Let T and T1 be semi-
rings of subsets of X, let # be a measure on T and #1 a measure on T1.

Ch. 2, 9 6J
EXTERIOR MEASURE
37
THEOREM 2. (X, T, #) and (X, T 1 , #1) generate the same exterior
measure in X if and only if #*#1 on T1 and #i# on T.
PROOF. We have #*==# on T and #i==#l on T1 by Theorem 1 (4).
Hence, if the exterior measures #* and #i are the same for all subsets of
X, then #*==#i==#l on T1 and #i#*==# on T.
Let now, conversely, #*#1 on T1 and #i# on T, and let 5 eX. We
wish to prove first that #*(5)#i(5); we may assume therefore that
#i(5)<oo. Then, given e>O, there exists a covering  Bn of 5 by sets
BnET1, such that
oo>#i(5) >  #l(Bn) -e?;:;  #*(B n ) -e.
Hence, all #*(B n ) are finite, and it follows that for n== 1, 2, . . . there
exist sets AnjET (f==l, 2, ...) such that Bne1Anj and
#*(B n »  #(Anj)-ef2n.
j
Then 5 e  Bn e nj Anj and
#i(5) >  #*(Bn)-e>  #(Anj)-2e,
nj
which implies that #i(5) >#*(5) -2e. This holds for all e>O, so #*(5)
#i(5). Similarly, using now that #i# on r, we obtain #i(5)#*(5). It
follows that #*==#i in X.
If X==R1 (the real line), Tthe semi-ring of all cells with #(A)b-a
for A==(a, bJET, and T 1 the semi-ring of all cells having integer end-
points with #l(A) ==b-a for A == (a, bJ ET1, then #* ==#1 on T 1 . If, how-
ever, A is for example the cell (0, !JET, then #*(A)==#(A)==!, but
#i(A)==I. Hence #i=l=-#*, which shows that the condition #*==#1 on T1
alone is not sufficient to entail equality of #* and #i in X.
We make one final remark. In Theorem 1 (2) it was proved that #*
is a-subadditive, and we will show now by an example that this cannot
be replaced by the statement that #* is always a-additive (although #*
may be a-additive in some particular examples). Let, once more, X ==R1
and let T 1 be the semi-ring of all cells having integer endpoints with
1l1(A)==b-a for A==(a, bJET1. Then, if 51==(0, !J and 5 2 ==(!, IJ, we
have #i(51) ==#i(5 2 ) == 1, and also #i(5 1 +52) == 1. Hence, although 51

38
MEASURE
[Ch. 2,  7
and 52 are disjoint, #i(5 1 +5 2 ) <#i(51) +#i(5 2 ). This shows that #i is
not even additive for a finite number of disjoint sets.
Exercises
THE SUBSEMI-RING OF ALL SETS OF FINITE MEASURE
6.1) Given the measure # on the semi-ring T, let T1 be the sub-
collection of all A ET satisfying #(A)<oo. Show that T1 is a semi-ring.
Denote the measure #, restricted to T1, by #1, and show that (X, T, #)
and (X, T1, #1) generate the same exterior measure.
7. Measurable Sets
In the preceding section, instead of extending the initial measure #
on the semi-ring T step by step (first to finite or countable unions of
sets of T, and in subsequent steps to more complicated sets), we have
immediately assigned to every subset 5 of X the exterior measure
#*(5). The set function #* is indeed an extension of # (in view of #*==#
on T), but #* cannot claim to be called a measure since it is in general
not additive but merely subadditive (as shown by the example in the
last paragraph of the preceding section). We have, therefore, been over-
shooting the mark, and the best we can do is to ask whether there
exists at least a collection of subsets of X, including T, such that #*
behaves like a measure on this collection. In addition, it is a mild re-
quirement that the collection shall be at least a ring. Assume, for the
moment, that such a ring exists, and let E be an arbitrary set in the
ring. Any A ETis also in the ring since the ring includes T, so any A Er
of finite measure is surely in the ring. Then the disjoint sets AE and
AEe==A-E are sets of the ring, hence #(A)===#*(AE)+#*(AEe) on
account of #*(A)===#(A) and the additivity of #* on the ring. It follows
that the ring is included in the collection A of all sets E satisfying the
condition that #(A)==#*(AE)+#*(AEe) holds for all A ET of finite
measure. We shall prove in the present section that the collection A
itself is a ring (even a a-field) such that #* is a measure on A, and it
will follow then from what we have now observed tht A is the largest
ring of subsets of X on which #* is a measure.

Ch. 2,  7J
MEASURABLE SETS
39
DEFINITION. The set E eX is called #-measurable (or, shortly, measur-
able if confusion is excluded) whenever
#(A) ==#*(AE) + #*(AEe)
for all A ET such that #(A) <00.
(1)
THEOREM 1. The set E is #-nteasurable if and only if
#*(5) ==#*(SE) +#*(SEe)
for all
SeX.
(2)
PROOF. Evidently, (2) implies (1). Assume, therefore, that (1) is
satisfied, and let SeX be arbitrary. Since #*(S)#*(SE)+#*(SEe) by
the subadditivity of #*, we have only to prove the inverse inequality.
For #*(5) ==00 the inverse inequality holds; we may assume, therefore,
that #*(5)<00. Then, given 8>0, there exists a a-set  An, covering
5, such that  #(An)<#*(S)+8. This shows first that all the sets AnET
are of finite measure, and furthermore, in view of SE e  A nE and
SEe e  An Ee , that
#*(SE)#*( AnE) #*(AnE)
and similarly
#*(SEe) #*(AnEe).
Hence,
#*(SE)+#*(SEe) {#*(AnE)+#*(AnEe)}
== #(An)<#*(S)+8,
where we have used that (1) holds for each AnET separately. Since
8>0 is arbitrary, the desired conclusion #*(SE) + #*(SEe)#*(s) follows.
The theorem shows that condition (2) is equivalent to the apparently
weaker condition (1), which serves as the definition of #-measurability
for the set E. In other words, we may as well use condition (2) as the
definition of #-measurability for the set E. This has advantages in the
following situation. If #*(5) is a set function, defined for all subsets 5
of X, and satisfying the first three conditions in sec. 6, Theorem 1 (i.e.
/1* is non-negative, a-subadditive and monotone, and #*(0)==0), then
/1* is called an exterior measure, irrespective of whether #* is generated
by a measure # on some semi-ring T or not. If #* is not generated by

40
MEASURE
[Ch. 2, 9 7
a measure #, the condition (1) is not available in order to select the
measurable sets from among all subsets of X, but now (2) is the "cor-
rect" definition for a set E to be measurable. The characterization of
measurable sets by means of condition (2) is due to C. CARATHEODORY
( 1 918, [1 J, p. 246).
In order to verify that a given set E e X is measurable, it is sufficient
to prove that #*(S)#*(SE)+#*(SEe) for all SeX, since the inverse
inequality is always true on account of the sub additivity of #*. Hence,
it is even sufficient to prove that #*(S)#*(SE)+#*(SEe) for ailS
such that #*(5)<00.
Let #* be an exterior measure in X, either generated by (X, T, #) or
not, and let A be the collection of all measurable sets.
THEOREM 2. If EEA, then EeEA. Furthermore, XEA and 0EA.
PROOF. Since the definition of measurability is symmetric in E and
Ee, the statements that EEA and EeEA are equivalent. If E==X, then
SE==SX==S and SEe===0. Hence, E==X is measurable,. since #*(5)==
#*(5)+#*(0). It follows that 0==xe is also measurable.
LEMMA cx. If E 1 EA and E2EA, then E1 +E 2 EA. If, in addition, E1
and E2 are disjoint, then #*{S(E 1 +E 2 )}===#*(SE 1 ) + #*(SE 2 ) for all 5 eX,'
in particular #*(E 1 +E2) =#*(E1)+#*(E2).
PROOF. Not yet assuming that E1E2==0, we have E1+E2==E1+
EE2 and (E1+E2)e==EE, so, using first that E 2 is measurable and
then that E1 is measurable,
#*(S)#*{S(E1 +E 2 )}+#*{S(E 1 +E 2 )e}
#*(SE1) +#*(SEE2)+#*{S(E1 +E 2 )e}
=#* (SE 1) + #*(SEE2) + # *(SEE)
== # * (SE 1) + # *(SE) ==# * (5).
Hence, all inequalities are equalities, and the first equality shows then
that E1 +E 2 EA.
One might believe for a moment that the second equality shows that
#*{S(E1 +E 2 ) }===#*(SE 1 ) +#*(SEE2).

Ch. 2, 9 7J
MEASURABLE SETS
41
This is not true, however, since #*{5(E 1 +E 2 )e} may be +00. Never-
theless, the stated relation itself is true, but this follows by observing
that the equalities above continue to hold if 5 is replaced by 5(E1 +E 2 ).
Hence, if E1E2==0, i.e. if EE2==E2, then
#*{5(E1 +E2)}==#*(5E1) +#*(5E 2 )
for all SeX.
COROLLARY. If En EA forn==l, "', p,then1 EnEAandill EnEA.
If, in addition, all En are disjoint, then #*(5 l En)== l #*(5E n ) for all
SeX,' in particular #*(l En)==l #*(E n ). The exterior measure #* is
therefore finitely additive on A.
PROOF. The only assertion which is not immediately evident is that
III En EA . This follows from II) En=(l E)C in view of Theorem 2.
THEOREM 3. If E 1 EA and E2EA, then E1-E2EA. Hence, combining
this result and the results of the preceding lemma, A is seen to be a ring.
PROOF. Follows from E1-E2=E1E.
THEOREM 4. If EnEA for n== 1, 2, . . ., then r EnEA. If, in ad-
dition, all En are disjoint, then #*(5 r En)==r #*(5E n ) for all SeX;
in particular
00 00
#*( En)== #*(E n ).
1 1
PROOF.
00
 En==E1+E2E+E3EE+...,
1
where the sets on the right are measurable and disjoint; we may, there-
fore, assume immediately that all En are disjoint. Then, for any 5 eX
and any index p, we have
p p 00 00
5( En)e==5 IT E::::>5 IT E==5( En)e,
111 1
hence
p p
#*(5)==#*(5  En)+#*{5( En)e}
1 1
p 00
 #*(5En)+#*{5( En)e}.
1 1

42
l\<IEASURE
[Ch. 2, 9 7
Letting Pi 00, and making use of the subadditivity of #*, we obtain
00
00
#*(5) #*(5En)+#*{5( En)e}
1 1
00
00
#*(S  En)+#*{5( En)e}#*(5).
1 1
It follows that the last inequalities are equalities, so r EnEA. Fi-
nally, replacing 5 by 5 r En, we obtain #*(5 r En)==r #*(5E n ).
It is evident now, from what has been proved in the Theorems 2,3,4,
that A is a a-field and that #* is a measure on A (indeed, #*(0) ==0, and
p* is monotone and a-additive on A).
In the remaining part of this section we assume again that #* is
generated by (X, r, #), and we show first that the semi-ring r, the
domain of definition of #, is a sub collection of A. Note that although
we have already proved that #*==# on r, this does not automatically
imply that reA.
THEOREM 5. If AoEr, then AOEA.
PROOF. We have to prove that #(A)==#*(AAo)+#*(AAg) for all
A Er of finite measure, and, on account of the sub additivity of #*, it
is sufficient to prove that
#*(AA o ) +#*(AAg)#(A)
for all A Er.
Now, AAoEr (so #*(AAo)==#(AAo)), and AAg==A-Ao==A-AA o is a
a-set  Bn with disjoint terms (so #*(AAg)==#*( Bn) #*(B n )==
 #(Bn)). It follows that
#*(AAo)+#*(AAg)#(AAo)+  #(Bn)==#(A)
by the a-additivity of # on r. This is the desired result.
Since A::) rand #*(A) ==#(A) for all A Er by sec. 6, Theorem 1 (4),
we may now draw the conclusion that the measure #* on A is an ex-
tension of the measure # on r. Furthermore, as already observed in the
beginning of this section, the a-field A is the largest ring (and hence the

Ch. 2, 9 7J
MEASURABLE SETS
43
largest a-field) on which #* is a measure. In view of these facts, no
confusion will arise if we denote, for any EEA, the number #*(E)
simply by #(E), and if we call this number simply the measure of the
set E. For any E not belonging to the a-field A we shall, however, con-
tinue to write #*(E), and continue to call this number the exterior
measure of the set E. Of course, the same can be done if #* is not gener-
ated by a measure # on some semi-ring T.
Given a semi-ring T1, satisfying Te T1 e A (it is possible, therefore,
that T1==A), the extended # is a measure on Tl, and we shall denote it
for a moment by #1. The question may be raised what will happen if
the extension procedure is repeated, beginning now with the measure
pIon T 1 . It would be rather unsatisfactory if the exterior measures #*
and #i, generated by (X, T, #) and (X, T1, #1) respectively, should turn
out to be different. We shall prove that this is not so.
THEOREM 6. Given the semi-ring T 1 such that Te T1 eA, and denoting
.the extended measure # on T1 by #1, the exterior measures #* and #i, gener-
ated by (X, T, #) and (X, T1, #1) respectively, are identical. Hence, the
,corresponding collections of measurable sets are also identical.
PROOF. We have #*==#1 on T1 (since T 1 eA, and since the extended
Jl equals #* on A), and also #i===#l on T1, so #i==# on T. Combining the
results that #*==#1 on T 1 and #i==# on T, it follows from sec. 6, Theo-
rem 2 that #i==#* in X.
We finally show that the question whether a given set E is measur-
able or not may already be settled by considering a class of simple
subsets of E.
THEOREM 7. The set E is measurable if and only if EA is measurable
jor all A E T of finite measure.
PROOF. It is evident that EEA implies EA EA for all A ET. Let now,
conversely, EA EA for all A ET of finite measure, hence
#(A) ==#*(AAE) +#*{A (AE)e}
by the definition of measurability. Observing that AAE=AE and

44
MEASURE
[Ch. 2,  7
A(AE)e==A(Ae+Ee)=AEe, we obtain
#(A) ==#*(AE) +#*(AEe)
for all A Er
of finite measure, and this shows that E is measurable.
Exercises
REGULAR EXTERIOR MEASURE
7. 1) Let # * be an exterior measure in X, and let A be the corre-
sponding collection of measurable sets; the measure #* on A will be
denoted by #1, and the exterior measure generated by (X, A, #1) by
#i. Show that #*#i, i.e., #*(S)#i(S) for all S eX. If #i=#*, i.e., if
# i (S) == # * (S) for all SeX, then # * is sometimes called a regular ex-
terior measure. Show that #* is regular if and only if there exists a semi-
ring r and a measure # on r such that #* is the exterior measure gener-
ated by (X, r, #).
7.2) Show that there exists a non-regular exterior measure.
CONTRACTED EXTERIOR MEASURE
7.3) Let #* be an exterior measure in X, and let the contracted ex-
terior measure #(S) be defined for all SeX by #(S)==sup #*(T) for
all TeS satisfying #*(T)<oc>. Show that # is an exterior measure in
X. The sets which are measurable with respect to #* or # are called
#-measurable or #e-measurable respectively. Show that E is #e-measur-
able if and only if E is #-measurable.
7.4) Let (X, r, #) generate the exterior measure #* and the con-
tracted exterior measure #, and let A be the collection of all #-measur-
able sets (hence, A is also the collection of all #e-measurable sets). The
measures #* and # on A will be denoted be # and #e respectively. The
definition of #e implies that, for any E EA, we have
#e(E)==sup #*(T) for all TeE
such that #*(T)<oc>. Show that
#e(E) ==sup #(F) for all measurable FeE
such that #(F)<oc>.

Ch. 2, 9 7J
MEASURABLE SETS
4S
7.5) Let # be a measure on the semi-ring T, and #1 a measure on the
semi-ring T 1 . Let (X, T, #) generate the exterior measure #* and its
contraction #, and, similarly, let (X, T1, #1) generate #i and #ie. As-
sume that #ie(A)#(A) for all A ET satisfying #(A) <00, and #(B)
#l(B) for all BET1 satisfying #l(B) <00. Show that #ie#* and ##i
in X. In particular, show that if #*(5) and #i(5) are both finite, the
#*(5) ==#(5) ==#ie(5) ==#i(5).
RELATIVE EXTERIOR MEASURE
7.6) Let #* be an exterior measure in X, let YcX be a #-measur-
able set, and let #0 be the restriction of #* to the subsets of Y (i.e.,
#0(5) ==#*(5) if 5 c Y, and #0(5) is not defined if 5 is not a subset of Y).
Show that #0 is an exterior measure in Y such that the #o-measurable
sets are exactly the #-measurable subsets of Y.
7.7) Let # * be an exterior measure in X, let Y c X be a set which is
not #-measurable, and let #0 be the restriction of #* to the subsets of
Y. Show that #0 is an exterior measure in Y such that any #-measur-
able subset of Y is #o-measurable. Show also that there exists a #0-
measurable set which is not #-measurable.
A MEASURE IN METRIC SPACE
7.8) Let A be the a-field consisting of all sets of the first category
and their complements in the complete metric space X; let #(E) ==0 if
E is of the first category, and #(E) = 1 if Ee is of the first category.
Show that # is  measure on A, #*(E)== 1 for all sets E of the second
category, and show also that th sets of A are the only #-measurable
sets.
CHARGES (FINITELY ADDITIVE MEASURES)
The following sequence of exercises is devoted to the theory of charges
(also called finitely additive measures or contents). Assume, in these exer-
cises, that T is a semi-ring of subsets of X such that, for A ET, BET
and Be A, the difference A - B is a a-set consisting of a finite number
of disjoint non-empty terms. Any a-set consisting of a finite number
of non-empty terms will be called an f-set (with respect to T). The set
function v on T is called a charge on T whenever v(0) ==0, v is monotone

46
l\IEASURE
[Ch. 2, 9 8
and finitely additive (Le., if AET, and A==l An, where AnET (n==
1, . . ., p) and all A n are disj oint, then v(A) == l v(A n)). Evidently any
measure on T is a charge on T.
7.9) Show that finite unions and finite intersections of I-sets are 1-
sets, and the difference of two I-sets is an I-set. Show also that any
I-set may be written as an I-set with disjoint terms.
7.10) Show that if v is a charge on T, and A e l An where A ET
and AnET (n== 1, . . " P), then v(A)l v(An).
7.11) If SeX is included in at least one I-set, then the exterior
charge v*(5) of S is defined by
v*(S)==inf  v(An)
for all I-sets  An::> S;
if S is not included in any I-set, then v*(5)==00. Show that v*(0)==0,
v* is monotone and finitely subadditive (Le., v*(l Sn)l v*(Sn))'
Show also that v* (A) ==v(A) for all A E T.
7.12) Show that (X, T, v) and (X, T1, VI) generate the same exterior
charge in X if and only if V*V1 on T1 and vi v on T.
7.13) If (X, T, v) generates the exterior charge v*, then the set E eX
is called v-measurable whenever v(A)==v*(AE)+v*(AEe) for all A ET
satisfying v(A)<oo. Show that E is v-measurable if and only if v*(S)==
v*(SE)+v*(SEe) for all SeX. If v* is an exterior charge in X which is
not generated by some triple (X, T, v), then the last condition is used
as the definition of v-measurability.
7.14) Let v* be an exterior charge in X. Show that the collection A
of all v-measurable sets is a field, and v* is a charge on A. If v* is gener-
ated by (X, T, v), show that TeA. Show also that if T1 is a semi-ring
of the same type as T such that TeT1 eA, and if the charge v* on T 1
is denoted by VI, then (X, T, v) and (X, Tl, VI) generate the same ex-
terior charge. Finally, show that E is v-measurable if and only if EA
is v-measurable for all A E T satisfying v(A) < 00.
7. 15) Show that the analogs of the statements in the Exercises 1, 2,
3, 4, 5, 6, 7 hold for charges.
8. Null Sets, and Sequences of Measurable Sets
Let #* be an exterior measure in X. Any set EeX, satisfying #*(E)
==0, is called a null set. Obviously, any subset of a null set is a null set.

Ch. 2,  8l
SEQUENCES OF MEASURABLE SETS
47
THEOREM 1. A ny null set is measurable.
PROOF. Let #*(E)==O and 5 eX. Then #*(5E) ==0, and #*(5)
#*(5Ee) since 5=>5Ee. Hence #*(5)#*(5E)+#*(5Ee), and this shows
that E is measurable.
COROLLARY. (1) If #* is generated by (X, T, #), and E eX has the
property that #*(EA) ==0 for all A ET of finite measure, then E is measur-
able, and either #(E) ==0 or #(E) ==00.
(2) If # is a measure on the semi-ring T, and either #(A)==O or
#(A) ==00 for any A E T, then every set E e X is measurable, and either
#(E) ==0 or #(E) ==00.
Proof. (1) Since, by Theorem 1, EA is measurable for all AET of
finite measure, the set E itself is measurable (by sec. 7, Theorem 7).
Let now F be measurable and of finite measure, so F e  A n where all
AnET and  #(An)<oo. Then
#(EF)#( EAn) #(EAn) ==0,
hence # (EF) ==0. If, therefore, #(E) is finite; the choice F ==E yields
fl(E) ==0.
(2) Any A E T of finite measure is now a null set, so it follows, if E eX
is arbitrary, that EA is a null set for all A ET of finite measure. Hence,
by (1), E is measurable, and either #(E) ==0 or #(E) ==00.
If all points of a set 5 c X, except possibly the points of a null set
Sl e5, have some property (P), we shall say that (P) holds almost
everywhere on 5, or also that almost every point of 5 has the property
(P). If the characteristic functions XE(X) and XF(X) of the sets E and F
are equal almost everywhere on X, we shall say that E and Fare
almost equal sets, and we denote this by E"'-IF.
'"rHEOREM 2. (1) E"'-IF if and only if #(E-F)==#(F-E) ==0.
(2) If E"'-IF, then #*(E)==#*(F)==#*(EF).
(3) If E "'-IF and F "'-IG, then E "'-IG; the relation "'-I is therefore an
equivalence relation on the collection of all subsets of X.
PROOF. (1) Follows immediately from the definition of the relation
EF.

48
MEASURE
[Ch. 2, 9 8
(2) If Ef'./F, then
#*(E)#*(EF) +#*(E -F) ==#*(EF)#*(E),
hence #*(E)=#*(EF). Similarly #*(F)==#*(EF).
(3) Let Ef'./F and Ff'./G, so
# (EFe) ==#(Ee F) = #(FGe) ==#(FeG) ==0.
The relation
E -G==EGe==EFGe+EFeGe cFGe+EFe
implies therefore that #(E -G) ==0. Similarly #(G- E) ==0. Hence E f'./G.
In the remaining part of this section we assume that # is a measure
on the a-ring A. If the extension procedure is applied to (X, A, #), we
obtain the a-field All of all #-measurable sets, and the measure #* on
All is again denoted by #. The results which follow for # on A are then
automatically valid for the extended # on All' We stress the fact, how-
ever, that we do not assume that A==AIl holds in what follows.
Given the sequence EnEA, the sets lim sup En and lim inf En are
also in A, since A is a a-ring. Furthermore, we know already that
lim En== En
if En is ascending,
and
lim En== II En
if En is descending.
THEOREM 3. (1) If all EnEA, En is ascending, and E=lim En, then
#(E)=lim #(En).
(2) If all EnEA, En is descending, E==lim En, and #(En)<oc> for at
least one value of n, then #(E)==lim #(En).
PROOF. (1)
#(E) ==#( En) ==#{E 1 + (E 2 -E 1 ) + (E 3 -E 2 ) + . . .}
=#(E 1 )+#(E 2 -El)+#(E 3 -E 2 ) + . . .
==lim{p(E1)+#(E 2 -E1)+. . . +#(En- E n-1)}
=lim #{E1+(E 2 -E1) + ... + (En- E n-1)}
=lim #(En).

Ch. 2,  8J
SEQUENCES OF MEASURABLE SETS
49
(2) If #(Em)<oo, then #(En)<oo for all nm, so #(E)<oo since
E cE n for all n. The sequence Em-En (m fixed such that # (Em) <00,
n variable) is then ascending, so (by (1))
#(Em) -#(E) ==#(Em -E) ==#{lim (Em-En)}
==lim #(Em-En)==#(Em)-lim #(En),
hence #(E) ==lim #(E n).
The example where En is the set {x: n<x<oo} in R1, and # is the
usual measure in R1, shows that the restriction #(En) <00 for some n
in part (2) of the theorem is essential.
THEOREM 4. Given the sequence EnEA, we have
#(lim inf En)lim inf #(En).
If, in addition, #(En+ E n+1+.. .)<00 for at least one value of n, then
#(lim sup En)lim sup #(En).
In particular, if the sequence of sets EnEA is convergent and #(En+
En+1+.. .)<00 for at least one value of n, then
#(lim En)==lim #(En).
PROOF. The set lim inf En= ;;=1 nr=n Ek is the limit of the as-
cending sequence P n== nr=n E k, and for each fixed n the inclusion
PncEk (kn) implies #(Pn)lim inf #(E k ) as koo. Hence
#(lim inf En)=lim #(Pn)lim inf #(En)
by part (1) of the preceding theorem.
The set lim sup En== rr;;=l r=n Ek is the limit of the descending
sequence Qn==r=nEk, and for each fixed n the inclusion Qn::::>Ek
(k):-n) implies #(Qn)lim sup #(Ek) as koo. Hence, if #(Qn)==
/-l(En+En+1+" .)<00 for at least one value of n, then
#(lim sup En)==lim #(Qn)lim sup #(En)
by part (2) of the preceding theorem.
The last statement follows easily by combining the inequalities
#(lim En)lim inf #(En)
and
#(lim En)lim sup #(En).

50
MEASURE
[Ch. 2,  9
If, in R1, the sets En are defined by
En==(0,2J for n== 1,3,5, . . .
and
En== (1, 3J for n==2, 4, 6, . . "
then lim inf En==(I, 2J and lim sup En==(O, 3J. Hence, if # is the usual
measure in R1, we have
# (lim inf En) <lim inf #(En) and # (lim sup En) >lim sup #(En),
showing that the inequalities may be proper.
Another example is as follows. Let, in two-dimensional space R2,
the set En (n==l, 2, ...) be the cell (0, n- 1 ; 0, nJ. Then limEn==0.
Furthermore, the usual measure #(En)== 1, and #(lim En) ==#(0) ==0, so
#(lim En) *lim #(En). This is in accordance with
#(En+En+1 +. . .)== 1 +(n+ 1)-1+(n+2)-1+. . . ==00 for all n.
Exercises
CHARGES
8.1) Show that the analogs of the statements in the Theorems 1
(with corollary) and 2 hold for charges (cf. the paragraph preceding
Exercise 7.9 for the definition of a charge).
9. Approximation Theorems, and the Sum Measure of two
Measures
We assume in this section that # is a measure on the semi-ring T,
and that (X, T, #) generates the a-field A of all #-measurable sets. The
extension of # onto A is again denoted by #. In the first theorem we
give still another characterization of measurable sets, in the form of an
inclusion theorem.
THEOREM 1 (INCLUSION THEOREM). The set E is measurable if and
only if, given any E>O, there exist measurable sets E 1e and E 2e such that
E 1e c E C E 2e and #(E 2e -E 1e ) <E.
PROOF. If E is measurable, we may choose E 1e ==E 2e ==E. Assume
now, conversely, that, given any 8>0, there exist measurable sets E 1e

Ch. 2, 9 9J
APPROXIMATION THEOREMS
51
and E 2e such that E 1e cEcE 2e and #(E 2e -E 1e )<8. Then, if SeX is
arbitrary, we have
#* (5) # * (5E) + # *(5Ee)# *(5E 2e ) + # * (5Ee)
# * (5E Ie) + # *{5(E 2e - E 1e)}+ # * (5Ee)
==#*(5) +#*{5(E2e-E1e)}#*(5) +8,
hence #*(5E) +#*(5Ee) ==#*(5), and this shows that E is measurable.
We shall give names to some special types of sets. Any set 0== r An,
where all AnET, has been called already a a-set, and by sec. 4, Theo-
rem 1 countable unions and finite intersections of a-sets are a-sets. Any
set 0 6 = nr On, where all On are a-sets, will be called a a6-set. Evi-
dently, all a-sets and a6-sets are measurable.
The measurable set E is said to be of a-finite rneasure if E c r En,
where En is measurable and #(En)<cx> for n= 1,2, . . . . In that case
E==r Fn, where all Fn==EEn are measurable and of finite measure.
We may even assume, if desired, that all F n are disjoint, since other-
wise  F n=F1 +F2F+F3FF+. . . . If E is of a-finite measure, then
Ecr Ak, where AkET and #(Ak)<cx> for k==l, 2, ... . Indeed, if
E c r En with #(En) <CX> for all n, then En c j Anj with AnjET and
#(A nj) < CX> for i == 1, 2, . . . . Hence E c nj A nj with #(A nj) < CX> for
n, i == 1, 2, . . . . If the whole set X itself is of a-finite measure, then the
measure # is called a a-finite measure. Similarly, the measure # is called
a finite measure whenever #(X)<cx>.
THEOREM 2 (COVERING THEOREM). (1) If the set 5 c X is sequential-
ly covered by T, then #*(5) ==inf #(0) for all a-sets 0 => 5.
(2) Given the set 5 eX, there exists a measurable set E => 5 such that
/1*(5)=#(E). Any set E satisfying these conditions is called a measurable
cover of 5, and, if #*(5)<cx>, the measurable cover of 5 is uniquely de-
termined modulo null sets (i.e., any two measurable covers of 5 are then
almost equal).
(3) Given the measurable set E of a-finite measure, there exists a a6-
set 06=rrr On such that the sequence of a-sets On is descending, E e0 6,
and #(06-E)=0 (the sets E and 06 are, therefore, almost equal).
PROOF. (1) Let 5c An=O, where all AnET. Since the a-set 0

52
MEASURE
[Ch. 2,  9
may be written as a a-set  Bn with disjoint terms, we have  #(Bn) ==
#(O) #(An), and it follows that in the definition of #*(5) which says
that #*(5)==inf  #(An) for all sequential coverings  An of 5 we may
restrict ourselves to disjoint sequential coverings  Bn==O. Hence
#*(5)==inf  #(Bn)==inf #(0)
for all a-sets 0  5.
(2) If #*(5) ==00, the whole set X is a measurable cover of 5, since
in this case #*(5)==#(X)==00. Assume, therefore, that #*(5)<00. By
(1) there exists a sequence of a-sets 0;", each 0;" covering 5, such that
#(0;")-#*(5)<n- 1 for n=l, 2, ... . Hence, if On==l1f=l 0; (so 5c
OncO;"), the sequence of a-sets On is descending, each On covers 5,
and 0#(On)-#*(5)<n-1. It follows that the a6-set 06== 111 On satis-
fies 5 C 0 6 C On for all n, and
0#(06) -#*(5)#(On) -#*(5) <n- 1
for all n,
so #(0 6 )==#*(5). This shows that 06 is a measurable cover of 5. Still
assuming that #*(5)<00, let E be another measurable cover of 5. Then
E0 6 is measurable, and the relations 5 C E0 6 C 0 6 and 5 C E0 6 C E im-
ply that
#* (5) ==#(E0 6 ) == #(0 6 ) ==#(E).
Hence #(06-E)==#(06)-#(E06)==0, and similarly #(E-0 6 )==0. The
sets E and 06 are, therefore, almost equal.
(3) Given the measurable set E of finite measure, the proof of (2)
shows the existence of a a6-set 0 6 == 111 On such that the sequence of
a-sets On is descending, Ec06, and #(06)==#(E). Hence #(06-E)==
#(0 6 ) -#(E) =0.
If #(E) ==00, but E is of a-finite measure, then E ==  Ej, where all
Ej are measurable and of finite measure. Corresponding to each Ej
there exists, therefore, a descending sequence of a-sets Ojn such that
EjcOjn and
# (Ojn -Ej) ==#(Ojn) -#(Ej) < (n. 2 i )-1
for n== 1, 2, . . . .
Then On== j Ojn is also a descending sequence of a-sets, such that
EcO n and
#(On-E)< (n'2 i )-1==n- 1
j
for n == 1, 2, . . .

Ch. 2, 9 9J
APPROXIMATION THEORE1\1:S
53
(observe that On-E ==OnEe== j OjnEe e j OjnEj== j (Ojn-Ej)). It
follows that the a6-set 0 6 === nl On satisfies E e 06 and #(06-E) ==0.
We conclude this section with a theorem on the sum measure of
two measures, which will be of importance in the discussion of the
Radon-Nikodym theorem (sec. 32).
THEOREM 3. If #1 and #2 are measures on the semi-ring T, then #=
#1 +#2 is a measure on T, and the corresponding exterior measures in X
satisfy #*==#i +#2. If the set E eX is #l-measurable as well as #2-
measurable, then E is #-measurable. The converse of this assertion holds
if, for all A ET, the numbers #l(A) and #2(A) are simultaneously finite
or infinite.
PROOF. It is evident that #(0)==0, and that # is monotone and a-
additive on T, i.e., # is a measure on T. Furthermore, #(0)==#1(0)+
#2(0) for any a-set O. In order to prove now that #i(5)+#2(5)#*(5)
for any set 5, we may assume that #*(5)<00. Then #*(5)=inf #(0)
for all a-sets 0:::> 5, hence
#*(5) ==inf {#1(0) +#2(0)}inf #1(0) +inf #2(0) ==#i(5) +#2(5).
For the proof of the inverse inequality we may assume that #i(5)+
#2(5) < 00. Then, given e >0, there exist a-sets 01 and O 2 , covering 5,
such that
#1(01)<#i(5)+e
and
#2(0 2 ) <#2(5) +e.
It follows that the a-set 0==0 1 0 2 covers 5; hence
#*(5)#(0) ==,U1(0) + #2(0)#1(01) + #2(0 2 ) <#i (5) + #2(5) + 2e,
and the desired inequality is obtained by observing that e>O is arbi-
trary.
If E is #l-measurable as well as #2-measurable, and if 5eX is arbi-
trary, then
#*(5E) +#*(5Ee) ==#i(5E) +#2(5E) +#i(5Ee) +#2(5Ee)
=#i(5) +#2(5) ==#*(5),
hence E is #-measurable.

54
MEASURE
[Ch. 2, 9 9
Finally, assuming now that for all A ET the numbers #l(A) and
#2(A) are simultaneously finite or infinite, we observe first that this
implies that, for all A ET, the three numbers #l(A), #2(A) and #(A) are
simultaneously finite or infinite. Given the #-measurable set E, we
shall prove that E is #l-measurable, i.e., we shall prove that #l(A)
#i(AE) +#i(AEe) holds for all A ET satisfying #l(A) <00. Subtracting,
for such a set A, the inequality
#2(A )#2(AE) +#2(AEe)
from the equality
#(A) ==#*(AE) + #*(AEe),
and observing that all numbers concerned are finite, we obtain the
desired inequality. The proof that E is #2-measurable is similar.
Exercises
LEBESGUE'S DEFINITION OF MEASURABILITY
9.1) Let #* be generated by (X, T, #). Show that if #(X) <00, then
EeX is #-measurable if and only if #(X) ==#*(E) +#*(Ee). This is the
original definition of measurability by H. LEBESGUE (1904, [IJ, p. 106)
for bounded sets in finite-dimensional number space; in this context T
is then the semi-ring of all cells contained in a fixed cell X=> E, and #
is the usual measure on T.
9.2) Let #* be generated by (X, T, #). Show that if there exists a
measurable set Y::::> E such that
#(Y) <00
and
#(Y)==#*(E)+#*(Y -E),
then E is measurable.
REGULAR EXTERIOR MEASURE
9.3) In Exercise 7.1 it was stated that the exterior measure #* in X
is regular if and only if #* is generated by some triple (X, T, #). Show
now that the exterior measure #* in X is regular if and only if each set
5 e X has a measurable cover (Le., denoting by A the collection of all

Ch. 2, 9 9J
APPROXIMATION THEOREMS
55
measurable sets, and writing #1==#* on A, there exists a set EEA
such that 5eE and #1(E)==#*(5)).
CONTRACTED MEASURE
9.4) Let (X, T, #) generate the exterior measure #* and the con-
tracted exterior measure #. Show that, for any 5 e X, we have #(5)
==sup #*(5B) for all f-sets B satisfying #(B) <00. We recall that an f-
set is a a-set consisting of a finite number of non-empty terms.
9.5) Let #1 and #2 be measures on the semi-ring T, and let, for all
A ET, the numbers #l(A) and #2(A) be simultaneously finite or infinite.
The contracted exterior measures in X, generated by #1, #2 and by the
sum measure #==#1 +#2, will be denoted by #1e, #2e and # respective-
ly. Show that #==#ie+#2e.
9.6) Given the measure # on the semi-ring T and the measure #1 on
the semi-ring T1, let # and #ie be the corresponding contracted ex-
terior measures. Assume that each A Er of finite #-measure is #1-
measurable, and #le(A)==#(A). Assume, similarly, that each BET 1 of
finite #l-measure is #-measurable, and #e(B)==#l(B). Show that the
collection A of all #-measurable sets is the same as the collection Al
of all #l-measurable sets, and #e==#le on A==A1. This is, for contracted
measures, an analog of sec. 6, Theorem 2 (a rather poor one, since the
hypotheses are stronger and the conclusion is weaker).
9.7) Let (X, T, #) generate the collection A of all measurable sets
with the contracted measure #e on A. Let T 1 be a semi-ring satisfying
TeT 1 eA, write #l==#e on T1, and let (X, T1, #1) generate the con-
tracted measure #le on the collection Al of all #l-measurable sets.
Show that A 1 ==A and #le==#e on A1==A. Observe, however, that in
the particular case that T 1 ==A, the second contraction is not necessary.
CHARGES
The next exercises are again devoted to the theory of charges. In all
these exercises, except in Exercise 9.12, we assume that (X, T, v) gener-
ates the exterior charge v* and the contracted exterior charge v in X;
the collection of all measurable sets is called A, and v* and v on A are
denoted by v and Ve respectively. We follow rather closely the expo-
sition of B. C. STRYDOM [IJ.

56
l\IEASURE
[Ch. 2, 9 9
9.8) Show that Theorem 1 (the inclusion theorem) holds for the
charge v. Show, in addition, that if E is measurable with v(E) <00, and
e>O is given, then there exist I-sets B1e and B2e such that B 1e cE cB 2e
and v(B 2e -B 1e )<e.
9.9) Show that if the set 5 is I-covered by T, then
v*(S) ==inf v(O) for all I-sets 0:::> S.
Show, in addition, that if E is measurable with v(E)<oo, then there
exists a descending sequence of I-sets On:::> E such that v( 0 n) ! v(E), the
set 0 6 = nl On is measurable, and V(06-E) ==0.
9.10) Show that Theorem 3 holds for charges.
9.11) Show that the assertions in Exercises 9.1,9.2 hold for charges.
9.12) Let v* be an exterior charge in X, let A be the corresponding
collection of all measurable sets, and write VI ==v* on A. Show that v* is
regular if and only if for each set 5 we have v*(S) ==inf v1(E) for all E EA
satisfying E:::> S.
9.13) Show that the assertions in Exercises 9.4, 9.5, 9.6, 9.7 hold for
charges.
CHARGES AND MEASURES
9.14) Let # be a measure on the semi-ring T (where T is assumed to
be of the type as used in the theory of charges). Then # is automatically
a charge on T, and so # may be extended in two different ways, either
by means of the extension procedure for measures or by means of the
procedure for charges. The exterior measure and its contraction, gener-
ated by (X, T, #), will be denoted by #: and #:e respectively; the ex-
terior charge and its contraction, generated by (X, T, #), will be de-
noted by #i and #ie respectively. The corresponding collections of
measurable sets are called Aa and Af respectively. Show that AfcA a ,
#a(E)==#f(E) for any EEAf satisfying #f(E) <00, and #ae==#fe on Af.
Note that, by this result, #fe is a measure on Af.
9.15) Let V be a charge on the ring J, and let, for each set EEJ,
the number T(E) be defined by T(E) ==inf {lim v(F n)} for all sequences
F n iE, F nEJ. Show that T is a measure on J (M. A. WOODBURY [IJ,
K. Y OSIDA and E. HEWITT [1 J).
9.16) Let v be a charge on the semi-ring T, and let (X, T, v) generate
the extended charge v and its contraction Ve on the collection A of all

Ch. 2, 9 9J
APPROXIMATION THEOREMS
57
v-measurable sets. The ring of all I-sets B satisfying v(B) < 00 will be
denoted by J. The measure T on J is defined as in the preceding exer-
cise by
T(B) ==inf {lim v(B n)}
for all sequences Bn i B, BnEJ.
Then V==T+Tp on J, where Tp is evidently a charge on J. The ex-
tension procedure for charges, with contraction included, is now ap-
plied to (X, J, v), (X, J, T) and (X, J, Tp) separately; the so obtained
contracted charges are Ve, Te and Tpe on the fields A, AT and A p re-
spectively. Show that A==ATnA p and Ve==Te+Tpe on A. Show also
that Te is a measure on A.
PURE CHARGES
9.17) Let v be a charge on the ring J such that v(B) < 00 for all
B EJ, and let A be a ring of v-measurable sets such that A  J (it is
not necessary that A is the ring of all v-measurable sets). The charge v
is called a pure charge on A whenever any measure # on A satisfying
O#(B)v(B) for all BEJ is identically zero on J, i.e. #(B) ==0 for all
BEJ. Show that v is a pure charge on A if and only if, given 8>0 and
any set EEA satisfying v(E)<oo, there exists a decomposition E==
r En into disjoint sets EnEA such that r V(En)<8. For the ex-
istence of a pure charge cf. Exercise 28.4.
9.18) With the notations of Exercise 9.16, Tp is a charge on the ring
J such that Tp(B)<oo for all BEJ, and A==ATnA p is a ring of Tp-
measurable sets satisfying AJ. Show that Tpe is a pure charge on A.
UNIQUE DECOMPOSITION
9.19) Once more with the notations of Exercise 9.16, we have now
Ve==Te+Tpe, where Te is a measure on A and Tpe is a pure charge on A,
and in addition Te(E) ==sup Te(EB) and Tpe(E) ==sup Tpe(EB) for all sets
BEJ. Show that if also Ve==T+Tc on A, where T is a measure on A
and Tc is a pure charge on A, and T(E)==sup T(EB) and Tc(E)==
sup Tc(EB) for all sets BEJ, then T==Te and Tc==Tpe' In other words,
show that the decomposition of Ve on A into a contracted measure and
a contracted pure charge is unique.

58
1\1:EASURE
[Ch. 2,9 10
INCREASING SEQUENCES OF MEASURES
9.20) Let fln (n== 1, 2, . . .) be arbitrary countably additive measures
on the semi-ring T of subsets of X, such that fln i fl on T. Show that fl
IS a measure on T.
9.21) Let the extension procedure for measures be applied to the
measures fln and fl of the preceding exercise, and let the extended
measure fl be a-finite. Show that the set E c X is fl-measurable if and
only if E is fln-measurable for all n simultaneously, and show that
fln(E) i fl(E) for any such set E.
9.22) The conclusion in Exercise 9.20 is no longer always true if the
convergence is not monotonically increasing. Indeed, let X be the set
of all positive integers, and let A be the a-field of all subsets of X.
Furthermore, let r Pi be a convergent series of numbers Pi satisfying
OPi 1. The numbers Pni (n, i== 1, 2, . . .) are defined by Pni==Pi for
in, and Pni== 1 for i>n, and the measures fln on A are defined by
fln(E) == iEE Pni. Show that fln converges (monotonically decreasing)
to a set function T which is not a measure.
10. Examples, in particular Lebesgue Measure and Stieltjes-
Lebesgue Measure
We return to the examples in sec. 5.
ExamPle 1 (Lebesgue measure). If X ==Rk (k-dimensional real number
space), and if Tis the semi-ring consisting of 0 and all cells A=(a1, b 1 ;
. . . ; ak, bkJ with fl(0) ==0 and fl(A)== rr=l (bi-ai), then fl is a measure
on T. The extension procedure yields the a-field A of all fl-measurable
sets E. Such a set E is called Lebesgue measurable, and its measure fl(E)
is called the k-dimensional Lebesgue measure of E. Lebesgue measure is
obviously a-finite.
The introduction of Lebesgue measure on the a-field of all Lebesgue
measurable sets is due to H. LEBESGUE (1902, [2J, [3J), but the notion
of Lebesgue measure on a a-field Ab of sets in finite-dimensional real
number space goes already back to E. BOREL (1898, [IJ). The a-field
Ab considered by Borel is the smallest a-field containing all cells (cf.
Exercise 4.1, and sec. 46), and it can be proved that Ab is a proper

Ch. 2, 9 10J LEBESGUE AND STIELTJES-LEBESGUE MEASURE
59
subcollection of the a-field A of all Lebesgue measurable sets (although
A, in a certain sense, is not much larger than Ab, since sec. 9, Theo-
rem 2 (3) shows that, given any EEA, there exists a set 06EAb such
that 0 6 ::::> E and the sets 0 6 and E are almost equal). More general
measures in finite-dimensional number space were introduced by J.
RADON (1913, [IJ), and the final step of considering a measure on a a-
field of subsets of an abstract point set was taken by M. FRECHET
(1915, [2J).
Evidently, any closed interval in Rk is the limit of a descending
sequence of cells, and any open interval is the limit of an ascending
sequence of cells. Hence, closed and open intervals are measurable and
they have the same measure as the cell with the same vertices. The
same is true then for any interval which is partially open. A degenerate
interval, i.e., a set [aI, b 1 ; . . . ; ak, bkJ such that an==b n for at least one
value of n, has measure zero. In particular, a set consisting of one
single point has measure zero, and the same is therefore true of any
set consisting of a finite or countable number of points, such as, for
example, the set of all rational points.
Since Rk is a metric space with respect to the familiar Euclidean
distance, we can slightly extend the inclusion and covering theorems
in the preceding section. We first observe that any open set OcR k is
a a-set. In order to prove this assertion, we cover Rk by a net A'O of
disjoint cubic cells of measure 1 (i.e., Rk is the union of a necessarily
countable number of disjoint cubic cells of measure 1; the collection
of all these cells is the net A'O; the cell (aI, b 1 ; . . . ; ak, bkJ is called
cubic whenever b1-a1 == . . . ==bk-ak); the net J1Ii of cubic cells of
measure 2- k is a refinement of A'O (i.e., each cell of J1Ii is entirely in-
cluded in some cell of A'O), the net A2 of cubic cells of measure 4- k is
a refinement of J1Ii, and so on. Given the open set 0, let Go be the
union of all cells of A'O which are included in 0, let G 1 be the union of
all cells of J1Ii included in 0 but not in Go, and so on. Then 0==  Gk,
and this may be written in the form O== An, where all AnET (the
sets An are even disjoint). It is not true that, conversely, every a-set
is open (any set, consisting of one cell, is not an open set). Hence, there
are less open sets than a-sets, and for this reason the first part of the

60
MEASURE
[Ch. 2,9 10
next theorem is a small extension of the covering theorem in the pre-
ceding section.
THEOREM 1. (1) For any set 5 c R k we have #*(5)==inf #(0) for all
open sets 0:::> S.
(2) The set E c Rk is Lebesgue measurable if and only if, given any
e>O, there exists an open set 0 and also a closed set F such that F c E c 0
and #(O-F)<e. Hence, if E is measurable, there is a set O(}, limit of a
descending sequence of open sets On, and also a set F (J, limit of an as-
cending sequence of closed sets F n, such that F (J c E c O(}, and the three
sets F (J, E and O(} are almost equal.
PROOF. (1) Given the a-set P== An with disjoint terms and given
e>O, there exist open intervals Bn such that AncBn and
#(Bn) <#(An) +e/2 n
Then O== Bn is open, O:::>P, and
#(O) #(Bn)< #(An)+e==#(P)+e.
for n = 1, 2, . . . .
Hence, the number inf #(0) for all open 0:::> 5 has the same value as
the number inf #(P) for all a-sets p:::> S. This shows that #*(5) =inf #(0)
for all open 0:::> 5.
(2) If there exist sets 0 and F, satisfying the conditions of the theo-
rem, then E is measurable by the inclusion theorem in sec. 9 (we use
then the obvious fact that open and closed sets are measurable). Let,
conversely, E be measurable, and let the open set 0:::> E satisfy #(O-E)
<!e (given any e>O, such a set 0 exists by (1); for #(E)<oc> this is
evident, for #(E) ==oc> we use a decomposition of E into sets of finite
measure as in the proof of sec. 9, Theorem 2 (3)). The set Ee is also
measurable, so there exists an open set 0':::> Ee satisfying #(0' -Ee)
<lEe Then F==(O')e is closed, FcE, and E-F=EFe==EO'==O' -Ee,
so #(E-F)<!e. It follows that FcEcO and #(O-F)<e.
We next consider the two variants, already introduced in sec. 5. If
the semi-ring r of all cells is replaced by the semi-ring rr of all rational
cells, then the generated exterior measure and the a-field of all measur-
able sets are unchanged. The situation is different, however, if r is
replaced by the semi-ring r i of all integer cells. In that case #*(E) is

Ch. 2, 9 10J LEBESGUE AND STIELTJES-LEBESGUE MEASURE
61
equal to the number of cells in T i of unit measure and having points
in common with E. Hence, if there exists a cell A E T i of unit measure
such that the intersection AE is neither empty nor the whole of A, we
have #(A) == 1, #*(AE) == 1 and #*(AEe) == 1, so #(A) <#*(AE) +#*(AEe),
and this shows that E is not measurable. If, however, E ==  A n, where
all AnETi, then E is evidently measurable. The conclusion is therefore
that for this variant the a-field of all measurable sets consists of all
sets E== An, where all AnETi (every measurable set is now a a-set).
The question may be raised whether there exist sets in Rk which fail
to be Lebesgue measurable. Such sets do exist, but so far it has not
been possible to exhibit one without making use of the axiom of choice.
We shall present an example in one-dimensional space R1, using the
fact that R1 is an additive group having the set of all rational numbers
as a subgroup.
THEOREM 2. There exists a set E e R1 which fails to be Lebesgue
measurable.
PROOF. The additive group R 1 is decomposed into cosets with re-
spect to the subgroupd of the rational numbers. If x is a point in one
of these cosets, then x+a (a rational) is in the same coset; it is evident,
therefore, that each coset has an infinity of points in common with the
cell (0, IJ. We choose one point in (0, IJ from each coset (axiom of
choice), and define E to be the set of these points. Hence Ee (0, IJ.
It is not difficult to see that if a1 and a2 are rational, a1 =Fa2, then the
sets
E1=={X: x==e+a1, eEE}
and
E2=={X: x==e+a2, eEE}
are disjoint (if e1 +a1 ==e2+a2, then e1-e2==a2-a1 Ed, so e1 and e2 are
in the same coset, i.e., e1==e2, and this would imply a1==a2). On the
other hand any point xER1 is in one of the cosets, so it may be written
as x===e+a, where eEE and aEd. In particular, if XE(O, IJ, then x==
e+a, where eEE and the rational number a satisfies -1 <a< 1 (since
O<e< 1). Let {aI, a2, . . .} be the set of all rational numbers in the
open interval (-1, 1), and En=={x: x==e+an, eEE} for n== 1,2, . . . .

62
MEASURE
[Ch. 2, 9 10
Then (0, IJ c r En by what we have just observed, and on the other
hand r En c .(-I, 2). Assume now that E is Lebesgue measurable,
and #(E) CX. Since it is easily seen that translation of a set in R 1 pre-
serves measurability and the value of the measure, it follows then that
all the disjoint sets En are measurable, and #(En)==cx for all n. In view
of r En c (-1,2), we obtain r #(En)#(r En)3, so #(En)==
cxo for all n. But (0, IJcr En; hence lr #(En)O, a contra-
diction. This shows that the set E is not Lebesgue measurable.
The first example of a non-measurable set in the sense of Lebesgue,
essentially the example reproduced here, was given by G. VITALI (1905,
[IJ).
ExamPle 2 (Discrete measure). If X is a non-empty point set, and r
is the semi-ring consisting of 0 and all sets A containing only one point
with #(0)==0 and #(A)==1 if A contains one point, then # is a measure
on T. If E c X and A E T, then EA is either empty or equal to A, so
EA is measurable. It follows that any E c X is measurable (by sec. 7,
Theorem 7); furthermore, #(E) is equal to the number of points in E.
This measure # is sometimes called the discrete measure in X.
ExamPle 3 (Discrete measure). If X is an infinite point set, and Tis
the field consisting of 0, X, all finite subsets of X and their comple-
ments with #(0)==0, #(A)n if A consists of n points, and #(A) ==00
for all other AET, then # is a measure on T. Given EcX and the set
A ET of finite measure, EA is either empty or finite, so EA is measur-
able. It follows that any E c X is measurable (by sec. 7, Theorem 7).
Furthermore, #(E) is once again equal to the number of points in E, so
# is the discrete measure in X.
ExamPle 4. If X is non-empty, and T is the a-field of all subsets of
X with #(0)0 and #(A)oo for all other A ET, then # is a measure
on T. Evidently the a-field A of all #-measurable sets satisfies AT.
ExamPle 5. If X is an uncountable point set, and T is the a-field of
all subsets of X with #(A)O if A is finite or countable and #(A) ==00
for all other A ET, then # is a measure on F. Once again A==T.

Ch. 2,9 10J LEBESGUE AND STIELTJES-LEBESGUE MEASURE
63
Example 6. If X :==R2, and r is the semi-ring consisting of 0 with
#(0)==0 and all horizontal linear cells A =={(x, y): a<xb, y==c} with
#(A):==b-a, then # is a measure on r. Denoting, for any EeX and
for any fixed real number Yo, the set {(x, Yo): (x, yo) EE} by E yo ' we
have E:==U y Ey. Furthermore Xy (an entire horizontal line) is measur-
able, and Ey==EXy; hence, if E is measurable, then each Ey is measur-
able. Conversely, the measurability of all Ey implies the measurability
of E (since in this case EA is measurable for all A Er), and then #(Ey)
is the one-dimensional Lebesgue measure of Ey. The final conclusion
is, therefore, that E:== U y E y is measurable if and only if all E y are
measurable, and in this case #(E) ==(X) if there is an uncountable
number of non-empty E y (even though perhaps #(Ey) ==0 for all these
Ey), and #(E)==y #(Ey) if the number of non-empty Ey is finite or
countable (the summation in y is only over those y for which E y is
non-empty) .
ExamPle 7. If X ==R2, and r is the semi-ring consisting of 0 with
,u(0) :==0, all horizontal linear cells A and all vertical linear cells B with
,u(A) and #(B) the one-dimensional Lebesgue measure of A and B re-
spectively, and all sets C consisting of one point with #(C):==O, then #
is a measure on r. If E eX, and Yo is fixed, we again define
Eyo=={(x, yo): (x, yo) EE},
and similarly
Exo=={(xo, y): (xo, y) EE} for any fixed Xo.
Evidently E==U x Ex==U y E y , and by a similar argument as in the
preceding example it may be seen that E is measurable if and only if
all Ex and Ey are measurable, and in this case #(Ex) and #(Ey) are the
one-dimensional Lebesgue measures of Ex and Ey respectively. Ob-
serve that the measurability of all Ey is not sufficient in order that E
be measurable. If E is measurable, and either the number of non-
empty Ex or the number of non-empty E y is finite or countable, then
/1(£) == x #(E x) or #(E) == y #(Ey) respectively; in all other cases
,u (E) == (X) for a measurable set E.
Example 8 (Stieltjes-Lebesgue measure). Let X==Rl, and let g(x) be
a real increasing function, defined and right continuous on R 1 (so g(X1)

64
MEASURE
[Ch. 2,9 10
g(X2) for Xl <X2, and g(x) !g(xo) as x!xo). If T is the semi-ring con-
sisting of 0 and all cells with #(0) ==0 and #(A) ==g(b) -g(a) for A == (a, b],
then # is a a-finite measure on T. The measure # is said to be generated
by the function g(x) , and it was proved in sec. 5 that any measure # on
T, satisfying #(A)<oo for all A ET, is generated by such an increasing
right continuous function g(x). Applying the extension procedure, we
obtain the a-field of all #-measurable sets E. Such a set E is called
5tieltjes-Lebesgue measurable (with respect to g(x)), and its measure
#(E) is called the 5tieltjes-Lebesgue measure of E (with respect to g(x)).
For g(x) ==x ordinary Lebesgue measure is obtained.
Stieltjes-Lebesgue measure is not due to Stieltjes as the name might
suggest; the name indicates that the integral based upon this measure
stands in a similar relation to the Lebesgue integral as the Stieltjes
integral (due to T. J. STIELTJES, 1894) stands to the Riemann integral.
Although g(x-)==limhtO g(x-h) exists for all x, there may be points
X for which g(x-) ¥=g(x); the measure g(b) -g(a) of the cell (a, b] may
be different, therefore, from the measure g(b) -g(a-) of the closed
interval [a, b]. Similarly, the measure g(x)-g(x-) of the set {x}, con-
sisting only of the point x, may be different from zero. Obviously, g(x)
is continuous at the point xOER1 if and only if # ({xo}) ==0. It is of some
interest, in this connection, to observe that the number of points where
g(x) is discontinuous is at most countable. Indication of the proof: The
number of points in (a, b], where g has a jump exceeding one, i.e., where
g(x) -g(x-) > 1, does not exceed g(b) -g(a); similarly, the number of
. h h . d 1 1 1 . f ,. H h b
pOInts were t e Jump excee s 2' 4' 8' . . " IS InIte. ence, t e num er
of points of discontinuity in (a, b] is at most countable, and then the
number of points of discontinuity on the whole of R1 is also at most
countable.
We finally mention one particular example. If g(x) has a jump of
value one at each integer, and g(x) is constant between consecutive
integers, then #*(5) is equal to the number of integer points contained
in the set S. It follows then immediately from the definition of measur-
ability that every set E cR1 is measurable. Note the similarity between
this particular example and the example of discrete measure.

Ch. 2,9 10l LEBESGUE AND STIELTJES-LEBESGUE MEASURE
65
Exercises
PROPERTIES OF LEBESGUE MEASURE
10.1) Let r be the semi-ring of all cells in Rk, let #1 be Lebesgue
measure on r, and #2 the measure on r defined by #2(A) ==00 for all
A *0. Finally, let ##1 +#2 on r. Show that there exists a set E cR k
which is #-measurable, but fails to be #l-measurable. Note the relation
to sec. 9, Theorem 3.
10.2) Show that a translation T in Rk preserves exterior Lebesgue
measure, i.e., if 5 c Rk and 5T is its image, then the exterior Lebesgue
measures #*(5) and #*(5 T ) satisfy #*(5T)#*(5). Show also that the
set ET is measurable if and only if E is so.
10.3) Let f(x) be non-negative and continuous on the closed one-
dimensional interval [a, b], and let F be its ordinate set, i.e.,
F=={(x, y): axb, O<yf(x)}.
Show that the two-dimensional Lebesgue measure of F is equal to the
Riemann integral
b
f f(x) dx.
a
The same holds, of course, if O<yf(x) is replaced by Oyf(x).
10.4) Show that the two-dimensional Lebesgue measure of a circular
sector of radius a and angle ex (0<ex2n) is lexa 2 .
10.5) Show that if (r, p) are polar coordinates in R2, and #1 is Lebes-
gue measure on the semi-ring r1 of all annular sectors
A [r1, r2; PI, p2)=={(r, p): 0r1r<r2, P1P<P2 with P2-P12n},
then #1 generates the same exterior measure, and therefore the same
measurable sets, as Lebesgue measure # initially defined on the semi-
ring r of all cells.
POLAR MEASURE
10.6) Let r 1 be the semi-ring of the preceding exercise, and let the
measure n on r1 be defined by n(A) == (r2-r1)(P2-P1). Show that every
n-measurable set is Lebesgue measurable, and conversely.

66
MEASURE
[Ch. 2,9 10
NON LEBESGUE MEASURABLE SETS
10.7) Show that if # is Lebesgue measure in Rk (kl) and F is a
#-measurable subset of Rk such that #(F) >0, then F contains a non-
#-measurable subset.
DIFFERENCE SET OF A SET
10.8) Let E be a Lebesgue measurable subset of Rk of finite positive
measure #(E), and let O<cx<l. Show that there exists a cell AcR k
such that #(EA) >cx#(A).
10.9) For EcRk, denote by J(E) the set of all points X-YERk with
xEE, YEE (the coordinates of x-yare the differences of the corre-
sponding coordinates of x and y). Show that if E is Lebesgue measur-
able and of positive measure #(E), then J(E) includes an open interval
around the origin. For the proof, we may assume that #(E)<oo (by
passing, if necessary, to a subset of E). In view of the result in the
preceding exercise there exists a cell A such that #(EA) >!#(A). We
multiply A with respect to its centre by a number greater than 1 such
that the so obtained larger cell A has its measure not exceeding !#(A).
It is evident now that there exists a point xOERk with all its coordi-
nates positive such that if xER k has all its coordinates smaller in abso-
lute value than the corresponding coordinates of Xo, then B==A+x
(that is, B is the set of all points a+x, aEA) is included in the larger
cell A, so #(AUB)!#(A). Let x be such a point, set P==EA and Q==
P+x. Show that the intersection PQ is not empty, and derive that
x E J (E). This will show then that J (E) includes the open interval of
all points x with coordinates smaller than the coordinates of Xo in
absolute value.
MEASURABLE SUBGROUPS OF THE ADDITIVE GROUP
10.10) Consider Rk as an additive group, and let G be a proper sub-
group. Show that if G is Lebesgue measurable, then it has measure
zero (such as e.g. the subgroup of all points with only rational coordi-
nates) .

Ch. 2, 9 11 ]
METRIC SPACE OF MEASURABLE SETS
67
ATOMS
10.11) Let # be a measure in X. The #-measurable set E of positive
measure is called an atom whenever for any #-measurable subset E 1 of
E we have either #(E1)0 or ,u(E-E1) ==0. If # is discrete measure,
then every point of X is an atom. If # is linear Lebesgue measure on
horizontal lines in the plane (cf. Example 6), then every vertical line
segment E satisfies #(E)==cx>, and any subset E1 of E of finite measure
satisfies #(E1)0, but E is not an atom since there are many subsets
E1 of E satisfying #(E1)==#(E-E 1 )==cx>. Show generally that if E is
a set of finite positive measure, and neither E nor any of its measur-
able subsets is an atom, then E contains subsets of arbitrarily small
positive measure.
10.12) Show by means of Zorn's lemma that if E is a set of finite
positive measure such that neither E nor any of its subsets is an atom,
and if O<cx<#(E), then there exists a subset E 1 of E such that #(E 1 ) cx.
10.13) Let # be a measure on the semi-ring T with the property
that, given any A ET of finite positive measure, there exists a set BEF
such that Be A and O<#(B) <#(A). Show that this does not imply
that the extended measure # is free of atoms of finite measure.
10.14) Let # be a measure on the semi-ring T, and assume that there
exists a number y (0< y<!) such that, given any A E T of finite positive
measure, there exists a set BET such that BcA and y#(A)<#(B) <
(l-y)#(A). Show that the extended measure # does not possess any
atom of finite measure.
10.15) Show that Lebesgue measure does not possess any atoms.
Show also that a Stieltjes-Lebesgue measure in the real line, generated
by a continuous and non-decreasing function g(x), does not possess any
atoms.
10.16) Let # be a measure in X, and E a set of a-finite #-measure.
Show that E ==E1 +E2 uniquely, where neither E1 nor any of its measur-
able subsets is an atom, and E 2 is a union of an at most countable
number of atoms of finite measure.
11. Metric Space of Measurable Sets
Given the measure # in X, we consider the collection A * of all sub-
sets of X of finite exterior measure, and we split up A* into classes of

68
MEASURE
[Ch. 2,  11
mutually almost equal sets (this can be done since by sec. 8, Theorem 2
the property to be almost equal is an equivalence relation). We shall
adopt the convention to call this collection of classes again A *, and to
speak about a set E of A* when, properly speaking, we mean the entire
class of sets almost equal to E. In other words, almost equal sets are
identified. With this convention, A * will now be made into a metric
space, i.e., we shall introduce a distance function p(E, F) for ailE, F EA *.
THEOREM 1. If p(E, F)=#*(E-F)+#*(F -E) for all E, FEA*, then
A* is a metric space with respect to the distance function p(E, F).
PROOF. Obviously p(E, F)==p(F, E), and p(E, F) ==0 if and only if E
and F are almost equal. Let now E, F, GEA*, and consider them for
the moment as individual sets, and not yet as equivalence classes of
sets. Since
E -F ==EFe==EGeFe+EGFe c EGe+GFe== (E -G) + (G-F),
we have
#*(E -F)#*(E -G) +#*(G-F).
Similarly
#*(F -E)#*(F -G)+#*(G-E).
Hence, by addition,
p(E, F)p(E, G)+p(G, F).
In particular, if E 1 is almost equal to E, the choice G==E1 shows that
p(E, F)p(E1, F). But then, since E and E 1 may be interchanged, we
have also p(E1, F)p(E, F); hence p(E, F)==p(E1, F). Finally, if F1 is
a set almost equal to F, a repeated application of this result shows that
p(E, F)==p(E1, F)==p(E1, F1),
justifying the definition of p(E, F) as the distance between the equiva-
lence classes of E and F.
As before, the collection of all #-measurable subsets of X is denoted
by A. Let Al be the subcollection of all #-measurable sets of finite
measure. Hence, Al is the intersection of A and A*. The following
theorem is now an immediate consequence of the preceding theorem:

Ch. 2, 9 11 ]
METRIC SPACE OF MEASURABLE SETS
69
THEOREM 2. If, for all E, FEA1, we define
p1(E, F)==#(E-F)+#(F-E),
then Al (with the same convention concerning identification of almost
equal sets) is a metric space with respect to the distance function PI (E, F).
The metric space Al is, evidently, a subspace of the metric space A* in the
preceding theorem (i.e., P1P on Al cA*).
The metric spaces A* and Al are complete; for Al (the more inter-
esting of the two spaces) this will follow from a more general result in
sec. 20 (for a direct, but rather tedious, proof we refer to Exercise 11.1).
If the metric space Al of #-measurable sets of finite measure is sepa-
rable, then the measure # is called a separable measure. Evidently, if #
is separable, Y a measurable subset of X, and #0 the restriction of # to
all measurable subsets of Y, then #0 is separable in view of the theorem
that any subspace of a separable metric space is separable.
THEOREM 3. If the semi-ring T, the initial domain of definition of #,
consists of a finite or countable number of sets, then # is separable.
PROOF. The collection Z of all finite unions l An, where p is vari-
able, all AnET and all An are of finite measure, is countable. It will be
sufficient to prove that Z is dense in AI. Let, for this purpose, EEA 1
and e>O be given. Then there exists a a-set S An=>E satisfying
#(S-E)<!e. Furthermore, if F==l An is a partial sum of S== An,
then FEZ and #(S-F)<!e for p sufficiently large. Hence
#(F-E)#(S-E)<!e
and
#(E - F) #(S - F) < !e,
so Pl(E, F) <c. This shows that Z is dense in AI.
THEOREM 4. Lebesgue measure in Rk and Stieltjes-Lebesgue measure
in R1 are separable.
PROOF. Lebesgue measure in Rk is generated by its restriction to the
countable semi-ring Tr of all rational cells, so it is separable by the pre-
ceding theorem. In order to prove the same for Stieltjes-Lebesgue
measure fl (with respect to the increasing and right continuous function

70
MEASURE
[Ch. 2, 9 11
g(x)) in R1, we observe first that the number of discontinuity points of
g(x) is at most countable. Let 5 be the subset of R1 consisting of all
rational points and all discontinuit:y points of g(x), and let r 1 be the
semi-ring consisting of 0 and all cells with endpoints in S. Then r 1 is
countable, and, given the arbitrary cell A and the number e>O, there
exists a cell A1Er1 such that A cA 1 and #(A 1 )-#(A)<e. This implies
that r1 generates the same Stieltjes-Lebesgue exterior measure (with
respect to g(x)) as the semi-ring of all cells, and therefore the same
measure. It follows that the measure is separable (since r1 is countable).
In Example 6 of the preceding section (the measure in R 2 generated
by the horizontal linear cells with linear measure) the measure # is not
separable, since any measurable E==U y Ey of finite measure contains
at most a countable number of non-empty E y ; hence, given any count-
able collection Z of such sets E, the collection Z also contains at most
a countable number of non-empty Ey, say E1, E2, . . . . It is impossible
now that Z is dense in AI, since there always exists a set E yo ' of unit
measure and disjoint to all En, and this implies
P1(E y ,E)=#(Eyo -E) +#(E -EYo)#(Eyo -E) =#(E yo )== 1 for all EEZ.
° .
Similarly, if X is an uncountable point set, and # is discrete measure
in X, then # is not separable.
Exercises
COMPLETENESS OF THE SPACES A* AND Al
11.1) Show that the metric spaces A * and AI, introduced in the
Theorems 1 and 2, are complete. Observe first that, given the funda-
mental sequence EnEA* (hence, p(En, Em)O as n, moo), there exists
a subsequence F n ==Emn such that
p(Fn, F n + p ) <2- n
for all n and all P>O,
and show that the set F ==lim sup F n satisfies
#*(F -F n)<2-(n-l)

Ch. 2, 9 11 J
METRIC SPACE OF MEASURABLE SETS
71
as well as
#*(F n-F)<2-(n-l)
for all n.
Show, finally, that this implies lim p(F, En) ==0, so that F is, therefore,
the limit element of the sequence En. For the completeness of AI, ob-
serve that whenever all En are measurable, then F is measurable.
METRIC SPACE OF ALL SETS OF FINITE CHARGE
11.2) Let v be a charge in X, let A * be the collection of all subsets
of X of finite exterior charge, and let Al be the subcollection of all v-
measurable sets of finite charge. Almost equal sets in A* and Al are
identified. Show that if
p(E, F) ==v*(E-F)+v*(F-E),
then A* is a metric space with respect to the distance function p(E, F),
and Al is a subspace.
JORDAN CHARGE
11.3) Let X==R1, and let r be the semi-ring of all cells A==(a, bJ
with v(A)==b-a. Applying the extension procedure for charges to
(X, r, v), we obtain a charge in R1 which is usually called Jordan
charge (Jordan content). Show that the corresponding metric spaces A*
and Al are not complete. For this purpose, let A==(O, 1J, and select
the subsets En (n== 1, 2, . . .) of A in the following way: For E 1 take
the cell (1, iJ, E 2 is obtained by adding to E 1 the middle ninth parts
(cells) of (O,lJ and (i, IJ, E3 is obtained by adding to E 2 the middle
27th parts of the cells of A -E 2 , and so on. For the complement F n==
A - En we have therefore
v (F n) == (1 -1) (1 -!) (1 --17) . . . (1 - 3 -n) ! fJ > 0
(for the benefit of the reader who is not familiar with the properties of
such products, a proof of fJ>O is indicated in the next exercise). Hence
v(En) i cx== I-fJ, where O<cx< 1. The sequence En is, evidently, a funda-
mental sequence in the metric space A* (as well as in the metric space
AI); assume now the existence of a set SEA* such that lim p(S, En) ==0.
Show that v*(S)cx; hence, there exists an I-set B  5 such that v(B) < 1.

72
MEASURE
[Ch. 2, 9 11
Show that, therefore, En-B is an ascending sequence of I-sets, non-
empty from a certain index no on, and so ot ascending positive charge
for nno. On the other hand En-B cEn-S, so v(En-B)v*(En-S)
O. This is a contradiction. Note the extension to Rk for k> 1.
11.4) Let the series of non-negative terms r an converge, and let
n
Pn= II (1 +ak)'
k=l
Show that the sequence Pn converges. For this purpose, observe that
since a finite number of factors may be deleted, it is no restriction of
the generality to assume that r an<t. Then
n
Pn<I+2ak
1
(proof by induction), and the convergence of Pn follows. If, in addition,
Oan< 1 for all n, and
n
qn== II (l- a k),
k=l
show that qn converges to a number q>O. For the proof we may as-
sume that Oan<t for all n. Then (l-an)-l 1 +2a n ; hence, q;;l con-
verges by what has just been proved, that is, qn converges to a number
q>O.

CHAPTER 3
DANIELL INTEGRAL
In the present chapter we introduce the abstract Daniell integral, generalizing
the Riemann integral (and the Stieltjes-Riemann integral) of classical analysis,
as well as the Lebesgue integral (and the Stieltjes-Lebesgue integral) dating from
the beginning of this century. It is one of the characteristic features of the Daniell
integral that there need not necessarily exist a measure in the underlying point
set X on which the integrands f(x) are defined.
12. Introduction of the Daniell Integral
Let # be a measure in X, and let A be a ring of #-measurable sub-
sets of X. The ring A is, therefore, a subcollection (proper or improper)
of the a-field Att of all #-measurable subsets of X. Any real function
I(x), defined at all points XEX, and assuming only a finite number of
different (finite) values, such that each value *0 among this finite
number is assumed on a #-measurable set of finite measure of A, will
be called a real step function (with respect to # and A). Such a step
function f(x) satisfies, therefore, either
f(x) ==0 for all XEX,
or it is of the form
q
f(x) ==  dnXFn (x),
n=l
where the numbers d n are finite, non-zero, and different for n== 1, . . " q,
and where the functions XFn(X) are the characteristic functions of the
disjoint non-empty sets F nEA of finite measure. We shall call this the
standard representation of the step function f(x). Obviously, the standard
representation is uniquely determined.
In order to avoid confusion, it is useful to observe that sometimes
in the literature the function f(x) == = 1 dnXFn (x) is called a step

74
DANIELL INTEGRAL
[Ch. 3, 9 12
function even though the condition #(F n) < 00 for n== 1, . . . , q is not
necessarily satisfied. We stress the fact, therefore, that in the present
discussion we adhere to the narrower interpretation where this con-
dition is satisfied.
LEMMA cx. A ny real function
v
f(x) ==  cnXEn (x),
n=l
where cn*=ioo, EnEA, and #(En)<oc> for n== 1, . . " p, is a step
function (note, therefore, that the numbers C n are not necessarily different
from each other and from zero, and the sets En are not necessarily disjoint).
PROOF. If 5 is an arbitrary non-empty subset of the index set
{I, . . " P}, and 5' is its complement, we define
F s == II (Ei-E j )== II EiEj.
iES,jES' iES,jES'
In particular F s== III E i for 5 =={1, . . ., P}. Then F sEA for all 5 (since
A is a ring), F S1 F s2 ==0 for 51*52, all Fs are of finite measure, and
the number of different F s is finite. If xEE i for some index i, then
either xEE j or xEEj for any index j *i, so x is a point of exactly one
of the sets F s for which i E5. Conversely, if x is a point of one of these
sets, then xEE i . Hence E i is the union of all Fs satisfying iE5, and
this shows that
f(x) ==  CiXEi(X) ==  dSXFs(x) ,
where d s ==  Ci.
iES
Deleting the F s for which F s == 0 or d s == 0, and joining terms with
equal coefficients, we obtain the standard representation of a step
function.
It is an immediate consequence of the lemma that the collection L
of all step functions (with respect to # and A) is a linear collection, i.e.,
if f, gEL and a, b are real constants, then af(x) +bg(x) EL. Also, given
f, gEL, it is not difficult to prove that the functions max (f, g) and
min (f, g) both belong to L (the function h==max (I, g) is the function
h(x), satisfying h(x)==max(f(x), g(x)) for each XEX). Indeed, if
f(x) ==  CiXEi (x)
1
and
n
g(x) ==  djXFj(X)
1
m

Ch. 3,9 12J INTRODUCTION OF THE DANIELL INTEGRAL
75
are the standard representations of j and g respectively, then j(x) ==Ci
and g(x)==d j on EiFj, j(X)==Ci and g(x)==O on E i -  Fj, j(x)==O and
g(x) ==d j on Fj-  Ei, and j(x) ==g(x) ==0 elsewhere. Hence, if h==
max(j, g), then h(x)==max(ci, d j ) on EiFj, h(x)==max(Ci, 0) on
E i -  Fj, h(x) ==max (0, d j ) on Fj-  Ei, and h(x) ==0 elsewhere. This
shows that h(x) is a step function.
In future applications we shall encounter a slightly different initial
situation yielding, however, the same final situation as described in
the preceding paragraph. Let, once more, # be a measure in X, and
let r be a semi-ring of #-measurable subsets of X, with the special
property that the difference of any two sets of r is a a-set with a
finite number of terms (such as, e.g., the semi-ring of all cells in R k ).
Then, as observed in sec. 4, the collection of all a-sets, containing only
a finite number of non-empty terms, is a ring A. Any real function
f(x) == l cnXEn(x), where Cn:F =:1=00, EnE rand #(En) <00 for n== 1, . . "
p, is now appropriately called a step function with respect to # and r.
It is evident that any step function with respect to # and the semi-ring
r is a step function with respect to # and the ring A, and conversely.
The collection L of all such step functions is, therefore, linear, and
j, gEL implies max (j, g) EL and min (j, g) EL. The main difference with
the situation in the preceding paragraph is that, in general, an arbi-
trary function JEL does not have its standard representation in terms
of characteristic functions of sets of the semi-ring r, but merely in
terms of characteristic functions of sets of the ring A.
THEOREM 1. I j, jor all
p
j==  cnXEn(X) EL,
1
we dejine
p
f(j)== cn#(En),
1
then f(j) is uniquely determined on L, and
(1) -oo<f(j) <+00,
(2) f(j+g) ==f(j) +f(g),
(3) f(aj) ==af(j) jor any real constant a:F=:I=oo,

76
DANIELL INTEGRAL
[Ch. 3, 9 12
(4) il I(x)g(x) on X, then f(f)f(g),
(5) il fn(x)! 0 for all XEX, then f(ln)! O.
PROOF. The uniqueness of f(f) is derived by observing that if (with
the notations as in the proof of Lemma ex) we have
p
f(x) ==  CiXEi(X)
1
and
E i ==  Fs,
iES
then
I(x) ==  dSXFs(X) ,
where d s ==  Ci.
iES
Hence
ci#(E i ) ==Ci  #(F s),
iES
so f(/) ==  ds#(F s), and this shows that if the standard representation
of I is i dnXFn(X) , then f(/)==i dn#(Fn). Since the standard repre-
sentation of I is unique, the same is true of f(f). Of the other proper-
ties of f(f) it is only (5) which needs a formal proof. We note first that
In!O on X implies In(x)O on X for all n, so f(fn)O. Given e>O, let
Pn=={X: fn(x)e}
for n = 1, 2, . . . ;
since In is a step function, the set P n is a set of A of finite measure.
Also, since In+1(X)/n(x) on X, we have Pn+1 cPn for all n. It follows
that P==lim P n== nl P n exists, and #(P) ==lim #(P n) by sec. 8, Theo-
rem 3. But P== nl Pn==0 on account of In(x)!O for all XEX, so
lim #(P n) ==#(0) ==0. Hence
#(P n) <e for nno(e).
Let M ==max 11(X) on X, let A be the measure of the set {x: 11(X) >O},
let nno(e), and In(x)==l CiXEi(X) where all E i are disjoint. Observing
now that some of the E i are entirely included in P n , and the remaining
E i are entirely included in P, we have
p p
In(x) ==  CiXEiPn(x) +  CiXEiPnC(X) ,
i= 1 i= 1
so
f(ln)M#(Pn)+eA<Me+eA==e(M +A).
It follows that f(ln)!O.

Ch. 3, 9 12J INTRODUCTION OF THE DANIELL INTEGRAL
77
The reader will have observed that, for a step function I(x), the
above f(/) is exactly what one intuitively feels that the integral of
I(x) with respect to the measure # should be. This is not the only
example, however, of a linear collection L of real functions f(x) with
the property that to each IEL corresponds a number f(f) such that
(1)-(5) of Theorem 1 are satisfied. We shall present a second example.
Let X==Rk (k-dimensional Euclidean space), and let L be the col-
lection of all real functions I(x), continuous on Rk, and such that the
set {x: l(x)=¥=O} is bounded for each IEL. It is customary to call the
closure C f of the set {x: I(x) =¥=O} the carrier of I(x); the collection L
may, therefore, be described as the collection of all real continuous
I(x) with bounded carriers C f (note that this does not mean, of course,
that the sets C f , IEL, are uniformly bounded). Evidently, L is a linear
collection, and I, gEL implies max (I, g) EL and min (I, g) EL. Any IEL
is Riemann integrable over the space Rk (in fact, if A is a closed inter-
val such that A:::> C f, then the integral of lover the whole space is
equal to the integral of lover A). Let f(/) be the Riemann integral of
IEL over X=R k . It is obvious that f(/) satisfies (1)-(4) of Theorem 1,
and we shall prove that (5) is satisfied as well. Let, for this purpose,
InEL and In!O on X==Rk. Given e>O, we introduce the sets
Pn=={x: In(x)e},
just as in the first example, where we proved that nr P n =0. Here
we shall obtain the even more striking result that Pno==0 for some
index no. Indeed, if
On==P=={x: In (x) <e},
then On is open, and On+1=>On for all n in view of In+1(X)/n(x) on X.
Furthermore, on account of In(x)!O for all XEX, every point XEX is
a point of On for all n?::-nx. Hence, r On==X, that is, r On covers
the carrier C h of 11(X). But then, by the Heine-Borel-Lebesgue covering
theorem, o On => Cfl for some index no, i.e., OnO=>C f1 , This means
that fno(x)<e for all XEC f1 , so (since Ino==O outside C f1 ) we have
fno(x)<e for all XEX. But then Ono==X, i.e., Pno==0. It follows now
immediately from fno(x)<e, holding on X, that In(x)<e for all n?::-no
and all XEX; in other words, In(x)! 0 holds uniformly on X. This result

78
DANIELL INTEGRAL
[Ch. 3, 9 12
is known as Dini's theorem. Finally, observing that all fn(x) vanish
outside Cfl' we obtain the desired property that f(fn)!O.
In order to obtain a common generalization of both examples, we
shall assume now that X is an arbitrary non-empty point set, and that
L is a non-empty linear collection of real finite functions f(x), defined
on X (if, therefore, fEL, then -oo<f(x) < +00 on X, and if f, gEL
and a, b are real constants, then af+bgEL). We assume furthermore
that if the functions f(x) and g(x) are in L, then the functions max (f, g)
and min(f, g) are also in L. It follows that f(x)E,L implies If(x)IEL,
since Ifl==max(f, O)-min(f, 0). Our next hypothesis is that on the col-
lection L there is defined a real linear functional f(f), i.e., given any
f(x) EL, there exists a real number f(f) such that
(1) -oo<f(f) <+00,
(2) f(f+g) ==f(f) +f(g),
(3) f(af)==af(f) for any real constant a=¥==l:oo.
We assume, furthermore, that the linear functional f(f) is non-nega-
tive, i.e., if f(x)O on X, then f(f)O (the term non-negative does not
mean, therefore, that f(f)O for all fEL). By the linearity of f(f),
this is equivalent to:
(4) if f(x)g(x) on X, then f(f)f(g).
Our final hypothesis is that f(f) is continuous under monotone con-
vergence to zero, i.e.:
(5) if fn(x)! 0 for all XEX, then f(fn)! o.
Any linear functional of this type is called an elementary integral on
the collection L; more precisely, f(f) is called the elementary f-inte-
gral of f(x) over X. If all confusion is excluded, we shall speak simply
about the integral of f, and we shall sometimes write ff instead of f(f).
I t will be our purpose, in the next two sections, to extend the ele-
mentary integral f(f) to a larger collection of functions (a collection,
therefore, which, if possible, contains L as a proper subcollection) in
such a manner that the properties (1 )-(5) are preserved as much as
possible. This extension procedure was introduced by P. J. DANIELL
(1918, [lJ, [3J), and for any f(x) in the extended collection the extended

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 79
f(f) is called the Daniell f-integral of f(x) over X. In the particular
case that L is the collection of all step functions
p
j(x) ==  cnXEn (x),
1
introduced in the beginning of this section, and
p
f(f)== cn#(E n ),
1
the extended integral is called a Stieltjes-Lebesgue-Radon integral (some-
times, these mathematicians' names are cited in a different order, or
only two out of the three names are used). In the still more particular
case that X is k-dimensional number space Rk, and # is Lebesgue
measure in Rk, the result will be the Lebesgue integral, due to H. LEBES-
GUE (1902, [2J, [3J).
Exercises
FURTHER EXAMPLES OF ELEMENTARY INTEGRALS
12.1) Show that if L is the collection of all real continuous functions
f(x) on the bounded closed interval A cR k , and f(f) is the Riemann
integral of f over A, then Land f(f) satisfy the conditions specified
above. This example served as an important illustration in Daniell's
first investigations. Note that Dini's theorem holds, i.e., given the
sequence fnEL such that fn!O on A, we have fn!O uniformly on A.
12.2) Let X =={x: xO}, let L be the collection of all functions f(x) ==
==ax (a real) defined on X, and let f(f)==a for f(x)==ax. Show that L
and f(f) satisfy the conditions specified above.
13. The Daniell Integral for Non-negative Functions
The approach to the extension problem for integrals by P. J.
DANIELL (1918, [IJ, [3J), and subsequent similar approaches by S.
BANACH (1937, [5J), H. H. GOLDSTINE (1941, [lJ) and M. H. STONE
(1948, [IJ), are very similar to the extension procedure for measures
from semi-rings to larger set collections. This is most marked in Stone's

80
DANIELL INTEGRAL
[Ch. 3, 9 13
version, and we shall show here that the problem for integrals is actu-
ally a particular case of the corresponding problem for measures, avoid-
ing thereby to a certain extent a repetition of the same arguments as
in the extension theory for measures.
Given two non-empty point sets X and Y, and the subsets A eX
and BeY, the set of all ordered pairs (x, y) such that xEA and YEB,
is called the Cartesian product of A and B, and denoted by A X B.
Hence A xBcXx Y. If A==0 or B==0, then A xB==0. The elements
(x, y) of Ax B are called the points of A X B.
Let X be a non-empty point set, and Rt the set of all finite positive
numbers. Given the real function l(x)O on X (where it is permitted
that I(x) assumes the value +00), we shall mean by the exterior ordinate
set F of I(x) the subset of X X Rt, defined by
F=={(x, y): XEX, O<y/(x) if I(x) <00, and O<y</(x) if I(x)==+oo},
and we shall mean by the interior ordinate set FO of I(x) the subset of
X X Rt, defined by
FO=={(x, y): XEX; O<y</(x)}.
Evidently po e F. Note that if I(xo) ==0, then the point (xo, 0) does not
belong to F (and then not to FO either). Hence, if I(x) ==0 for all XEX,
then F == 0. Generally, if I(x), g(x), . .. is a collection of non-negative
functions on X, we shall without further specification denote the corre-
sponding exterior ordinate sets by F, G, . . . .
As in the preceding section, we assume now that L is a non-empty
linear collection of real finite functions on X, such that I, gEL implies
max (I, g) EL and min (I, g) EL. The subcollection of all l(x)O in L will
be denoted by L +. Note that L + is not empty, since L + contains at
least the function which is identically zero on X.
THEOREM 1. The collection r 01 all dillerences F-G, where I, gEL+,
. ..
s a s em-rng.
PROOF. We observe first that although the set F -G in X X Rt is
uniquely determined by the non-negative functions 1 and g, the converse

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 81
is not true. Indeed, F -G==F 1-G 1 implies merely that I(x) ==/1 (x) and
g(X)==gl(X) at the points XEX where I(x»g(x), and that 11(X)gl(X)
at the points XEX where I(x)g(x). It follows that if F -GE r is given,
we may assume without loss of generality that g(x)/(x) on X, since
F-G==F-G 1 , where gl==min(/, g)EL+.
(a) Since L + contains at least one function I with corresponding
exterior ordinate set F, we have 0==F -FE r .
(b) Given A==F-GE r and B==F 1 -G 1 E r, we have AB==F2-G 2 ,
where 12==min(/, 11)EL+ and g2==max(g, gl)EL+, so ABE r.
(c) Let A==F -GE r and B==F1-G1E r, where B cA. Then we may
assume that 1/lglg on X (otherwise, assuming already that Ig
and 11gl, we may first replace 11 and gl by 12==min (I, 11) and g2==
min (I, gl) respectively, and then 12 and g2 by 13max (g, 12) and g3==
max(g, g2) respectively), so A-B==(F-F1)+(G1-G) is the union of
two sets of r.
It follows that r is a semi-ring.
Fronl now on we shall assume, unless stated otherwise, that if F-G
E r, then I(x)g(x) on X.
THEOREM 2. II r is the semi-ring 01 the preceding theorem, and f(/)
is an elementary integral on the lunction collection L, then the set lunction
il on r, delined by il(F-G)==f(/-g), is a measure on r.
PROOF. We note first that il(F -G) is uniquely determined by F -G,
for if F -G==F 1-G1, then 1== 11 and g=gl for all x satisfying I(x) >g(x) ,
and 11==gl for all x satisfying I(x)==g(x), so l-g==/1-g1 for all XEX.
Evidently il(0) ==0, and Oil(A)<oo for any A E r. Furthermore, il is
monotone. It will be sufficient, therefore, to prove that il is a-additive
on r. Let, for this purpose, F-G== (Fn-Gn), where F-GE r, all
Fn-GnE r, and all Fn-Gn are disjoint. Then, by the properties of
Lebesgue measure in the real line,
n
r n(x) =={/(x) -g(x)}-  {Ii (x) -gi(X)}! 0
i=l
for all XEX,
so lim f(rn) ==0 by property (5) of the elementary integral. Hence,
n
f(f-g)==lim  f(/i-gi),
i=l

82
DANIELL INTEGRAL
[Ch. 3, 9 13
1.e.,
n 00
fi(F-G)==lim  fi(Fi-Gi)== fi(Fi-G i ).
i=l 1
This shows that fi is a-additive on r.
Applying the extension procedure for measures to (X X Rt, r, fi),
the measure fi on r is now extended, and we obtain first the exterior
measure fi* in X X Rt, and then the extended measure fi on the a-
field A of all fi-measurable subsets of X X Rt. For any function f(x)O
whose exterior ordinate set F is ,a-measurable, we define f(f) by
f(f) ==fi(F) , and (f) is then called the Daniell integral of f. The col-
lection of all f of this kind will be denoted by M+, and any fEM+ will
be called a non-negative f-measurable function. Obviously, any fEL +
is f-measurable, and the old and new values of f(f) are the same on
L + (that is, the new f(f) is an extension of the initial elementary
integral on L +).
In principle we have now completed our task (at least for non-nega-
tive functions), and it seems at first as if all that remains to be done
is the reformulation for the present case of the theorems of measure
theory. This is true to a certain extent, but it is not the whole truth.
Indeed, the sets of A are subsets of X X Rt, and in X X Rt there exists
an additive structure along the Rt-axis. Consequently, there will also
be an additive structure in M+, and this is a feature which falls outside
the immediate scope of measure theory. Hence, it is to be expected
that the linearity properties of M+ will not follow quite automatically
from the known properties of fie
THEOREM 3. (1) If f, gEM+, then max(f, g)EM+ and min(f, g)EM+.
If, in addition, f(x)g(x) on X, then f(f)f(g).
(2) If f(x)O on X, then fEM+ if and only if min (f, P) EM+ for all
P(x) EL +.
(3) If fn!fo on X, where all fnEM+ and f(fn)<OC> for at least one
value of n, then foEM+ and f(fn)!f(fo). In particular, if fn!O on X,
where all fnEM+ and f(fn)<OC> for some n, then f(fn)!O.
PROOF. (1) The first assertion is derived by observing that max (f, g)
and min (f, g) have the exterior ordinate sets F +G and FG respectively.

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 83
If, moreover, I?;::-g on X, then F => G, so
f (I) === il (F)  il ( G) == f (g) .
(2) If IEM+ and PEL+cM+, then min(/, P)EM+ by (1). Let, con-
versely, IO on X and min(/,p)EM+ for all PEL+. This means that
the ordinate sets F and P satisfy FPE A for all PEL+. But then, if
A==P-Q (PEL+, qEL+) is an arbitrary set of T, we have also FAE A
for any A E T , and it follows by sec. 7, Theorem 7 that FE A .
(3) If In! 10, the sequence of sets F n is descending and has F 0 as
its limit. Hence, if all InEM+ and f(ln)<(X) for some n, then the set
F 0 is measurable and il(F n) ! il(F 0) by sec. 8, Theorem 3. In other
words, 10EM+ and f(ln) !f(/o).
Our next aim is to show that the collection M+ is closed under
multiplication by non-negative constants. For this purpose we intro-
duce the following notation. Given the set E c X X Rt and the constant
a>O, we shall denote by aE the set of all points (x, ay) for which
(x, Y)EE. Note that aE 1 E 2 ===(aE 1 )n(aE 2 ) and aEe==(aE)e.
THEOREM 4. II IEM+ and a is a non-negative constant, then aIEM+
and f (al) === af (I). I I E is an arbitrary subset 01 X X Rt and a is a posi-
tive constant, then il*(aE) ==ail*(E).
PROOF. In a certain sense the theorem is trivial since, for a>O, the
mapping I-+al of L + onto the whole of L + is one-one; it generates
therefore a one-one mapping of T onto the whole of T which may be
extended in a natural manner into a one-one mapping of A onto A
with the property that all measures are multiplied by a. Restricting
the mapping to ordinate sets in A , we obtain the first assertion, and
the second assertion is obtained by taking greatest lower bounds for
all il-measurable sets covering the given set E. The following lines
supply the details.
For a===O the first statement is evident (also in the case that f(/) ===(X);
we recall that O.(X)===O by definition), let therefore a>O. The set A c
XxRt satisfies AE T if and only if aAE T, and then il(aA)===ail(A) by
the linearity of f(/) on L. Hence, the set BcXxRt is a a-set if and
only if aB is a a-set, and then il(aB)==ail(B). It follows that the arbi-
trary set E c X X Rt is covered by the a-set B if and only if aE is

84
DANIELL INTEGRAL
[Ch. 3, 9 13
covered by the a-set aB; in particular, E fails to be sequentially
covered by r if and only if aE fails to be so. If E and aE are sequential-
ly covered, we obtain ii*(aE) =aii*(E) by taking greatest lower bounds
for the measures of all a-sets B :::> E and aB:::> aE respectively; if E and
aE fail to be sequentially covered, then ii*(E) =ii*(aE) =00, so ii*(aE) ==
aii* (E) .
Let now EE A . We wish to prove that aEE A ; it will follow then
from the already established result that ii(aE) ==aii(E), and in particu-
lar it will follow that fEM+ implies afEM+ with f(af)==af(f). In
order to prove that aEE A , it is sufficient to show that
ii*(5) =ii*(5naE) +ii*{5n(aE)e}
for any
5 c XxRt.
Now, since EE A , we have
ii*(a- 1 5) =ii*(a- 1 5nE) + ii*(a- 1 5nEe);
hence, multiplying by a,
ii*(5) ==ii*(5naE) +ii*(5naEe) ==ii*(5naE) +ii*{5n(aE)e}.
This shows that aEE A .
Next, we shall consider the interior ordinate set FO of a function
f(x)O. The difference G==F -Fo is sometimes called the graPh of f(x);
properly speaking, G is that part of what is usually called the graph of
f where O<f(x) < 00.
THEOREM 5. (1) If f(x)O on X, if FOcHcF, and H is ii-measur-
able, then F and FO are ii-measurable, so jEM+. In particular, F and FO
are ii-measurable or non-ii-measurable simultaneously.
(2) ii(FO) ==ii(F) ==f(f) for any fEM+ (although ii(F -FO) ==0 does not
necessarily hold,' cf. ExamPle 4 at the end of the section).
PROOF. (1) Since HE A , we have
(l+n-1)HE A
and
(l-n- 1 )HE A
for n==2, 3, . . .
by the proof of the preceding theorem. The desired result that FE A
and FOE A is derived by observing that
F ==lim.(1 +n-1)H
and
FO=lim(1-n-1)H
as n-+oo.

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 85
(2) If il(F -FO) ===ex>O, then
il{n- 1 (F -FO)} == ex/n
for n===2, 3, . . "
so
il(FO)lex+lex+ . . . ===00,
SInce
FOl(F-FO)+l(F-FO)+... .
It follows that il(F) ===il(FO) ==00.
If il(F-FO) ==0, then
il(F) === il(FO) + il(F - FO) == il(FO).
The property that il(FO)==f(/) holds for all IEM+ will enable us now
to handle not only decreasing sequences of functions as in Theo-
rem 3 (3), but also increasing sequences (this could not be done so
easily before, because it is not true in general that O/n i I implies
F n iF; it is true, however, that OI nil implies F i FO).
THEOREM 6 (INTEGRATION OF INCREASING SEQUENCES). II In i I on
X, and all InEM+, then IEM+, and f(ln) i f(/).
PROOF. Since all FE A and Fi FO, we have FOE A and il(F) i il(FO)
by sec. 8, Theorem 3 (1), so IEM+ and f(ln) i f(/).
The theorem on the integration of increasing sequences, for the par-
ticular case of Lebesgue integration, is due to B. LEVI (1906, [1 J). He
assumed, in addition, that the sequence of numbers f(ln) is bounded.
For the case that X is a bounded interval and the sequence In (x) is
uniformly bounded, the theorem was established already earlier by
H. Lebesgue.
THEOREM 7. Let In (x) EM+ lor n=== 1, 2, . . " and let
h(x) ===sup I n(x),
P(x) ==lim sup I n(x)
and
k(x) ==inf I n(x),
q(x) ==lim inf I n(x) ;
their exterior ordinate sets will be denoted by H, K, P and Q respectively.
Then H, K, P and Q are measurable (i.e., h, k, p, qEM+), and they have

86
DANIELL INTEGRAL
[Ch. 3, 9 13
the same measure as
00 00
 F n, II F n, lim sup F nand lim inf F n
1 1
respectively. In particular, il In EM + (n=== 1,2, . . .) and lo==lim In exists,
then IOEM+.
PROOF. Since Ho==r F and K==nr Fn, we have immediately
the result that h, kEM+ and fi(K)===fi(nr Fn). Furthermore, if (x, Y)E
r F n, then y/n(x) for some n, hence yh(x), i.e. (x, y) EH. This
shows that r FncH. It follows then from
00 00
HO== Fc FncH
1 1
and
fi(HO) ==fi(H)
that fi(H)=fi(r F n).
Writing hn==sup (In, In+1, . · .) and kn===inf (In, In+1, . . .) for n== 1,2,
. . " we have hnEM+, knEM+ for all n, and hn!P, k n jq on X, so
p, qEM+. Furthermore, if (x, y) EPO, then y<P(x), i.e., y<ln(x) for
infinitely many n; hence (x, y) Eljm sup F. This shows that po c
lim sup F. Conversely, if (x, y) Elim sup F n, then y/n(x) for infi-
nitely many n, Le., yP(x); hence (x, y) EP. This shows that lim sup F n
c P. It follows then from
POclim sup Fclim sup FncP
and
fi(PO) ==fi(P)
that fi(P)==fi(lim sup Fn). The proof for Q is similar.
THEOREM 8 (FATOU'S LEMMA; P. FATOU, 1906, [lJ, FOR LEBESGUE
INTEGRATION). II In EM + lor n== 1,2, . . " then
(lim inf In)lim inf (In)'
PROOF. Writing again q(x)==lim inf In(x), we have (by the preceding
theorem and sec. 8, Theorem 4)
f(lim inf In)==f(q) ===fi(Q) ==fi(lim inf F n)
lim inf fi(Fn)==lim inf f(fn).

Ch. 3,  13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 87
THEOREM 9. If fnEM+ for n==l, 2, "', and fn(x)g(x) for all nand
all XEX, where gEM+ and f(g) <00, then
f(lim sup fn)lim sup f(fn).
If, in addition, f(x)==lim fn(x) exists, then f(f)==lim f(fn).
PROOF. Since F n c G for all n, we have  F n c G, so
ii(F 1 +F2+' · . )ii(G)==f(g)<oo.
Hence (by sec. 8, Theorem 4),
ii(lim sup F n)lim sup ii(F n),
so that by Theorem 7
f(lim sup fn) ==ii(lim sup F n)lim sup ii(F n) ==lim sup f(fn).
If, in addition, f(x) ==lim fn(x) exists, then it follows, by combining this
result and Fatou's lemma, that
f(f)lim sup f(fn)1im inf f(fn)f(f),
so f(f)==limf(fn).
The final part of this section will be devoted to the proof that f(f)
is linear on M+. We first note some auxiliary facts
If any function s(x)O on X, having the property that there exists
a sequence In(x) EL + satisfying In i S, is called a a-function for the
present purposes, then it is evident by Theorem 6 that such a a-
function s(x) is in M+, and f(ln) i f(s). If Sl(X) and S2(X) are a-functions,
then s3==min (Sl, S2) is also a a-function (if In i Sl, k n i S2, then min(ln, k n )
i S3). Next, if any function f(x) on X, having the property that sn!f
for some sequence of a-functions sn(x), is called a a6-function, then it is
evident that such a a6-function f is in M+, and Theorem 3 (3) shows
that f(sn) !f(f) if f(sn)<oo for some value of n. Obviously, if f1(X)
and f2(X) are a6-functions, then f3==min (f1, f2) is also a a6-function.
Given now any ii-measurable subset E of X X Rt of finite ii-measure,
there exists by sec. 9, Theorem 2 (3) a descending sequence of a-sets Bn
such that E is the difference of the limit of the sequence Bn and a set
of /i-measure zero. This holds in particular if E is the exterior ordinate
set of a function of M+ with a finite integral, but it is not true in general

88
DANIELL INTEGRAL
[Ch. 3,  13
that the corresponding a-sets Bn are then automatically ordinate sets
of a-functions. The following lemma supplies more precise information
concerning this point.
LEMMA cx. Given fEM+ with f(f)<oc>, there exists a decreasing se-
quence of a-functions sn(x)f(x), such that all f(sn) are finite, and the
G6-function f1(x)==lim sn(x) satisfies f(sn)!f(f1)==f(f). In addition,
f1-fEM+ and f(f1-f) ==0.
PROOF. Given E>O, there is (by sec. 9, Theorem 2 (3)) a a-set 0==
 (Pn-Qn) with disjoint terms in XxRt, covering the ordinate set F
of f, such that
il(O) -il(F) ==il(O- F) <E.
The sets Pn and Qn in O== (Pn-Qn) are the ordinate sets of the
functions Pn, qn EL + (Pnqn on X), and it follows from  (Pn-Qn) => F
that
s(x) ==  {Pn(x) -qn(x)}f(x)
for all XEX'
,
hence, the a-function s(x) exceeds f(x), and
k k 
f(s)==lim f{ (Pn-qn)}==lim  f(Pn-qn)== f(Pn-qn)
k 1 1 1
by Theorem 6 and the linearity of f(f) on L +. The ordinate set 5 of s
satisfies, therefore, 5 => F and
so
il(5)==f(s)== f(Pn-qn)== il(Pn-Qn)==il(O),
Of(s) -f(f) ==il(5) -il(F) ==il(O) -il(F) <E.
Note that replacing 0 by 5 has the effect that the straggling threads
(or, more-dimensionally speaking, the straggling sheets) of 0 outside F
are now neatly put on top of each other. It is an immediate conse-
quence of the foregoing that there exists a sequence of a-functions
sn(x), all of them exceeding f(x), such that f(sn)-f(f) is finite and
tends to zero as noc>. We may even assume that Sn+1Sn on X for all
n (otherwise, we replace S2 by s2==min (Sl, S2), S3 by s3==min (S2, S3),
and so on). Hence, there is a a6- f unction f1 such that Sn!f1f, and
f(sn) !f(f1) since all f(sn) are finite. But we have also f(sn) !f(j),

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 89
so f(f1)==f(f), i.e., the set F 1 -F is of ,a-measure zero. This implies,
finally, that f1-fEM+ and f(f1-f) ==0. Indeed, given any E>O, there
is a a-set O'== (P-Q) with disjoint terms, covering F 1 -F, and
such that ,a(0') <E. Then
s' (x) ==  {p(x) -q(x)}
is a a-function exceeding d(x) ==f1(X) - f(x), and f(s') ==,a(0') <E. Hence,
the ordinate set D of d(x) satisfies ,a*(D)f(s') <E. This holds for all
E>O; it follows that D is ,a-measurable and ,a (D) ==0, so d==f1-fEM+
and f(f1-f)==f(d)==0.
The lemma shows that, given any fEM+ with f(f) <00, there exists
a a6-function f1f such that f(f1-f) ==0. This suggests that it may be
useful for our purposes to consider more closely the functions h(x) EM+
satisfying f(h)==O. Such a function h(x) is called a null function; it
follows by comparison of the ordinate sets that any function h 1 (x),
satisfying 0h1h, is also a null function. Any subset E of X, whose
characteristic function XE(X) is a null function, is called a null set. If E
is such a null set, then f(nXE)==nf(XE) ==0 for n== 1,2, .. " so f(oo' XE)
==0 by the theorem on integration of increasing sequences. It follows
that the ordinate set E X Rt of 00' XE is of ,a-measure zero. Hence, any
non-negative function h(x), vanishing outside a null set EeX, is a
null function. Conversely, if h(x)O is a null function, then E ==
{x: h(x) >O} is a null set. Indeed, setting
En=={x: h(x)n-1} for n== 1,2, .. .,
we have n-1XEnh on X, so n- 1 XEn is a null function, i.e., f(XEn)==O.
Hence, since XEnixE, we have also f(XE)==O, so E is a null set. The
following lemma is now an easy consequence.
LEMMA fJ. If fEM+, and if the non-negative function f1(X) has the
property that {x: f1(X) =l=-f(x)} is a null set, then f1 EM + and f(fl)==f(f).
Furthermore, the union of two null sets is a null set.
PROOF. Let E =={x: f1 =l=-f}. Since E is a null set, we have ,a(E X Rt)
==0. Hence, if F is the ordinate set of f, the sets F + (E X Rt) :and
F - (E X Rt) are ,a-measurable, and of the same ,a-measure as F. Further-
more, the ordinate set F 1 of f1 is included between F - (E X Rt) and

90
DANIELL INTEGRAL
[Ch. 3,  13
F + (E X Rt). Hence, F 1 is fi-measurable, and fi(F 1) ==12 (F) , i.e., 11 EM+
and f(/1)==f(/).
Next, if E1 and E 2 are null sets, then the ordinate sets of XEI and
XE 2 are of fi-measure zero. It follows that their union is also of 12-
measure zero; this union, however, is exactly the ordinate set of
XE 1 +E 2 . Hence, f(XE 1 +E 2 )==0, i.e., E1+E2 is a null set.
THEOREM 10. (1) II I, gEM+, then l+gEM+ and f(/+g)==f(/)+
J(g).
(2) II I, gEM+, and g1 on X, then l-gEM+.
PROOF. (1) If I, gEL+, then l+gEL+ and f(l+g)==f(/)+f(g).
Hence, if I and g are a-functions, then I+g is a a-function, and
J(/+g)==f(/)+f(g). Indeed, if lnil and knig, where In, knEL+, then
In+kn i I+g, so I+g is a a-function, and
f(/+g)==lim f(ln+kn)==lim f(ln)+lim f(kn) ==f(/) +f(g).
Next, let I and g be a6-functions such that f(/) and f(g) are finite.
Then there exist sequences of a-functions sn!1 and tn!g such that
f(sn) !f(/) and f(t n ) !f(g). Since sn+tn!/+g, Theorem 3 (3) shows
that l+gEM+, and
f(/+g)==lim f(sn+tn)=lim f(sn)+lim f(tn)=f(/)+f(g).
If I, gEM+ with f(/) and f(g) finite, it follows from Lemma ex that
there exist a6-functions 11 and gl such that 11==1 except on a null set
E1, and gl==g except on a null set E 2 . Then l+g==/1 +gl except on the
null set E1 +E2, and since 11 +gl EM + with f(/1 +gl)==f(/1)+f(gl)
by what has already been proved, a repeated application of Lemma fJ
shows now that l+gEM+ with
f(/+g) ==f(/1 +gl) ==f(/1) +f(gl) =f(j) +(g).
The case that one at least of the numbers (/) and f(g) is infinite
is left to investigate. In order to show that l+gEM+, it will be suf-
ficient to prove that min (I+g, P) EM+ for any pEL + (by Theorem 3 (2)).
Writing j1==min(l, P) and gl==min(g, P), we have min(/+g, P)==
min (11 +gl, P); hence, it is sufficient to prove that min (11 +gl, P) EM+.
But 11, gl EM+, and the integrals f(j1) and f(gl) are finite (they do

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 91
not exceed f(P)), so 11 +glEM+ by what has already been proved. It
follows that min (11 +gl, P) EM+, so l+gEM+. Finally, since j+gj
and I+gg, the equality f(l+g)==oo==f(/)+J(g) is evident. This
completes the proof of the first part.
(2) We assume first that gjP on X for some fixed P(x) EL +. Then,
if I, gEL +, the assertion is true (i.e., j-gEM+), and the same holds if
I and g are a-functions, since In i j, k n i g, where In, knEL + and knln,
implies I-g==lim (In-kn) EM+ by Theorem 7 (it is permitted to as-
sume that knln on X, since otherwise we replace k n by min(kn, In)).
If I and g are a6-functions, the assertion is also true, since sn!/, tn!g,
where Sn and t n are a-functions and tnsnp on X, implies I-g==
lim(sn-tn)EM+ by Theorem 7 (it is permitted to assume that tn
snP on X, since otherwise Sn may be replaced by s==min (sn, P) and
t n by min (s, t n )).
Next, assuming only that I, gEM+ satisfy g/P on X, we may
conclude again by Lemma ex that there exist a6- f unctions 111 and
glg such that 11==1 and gl==g, except on a null set. We may assume
that gl/1P, since otherwise 11 may be replaced by ji==min(/1, P)
and gl by min(gl, Ii). Then l-g==/1-g1 except on a null set, and
since 11-g1 EM+ by what has already been proved, an application of
Lemma fJ shows that l-gEM+.
We now remove the restriction that IP on X for some fixed
P(x) EL +, i.e., we assume only that I, gEM+ and g1 on X. Choose an
arbitrary PEL+, and let In==min(l,nP), gn==min(g,np) for n==I,2,
. . . . Then In, gn EM + and gnjnnp on X, so In-gnEM+ by what
has already been proved. Furthermore In-gn i k, where k(x)==/(x) -g(x)
at all x where P(x) >0, and k(x) ==0 at all x where P(x) ==0. It follows
that k(X)EM+, so that, since min(k, p)==min(/-g, P) for all XEX, we
have finally min(/-g, P)EM+ for all PEL+. But then l-gEM+ by
Theorem 3 (2).
It is evident that, once it has been proved that j-gEM+, we obtain
as an immediate consequence the result that f(/) ==f(g) +f(j-g), and
this implies f(/-g)==f(j)-f(g) if f(g) <00. Note, concerning this
point, that j(x) ==g(x) + {/(x)-g(x)} holds for all XEX, also if j(x)==+oo
or j(x) ==g(x) == +00.

92
DANIELL INTEGRAL
rCh. 3, 9 13
We make one final remark. In view of the linearity of f(f) on M+,
the theorem on the integration of increasing sequences may also be
formulated as follows:
If g(x)==r gn(x) on X, and all gnEM+, then gEM+ and f(g)==
r f(gn).
Examples. (1) X is the interval (0, 2J on the real line ; L consists of
all functions f(x) ==ax, where a is a real constant, and f(f) ==2a. Then
L + consists of all f(x) ==ax with aO; the ordinate set of a function
fEL + is therefore a triangle, and f(f) is the ordinary area of the trian-
gle. If the a-function s(x) is the limit of the increasing sequence Pn(x) ==
anxEL +, then s(x) is either identically +00 or s(x) ==ax, where a==
lim an<oo. In the second case, therefore, s(x)EL+. Similarly, if f(x) is
a a6-function not identically +00, then again fEL+. Next, if fEM+
and f(f) ==0, there exists a a-function s(x) ==axf(x) with arbitrarily
small a; hence, f(x) ==0 on X. It follows by Lemma ex that any fEM+,
satisfying f(f) <00, is of the form f(x) ==ax with aO, i.e. fEL +. Finally,
in view of the fact that fEM+ if and only if min (f, P) EM+ for all pEL +, it
is not difficult to verify now that any fEM+ is either identically +00 or
fEL +. The extension procedure, therefore, adds to the functions of L +
only the function which is +00 for all XEX.
(2) Let X be the union of the intervals [-1,0) and (0, IJ, and let
{ 0 on [-1, 0), { X on [-1, 0),
q?1 (x) == q?2(X) ==
x on (0, IJ, 0 on (0, IJ.
Let L consist of all linear combinations f(x) ==aq?l(x) +bq?2(X) with
f(f)==a. Then the collection M+ consists of all f(x)O which are arbi-
trary on [-1,0) and of the form aq?l(x) with Oaoo on (0, 1J. This
example shows that f(f)==O as soon as f(x) ==0 on (0, IJ; hence, there
are many nonzero functions tor which f(f) ==0.
(3) Let X and qJ1(X) be the same as in the preceding example, and
let L consist of all f(x) == aqJ1 (x) with f(f)==a. If E is any non-empty
subset of [ - 1, 0) X Rt, then E is not sequentially covered by the semi-
ring T, and EA==0 for all AE T, so E is ,a-measurable and ,a (E) ==00.
The collection M+ is, therefore, the same as in the preceding example,
but f(f)==oo for fEM+ as soon as f(x) >0 for at least one XE[ -1,0).
(4) Let X be a non-empty point set, and let #(E), for EeX, be the

Ch. 3, 9 13J DANIELL INTEGRAL FOR NON-NEGATIVE FUNCTIONS 93
number of points in E (i.e., # is the discrete measure in X). Further-
more, let L be the collection of all step functions f(x) == l cnXEn (x)
with f(f)==l cn#(En). This may also be described by saying that L
consists of all f(x) satisfying f(x) ==0 except at a finite number of points
Xl, . . " x q , and f(f) == f f(x n ) for such a function fEL. It is easily
verified that, for any non-negative function f(x) o X and any P(x) EL +,
the function min(f, P) is a step function; hence, M+ consists of all non-
negative functions on X. If fEM+, and f(x) >0 at an uncountable
number of points x, then f(f)==oc>; if the number of points x at which
f(x) >0 is at most countable, say Xl, X2, . . " then f(f)==  f(x n ). Note
that if the total number of points in X is finite, then any fEM+ satis-
fying f(f)<oc> is already in L+. Note also (in connection with Theo-
rem 5) that if the total number of points in X is uncountable, then the
graph G of the function xx(x) satisfies il(G) ==oc>, since G is not se-
quentially covered by r.
Exercises
"THE GRAPH OF A FUNCTION
13.1) Show that the graph G==F -FO of the function fEM+ satisfies
either il(G) ==0 or il(G) ==oc>.
AN EQUIVALENCE THEOREM
13.2) Let Land f(f) satisfy the conditions specified in this section,
and assume that, with respect to the same point set X, the pair L1
and f1(f) satisfies the same conditions (i.e., f1(f) is an elementary
integral on the linear function collection L 1 ). Show that the corre-
sponding classes M+ and Mt of non-negative measurable functions
. are identical, and the extended integrals f (f) and f 1 (f) coincide on
M+==Mt, if and only if L+cMt with f 1 (f)==f(f) on L+ and LtcM+
with f(f)==f 1 (f) on Lt.
THE CONTRACTED INTEGRAL
13.3) Let il* be the exterior measure in X X Rt introduced in this
section, and let il be the corresponding contracted exterior measure.

94
DANIELL INTEGRAL
[Ch. 3, 9 13
Since any fi-measurable set is fie-measurable and conversely by Exer-
cise 7.3, the functions of M+ are exactly those with fie-measurable ordi-
nate sets. For any fEM+, we write fe(f)fie(F), where F is the ex-
terior ordinate set of f. Show:
(a) if E c X X Rt is fi-measurable, then fie(E) sup fi(EA) for all
AE T '
,
(b) if fEM+, then fe(f)sup f{min(f, P)} for all pEL+;
(c) if fEM+, then fe(f)sup f(g) for all gEM+ satisfying gf and
f(g) <00.
FINITELY ADDITIVE INTEGRAL
In the next exercises, and in further exercises on finitely additive
integrals in subsequent sections, we follow closely the exposition of
B. C. STRYDOM [1].
13.4) Let the linear collection L of functions satisfy the conditions
of this section, and let f(f) be a positive linear functional on L (i.e.
f(f) satisfies the conditions (1)-(4), but not necessarily (5), of the
preceding section). As shown in the proof of Theorem 1, the collection
T of all differences A ==P-Q, where P, Q are the exterior ordinate sets
of p, qEL + (Pq on X), is a semi-ring with the property that any differ-
ence A-B (A, BE T ) is an f-set. Write v(A)==f(P-q) for A==P-Q
E T, and show that v is a charge on ]' . The extension procedure for
charges yields the collection A of all v-measurable subsets of X X Rt
and the collection M+ of all f(x)O with v-measurable exterior ordinate
sets. Show that if f, gEM+, then max(f, g)EM+ and min(f, g)EM+. If,
in addition, fg on X, then f(f)f(g). Show also that if f(x)O on X,
then fEM+ if and only if min(f, P)EM+ for all P(x)EL+. Finally, show
that if fEM+ and a is a non-negative constant, then afEM+ and
f(af)af(f). More generally, if E is an arbitrary subset of XxRt
and a is a positive constant, then v*(aE)av*(E).
13.5) With the same notations as in the preceding exercise, show
that if fEM+ satisfies f(f) <00, and if 8>0, then there exist functions
f1, f2 EL + such that f1ff2 and f(f2-f1)<8.
13.6) With the same notations as in the preceding exercises, show
that if f, gEM+, then f+gEM+ and f(f+g)==f(f) + f(g).
13.7) With the same notations as in the preceding exercises, show

Ch. 3, 9 14J THE DANIELL INTEGRAL FOR REAL FUNCTIONS 95
that if f, gEM+, where fg on X and f(g)<oo, then f-gEM+ and
f(f-g)=f(f)- f(g).
14. The Daniell Integral for Real Functions
We first consider non-positive functions. If f(x)O on X, and
-f(X)EM+, we define f(f)=-f(-f). In the particular case that
f(x)O and fEL (i.e. -fEL+), the old and new values of f(f) are the
same.
Next, we consider real functions on X, assuming positive values as
well as negative values. For any real f(x) on X, the functions f+(x)
and f-(x) are defined by f+=max (f, 0) and f-==min (f, 0). The functions
f+(x) and f-(x) are called the positive part and negative part of f(x) re-
spectively. Evidently f==f++f- and Ifl=f+-f-. Any f(x) satisfying
f+EM+ and -f-EM+ is called an f-measurable function. We shall not
define the integral f(f) for all f-measurable functions f, but merely for
the f-measurable functions f for which one at least of the numbers
f(f+) and f(f-) is finite. For such a function f the integral f(f) is then
defined by f(f) =f(f+) +f(f-), and f is called an f-integrable functi01.
If f is f-measurable, and f(f+)==+oo, f(f-)=-oo, then f(f) is left
undefined (although it is true that f(f+) +f(f-) ==0 by our previous
definitions; this implies therefore, that the formula f(f) ==f(f+) +f(f-)
does not hold automatically for all f-measurable functions). Any f-
integrable f(x), satisfying -oo<f(f) < +00, is called an f-summable
function. Obviously, any fEL is f-summable, and the initial and new
values of f(f) are the same. We repeat the main points.
DEFINITION. If f+EM+ and -f-EM+, the function f(x) is called f-
measurable on X. The collection of all f-measurable functions will be
denoted by M.
If fEM, and one at least of the numbers f(f+) and f(f-) is finite, the
function f(x) is called f-integrable over X, and its integral f(f) is defined
by f(f) =f(f+) +f(f-) ==f(f+) -f( -f-).
If f is f-integrable, and -oo<f(f) < +00, the function f(x) is called
.f"-summable over X. The collection of all f-summable functions will be
denoted by L 1 .

96
DANIELL INTEGRAL
[Ch. 3,  14
In order to avoid confusion, it is perhaps useful to remark that some
mathematicians use the term "integrable" where we use "summable";
in that case they usually do not have a special name for the more
general notion of integrability which is used here.
We collect some simple consequences of the definitions in one theorem.
THEOREM 1. (1) II g is integrable over X with f(g) >-00, and il the
measurable lunction I(x) satislies I(x)g(x) on X, then I is integrable,
and f(/));f(g).
(2) I I I and g are measurable, then max (I, g) and min (I, g) are measur-
able. I I I and g are summable, then max (I, g) and min (I, g) are summable.
(3) II I is summable, then III is summable, and If(/)If(l/l). II I is
measurable, g is summable, and 1/1lgl on X, then I is summable.
(4) II I is integrable, and a*=l:oo is a real constant, then al is inte-
grable, and f(al)==af(/).
PROOF. (1) f(/+) and f(/-) exist since IEM, and in view of I+g+,
I-g- we have f(/+)f(g+) and f(/-)f(g-). In addition, (g-)
>-00 on account of f(g) >-00. Hence f(/-) >-00, i.e., I is inte-
grable. Finally,
f(/) ==f(/+) +f(/-)f(g+) +f(g-) ===f(g).
(2) We note first that
{max (I, g) }+=max (/+, g+),
{min (I, g) }+===min (/+, g+),
{max(/, g)}-==max(/-, g-),
{min (I, g)}-==min (/-, g-).
It follows immediately that I, gEM implies max (I, g) EM and min (I, g)
EM. Let now I, gEL 1 , and set h==max(/, g). Then h+/++g+, so
f(h+)f(/++g+) =f(/+) +f(g+) <00,
and
h- / -
7 ,
so f(h-));f(/-) > -00. It follows that f(h) is finite, i.e., hELl. The
proof for min (I, g) is similar.
(3) If I is summable, the numbers f(/+) and f(f-) are finite, so
f(I/I)==f(/+)+f(-I-) is finite. Hence, III is summable. Comparing
the formulae f(/)==f(I+)-f(-I-) and f(I/I)==f(/+)+f(-I-) and
observing that f(/+));O and f(-I-));O, we find If(/)If(I/I).

Ch. 3,9 14J THE DANIELL INTEGRAL FOR REAL FUNCTIONS 97
Next, let IEM, gEL 1 , and 1/1lgl on X. It follows from IEM that
1+, -I-EM+, so 1/1:::=1++(-I-)EM+, i.e., f(l/l) exists. Then, on ac-
count of Igi ELI and 1/1lg/, we have (I/I)f(lgl)<oo, and this im-
plies that f(/+) and f( -1-) are finite. Hence, I is summable.
(4) Evident.
Although the summability of I implies the summability of III, it is
not always true that summability of III implies the same for I; this is
due to the fact that I may fail to be measurable (cf. Exercise 17.3).
However, if I is measurable and III is summable, then I is evidently
summable. In other words, the class L 1 of all summable functions
consists exactly of those measurable I for which f(I/D is finite.
It will be our next task to show that (/) is linear on the collection
L1 of all summable functions. This is not trivial, since it is not always
true that (/+g)+==I++g+ and (/+g)-==I-+g-. In the proof an es-
sential role will be played by Theorem 10 (2) in the preceding section,
asserting that if I, gEM+ and g1 on X, then l-gEM+.
THEOREM 2. II I and g are measurable, then I+g is measurable. II I
and g are summable, then I+g is summable, and f(/+g)==f(/)+f(g).
PROOF. We first state and prove a particular case. If 11 and gl are
non-negative and measurable, then 11 -gl is measurable. Also, if 11
and gl are non-negative and summable, then 11-g1 is summable and
f(/1-g1)==f(/1)-f(gl)' In order to prove these assertions, let h==
min (11, gl). Then hEM+ and h/1, hgl on X, so 11-hEM+ and gl-h
EM+ by Theorem 10 (2) in the preceding section. In addition, at any
point XEX one at least of the functions 11-h and gl-h vanishes.
Moreover, if 11,glEL1, then hELl, and also II-hELl and gl-hELl.
It follows that, in any case, 11-h and -(gl-h) are the positive and
negative parts of the function (/1-h)-(gl-h)==/1-g1. Note that the
last equality holds at any point XEX, in particular at any point x
where 11 (X) ==gl(X) ==00. Hence, 11-g1 is measurable, and in the case
of summability
f(/1-g1):::=f(j1-h)-f(gl-h) =={f(/1) -f(h)}-{f(gl) -f(h)}
=f(/1) -f(gl).

98
DANIELL INTEGRAL
[Ch. 3, 9 14
Let now I and g be measurable, but arbitrary otherwise. Then
I+g= (/++g+) - (-I--g-),
also at the points x where I and g are infinite of opposite sign, so that,
setting 11=1++g+ and gl==-I--g-, we have the case already con-
sidered. It follows that l+g==/1-g1 is measurable, and in the summa-
bility case
f(/+g) ==f(/1-g1)==f(/1)-f(gl)
==f(/+) +f(g+) -{f( -1-) +f( -g-)}==f(/) +f(g).
Of course, there exist slight generalizations, such as: If I and g are
integrable with f(/) >-00, f(g»-oo, then I+g is integrable, and
f(/+g) ==f(/) +f(g).
If In is a sequence of functions on X, the following formulae, gener-
alizing the formulae in the proof of Theorem 1 (2), are easily derived:
(sup In)+=sup I;;, (sup In)-==sup I;; (inf In)+==inf I;;, (inf In)-==inf I;.
Hence, if all In are measurable, the same is true of sup In and inf In
(by Theorem 7 in the preceding section). Furthermore, if
hn==sup (In, I n+1, . · .)
and
kn==inf (In, In+1, · · .)
for all n,
we have
hn!lim sup In==P
and
k n flim inf In==q
(i.e., p==inf h n and q==sup k n ). Hence, if all In are measurable, then
all h n and k n are measurable by what we have already observed, so
that p and q are also measurable. In addition,
p+== (inf hn)+==ini h;; =inf {sup (I;;, t:+ l' · · .) }==lim sup I;;,
and similarly p-==lim sup I;, q+==lim inf I; and q-==lim inf I;.
THEOREM 3 (LEBESGUE'S THEOREM ON DOMINATED CONVERGENCE;
H. LEBESGUE, 1910, [4J, FOR LEBESGUE INTEGRATION). II In is a
sequence 01 measurable lunctions on X, and il there exists a summable
tunction g(x) such that Iln(x) Ig(x) lor all n and all XEX, then lim inf In(x)

Ch. 3,  14J THE DANIELL INTEGRAL FOR REAL FUNCTIONS 99
and lim sup In(x) are summable, and
-f (g) f (lim inf 1 n) lim inf f (I n) lim sup f (I n) f (lim sup 1 n) f (g).
In particular, il in addition I(x)==lim In(x) exists on X, then 1 is sum-
mable, and f(/)==1im f(ln).
PROOF. The functions P(x)==lim sup In(x) and q(x)==lim inf In(x)
are measurable, and since their absolute values do not exceed g(x) ,
they are also summable. It follows then from the Theorems 8, 9 in the
preceding section that
Of(lim inf I;)lim inf f(/)lim sup f(/;)f(lim sup I;;)f(g) ;
hence, by the remarks above,
Of(q+)lim inf f(/;)lim sup f(/;)f(P+)(g). (1)
Similarly,
Of{lim inf (-I;)}lim inf f( -I;)lim sup f( -I;)
f{lim sup (-I;)}f(g),
so that, multiplying by -1 and observing that
-lim inf(-I;)==lim sup I;
and
-lim sup (-I;)==lim inf I;,
we obtain
-f(g)f(lim inf I;)lim inf f(/;)lim sup f(/;)f(lim sup I;)O,
that is,
-f(g)f(q-)lim inf f(/;)lim sup f(/;)f(P-)O. (2)
Addition of (1) and (2) gives the two relations
-f(g)f(q)lim inf f(/;;)+lim inf f(/;)=A,
B==lim sup f(/;)+lim sup f(/;)f(P)f(g).
Since it follows from the well-known inequalities
lim inf cxn+lim inf fJnlim inf(cxn+fJn)
and
lim sup cxn+lim sup fJnlim sup (cxn+fJn)

100
DANIELL INTEGRAL
[Ch. 3,  14
that
Alim inf f(fn)lim sup f(fn)B,
\
the desired result
-f(g)f(q)lim inf f(fn)lim sup f(fn)f(P)f(g)
follows.
COROLLARY. If fn is a sequence of measurable functions on X such
that the limit function f(x)=='L/f fn(x) exists on X, and if there exists a
summable function g(x) such that Il fn(x)lg(x) for all p and all XEX,
then f is summable, and f(f)==r f(fn).
Since the functions fn may assume the values :1:00, the reader is
referred to sec. 5 for the definition of l fn(x) and r fn(x).
We recapitulate the properties of the collection LV) of all finite valued
summable functions:
The collection L) is a linear collection, and f, gELf) implies max (f, g)
ELV) and min (f, g) ELV). Furthermore, the extended integral f(f) has
on Lf) the same properties as the initial elenlentary integral f(f) on
the initial domain of definition L (i.e., f(f) is finite and linear on LV),
f(f)f(g) for fg, and fnto implies (fn) to by the theorem on domi-
nated convergence). Hence, f(f) on L) is an extension of f(f) on L
such that the characteristic properties of f(f) are preserved. It is evi-
dent, from the properties of the extension procedure for measures (cf.
sec. 7, Theorem 6), that a repeated application of the procedure, con-
sidering now f(f) as an elementary integral on LV), will give no further
extension of f (f) .
Finally, we return to some of the examples at the end of the pre-
ceding section. In the first example X == (0, 2J, and L consists of all
f(x)==ax (a constant) with f(f)==2a. We have already proved that M+
is obtained by adding to L + the function which is identically +00 on
X. Hence, it is easily seen that L 1 ==L; the extension, therefore, is
Improper.
In the fourth example X is an arbitrary non-empty point set, and
L consists of all f(x), vanishing except at a finite number of points

Ch. 3, 9 14J THE DANIELL INTEGRAL FOR REAL FUNCTIONS 101
Xl, . . " x q (these points and the number of them varying with I), with
f(/) == f f(xn). We have already proved that M+ consists of all/(x);?;O,
and f(/) == XEX I(x) for fEM+. It is easily seen now that any f(x) on
X is measurable, and L 1 consists of alii satisfying XEX If(x) 1<00. In
addition, f(f)=XEX f(x) for fELl. Hence, if X contains an infinite
number of points, the extension is proper.
Exercises
FURTHER EXAMPLES
14. 1) Consider the second and third examp les at the end of the pre-
ceding section, and show that in the second example L1 consists of all
fwhich are arbitrary on [-1,0) and of the formf{x)==ax (-oo<a<oo)
on (0, IJ with f(f)=a, whereas in the third example L 1 ==L.
THE CONTRACTED INTEGRAL
14.2) We resume the investigation of the" contracted integral" fe (f) ,
introduced on M+ in Exercise 13.3. It was proved there that, for any
fEM+, we have
fe(f)===sup f{min(f, P)} for all PEL+.
Show by means of this relation that, for I, gEM+, we have fe(f+g)==
fe(f)+Fe(g). Hence, if f, gEM+ with fe(g) <00, and gf on X, then
fc(f-g)===fe(f)-fc(g). Develop the theory of fe(f) for real measur-
able functions, corresponding to the theory for f(f) in the present
section. Show in particular that a repeated application of the ex-
tension procedure gives no further extension (repeated contraction is
not necessary).
ABSTRACT RIEMANN INTEGRAL
14.3) We resume the investigation of the positive linear functional
J(/), initially given on L and extended onto M+ in the Exercises 13.4-
13.7. If 1 is real on X, and f+, -f-EM+ with /(/+) and /( -1-) finite,
then 1 is called / -summable, and the abstract Riemann integral / (I)
is defined by /(/)===/(f+)-/(-f-). Show that if f and g are /-

102
DANIELL INTEGRAL
[Ch. 3,9 15
summable, then max (f, g) and min (f, g) are f -summable. Show also
that if f is f-summable and a is a finite real constant, then If I and af
are f-summable, and If(f)lf(I/I). Furthermore, if f and g are f-
summable, then f+g is f-summable, and f(f+g)==f(/)+f(g).
Hence, f(f) is a finite positive linear functional on the collection of
all f-summable functions. Show that f is f-summable if and only if,
given E>O, there exist 11, f2 EL such that f1ff2 on X andf(f2-f1)
<E. Show, finally, that a repeated application of the extension pro-
cedure gives no further extension.
RIEMANN INTEGRAL
14.4) Let X=Rk (k-dimensional number space), let # be Lebesgue
measure in X, and let L be the collection of all step functions f(x) ==
) CnXAn(X), where all cn=l=-:l:oo and all An are cells. Finally, let f(f)==
l cn#(A n ) for f==) cnXAn(x) EL. Then f(f) is an elementary integral
on L; hence, f(f) is surely a positive linear functional on L. We apply
to f(/) the extension procedure described for non-negative functions
in Exercise 13.4, and completed for real functions in the preceding
exercise. Show that the thus obtained abstract Riemann integral f (f)
is exactly the Riemann integral from classical analysis, i.e., f is f-
summable if and only if there exists a cell AI, depending on f, such that
f is Riemann integrable over AI and vanishes outside AI, and in that
case f(f) is the Riemann integral of f over AI'
15. Null Functions and Null Sets
In sec. 13 we have introduced certain non-negative functions which
were called null functions. In the present section we shall introduce
the corresponding notion for real functions.
Any function f on X, such that f(lfl) exists and satisfies f(lfl) ==0,
is called an f-null function. The function f(x) is, therefore, a null
function if and only if the exterior ordinate set of If(x) I is of ,a-measure
zero. By comparing exterior ordinate sets it is evident that if f is a null
function, and Ig(x) 1lf(x) I on X, then g is a null function. In particular,
since Of+lfl and O-f-Ifl, it follows that if I is a null function,
then f+ and -f- are null functions, i.e., f(/+)==f( -1-)==0. Hence,
any null function f is measurable, and f(/) ==0.

Ch. 3, 9 15J
NULL FUNCTIONS AND NULL SETS
103
Obviously, for non-negative functions the present notion of null
functions coincides with the notion introduced in sec. 13.
Any set E eX, for which the characteristic function XE(X) is an f-
null function, is called an f-null set. If E is such a null set, then
f(nXE)==O for n==1,2, "', so f(OO'XE)==O by the theorem on in-
creasing sequences. It follows that any function f, vanishing outside a
null set E, and assuming arbitrary values on E, is a null function. If
En is a null set for n== 1, 2, . . " then the union E ==  En is a null set
(indeed, if Sn== Ei, then XS n iXE).
Finally, any two functions f1, f2 on X, for which f1-f2 is a null
function, will be called f-almost equal functions. Evidently, the proper-
ty to be almost equal is an equivalence relation (if f1-f2 and 12-f3
are null functions, then If1-f21 + If2-f31 is a null function, so that
f1-f3 is a null function on account of If1-f31lf1-f21+lf2-f31; note
that the triangle inequality holds also if one or more of f1, f2, f3 are
=1:00).
THEOREM 1. The function f is a null function if and only if E ==
{x: f(x) *O} is a null set.
PROOF. Let E =={x: f(x) *O} be a null set. As remarked already,
00' XE(X) is then a null function, and I/(x) Ioo' XE(X) on X. Hence, f is
a null function.
Let now, conversely, f be a null function, and let E =={x: f(x) *O} and
En=={x: If(x)ln-1} for n== 1,2, . . . . Since f is a null function and
n-1XEnlfl on X, the function n- 1 XEn is a null function, so f(XEn)=O
for n==l, 2, ... . Hence, since XEnixE on X, we have also f(XE)==O,
i.e., E is a null set.
COROLLARY. The functions f1 and f2 are almost equal if and only if
E ={x: f1(X) *f2(X)} is a null set.
Obviously, if f1 is measurable and h is a null function, then f2==
f1 +h is measurable. It would be tempting to conclude now immedi-
ately that if f1 and f2 are almost equal, and f1 is measurable, then f2
is also measurable. The proof would follow from f2==f1 + (f2-f1)' Un-
fortunately, the last equality need not be true for all x (if f1 == +00, and
f2 is finite and *0 at some point x, the equality is false at that point).

104
DANIELL INTEGRAL
[Ch. 3, 9 15
The stated fact, however, is true, and a correct proof is derived by ob-
serving that although 12==/1+(/2-/1) may not be true for all x, the
equality 12=/1+ (-/1XE)+/2XE, where E is the null set {x: 11=1=-12},
holds at all points XEX. Since the null functions -/1XE and 12XE are
measurable, 12 is seen to be measurable. In addition, we have evi-
dently It/t+oo'XE and It/t+oo'XE on X; hence f(lt)f(/t)+
f(oo' XE)==f(/!), and similarly f(/!)f(lt). It follows that f(lt)==
f(/t). Similarly f(/;;)==f(/l)' This leads to the result that if 11 and
12 are almost equal, and 11 is integrable, then 12 is integrable, and
f(/2)==f(/1) as well as f(1/21)==f(1/11). It follows now also that if 11
and 12 are almost equal, gl and g2 are almost equal, and 11 and gl are
summable, then all +bg 1 and al2+bg2 are summable and almost equal
for any pair of finite real constants a, b; hence
-00<f(aI2+bg2) =f(a/1 +bg1) <00
and
0f(laI2+bg21) ==f(l a /1 +bg11) <00.
In view of the fact that we have proved now that any change in
the values of a function on a null set affects neither its measurability
or integrability, nor the value of the integral if it exists, we are able
to state small generalizations of the theorems on increasing sequences
and dominated convergence.
THEOREM 2. (a) II O/n i I on X, except possibly on a null set, and
il all In are measurable, then I is measurable, and f(ln) i f(/).
(b) II I=lim In exists on X, except possibly on a null set, and il there
exists a summable lunction g(x) such that Ilnllgl lor all n, except possi-
bly on a null set which may depend on n, then I is summable, and f(/)==
lim f(ln).
PROOF. The theorem is reduced to the already familiar case by intro-
ducing functions I(x) and 1* (x) , zero on the exceptional null set E (note
for part (b) that the union of a countable number of null sets is a null
set), and equal to In(x) and I(x) respectively on X -E.
THEOREM 3. II I is summable, then the set E=={x: I/(x)l==oo} is a null
set.

Ch. 3,  15J
NULL FUNCTIONS AND NULL SETS
105
PROOF. We have XEn-1Ifl for n== 1,2, . . . . Hence, denoting the
exterior ordinate sets of XE and If I by P E and IFI respectively, we have
P E c n- 1 IFI, so il*(PE)n-1il(IFI) for n== 1,2, . . . . Since il(IFI)==
f(lfl) <00 by hypothesis, it follows that il*(P E ) ==0, i.e., f(XE) ==0, and
this shows that E is a null set.
THEOREM 4. If Ofn i f on x, except possibly on a null set, if all fn
are summable, and if there exists a finite constant C such that f(fn)C
for all n, then f is finite on X, except possibly on a null set.
PROOF. f(fn) i f(f) by Theorem 2, and f(fn)C for all n, so f(f)C.
It follows that f is summable, and therefore, by the preceding theorem,
f is finite on X, except possibly on a null set.
COROLLARY. If fn is a sequence of non-negative summable functions
on X such that r f(fn) <00, then f(x)==r fn(x) is finite on X, ex-
cept possibly on a null set. In other words, the series r fn(x) converges
in the classical sense for all XEX, except possibly for the points x in a
null set.
I t may very well happen in some examples that the only null fuction
is the function which is identically zero on X. Evidently, this is so if
and only if the only null set is the empty set. We have already met
such examples in the preceding section. If X == (0, 2J, and L consists of
all f(x)==ax (a constant) with f(f)==2a, then the collection L1 of all
summable functions coincides with L, and any null function is, there-
fore, identically zero. Also, if X is an arbitrary non-empty point set,
and f(f) is the integral with respect to discrete measure in X, then L 1
consists of alii satisfying XEX If(x) 1<00 with f (f) == XEX f(x). Hence
f(lfl) ==0 if and only if f(x) ==0 on the whole of X. Fortunately for the
contents of the present section, there exist also examples where not
every null function vanishes identically. The most well-known example
is the Lebesgue integral in Rk. According to the definition at the end
of sec. 12, the collection L consists then of all step functions f(x) ==
l cnXEn(x), where all En cR k are Lebesgue measurable and #(En) <00
(# is here Lebesgue measure), and f(f)==l cn#(En). Since any set E,
consisting of a finite or countable number of points, satisfies #(E) ==0,
its characteristic function XE is a null function.

106
DANIELL INTEGRAL
[Ch. 3, 9 15
Exercises
THE THEOREM ON DOMINATED CONVERGENCE
15.1) In the proof of the theorem on dominated convergence (Theo-
rem 3 of the preceding section) we have decomposed all functions In
into I; and I;;, and the result was derived for the sequences I; and I;;
separately. One may ask whether it is not simpler to consider instead
the sequence of non-negative functions hn/n+g, derive that
Of(lim inf hn)lim inf f(hn)lim sup f(hn)f(lim sup hn)2f(g),
and then subtract f(g). Unfortunately, this gives difficulties since In==
(In+g) -g does not necessarily hold at points x where g(x) == +00. Show
that these difficulties may be overcome by means of Theorem 3 in the
present section; note, of course, that the proof of Theorem 3 does not
make use of the theorem on dominated convergence.
THE CONTRACTED INTEGRAL
15.2) Develop the theory of null functions and null sets for the
contracted integral fe(/), defined generally in Exercise 14.2.
ABSTRACT RIEMANN INTEGRAL
15.3) Let f (I) be the abstract Riemann integral, defined in Exer-
cise 14.3. Any I on X, satisfying f(I/I)==O, is called a f-null function,
and any set E c X, such that XE is a f -null function, is called a f -null
set. Show:
(a) if Igl1/1 on X, and I is a null function, then g is a null function;
(b) if I is a null function, then I is f-summable, and f(/)==O;
(c) any I, bounded on X, and vanishing outside a null set, is a null
function;
(d) any finite sum of null functions is a null function, and any
finite union of null sets is a null set;
(e) if I is a null function, then {x: I/(x) I > ex} is a null set for any ex>O;
(f) if any two functions 11, 12 on X, for which 11 - 12 is a f -null
function, are called f -almost equal, then the property to be alnlost
equal is an equivalence relation.

Ch. 3,9 16J METRIC SPACE OF ALL SUMMABLE FUNCTIONS 107
16. The Metric Space of all Summable Functions
We assume, as before, that L 1 is the collection of all f-summable
functions on the point set X. By the results in the preceding section,
L1 may be decomposed into classes of mutually almost equal functions,
and (in a similar manner as we did with almost equal sets in sec. 11)
we shall denote the collection of these classes again by L1, and speak
about a function IEL1 when, properly speaking, we mean the entire
.class of all functions which are almost equal to I. Almost equal functions
are therefore identified. Introducing then, for IEL1, the notation 11/11==
.f(l/l), the number 11/11 is finite, unambiguously defined (i.e., all I of
the same class have the same f(I/I)), and
11/-gII==O if and only if I==g (almost),
11/-gII"h- III + Ilh-gll,
'SInce
f(l/-gl)f(l/-hl + Ih-gl) ==f(lh- II) +f(lh-gl).
This shows that L 1 is a metric space with respect to the distance
function p(/, g) == II/-gli. The distance 11111 between I and the null function
is called the L1-norm of I (note that we may say now "the" null function).
If all functions in L 1 are multiplied by the real constant a, then all
distances are multiplied by lal; furthermore, all distances are invariant
under translations (since 11(/+h)-(g+h)II==II/-gll).
THEOREM 1. The metric space L1 is comPlete.
PROOF. We have to show that, given the sequence InEL1 satisfying
limlllm-lnll==O as m, noo, there exists a function IEL1 such that
1imil/-Inll==0 as noo. Since limlllm-lnll==O, there exists a sub-
sequence gn of In satisfying ;:'=1 IIgn+1-gnll<oo (choose gl==ln 1 , where
n1 is the smallest index n such that III n - 1m II <! for all m >n; generally,
choose gp==lnp' where np is the smallest index n>n p -1 such that
Jlfn-Imll< 1/2 P for all m>n). Then, if
00
g(X)==lg1(X)I+  Ign+1(X)-gn(x) I,
n=l

108
DANIELL INTEGRAL
[Ch. 3, 9 16
we have
p
f(g)==lim f{lg11+ Ign+1-gnl}
p-+oo 1
p 00
==lim (1Ig111+ l[gn+1-gnll)==llg111+ IIgn+1-gnll<oo,
p-+oo 1 1
so g is summable, and it follows that the set E =={x: g(x) ==oo} is a null
set. In other words, except at the points of the null set E, the series
Ig1(X) I + r Ign+1(X) -gn(x) I converges in the classical sense; it follows
by a well-known theorem that the series gl(X) + r {gn+1(X) -gn(x)}
converges in the classical sense for x EX - E. Hence, if
! 0 on E,
I(x) == 00
gl(X) + 7 {gn+1(x)-gn(X)} on X-E,
then the function I(x) is finite on X, and IEL1 (the theorem on domi-
nated convergence may be applied, since I==lim gp and
p-l p-l
Igpl== Ig1 +  {gn+1-gn}Ilg11+  Ign+1-gnlg
1 1
on X, except on a null set). Furthermore,
00
00
I/-gpl == I {gn+1-gn}I Ign+1-gnl,
p p
except on a null set, so
00
00
11/-gplIll Ign+1-gnlll==  Ilgn+1-gnll0
p p
as poo.
Consequently, we may draw the conclusion that
11/- I nll II/-gpil + IIgp- I nil O
as n (and p) oo.
This completes the proof. Note that since in a metric space the limit
point of a fundamental sequence, if it exists, is unique, the limit
function I(x) ELI is uniquely determined.
COROLLARY. II In is a lundamental sequence in L1 with limit lunction
I, then lim III nil == 11/11. Furthermore, In contains a subsequence I nk such that

Ch. 3, 9 16J METRIC SPACE OF ALL SUMMUABLE FNCTIONS 109
f nk(x) converges pointwise on X to f(x) as k-+oc>, except on a null set. It
is even so that every subsequence of fn contains a subsequence converging
pointwise on X to f(x), except on a null set.
PROOF. The first assertion follows from "lfll-llfnllllIf-fnil. The
second assertion is evident from the proof of the theorem above, and
the last assertion follows by observing that any subsequence of a
fundamental sequence in the metric space L 1 is itself a fundamental
sequence, having the same limit as the whole sequence.
THEOREM 2. The initial domain of definition L of f(f) is dense in
the metric space L 1 .
PROOF. Given fELl and E>O, we have to show the existence of a
function f*EL such that IIf-f*II<e. Let first fO on X. On account of
sec. 13, Lemma cx, there exists a a-function s(x)f(x) such that f(s-f)
<!E. By the definition of a a-function there exists a sequence SnEL+
such that Sn i s on X. Hence, since f(sn) i f(s), we have f(s-sp)===
f(s)-f(sp)<!E for a sufficiently large p. It follows that, for this
value of p, the function spEL + satisfies
Ilf-splI==f(lf-spl)f (If-sl) +f(ls-spl) ==f (s- f) +f (s-sp) < tE.
If fELl, then f===f+-(-f-), and by what has already been proved,
there exist functions s, tEL + such that
f(lf+-sl) <t e
Then f*==S-tEL, and
and
f(l-f--tl)<t E .
IIf- f*ll==f(lf- (s-t) I) ==f(1 (f+ -s) - (-f--t) I) <E.
The final conclusion is, therefore, that the metric space L 1 of all
summable functions is the closure of the initial collection L. Obvi-
ously, the metric space L1 is separable if and only if the initial col-
lection L, as a metric space with respect to the same distance function
IIf-gll, is separable (indeed, note that any dense subset of L is also
dense in L 1) .

110
DANIELL INTEGRAL
[Ch. 3,  16
Exercises
THE CONTRACTED INTEGRAL
16.1) Let fe(f) be the contracted integral corresponding to f (f).
Show that, under identification of e-almost equal functions, the col-
lection LIe of all fe-summable functions is a complete metric space
with respect to the di.stance function fe(lf-gl), such that the initial
domain of definition L of f(f) is lying dense in this space. Anyequiva-
lence class of functions fELl is included entirely in some equivalence
class of LIe, and fe(f)==f(f) for all these f. Show that, conversely, any
equivalence class of functions gEL 1e contains an entire equivalence
class of functions from L 1 ; hence, LIe and L 1 may be identified as
metric spaces.
16.2) The preceding exercise shows that the distinction between L 1
and LIe lies only in the fact that there may be more fe-null functions
than f-null functions. Show, by means of Example 6 in sec. 10, that
this may actually occur.
RIEMANN INTEGRAL
16.3) Let f(f) be an abstract Riemann integral. Show that, under
identification of f -almost equal functions, the collection of all f-
summable functions is a metric space with respect to the distance
function f(lf-gl). The metric space is not always complete; show, in
particular, that it fails to be complete if f(f) is the classical Riemann
integral in R1. Show also that the initial domain of definition L of
f (f) is dense in the metric space.

CHAPTER 4
STIEL T JES-LEBESGUE INTEGRAL
It will be shown in sec. 17 that in the presence of a rather mild additional
condition on the linear collection L, any Daniell integral which is the extension
of an elementary integral on L turns out to be a Stieltjes-Lebesgue integral,
Le., the same integral may be obtained by extension of an elementary integral
defined on the collection of all step functions with respect to an appropriate
measure in X. In sec. 18 (and in the exercises belonging to that section) the
reader will find a number of useful properties of the Stieltjes-Lebesgue integral;
the next two sections are devoted to integrals with complex integrands, and
sec. 21 contains the theorem that if the extension procedure is applied to the
Riemann integral (for continuous functions in Euclidean space), then the result
is exactly the Lebesgue integral. As an immediate generalization we shall ob-
tain the famous Riesz representation theorem for positive linear functionals on
the space of continuous functions. In the exercises of sec. 21 some classical inte-
grals are computed explicitly.
17. The Induced Measure in X, and the Representation Theorem
In the preceding chapter the theory of the general Daniell integral
was developed, and in many respects the so obtained results have a
final form. In some special cases, however, new problems arise. Among
these special cases one of the most important is the case that the linear
function collection L, which serves as the domain of definition of the
elementary integral f(f), consists of all step functions with respect to
. a given measure # on a ring A of subsets of X, and f(f) is defined on
L in the natural manner (i.e., any fEL is of the form f==) CnXE n , where
cn=F =1:00, EnEA and #(En) <00 for n== 1, . . " p, and f(f) ==) cn#(En)
for this f). As defined in sec. 12, the extended integral f(f) is then
called a Stieltjes-Lebesgue integral. The following is now a plausible
conj ecture.
If the extension procedure for measures is aPPlied to (X, A, #), and All

112
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
is the so obtained a-field of all #-measurable subsets of X, then the charac-
teristic function XE of any set EEAp, is f-measurable, and f(XE)==#(E).
Conversely, if XE is f-measurable, then EEAp,.
Next, even though L is not always a collection of step functions,
the question may be raised whether, nevertheless, it is so that the
corresponding extended Daniell integral f(f) is always a Stieltjes-
Lebesgue integral. In other words, we ask if there exists a measure v,
defined on some ring T of subsets of X, such that the same Daniell
integral f(f) may also be obtained by extension of the elementary
integral corresponding to the step functions with respect to v and T.
A natural method for obtaining such a measure v seems to present
itself in view of the following facts and definitions.
If E is a subset of X, then the characteristic function XE is f-
measurable if and only if XE is f-integrable (for non-negative functions
measurability and integrability are identical notions). Any EeX for
which XE is measurable is called an f-measurable or f-integrable subset
of X. If EeX is measurable, and f(XE) <00, then E is said to be an
f-summable subset of X. The empty set 0 e X is always summable.
Given the measurable subsets E1 and E 2 of X, the sets E1 +E2, EIE2
and E1-E 2 are also measurable, since
XE 1 +E 2 ==max (XE1' XE 2 ),
X E 1E2==min (XE 1 , XE 2 )
and
XE1-E2==XE1-XE1E2'
Furthermore, if En (n==l, 2, ...) is a sequence of disjoint measurable
sets, then the set  En is measurable, and f(XEJ== f(XEn)' by the
theorem on integration of monotone sequences. It follows that the col-
lection of all measurable sets EeX is a a-ring T, and v(E)==f(XE) is a
measure on T. The measure v on T is said to be induced by the integral
f(f). It follows now that the step functions s(x)==l cnXEn(x), where
Cn:F=ioo, EnET and v(En)<oo for n== 1, .. " p, form a linear col-
lection Ls (such that s, tEL s implies max(s, t)ELs and min(s, t)EL s ),
and f(s) == l cnf{xEn) == l cnv(En) is an elementary integral on Ls.
More explicitly as above, the question may be raised, therefore, if the

Ch. 4,  17J
INDUCED MEASURE IN X
113
extension procedure for the integral, applied to f(s) on Ls, will gener-
ate again the same integrable functions with the same value of the
integral as the procedure with L as initial domain of definition. It
turns out that the answer to this question is not always affirmative;
in the example where X == (0, 2J, and L consists of all functions f(x) =ax
with f(j)==2a, the collection M+ of all non-negative measurable
functions contains, besides the function 00' Xx (x) , only the functions
of L +, so the empty set 0 is the only measurable subset of X, and Ls
consists only of the null function.
It is evident, therefore, that in the examples where the answer is
affirmative, a further condition on L and (or) f(j) should be satisfied.
In order to obtain such a condition, assume that f(j) is a Stieltjes-
Lebesgue integral generated by an elementary integral on the col-
lection Ls of step functions s(x). Obviously, the characteristic function
xx(x) of the whole set X satisfies min(Xx, s)EL;- for any sEL;-; hence,
XXEM+ by sec. 13, Theorem 3 (2). This shows that the measurability of
the function xx(x) is a necessary condition in order that f(j) should be
a Stieltjes-Lebesgue integral. The main result in the present section
will be that the condition is also sufficient. More explicitly, we shall
prove the following representation theorem.
Ij the elementary integral f(j) on the initial domain oj dejinition L
(where L does not necessarily consist oj step functions) has the property
that, upon extension oj f(j), the function xx becomes f-measurable, then
the extended Daniell integral f(j) is a Stieltjes-Lebesgue integral with re-
spect to the measure 'V induced in X by the integral tselj.
Evidently, once more by sec. 13, Theorem 3 (2), the condition XX EM +
is satisfied if and only if min(f,xx)EM+ for any jEL+, i.e., since
xx(x)==1 for all XEX, we have
XX EM + if and only if min (f, 1) EM+ for any fEL +.
We shall assume from now on, unless the contrary is explicitly
stated, that XXEM+. Note that the condition is surely satisfied if the
more restrictive condition
min(j,I)EL+ for any fEL+
holds. Both conditions were introduced by M. H. STONE ([IJ, Note II).
The more restrictive condition is simpler in so far as it is merely a

114
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
condition on the collection L, and it is satisfied in some of the most
important examples (e.g. if L is a collection of step functions, or if L
is the collection of all continuous functions with bounded carrier in Rk).
Some simple consequences of the condition that xxEM+ are col-
lected in the following theorem.
THEOREM 1. (a) Any lunction, which is constant on X, is measurable.
(b) II I is measurable, and -oo<a<oo, then the lunctions max(/, a)
and min (I, a) are measurable.
PROOF. (a) Evident.
(b) Follows from the theorem that if I and g are measurable, then
max (I, g) and min (I, g) are measurable.
THEOREM 2. II I is measurable, and -oo<a<oo, then the set A==
{x: I(x) >a} is measurable. II I is summable, and O<a<oo, then the set
A =={x: I(x) >a} is summable.
PROOF. If In(x)=n{/-min(/, a)} for n==l, 2, "', then In(x)joo on
A and In(x)==O on X-A. Hence, min(ln, l)jxA on X. Furthermore,
in view of the preceding theorem, all functions In are measurable; the
same holds, therefore, for the functions min (In, 1). It follows that XA
is measurable, i.e., A is a measurable set.
If I is summable (so f(/+)<oo) and O<a<oo, then f(XA)<OO on
account of OXAJ+/a, so A is a summable set.
COROLLARY. (a) II I is measurable, and -oo<b<oo, then the set
B=={x: I(x)<b} is measurable. II I is summable, and -oo<b<O, then
the set B=={x: I(x)<b} is summable.
(b) II I is measurable, and -oo<a<oo, then the sets {x: I(x) >a},
{x: I(x) <a} and, by comPlementation, the sets {x: I(x)a} and {x: I(x)a}
are measurable. Hence, by subtraction, the set {x: I(x) ==a} is also measur-
able. Furthermore, the sets {x: I(x) ==oo}, {x: I(x) == -oo} and, by comPle-
mentation, the sets {x: I(x) < oo} and {x: I(x) > -oo} are measurable.
Hence, by subtraction, the set {x: -oo</(x)<oo} is measurable.
PROOF. Note that complementation is permitted since the set X is
measurable. In order to show that {x: I(x) ==oo} is measurable, we ob-
serve that {x: l(x)==oo}==II;;'=l {x: I(x»n}, and we recall that the col-

Ch. 4, 9 1 7J
INDUCED MEASURE IN X
115
lection of all measurable sets is a a-ring (since X is measurable, the a-
ring is even a a-field).
It is important that the converse of Theorem 2 is also true.
THEOREM 3. If f(x)O on X, and the set A =={x: f(x) >a} is measurable
for all positive a, then fEM+.
PROOF. The set
00
R=={x: f(x)==oo}== II {x: f(x) >n}
n=l
is measurable by hypothesis, so g(x) ==00' XR(X) EM+. Let 1 <l5<00, and
let
A =={x: l5m<f(x)l5m+ I}
for m==O, =I: 1, =1:2, . . . .
By hypothesis A is measurable (since A is the difference of {x: f>l5 m }
and {x: f>l5 m + 1 }), so the characteristic function Xn(x) of A is measur-
able, and it follows that
00
f6==  l5mX+gEM+.
m= -00
If l5! 1 through a suitable sequence of values l5n (choose l5 1 > 1 arbitra-
rily, and then e.g. l5n+1 ==l5; this ensures that each partition on the
Rt-axis is a refinement of the preceding partition), then f6 n i f, so fEM+.
COROLLARY. If f(x) is real on X, and the sets {x: f(x) >a} and {x:
f(x) < b} are measurable for all positive a and all negative b respectively,
then f is measurable.
It is of interest to note that, by modifying the functions f6 somewhat,
we obtain the following approximation theorem.
THEOREM 4 (ApPROXIMATION THEOREM). If f(x) EM+, there exists a
sequence of functions fn(x) EM+, each fn assuming only a finite number
of finite values, such that fn if on X. If, in addition, f is summable, then
all fn are step functions (i.e., fn assumes its positive values on summable
sets) .

116
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
PROOF. Choose n and the set R as in the preceding theorem (i.e.,
R=={x: f(x) ==oc>} and n+1 == for n== 1, 2, . . .), and let, for ==n, the
function fn(x) be defined by
n'2 n
fn(x)==  mX(x)+nXR(x).
m= -n° 2 n
If fO is summable, and fn assumes only a finite number of different
values and satisfies Ofnf, then fn is necessarily a step function (cf.
Theorem 2).
THEOREM 5. If fEM+, and -oc><P<O or O<P<oc>, then fPEM+.
PROOF. If O<P<oc> and a>O, the set {x: f>a 1 / P } is measurable by
Theorem 2 (since f is measurable). Hence, since {x: fp >a}=={x: f>a 1 / P },
the set {x: fP>a} is measurable. This holds for all a>O, so fp is measur-
able by Theorem 3.
If -oc><P<O and a>O, the set {x: f<a 1 / P } is measurable by Theo-
rem 2, Corollary. Hence, since {x: fp >a}=={x: f<a 1 / p }, the set {x: fP>a}
is measurable. This holds for all a >0, so fp is measurable by Theorem 3.
THEOREM 6. If f and g are measurable functions, then fg is a measur-
able function.
PROOF. If f and g do not assume the values :l:oc>, the proof is simple.
Indeed, f+g and f-g are measurable by sec. 14, Theorem 2; hence,
If+gl and If-gl are measurable. It follows from the preceding theorem
that If+gl2 and If-gl 2 are measurable, and the desired result is then
derived by observing that
fg==!{lf+gI 2 -lf-gI 2 }.
The proof fails, however, if the functions assume the values :1:0C>, since
then the last equality does not hold at all points XEX (if e.g. f==oc> and
O<g<oc> at a point x, then fg==oc> and If+gI 2 -lf-gI 2 ==0 at that
point). The difficulty may be overcome by observing that
fg== (f+ + f-) (g+ + g-) == f+g+ + f+g- + f-g+ + f-g-
holds at all points XEX, so that we may assume \vithollt loss of gener-
ality that f, gEM+. Let fn==min(f, n) and gn==min(g, n) for n== 1,2,

Ch. 4, 9 17J
INDUCED MEASURE IN X
117
. . . . Then fngn EM + by what has already been proved, and fngn i fg
on X. Hence fgEM+.
As noted earlier in this section, the collection of all f-measurable
subsets of X is a a-ring r, and v(E) ===f(XE) is a measure on r. Since,
by hypothesis, the set X is measurable, the a-ring r is a a-field. If
Ls is the collection of all corresponding step functions f(x) === l cnXEn (x),
where cn=¥==l:oo, En Er and v(En)<oo for n===l, "', p, then f(f)==
l cnv(En) is an elementary integral on Ls. By Theorem 2 all sets
{x: f(x) >a} are summable for any fEL and any a>O, so Ls does not
exist exclusively of the null function (except in the trivial case that L
consists exclusively of the null function). Once more, the question is
raised now if the extension procedure, using f(f) as an elementary
integral on Ls, generates the same integrable functions with the same
value of the integral as the procedure with L as initial domain. The
answer will follow from the following lemma.
LEMMA cx. Let f(f) and f(f) be elementary integrals on L(f) and
L(f) respectively, where the linear function collections L() and L(f),
both on the same point set X, have the property that if f and g are in one
of them, then max(f, g) and min(f, g) are in the same collection. Then the
extended Daniell integrals f(f) and f(f) are identical (i.e., the collections
of integrable functions are the same with the same value of the integral) if
and only if any fEL+(f) is f-measurable with f(f) ==f(f) , and any
fEL+(f) is f-measurable with f(f)==f(f).
PROOF. For the extension of f(f) we have introduced the semi-ring
r J of all sets F -G (F and G ordinate sets of f, gEL +(f); fg), and
ilJ(F -G) ==f(f-g) is then a measure on r J ' The measure ilJ is ex-
tended, whereby the a-field AJ of all ilJ-measurable subsets of X X Rt
is generated. Similarly, for the extension of f(f) the semi-ring r" of
all sets F -G is introduced (F and G ordinate sets of f, gEL +(f) ;
f'?g) , and il,,(F-G)==f(f-g) is a measure on r" , generating by ex-
tension the a-field A " of all il ,,-measurable subsets of X X Rt. Obvi-
ously, the extended integrals f(f) and f(f) are identical if and only
if AJ === A ", and ilJ===il" on AJ == A ". It follows easily from sec. 6,
Theorem 2 that this occurs if and only if any A E r J is il ,,-measurable
with ilJ(A)==ilJ(A), and any BE r" is ilJ-measurable with ilJ(B)==

118
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
il ,,(B). In view of the definitions of r oF ' r " , iloF and il" the last con-
dition is equivalent to the condition that any fEL +(f) is f-measur-
able with f(/)===f(/), and any fEL+(f) is f-measurable with (f)===
f(f).
In order to apply the lemma in the present situation, we denote the
elementary integral f(f) on the collection Ls of step functions by fs(f) ,
and we shall prove that the extended integrals f(f) and fs(f) are
identical. Evidently, by the definition of Ls, any fEL;- is f-measur-
able, and f(f) ===fs(f) for fEL;-. For the proof of the second part, let
fEL + be given. Since f is non-negative and f-summable, there exists
by Theorem 4 a sequence of functions fn EL ;- such that fn if on X. It
follows then by the theorem on integration of increasing sequences
that f is Js-measurable, and fs(f n) i fs(f). However, fs(f n) ==f (f n)
since fnELs, and f(fn) i f(f), once more by the theorem on integration
of increasing sequences. Hence fs(f) ==f(f) for any fEL +. In view of
the lemma, the extended integrals fs(f) and f(f) are now identical.
We recall that if the initial function collection L is a collection of
step functions, corresponding to the measure fl in X, then the gener-
ated Daniell integral (f) is called a Stieltjes-Lebesgue integral. This
is usually indicated by the notation f(f)==/xfdfl.
The obtained result may be summarized, therefore, as follows.
THEOREM 7 (REPRESENTATION THEOREM). In order that the Daniell
integral f(/) should be a Stieltjes-Lebesgue integral /xfdv with respect to
the measure v(E) ==f(XE) induced in X by the integral itself, it is neces-
sary and sufficient that the characteristic function xx of X is f -measur-
able.
There are several points which deserve further attention. Denoting,
as before, the a-field of all f-measurable subsets of X by r, the set
function v(E)==f(XE) is a measure on r, and we may apply, therefore,
the extension procedure for measures to (X, r, v), obtaining thereby
the a-field A of all v-measurable subsets of X. Evidently reA, and
we shall prove that, fortunately, r==1.

Ch. 4, 9 17J
INDUCED MEASURE IN X
119
THEOREM 8. If r is the a-field of all f-measurable subsets of X, and
the extension procedure for measures is aPPlied to the measure v(E) ==f(XE)
on r, then no extension of v is obtained.
PROOF. Denoting, as above, the a-field of all v-measurable sets by
A, we have to prove that every EEA is already contained in r.
(a) We first give the proof for sets in A of measure zero. Let, there-
fore, EEA and v(E) ==0 (observe that the symbol v in v(E) is already
the extended v; it is not immediately permitted, therefore, to write
v(E)==f(XE))' Then, given e>O, there exists a a-set 0 with respect to
rsuch that OE and v(O)<e. But OEr (since ris a a-ring); hence,
xo is f-measurable and v(O)==f(xo). It follows that XEXO and f(xo)
<c. Choosing now a suitable decreasing sequence On  E such that
f(Xon)!O, we see that XE is an f-null function, i.e., EEr.
(b) Next, we consider sets in A of finite measure. Every a-set  An
(all AnEr) is contained in r, since r is a a-ring, so it follows that the
limit of any decreasing sequence of a-sets is also contained in r. Hence,
since any E EA of finite measure is the difference of such a limit and
a set of measure zero, we may conclude by (a) that EEr.
(c) We now consider the general case, and we begin by proving that
if E is a subset of X, satisfying EA Er for all A Er of finite measure,
then EEr. The hypothesis concerning E may be formulated alterna-
tively as: min(XE,XA) is f-measurable for all f-summable sets A.
Observing that, for Oc<oo, we have
. _ J min (XE, XA)
mln(XE, cXA)- l . ( )
c mln XE, XA
for
for
cl,
Oc< 1,
it follows that min (XE, CXA) is also f -measurable for all f -summable
setsA. Hence,min(XE, s) isf-measurableforanystepfunctions(x)EL;-
(since s(x) is a finite linear combination with non-negative coefficients
of characteristic functions of disjoint f-summable sets). But then,
since the integral is generated by its own restriction to Ls, the col-
lection Ls may play the role of L in sec. 13, Theorem 3 (2), and it
follows that XE is f -measurable, i.e., E E r.
Let now EEA. Then EAEA for all AEr of finite measure; hence,
EA Er for all A Er of finite measure (since EA is of finite measure,
so that (b) may be applied to conclude from EA EA that EA Er). It
follows by the preceding paragraph that E E r.

120
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4,  17
We finally consider the classical case that f(f) is, already from the
beginning on, a Stieltjes-Lebesgue integral. Let # be a measure on the
ring A of subsets of X, and let the elementary integral f(f) == l cn#(E n )
correspond to the step functions f(x) == l cnXEn(x) with respect to #
and A. The extended integral f(f) induces in X the measure v; let Av
be the a-field of all v-measurable subsets of X (by the last theorem
Av is also the a-field of all f-measurable subsets of X). Finally, let AJl
be the a-field of all #-measurable subsets of X, obtained by applying
the extension procedure for measures to (X, A, #). It is a reasonable
conjecture that Av==AJl, and v==# on Av==AJl. We shall prove that the
conjecture is true; consequently, the concepts of f-measurable, #-
measurable and v-measurable subsets of X are equivalent, and the
same applies to the notations f(f), Ixfd# and Ixfdv.
THEOREM 9. Let the Stieltjes-Lebesgue integral f(f) be generated by ex-
tension of the elementary integral corresponding to the step functions with
respect to the measure # on the ring A of subsets of X, and let v(E) ==f(XE)
be the measure induced by f(j) on the a-field Av of all f-measurable
subsets of X. Finally, let AJl be the a-field of all #-measurable subsets of
X. Then Av==Att, and v==# on Av==AJl.
PROOF. (a) We show first that any EEAJl of finite #-measure is
contained in Av, and v(E)==#(E). This is trivial if EEA and #(E)<oc>.
Hence, if E= An is a a-set (all AnEA and disjoint) of finite #-
measure, then EEAv (since Av is a a-ring), and v(E) ==  v(An)==  #(An)
==#(E). It follows then immediately that any set E1EAJl of #-measure
zero is also of v-measure zero, and also that any set E2EAJl' which is
the limit of a decreasing sequence of a-sets of finite measure, is in Av,
and v(E 2 )==#(E 2 ). Hence, since any EEAJl of finite #-measure is of the
form E 2 -E1, the set E is in Av, and v(E)==#(E).
(b) Let fn be a sequence of non-negative step functions with respect
to # and A (i.e. fnEL+) such that fntf on X (the function f is, there-
fore, what we have called a a-function), and let O<a<oc>. Then the
set
00
{x: f>a}==  {x: jn>a}
n=l
is #-measurable, since AJl is a a-ring and all {x: fn>a}, being sets in A,

Ch. 4, 9 17J
INDUCED MEASURE IN X
121
are #-measurable. It follows that the set
00
{x: fa}== II {x: f>a-n- 1 }
n=l
is also #-measurable. If, in addition, f(f) <OC>, and we denote the set
{x: fn>a}EA by A n, then #(An) ==..jf(XAn)a-1f(f) since XA n <f n/af/a.
Hence, if A=={x: f>a}, then #(A)===lim #(An)a-1f(f)<0c>, so that
the set {x: fa} is also of finite #-measure.
(c) We prove now that any EEAv of finite v-measure is contained
in Ajl, and #(E)===v(E). Let, therefore, EEAv and v(E) ==f(XE) <OC>.
Given e>O, there exists (by sec. 13, Lemma ex) a a-function S(X)XE(X)
such that f(s)-f(x)<e. Consider the set A=={x: s(x)I}. By (b) the
set A is #-measurable and of finite #-measure; by Theorem 2, Corollary,
the set A is also f-measurable, and Of(XA)-f(XE)<e on account of
XEXAS. We may conclude, therefore, that for any e>O there exists
a set A ::::>E, #-measurable as well as v-measurable, such that #(A)<oo
and v(A)-v(E)<e. If, in particular, v(E) ==0, then v(A)<e; hence
since #(A)==v(A) by (a) on account of #(A)<oc>, we have #(A)<e. It
follows that #*(E)<e for all e>O, so E is #-measurable, and #(E)==
O==v(E) .
If, more generally, O<v(E)<oc>, there exists a descending sequence
of sets An, #-measurable as well as v-measurable, such that An::::> E 
#(An)<oc> and v(An)-v(E)<n- 1 . It follows that A==lim An is #-
measurable as well as v-measurable, A ::::>E, #(A)<oc> and v(A)-v(E)
===0, so v(A-E)==O. Then A-E is #-measurable, and #(A-E)==O by
what has already been proved. Hence, E ==A - (A - E) is #-measurable
and #(E)===#(A)==v(A) ===v(E) , where #(A)==v(A) follows again by (a)
This completes the proof for sets of finite measure.
(d) E EAjl if and only if EA EAjl for all A EAjl of finite #-measure.
The last condition, by (a) and (c), is equivalent to EA EAv for all A EAv
of finite v-measure, and finally, this is equivalent to E EAv.
In (a) and (c) we have already proved that if either v(E)<oc> or
/1(E)<oc>, then v(E)==#(E); it follows, therefore, that v(E)===#(E) for
all EEAv==Ajl.
If X ==Rk (k-dimensional real number space), and # is Lebesgue
measure on Rk, then the Lebesgue integral f(f) is obtained (by defi-

122
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
nition) by extension of the elementary integral J(f) == 'f cn#(En), in-
itially defined on the collection of all step functions f(x) == 'f cnXEn (x),
where all En are #-measurable and of finite #-measure. The last theo-
rem shows that the same is true if we restrict ourselves to the col-
lection of all step functions of the particular form f(x) == 'f CnX.An(x) ,
where all An are cells (A is then the ring of all finite unions of cells).
Exercises
LEBESGUE'S DEFINITION OF THE INTEGRAL
17.1) Let J(f)==Jxfd#, where #(X) <00, and let the measurable
function f(x) satisfy Of(x)M on X, where M is a finite constant.
For the partition P=={O==ll <l2<' . . <In+1==M} of the interval [0, MJ,
let Ek=={X: lk<f(x)lk+1} and s(f; P)== lk#(E k ). Show that f(f)==
sup s(f; P) for all possible partitions P (n and the partition points
l2, "., In variable). Almost equivalently, show that J(f)==lims(f; P)
as (P) ==max (lk+1-lk) O. This, for the particular case of the Lebes-
gue integral over a bounded interval, was the original definition of H.
LEBESGUE (1904, [IJ, p. 112). For the case of an unbounded fEM+
Lebesgue introduced the functions fn==min(f, n) for n==l, 2, "', and
definedJ(f)==lim(fn). Show that, without any restrictions concerning
p(X) and boundedness of fEM+, we have J(f)==sup s(f; P) for all par-
titions P=={O==ll <l2< . . . <In} (n and l2, . . ., In variable).
SAKS' DEFINITION OF THE INTEGRAL
17.2) Let (f)==Jxfd# and, for a given fEM+, let mA==inf f on the
measurable set A. Show that J(f)==sup  m.Ak#(A k ) for all decompo-
sitions of X into disjoint measurable sets AI, . . " An (n and the sets
AI, . . " An variable). This is the definition of (f) which occurs in the
book by S. SAKS [IJ.
AN EXAMPLE
17.3) Show the existence of a function f(x), defined on R1, such that
If(x) I is Lebesgue summable but f(x) fails to be Lebesgue summable.

Ch. 4, 9 17J
INDUCED MEASURE IN X
123
STONE'S CONDITION
17.4) On X == [0, IJ, let L be the collection of all real continuous
functions I(x) on X such that there exists a subinterval [a, IJ of X,
where a may depend on I, with I(x)==kx on [a, 1]. In other words, the
"last" part of the graph of I(x) js a line segment lying on a line passing
through the origin. For each IEL the number f(/) is defined as the
Riemann integral of lover X. Show that f(/) is an elementary integral
on L. Show also that not every IEL + satisfies the condition that
min(/, I)EL+. Note, however, that min(/, I)EM+ for any IEL+, so
that the extended f(/) is a Stieltjes-Lebesgue integral.
ANOTHER NECESSARY AND SUFFICIENT CONDITION FOR A DANIELL
INTEGRAL TO BE A STIELTJES-LEBESGUE INTEGRAL
17.5) Let f(/) be a Daniell integral, initially defined on L. It has
been observed by T. A. SPRINGER (lecture notes) that a sufficient con-
dition for f(/) to be a Stieltjes-Lebesgue integral is that I, gEL implies
IgEL (the collection L is then what is usually called an algebra of
functions). The example in the preceding exercise shows that the con-
dition is not necessary. A condition which is necessary as well as suf-
ficient is that for any IEL+ we have 1 2 EM+. Assume this condition
satisfied, and show first that if I, gEL, then IgEM. Next, show that if
I and g are summable, then also IgEM. Finally, show already that
IEL + implies that InEM+ for n== 1, 2, 3, . . . .
17.6) On {y: OyI}, let Po(Y) ==0, P1(y)==!(I-y), generally Pn+1(Y)
==!{I-y+p(y)} for n==O, 1, 2, . . . . Show that qn(Y) == I-Pn(Y) !y!
uniformly on [0, IJ, so that, denoting qn(y)-qn(O) by Qn(Y), we have
lim Qn(Y)==Y! uniformly on [0, IJ. Derive from this result the existence
of a sequence of polynomials P n(Y) such that P n(O) =0 for all nand
lim P n(Y) ==y i on [0, 00). Finally, show that if the finite-valued function
I on X satisfies InEM+ for n== 1, 2, . . " then It, 1 1 , Ii, . . . EM+.
17.7) Let f(/) be a Daniell integral, initially defined on L, and let
1 2 EM+ for all IEL+. It follows then from the results in the preceding
exercises that Ii, 1 1 , Ii, . . . EM+. Show that this implies immediately
that min(/, I)EM+. Hence, f(/) is a Stieltjes-Lebesgue integral.

124
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 17
THE CONTRACTED INTEGRAL
17.8) Let fe(l) be the contracted integral corresponding to the Stiel-
tjes-Lebesgue integral f(/). Hence, since f-measurable and e-measur-
able functions are the same (so that, also, f -measurable and fe-measur-
able subsets of X are the same), the function xx(x) is measurable. Show:
(a) If I is fe-summable and O<a<oo, then the set {x: I(x) >a} is e-
summable. (b) If IO is e-summable, there exists a sequence of non-
negative fe-step functions In such that In i I. (c) If fe(/) induces in X
the measure Ve, and the extension procedure is applied to the elementa-
ry integral on the ve-step functions (contraction is now not necessary),
then the result is again fe(/). (d) If v is the measure induced in X by
f(/), then Ve is the contracted measure corresponding to v. Hence,
since it follows from Theorem 9 that V==/-l whenever f(/)==Jld/-l, we
have also Ve==/-le.
ABSTRACT RIEMANN INTEGRAL
17.9) Let f (I) be an abstract Riemann integral. Any set E c X, satis-
fying XEEM+, is called a f-measurable subset of X. If, in addition,
f(XE) <00, then E is called f-summable. Show that the collection of all
f -measurable subsets of X is a ring A, and v(E) == f (XE) is a charge on
A. Show by an example that A may consist exclusively of the empty set.
17.10) Let v be a charge on the ring A of subsets of X, and let Ls be
the collection of all step functions with respect to v and A. Further-
more, let the abstract Riemann integral f(/) be obtained by extension
of the positive linear functional corresponding in the natural manner to
v and Ls. Show that xx EM +. The integral f(/) is called a Stieltfes-
Riemann integral, and denoted by Ildv.
17.11) Let f (I) be an abstract Riemann integral, and assume that
xxEM+. Show that any function which is constant on X, is f-measur-
able. Show also that if I is f-measurable, and -oo<a<oo, then
max(f, a) and min(/, a) are f-measurable.
17.12) Let f (I) be an abstract Riemann integral, and assume that
xxEM+. Show that, for any non-negative f-summable I, the set {x:
I(x»a} is f-summable for all a satisfying O<a<oo, except for at
most a countable number of values of a. This result is due to L. H.
LOOMIS [IJ.

Ch. 4,9 18J
FURTHER PROPERTIES OF THE INTEGRAL
125
17.13) Let f(/) be an abstract Riemann integral, let XX EM +, and
let v be the charge induced in X by f (I). Assume now that j(x) is non-
negative and bounded on X, vanishing outside a f -summable subset
of X, and such that {x: I(x»a} is f-summable for all positive a with
at most a countable number of exceptional values. Show that I is f-
summable, and show that I is also summable with respect to the Stiel-
tjes-Riemann integral fldv. Show, in addition, that f(/)==fldv for
this function I.
17.14) Let f(/) be an abstract Riemann integral generated by ex-
tension of a positive linear functional on the collection L, and let each
jEL be bounded and vanish outside a f-summable subset of X (the
subset may depend on I). Finally, let f(/) induce the charge v in X.
Show that in order that f(/)==fldv for all I, it is necessary and suf-
ficient that XXEM+.
17.15) Show that if f(/) is a Stieltjes-Riemann integralfldv, and
j is non-negative and f-summable, then Ip js f-summable for all P
such that O<P<oc>. Show that if I and g are f-summable, then Ig is
f -summable.
17.16) Let v be a charge on the ring A, and f(/)==/ldv. The inte-
gral f(/) induces the charge VI on the ring Al of subsets of X. Obvi-
ously, A c Al and VI ==v on A. Applying the extension procedure for
charges to (X, A, v), we obtain the extended charge v on the ring Av of
all v-measurable subsets of X. Show that Al ==Av, and VI ==v on Al ==Av.
18. Further Properties of the Integral
In the present section we shall derive some further properties of the
Stieltjes-Lebesgue integral f(/)==/xldfl, corresponding to the measure
it in X. As defined in sec. 15, a null function is a function I(x) satisfying
f(I/D == IIII dfl==O,
x
and a null set is a set E c X satisfying
f(XE)== I X E d/1==O.
x
In the present case fxXEdfl==fl(E) , so the null sets are exactly the sets

126
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4,9 18
of fl-measure zero. The theorem that 1 is a null function if and only if
the set {x: I(x) =¥=O} is a null set may be stated, therefore, in the equiva-
lent formulation that 1 is a null function if and only if I(x) == 0 almost
everywhere on X. The metric space L 1 is obtained by identification of
summable functions which are equal almost everywhere on X, and the
distance II/-gil between I, gEL 1 is then fx I/-gi dfl.
THEOREM 1. For any summable I, the set {x: I(x) =¥=O} is 01 a-linite
measure (i.e., a countable union 01 sets ollinite measure).
PROOF. We have
00
00
{x: I=¥=O}==  {x: l>n- 1 }+  {x: 1< -n- 1 },
n=l n=l
and each of the sets in these countable unions is of finite measure by
Theorem 2 in the preceding section.
Given the integrable function 1 and the measurable set E eX, the
product IXE is a measurable function, satisfying (IXE)+/+ and (IXE)-
. /-. It follows thatfx(IXE)+dfl andfx(IXE)-dfl exist, and one at least
of them is finite. HencefxlxEdfl exists. The jntegralfxlxEdfl is called
the integral oll(x) over the set E, and denoted by fEldfl. We extend the
definition of integration over a fl-measurable subset E of X to all cases
in which IXE is defined on X and integrable over X (where it is under-
stood that IXEO on X -E, even though I(x) itself is perhaps not
defined on X -E). We finally note that the argument used above shows
that if 1 is integrable over E, then 1 is integrable over any measurable
subset of E.
THEOREM 2 (MEAN VALUE THEOREM). III is integrable over the measur-
able set E, and -(X)a/(x)b(X) on E, then
afl(E)f Idflbfl(E).
E
PROOF. The desired result is immediately derived by observing that
fEadfl==afl(E) for any constant a satisfying -(X)a(X) (we recall that
(=l:(X)) . 0==0. (=l:(X)) ==0).

Ch. 4, 9 18J
FURTHER PROPERTIES OF THE INTEGRAL
127
TI-IEOREM 3. If En is a sequence of disjoint measurable subsets of X,
and f(x) is integrable over E ==  En, then
I fd#==  I fd#.
E En
PROOF. If the theorem is true for f+ and f-, then it is true for f.
We may assume therefore that fO on E. Then fXEnO for all n, and
fXE==  fXEn' so
I fd#== I fXEd#==  I fXEn d #==  I fd#
E x X En
by the theorem on integration of increasing sequences.
The last theorem shows that, given the non-negative #-measurable
function f(x) on X, the set function V(E)JEfd# is a measure on the
a-field All of all #-measurable sets E eX. We may apply, therefore, the
extension procedure for measures to (X, All, v), and it is of some im-
portance to observe that the a-field Av of all v-measurable sets may be
properly larger than All' If E is a #-measurable set, containing a subset
Eo which is not #-measurable, and if f==O on E, then E is a v-null set,
and so Eo is a v-null set. Hence EOEAv, but EOEAIl does not hold.
It is a rather trivial task to prove that if f is measurable on X, and
JEfd#O for all measurable E c X, then f==O almost everywhere on
X. It is sufficient for the proof to select for E the sets {x: f(x)O} and
{x: f(x) <O} respectively; the result is then that f+ and f- are null
functions. The question may be raised, however, if it is not possible to
deduce a similar result from a weaker hypothesis. An answer is con-
tained in the following theorem.
THEOREM 4. Let the measure # in X be generated by aPPlying the ex-
tension procedure to (X, r, #), where r is a semi-ring, and let the measur-
able function f(x) satisfy JAfd#==O for all A Er. Then f(x) o almost
everywhere on any set which is sequentially covered by r.
PROOF. It is sufficient to prove that f(x)O almost everywhere on
any set A Er.
(a) Let first #(A)<oo. By the remark preceding the present theo-

128
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 18
rem, the proof may be reduced to the proof that JEfd#O for any #-
measurable E c A (the summability of f over any #-measurable E c A
is guaranteed, since f is summable over A). If E c A is a a-set with
respect to T, we have JEfd#==O by the preceding theorem; hence, if
E c A is the limit of a descending sequence of a-sets Ene A, then also
JEfd#==O by the theorem on dominated convergence (lim fXEn==fXE
and IfXEnllfXEll for all XEX). Since any #-measurable E cA is the
difference of such a limit and a #-null set, it follows that JEfd#==O for
any #-measurable E cA.
Note that we may conclude already from what has been proved so
far that f==O almost everywhere on any set of finite measure.
(b) Let A ET (without restriction on #(A)), and let O<a<oo. Since
j is summable over A, the set Aa=={x: xEA, f(x) >a} is of finite measure
by Theorem 2 in the preceding section, so f==O almost everywhere on
Aa by (a). On the other hand f>a>O on Aa. Hence #(Aa) ==0 for all
positive a. Similarly, #(Ab)==O for all negative b, where Ab=={x: xEA,
j(X) <b}. It follows that fO almost everywhere on A.
I t is a consequence of the last theorem that if # is Lebesgue measure
in Rk, and the integral ot f(x) over any cell vanishes, then f(x) ==0
almost everywhere on Rk. Note, however, that in the general case it
does not always follow from JAfd#==O for all A ET that f==O almost
everywhere on X. A counterexample is obtained by considering the
example that X ==R2 and # is one-dimensional Lebesgue measure on
the semi-ring T of all horizontal cells A =={(X1, X2) : a<x1b, X2==C}. If
j is the characteristic function of a vertical line, then JAfd#==O for all
A ET, but f== 1 on a set of infinite measure.
We conclude the section with a theorem on differentiation with re-
spect to a parameter under the sign of integration. Before doing so,
the theorems on integration of monotone sequences and on dominated
convergence will be generalized to a certain extent; we replace the
discrete parameter n by a continuous parameter A.
LEMMA cx. (a) Let f(x, A) be a #-measurable function on X for all
values of the real parameter A satisfying A1A<Ao, and let Of(x, A) i f(x)

Ch. 4,  18J
FURTHER PROPERTIES OF THE INTEGRAL
129
on X as Ai Ao, except on a null set. Then f(x) is fl-measurable, and
I f(x, A) dfl i I f(x) dfl
x x
as A i Ao.
(b) Let f(x, A) be fl-measurable on X for all values of the parameter A
satisfying A1A<Ao, and let lim f(x, A)==f(x) on X as Ai Ao, except on a
null set. Furthermore, let g(x) be a summable function such that If(x, A) I
g(x) on X for all A, except on a null set which may depend on A. Then f
is summable, and
lim I f(x, A) dfl== I f(x) dfl
x x
as A i Ao.
Similar assertions hold if Ao<AA1 and A!Ao.
PROOF. Follows immediately by observing that, for any sequence
An satisfying An i Ao, the desired conclusions hold for the sequence
f(x, An).
THEOREM 5 (DIFFERENTIATION UNDER THE SIGN OF INTEGRATION).
Let f(x, A) be summable over X for all values of the parameter A satisfying
A1<A<A2, and let f;.(f)=/xf(x, A)dfl. Furthermore, let for the fixed value
Ao the partial derivative of (x, AO)/OA exist on X, except on a null set. Final-
ly, let g(x) be a summable function such that, for all A in a neighbourhood
of Ao,
f(x, A) - f(x, Ao) (
g x)
A-Ao
on X, except on a null set which may depend on A. Then of (x, AO)/OA is
summable over X, and
( df;.(f) ) == f !L (x, AO) dft.
dA A=AO OA
x
PROOF. Follows immediately from part (b) of the preceding lemma,
replacing f(x, A) in the lemma by {f(x, A)-f(x, Ao)}/(A-Ao).
Note that if the hypotheses of the theorem are satisfied in the closed
A-interval {A: A1AA2}, and Ao coincides with Al or A2, then the con-
clusion holds provided the derivatives with respect to A in the point Ao

130
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 18
are understood to be one-sided derivatives. Finally, the following par-
ticular case is of importance for the applications.
Let f(x, A) and g(x, A) be summable over X for {A: Al <AA2}, and let
f;.(f)==/xf(x, A)d#. Furthermore, let of (x, A)joA exist on X for these values
of A. Finally, for all A' in a neighbourhood of A and all XEX, let
f(x, A') - f(x, A)  ( A )
A'-A gx,.
Then df;.(f) jdA== /x{of(x, A)joA}d# for all A.
Exercises
SUMMABILITY
18.1) Let #(X) <00, let f be real, finite and measurable on X, and
let En=={x: n-l f(x) <n} for n==O, :l: 1, :l:2, . . . . Show that f is
summable over X if and only if oo Inl#(En)<oo.
18.2) Let the non-negative function f be #-measurable on X, and
let, for n==l, 2, "', fn(x)==f(x) on {x: f(x)n} and fn(x)==n on {x:
f(x) >n}. Show that f(fn) t f(f).
18.3) Let the non-negative function f be summable over X, and let,
for n== 1, 2, . . " fn(x) ==f(x) on {x: f(x)n} and fn(x) ==0 on {x: f(x) >n}.
Show that f(fn) i f(f). Show also that this may be false if f is only
assumed to be non-negative and #-measurable on X, but that it is
again true if f is non-negative, #-measurable and finite almost every-
where on X.
18.4) Show that if f is summable over X, and En=={x: f>n} for
n== 1, 2, . . " then lim n#(En) ==0 as noo. Show also that the converse
is not always true.
UNIFORM CONVERGENCE
18.5) Let #(X) <00, and let fn(x) be a sequence of summable
functions, converging uniformly on X to f(x). Show that f is summable,
and lim f(lf-fnl)==O. This implies then that lim f(fn)==f(f). In other
words, if #(X) <00, then uniform convergence implies convergence in
the metric space L 1 of all summable functions. Show also that this does

Ch. 4, 9 18J
FURTHER PROPERTIES OF THE INTEGRAL
131
not necessarily hold if #(X)==oo, but that it is again true if Ifnlg
almost everywhere on X for some summable g, since in this case ordi-
nary pointwise convergence of In to I almost everywhere on X already
implies that lim f(I/-lnl) :::=0.
POINTWISE CONVERGENCE
18.6) If a sequence of functions In on X converges pointwise on X
as well as in the metric space L1, then the limits are the same. More
precisely, show that if In (x) and I(x) are summable over X, limf(I/-lnl)
==0, and lim In(x)==g(x) almost everywhere on X, then I(x)==g(x)
almost everywhere on X.
18.7) Let En be the sequence of closed intervals [0, lJ, [!, 1J, [0, -lJ,
[1, iJ, [I, IJ, [0, 1J, · . " let In (x) == XE n (x) , and let f(ln) be the Lebesgue
integral of In over [0, IJ. Show that In converges to zero in the metric
space L 1 of all functions which are Lebesgue summable over [0, IJ, but
there exists no point x where In (x) tends to zero in the ordinary
pointwise sense. Hence, even though #(X)<oo and Ilnlg on X for
some summable g, it is not always true that convergence in L1 implies
pointwise- convergence almost everywhere on X. Note, however, that
by the corollary of the completeness proof for L1 there exist many
subsequences of In converging pointwise to zero almost everywhere
onX.
Show also that, even though # (X) <00, pointwise convergence almost
everywhere on X of the sequence of summable functions gn to the
summable function g does not always imply convergence of gn to g in
L 1 . In order to obtain convergence in L1 it is, therefore, essential that
an additional condition, such as e.g. in the theorem on dominated
convergence, should be satisfied.
CONVERGENCE IN MEASURE
18.8) Given the sequence of measurable functions In, fjnite almost
everywhere on X, and the measurable function 10, finite almost every-
where on X, it is said that In converges on X to 10 in measure if limnoo #{x:
Ifn-/ola}==O for every a>O. Show that if In and gn converge in
measure to I and g respectively, then In+gn converges in measure to
f+g.

132
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 18
18.9) Let In be a sequence of measurable functions on X, let g be a
summable function such that Ifnlg on X for all n, and let lim In==1
almost everywhere on X. Show that if Dp==;:=p {x: Ifn-/la} for
some a>O, then lim #(Dp) ==0 as P-+oc>. Note that it follows in par-
ticular that In converges in measure to I.
18.10) Let #(X) < oc>, let In and I be measurable and finite almost
everywhere on X, and let lim fn==1 almost everywhere on X. Show
that if Dp has the same meaning as in the preceding exercise, then
lim #(Dp) ==0. In particular, it follows again that In converges in
measure to I.
18.11) Let I n(x) be defined on X =={x: Ox< oc>} for n== 1, 2, . . . by
In(x)==O on [0, n) and (n+ 1, oc», In(x)== 1 on en, n+ IJ. Show that In
converges pointwise to zero on X, but not in measure.
EGOROFF'S THEOREM
18.12) Let the hypotheses of either Exercise 18.9 or Exercise 18.10
be satisfied. Show that, given 8>0, there exists a set De of measure not
exceeding 8 such that In converges to I uniformly on X -De. With the
hypotheses of Exercise 18.10 this is EGOROFF'S theorem (1911, [1 J).
18.13) Use the example in Exercise 18.7 to show that if In converges
to I in measure, then In does not necessarily converge pointwise to I
almost everywhere on X, even if #(X)<oc> and Ilnlg on X for a
summable function g.
CONVERGENCE IN MEASURE AND POINTWISE CONVERGENCE
18.14) Show that convergence in measure and pointwise convergence
almost everywhere on a set X of a-finite measure are related in the
following way. If all fn(x) and I(x) are measurable and finite almost
everywhere on X, then In converges in measure to I on every subset
of finite measure if and only if every subsequence of In contains a
subsequence which converges pointwise to I almost everywhere on X.
CONVERGENCE IN MEASURE AND IN NORM
18.15) Show that if In (n==l, 2, ...) and I are summable over X,
and limJxl/-Inld#=O, then In converges on X to I in measure. Show
that the converse does not always hold, even though #(X)<oc>, but

Ch. 4, 9 18J
FURTHER PROPERTIES OF THE INTEGRAL
133
that it does hold whenever there exists a summable g(x) such that
Ifnlg almost everywhere on X.
DISTRIBUTION FUNCTIONS
18.16) Let O</-l(X) <00, and let f(x) be finite and /-l-measurable on
X. Show that if the real function P(t) of one variable is defined by
P(t)==/-l{x: -oo<f(x)t},
then P(t) is increasing and right continuous such that
P(-oo)=O
and
P(+oo)==lim P(t)==/-l(X),
tt oo
The function P(t) is called a distribution function of f(x) ; sometimes the
use of this name is restricted to the case that /-l(X) == 1 (then P(t) is
the "probability" that f(x) assumes a value t).
18.17) Let /-lp be the Stieltjes-Lebesgue measure on the real line R1,
generated by the function P(t) of the preceding exercise (cf. sec. 10,
Example 8). Show that if E is any /-lp-measurable set in R!, then
{x: f(x) EE} is /-l-measurable, and
/-l{x: f(x) EE}==/-lp(E).
Note that /-l(X) ==/-lp(R1) , and that if E consists of one point to, then
/-l{x: f(x) == to} ==/-lp({to}) .
18.18) Let X, /-l, f and /-lp have the same meaning as in the preceding
exercise. Show that if F(t) is /-lp-integrable and G(x) is defined on X by
G(x) ==F(t) on {x: f(x) ==t}, then G(x) is /-l-integrable, and
fGd/-l== fFd/-l p ,
x Rl
Note, in particular, that if f is /-l-integrable, then t is /-lp-integrable, and
ffd/-l== ftd/-l p ,
x Rl
18.19) Let y==P(t), defined on -ootoo, be increasing and right
continuous, and let
P(-oo) ==0, P(+oo)==lim p(t)==a
tt oo
with O<a<oo.

134
STIEL TJ ES- LEBESG UE INTEGRAL
[Ch. 4, 9 19
Show that its inverse function f*(y), defined on Oya by f*(y)
inf{t: P(t) >y} for Oy<a, f*(a)==lim yta f*(y), is increasing and right
continuous.
18.20) With the notations of the preceding exercise, and flL Lebes-
gue measure on [0, a], show that
flL{Y: -oo<f*(y)t}==P(t) for -oo<t<oo.
Use Exercise 18.18 to show that any Stieltjes-Lebesgue integralfRl F dflp
(where flp is the measure generated by P(t)) may be expressed as a
Lebesgue integral over [0, a]. Consider in particular the extreme case
that P(t) ==0 for t<O and P(t) ==a for tO.
18.21) Let O<fl(X)==a<oo, let f(x) be finite and fl-measurable on
X, and let P(t) be the distribution function of f on R1 according to
Exercise 18.16. Furthermore, let f*(y), defined on [0, a], be the inverse
function of y==P(t) according to Exercise 18.19. Then f(x) on X and
f*(y) on [0, a] are equidistributed, i.e.,
fl{X: -oo<f(x)t}==flL{Y: -oo<f*(y)t} for every t.
Show that if fn(x) is a sequence of finite fl-measurable functions on X,
tending on X to the finite function f(x), then lim f(y)==f*(y) on [0, a],
except on an at most countable subset of [0, a].
18.22) Let O<fl(X)==a<oo, let f(x) be finite and fl-integrable over
X, and let P(t) be the distribution function of f on R 1 . Show that
o 00
j fdfl== - j P(t) dt+ j {a-p(t)}dt,
x -00 0
where the integration is Lebesgue integration over (-00, OJ and [0, 00)
respectively.
19. The Integral for Complex Functions
Given the real fl-measurable functions g(x) and h(x) on the fl-measur-
able set EeX, the complex function f(x)g(x)+ih(x) is called a fl-
measurable comPlex function on E. If g and hare fl-summable over E,
then f==g+ih is said to be fl-summable over E, and the integral fEfdfl
is defined by
j fdfl j gdfl+ij hdfl.
E E E

Ch. 4, 9 19J
INTEGRAL FOR COMPLEX FUNCTIONS
135
The collection of all complex functions f such that f is summable over
E is a complex linear collection, and !Efdfl is on this collection a
comPlex linear functional (i.e., if f1 and 12 are summable over E, and
ex, fJ are complex constants, then exf1 +fJf2 is summable over E, and
I (exf1 +fJf2) dfl==ex f f1 dfl+fJ I f 2 dfl).
E E E
THEOREM 1. The comPlex fl-measurable function f is fl-summable over
E if and only if If I is fl-summable over E, and in that case
If fdfll/lfl dfl.
E E
PROOF. The measurability of f==g+ih (g and h real) is by definition
equivalent to the measurability of g and h, and implies therefore (by
sec. 17, Theorems 5 and 6) the measurability of Ifl==(g2+h2)l. Note
that we make use here of the hypothesis that the integral is a Stieltjes-
Lebesgue integral, and not merely a general Daniell integral (cf. how-
ever, sec. 22, Theorem 3).
Let now f==g+ih be summable. Then g and hare summable, so Igl
and Ihl are summable. It follows that If I is summable, since If I is measur-
able and 1/1lgl+lhl.
Let, conversely, I be measurable and If I summable. Then Igi and Ihl
are summable, since they are measurable, and Igllfl, Ihllfl. It follows
that g and hare summable, so f==g+ih is summable.
Assuming now that f is summable over E, let !Efdfl==reiq;, where
r== I!Efdfll and cp is real. Then f1(X) == f(x) e-iq; satisfies !Ef1dfl==r. Hence,
writing f1==gl+ih1 (gl and hI real), we have r==!Eg1dfl+i!Eh1dfl, so
r=!Eg1dfl. It follows that
II fdfll==r== I gldfl/lg11 dfl/lf11 dfl== Ilfl dfl.
E E E E E
Observe that it has been used in the last part of the proof that f, f1
and gl are finite almost everywhere on E. Note also that I!Efdfll==
IE III dfl<=>glO and hI ==0 almost everywhere on E <=>f10 almost every-
where on E<=>f(x) == If(x) I eiq;, with cp fixed, holds almost everywhere on E.
The definition of a comPlex null function is the same as in the real
case: I==g+ih is a null function whenever f(lfl) ==0. It follows immedi-

136
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 20
ately that I==g+ih is a null function if and only if 1==0 almost every-
where (i.e., ghO almost everywhere). Also, 11 and 12 are defined to
be almost equal whenever 11 - 12 is a null function, i.e., whenever 11 == 12
almost everywhere. Hence, if 11 and 12 are almost equal, and 11 is
summable, then 12 is summable, and f(/l)f(/2) as well as f(1/11)==
f(1/21). We finally note that the theorem on dominated convergence
holds for complex functions.
II Ilim In almost everywhere, il all In are measurable, and il there
exists a summable lunction g such that Ilnlg almost everywhere lor all n,
then I is summable and f(/)lim f(ln).
The proof is derived by considering the real and imaginary parts
separately.
20. The Metric Space of all Complex Summable Functions
We shall from now on denote by L 1 the complex linear collection of
all complex functions which are fl-summable over X, and the real linear
subcollection of all real fl-summable functions (which was called L 1 in
the preceding sections) will then be denoted by Lr). In sec. 16 it was
shown that if almost equal functions are identified, then Lr) is a com-
plete metric space with respect to the distance function II/-gll==
f(l/-gl), and the initial function collection L (i.e., the initial domain
of definition of the integral) is dense in Lr). Since we have here a
Stieltjes-Lebesgue integral, the distance function may be written as
11/-glI==Jxl/-gldfl, and it follows in particular that the collection of
all real step functions (corresponding to the measure fl) is dense in Lr),
since the integral may be considered as the extension of its own re-
striction to the fl-step functions. Obviously, identifying almost equal
complex functions, the complex collection L 1 is a metric space with
respect to the same distance function II/-gll== Jx I/-gi dfl, and it is not
difficult to prove that the metric space is complete. This may be done
in two different ways; we can either repeat the proof for the real case
(the proof is unchanged), or we can observe that if In==gn+ihn (with
gn and h n real) is a fundamental sequence in L1, then gn and h n are
fundamental sequences in Lr), converging therefore (in the sense of
the metric) to functions gELr) and hELr) respectively, and In converges
then to l==g+ihEL 1 .

Ch. 4, 920J SPACE OF ALL COMPLEX SUMMABLE FUNCTIONS 137
We recall that, by the results in sec. 11, the collection Al of all #-
measurable subsets of X of finite measure is a metric space with respect
to the distance function p1(E, F) ==#(E -F) +#(F -E), provided almost
equal sets are identified. Since the expression for p1(E, F) may be
written as
p1(E, F) == f IXE-XFI d#== IIXE-XFII,
x
it is evident that the metric space Al may be regarded as a subspace
of Li r ) (more precisely: the one-one mapping of any E EA 1 onto its own
characteristic function XEELi r ) preserves distances). This fact enables
us to prove easily that Al is complete.
THEOREM 1. The metric space Al of all sets of finite measure, with the
distance function
p1(E, F) == f IXE-XFI d#,
x
is comPlete.
PROOF. Let limlixEm -xEnll==O as m, noo. Then, since Li r ) is com-
plete, there exists a function fELi r ) such that limllf-xEnll==O as noo,
and by the corollary of the completeness theorem for Li r ) there is a
subsequence XE p (pn1, n2, · . .) such that XE p converges pointwise to
f almost everywhere on X. Since each XE p assumes only the values zero
and one, the same is then true of f (almost everywhere), so f==xE for
some #-measurable set E of finite measure (#(E) <00 on account of
#(E)== f XEd#== ffd#==lIfll<oo).
x x
If the collection of all functions g+ih, with g, hEL, is symbolically
denoted by L+iL, then it follows immediately from the property that
L is dense in Li r ) that L+iL is dense in L1. In particular, the col-
lection of all complex step functions (corresponding to the measure #)
is dense in L 1 .
THEOREM 2. The metric space L1 is separable if and only if the measure
/1 is separable.
PROOF. (a) Let L 1 be separable. It follows then (by the theorem
that any subspace of a separable metric space is separable) that the

138
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 20
subspace of all characteristic functions of sets of finite measure is sepa-
rable. This is equivalent to the assertion that Al is separable, i.e., that
# is separable.
(b) Let # be separable, Le., let there exist a countable collection Z
of subsets of X such that Z is dense in the metric space Al of all sets
of finite measure. The collection S(Z) of step functions =l YnXFn(X),
where Yn is rational complex and FnEZ for n==l, ..., p, is then also
countable (note that F1, . . " F p are not necessarily disjoint). We shall
prove that S(Z) is dense in L 1 . Let fELl and e>O. Since the col-
lection of all step functions is dense in L1, there is a step function
sl(x)==fcxnXEn(x) such that IIf-s111<le. Choosing the rational com-
plex numbers Y1, . . " Yp such that f Icxn -Ynl#(En) < 1E, and setting
s2(x)==f YnXEn(X) , we have
p
Ils1-s211 Icxn-Ynl#(En)<le.
1
Next, choosing for n== 1, . . ., p the sets F n EZ such thatJx IXEn -XFJ d#
<ef(3PIYnl), and writing S3(X) == f YnXFn(X) , we have S3 ES (Z) and
p
IIs2-s31/ IYnif IXEn -xFnl d#< lEe
I X
Hence, S3ES(Z) and Ilf-s31/I/f-s1Ir+ I/s1-S21/+ Ils2-s31/ <E. This shows
that S(Z) is dense in L1, and since S(Z) is countable, it follows that L 1
is separable.
Exercises
A NECESSARY AND SUFFICIENT CONDITION FOR MEASURABILITY
20.1) Let # be a measure on the ring A of subsets of X, let L be the
collection of all real step functions with respect to # and A, and let
J fd# be the corresponding extended Stieltjes-Lebesgue integral. Show
that the real function f(x) on X is #-measurable if and only if the
following condition is satisfied: Corresponding to any set E E A of
finite measure, there exists a sequence sn(x) EL converging pointwise
almost everywhere to f(X)XE(X),

Ch. 4,9 21J RIEMANN INTEGRAL AND LEBESGUE INTEGRAL 139
21. Riemann Integral and Lebesgue Integral, and the Riesz
Representation Theorem
The reader will have noted that we have not yet evaluated any
integrals, except in rather trivial cases. One of the means to improve
the situation will be a theorem asserting that any bounded function,
Riemann integrable over a bounded interval in finite-dimensional real
number space, is Lebesgue summable over the same interval, and the
two integrals have the same value. As soon, therefore, as we shall also
be familiar with the rules which govern the transformation of a multi-
ple integral into a repeated integral, the machinery for computing
ordinary Riemann integrals of one variable is available in order to
handle Lebesgue integration in an efficient manner.
THEOREM 1. Let fl be Lebesgue measure in k-dimensional real number
space Rk. If the bounded function f(x) is Riemann integrable over the
bounded interval JeRk, then f(x) is Lebesgue summable over J, and the
values of the Lebesgue integral (L) fL1fdfl and the Riemann integral
(R) jL1 f are equal.
PROOF. We assume first that f is real-valued. There is no loss of
generality in assuming also that J is a cell, since this affects neither
the Riemann nor the Lebesgue integral. Let No be a net, consisting of
a finite number of disjoint cells covering J exactly, i.e., if J1, . . " J p
are the cells of the net No, then l Lli==J. Let mi==inf f(x) on J i , and
Mi==sup f(x) on J i for i=== 1, . . " p. The functions go(x) and ho(x) are
defined on J by gO(x)==mi on LJ i and ho(x)==Mi on J i (i==l, "', P).
Then
p
 mifl(Ji)==(L)fgo(x)dfl,
1 
p
 Mifl(Ji)==(L)fho(x)dfl.
1 
We consider now a sequence No, N 1, . .. of nets of the same kind as
No, such that:
(a) the net No, which was introduced above, is the first member of
the sequence,
(b) the net N n is a refinement of the net N n-1 (n=== 1, 2, . . .), Le.,
each cell of N n is entirely included in some cell of N n-1,
(c) the maximal diameter of the cells of N n tends to zero as n-+oo.
Corresponding to the net N n the functions gn(x) and hn(x) are defined

140
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 21
in the same way as go(x) and ho(x) were defined for No. For any point
XEL1, the numbers gn(x) form an increasing sequence and the numbers
hn(x) form a decreasing sequence. Hence, gn(x) i m(x) and hn(x)! M(x) ,
where m(x)M(x), and the functions m(x) and M(x) are fl-measurable.
It follows that
(L) J gndfli (L) J mdfl,
L1 L1
(L) J hndfl! (L) J M dfl
L1 L1
by the theorem on dominated convergence. Consequently,
limmi,u(L1i)===(L)Jm(x)dfl as noo,
L1
lim  Mifl(L1i)==(L)JM(x) dfl as noo,
L1
and, since f(x) is Riemann integrable over LI by hypothesis, the limits
on the left are both equal to (R)JLtI. But then
(R) J 1=== (L) J m dfl== (L) J M dfl,
L1 L1 L1
so (L)JLt (M -m) dfl===O. It follows that m(x)==M(x) almost everywhere
on LI, so that, in view of m(x)f(x)M(x), we have also m(x)==/(x)===
M(x) almost everywhere on L1. This shows that f(x) is Lebesgue sum-
mable over L1, and
(L) J I d,u== (L) J m dfl== (R) J f.
L1 L1 L1
If I==g+ih (g and h real), the above proof is valid for g and h separately,
and the same result is obtained.
Remarks. (a) The interval L1 may be replaced by any bounded Le-
besgue measurable set E cR k provided (R) JEf is defined, that is, pro-
vided fXE is Riemann integrable over any interval L1 covering E.
(b) Note that the characteristic function of the set of all rational
points in [0, IJ is bounded and Lebesgue summable, but not Riemann
integrable over [0, 1 J.
(c) If f(x)O on axb, and f(x) is (properly) Riemann integrable
over a+15xb for all15 satisfying O<15<b-a (this implies, therefore,
that f is bounded on [a+15, bJ), then the Lebesgue integral of f over
[a, bJ exists and equals lim (R) J:+tJI as 15! 0 (it is not excluded, however,

Ch. 4,9 21J RIEMANN INTEGRAL AND LEBESGUE INTEGRAL 141
that the limit is +00). In particular, if the improper Riemann integral
of the non-negative function lover [a, bJ exists as a finite number,
then I is Lebesgue summable over [a, bJ, and the Lebesgue integral
equals the improper Riemann integral. The proof is immediately de-
rived by means of the theorem on integration of monotone sequences.
Similarly, if l(x)O on [a, 00), and f is Riemann integrable over [a, bJ
for all finite b>a, then the Lebesgue integral of lover [a, 00) exists
and is equal to lim (R) J:f as b joo. If, however, I(x) has an infinite
number of changes of sign, then it may happen that the improper
Riemann integral exists, and the Lebesgue integral fails to exist (cf.
Exercise 21.2).
Let L be the collection of all real continuous functions I on Rk, such
that the set {x: I(x) *O} is bounded. In other words, L is the col-
lection of all real continuous functions on Rk having a bounded carrier
(we recall that the carrier of I is the closure of the set {x: I(x) *O}).
Then L is a real linear collection, and I, gEL implies max (I, g) EL and
min(/, g)EL. If, for IEL, the number f(/) is defined as the Riemann
integral of f over Rk, then f(/) is an elementary integral on L (cf.
sec. 12) . We shall prove now that the extension procedure for integrals,
applied to f(f) on L, generates the Lebesgue integral.
THEOREM 2. The integral over Rk, generated by extending the Riemann
integral f(f) lor continuous functions I possessing a bounded carrier, is
the Lebesgue integral.
PROOF. By sec. 17, Lemma ex (which is a reformulation of the theo-
rem that (X, r, #) and (X, r1, #1) generate the same measure if and
only if #1==# on rand #==#1 on r1), it will be sufficient to prove that,
for any continuous I with a bounded carrier, the Lebesgue integral
equals the Riemann integral, and that, for any step function 1==
l cnX.An(x) where all An are cells, the extended integral f(f) exists
and equals the Lebesgue integral of I (note that we use here that the
Lebesgue integral is generated by extension of its own restriction to
this particular class of step functions; cf. the remark at the end of
sec. 17). The first assertion follows immediately from the preceding
theorem, and in the proof of the second assertion we may, evidently,
restrict ourselves to the case that I==XA, where A is a cell. Selecting

142
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 21
the cell B such that A is included in the interior of B, there exists a
sequence fn of continuous functions, all of them vanishing outside B
and satisfying Ofnl, such that lim fn==XA and lim f(fn) ==fl(A) ,
where fl denotes Lebesgue measure in Rk. Then, by the theorem on
dominated convergence, XA is f-summable and lim f(fn)==(XA)'
Hence f(XA) ==fl(A) ==j XAdfl. This completes the proof.
COROLLARY. The integral over the bounded closed interval L1 c Rk,
generated by extending the Riemann integral f(f) over L1 for continuous
functions, is the Lebesgue integral over J.
PROOF. The proof is similar to the proof of the above theorem if the
following points are taken into account:
(a) The Lebesgue integral over J is the same as over L1 e , where J e
is the cell having the same vertices as J.
(b) The Lebesgue integral over J e is obtained by extension of its
own restriction to the collection of all step functions l cnXAn(X), whee
all A n are cells included in J e .
(c) The proof that, for any cell A included in J e , its characteristic
function XA is f-summable, and f(XA)==fl(A), is even simpler than
above since the functions fn satisfying lim fn==XA are merely required
to be continuous on J (i.e., if fn is defined to be zero outside J, then
fn may be discontinuous on the boundary of J).
There exists an important generalization of the last theorem which,
for k== 1, is essentially F. RIESZ'S representation theorem for positive
linear functionals on the collection of continuous functions of one vari-
able (1 909, [1 J) .
THEOREM 3 (RIESZ REPRESENTATION THEOREM FOR POSITIVE LINEAR
FUNCTIONALS). Let L be the collection of all real functions f(x), defined
and continuous on Rk, and having a bounded carrier. Then any non-
negative linear functional f(f) on L is an elementary integral on L.
Hence, there exists a measure v in Rk such that
f(f) == I fdv
Rk
for all fEL.
The a-field Av of all v-measurable subsets of Rk contains all open sets, and

Ch. 4, 921J RIEMANN INTEGRAL AND LEBESGUE INTEGRAL 143
therefore all closed sets and all cells. In addition, every bounded set E E Av
is of finite v-measure.
Conversely, if Av is a a-field of subsets of Rk, containing all open sets,
and if v is a measure on Av such that v(E) <00 for all bounded EEAv,
then JRkfdv exists as a finite number for each fEL, and
/(f) == I fdv
Rk
is a non-negative linear functional on L.
PROOF. Let f(f) be a non-negative linear functional on L. Given
the sequence fnEL, satisfying fn!O on Rk, we first select a bounded
closed interval J covering the carrier C /I of f1, and then we select the
function gEL + such that g(x) 1 on J. Denoting now maxx fn(x) by
IIfnll, we have fn(x)lIfnllg(x) for all n and x, so f(fn)"fn"f(g)! 0,
since Ilfnll!O by Dini's theorem (cf. sec. 12). This shows that f(f) is an
elementary integral on L, and, since L has the additional property that
fEL implies min (f, 1) EL, the (extended) integral f(f) is a Stieltjes-
Lebesgue integral, so f(f)==JRkfdv, where v is the measure induced in
Rk by the (extended) integral f(f). If E is an arbitrary bounded open
subset of Rk with complement Ee==Rk-E, and f(x)==d(x, Ee) is the
ordinary Euclidean distance in Rk between the point x and the set Ee,
then the function f is continuous and vanishes outside of E, so fEL.
I t follows by sec. 17, Theorem 2 that the set {x: f(x) >O} is v-measur-
able. This set, however, is exactly E. Hence, any bounded open set E
is v-measurable, and the same is then true for any open set, either
bounded or unbounded. In order to prove that v(E) <00 for any bounded
v-measurable E cRk, it will be sufficient to give the proof for the case
that E is a bounded closed interval. In this case, selecting gEL + such
that g 1 on E, we have
v(E)== J XEdv1 gdv==f(g) <00.
Let now, conversely, the a-field Av contain all open sets, and let l'
be a measure on Av satisfying v(E) <00 for all bounded EEAv. Then,
if fEL, the set {x: f+>a} is open for any a>O (since f is continuous),
so {x: f+>a}EAv for all a>O. It follows by sec. 17, Theorem 3 that f+
is v-measurable, so J f+dv exists. Furthermore,
J f+dvllf+llv(Cf)<oo.

144
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 21
Similarly, II-dv exists as a finite number. This shows that f(/) ==1 Idv
exists as a finite number for any IEL. Obviously, f(/) is a non-negative
linear functional on L.
The problem of the uniqueness of v in the first part of the theorem,
and the extension to linear functionals which are not necessarily non-
negative, will be discussed in sections 47, 49.
ExamPle. If the leal function g(x) is increasing and right continuous
on R1, and fly is the Stieltjes-Lebesgue measure gel1erated by g(x) , then
any open subset of R1 is fly-measurable in view of the fact that any
open set is a a-set with respect to the semi-ring of all cells (for the
proof, cf. sec. 10, Example 1). Hence, by the second part of the Riesz
representation theorem, I Rl I dfly exists as a finite number for any con-
tinuous I which vanishes outside a bounded subset of R1. In particular,
if J is a bounded closed interval, then/L1ldfly exists as a finite number
for any I which is continuous on J.
"-
Note that if h(x) is a second function which is also increasing and
right continuous on R1, then h is fly-measurable, since for any a>O and
b<O, the sets {x: h(x»a} and {x: h(x)<b} are of the form [xo, 00) or
(xo, 00) and (-00, Xl] or (-00, Xl) respectively. It follows that h+ and
h- are fly-integrable, and h itself is fly-summable over any bounded
in terval.
Exercises
RIEMANN INTEGRABILITY
21.1) Let I(x) be bounded on the bounded interval LI c Rk. Show
that I(x) is Riemann integrable over J if and only if I is continuous on
J, except at most on a Lebesgue null set.
EXAMPLES IN LEBESGUE INTEGRATION
In the following sequence of exercises the Lebesgue integral of lover
the interval from a to b will be denoted by I: Idx.
21.2) Show that the function I(x) ===sin x/x is not Lebesgue inte-
grable over [0, 00), although I(x) is continuous on [0, b] for every b>O,
and lim It Idx exists as bi 00.

Ch. 4,  21J RIEMANN INTEGRAL AND LEBESGUE INTEGRAL 145
21.3) Show that the function
d ( . 1 ) . 1 2 1
f(x) == - x 2 sIn - == 2xsIn- - - cos-
dx x 2 x 2 X x 2
is not Lebesgue integrable over [0, IJ, although f(x) is continuous on
[8, IJ fo every 8 satisfying 0<8<1, and limfe 1 fdx exists as 8!0.
21.4) Show that if p and q are positive, then
1
f Xp-l 1 1 1 1
l+x q dx=p- p+q + P+2q - p+3q +....
o
Consider, in particular, the cases that p== 1, q== 1 and p== 1, q==2.
21.5) There exist many methods for computing the value of
b
cx==lim I (sin x/x) dx.
bt oo 0
The following method is elementary, and yields at the same time the
sum of the series
 n- 1 sin nx
for
0<x<2n.
Show first, by summing the finite geometrical series = -n e ikt and
taking real parts, that
i+cos t+cos 2t+ . . . +cos nt==sin (n+-l)t/2 sin it
for 0<t<2n. Next, show that, for 0<x<2n,
x/2
1 . 1 . 1. _ f sin (2n + 1 ) t
"2X + SIn x + "2 sIn 2x + . . . + - sIn nx - . dt.
n sInt
o
(1)
Furthermore, show that if g(t) is now defined on [0, ixJ by g(O) ==0,
g(t) == (sin t)-1-t- 1 , then g has a continuous derivative on [0, -lxJ, so
that it follows by partial integration that ft/ 2 g(t) sin ktdt tends to zero
as k tends to 00. Note, for later purposes, that the convergence is uni-
form in x for Oxn. Show now that, for 0<x<2n, the expressions in
(1) have a limit as n-+oo if and only if ft/ 2 {sin (2n+ 1 )t/t}dt has a limit,
and these limits are then the same. Observe that the last integral has

146
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4,  21
indeed a limit for n-+oo, namely the required number ex. Finally, set
x==n in the so obtained equality r n- 1 sin nx==ex-!x. This shows
that ex===!n and r n- 1 sin nx===!(n-x) for 0<x<2n.
21.6) Show that the partial sums sn(x) of  n- 1 sin nx are unI-
formly bounded on [0, 2nJ, and show also that
00
00
 n- 2 cos nx=  n- 2 -!nx+!x 2
1 1
for
0x2n.
Deduce from this formula that oon-2-1n2 and oo ( _I ) n-l n -2- 1 n 2
1 - 6 1 -12 ·
21.7) Show that
1 1
J o: dx= - f log(+x) dx= - l2 n2 ,
o 0
1 1
f log x f log ( X l-X)
dx == dx = - in 2 ,
I-x
o .0
"
and
1
f logxlog(l-x)dx=  1 =  {  _ 1 - }
1 n(n+I)2 1 n n+l (n;I)2
o
- 2 - 1"7T2
- 6 0JfI ·
21.8) Show that
00
 n- 3 sin nx==!n 2 x-!nx 2 + l 2 x 3
1
for
0x2n,
and deduce from this formula that 1-3-3+5-3-7-3+...= l2 n3.
Show now that
n/2 00 1
f (log tg cp)2 dcp = f (log X)2 dx === 2 f (log x)2 dx = 1.n 3 ,
1 + x 2 1 + x 2 8
000
21.9) Show that
n/2 n/2
I = flog sin xdx= flog cos xdx== -in log 2.
o 0

Ch. 4,9 21J RIElVIANN INTEGRAL AND LEBESGUE INTEGRAL 147
21.10) Show, by considering 21 1 + 21 2 and 21 1 - 21 2 , and by using
the results in Exercises 21.8 and 21.9, that
n/2 n/2
1 1 == 1 (log sin x)2dx== 1 (log cos x)2dx== l4 n3+!n(log 2)2,
o 0
n/2
1 2 == 1 (log sin x) (log cos x) dx==!n(log 2)2- 41s n3.
o
21.11) Show that if
x
F(x) == (I e- t2 dt)2,
o
1
f e-x2(1+t2)
G(x) == dt
l+t 2
o
for xO, then the derivative of F(x)+G(x) vanishes for all xO. Show
that F(x) +G(x) ==!;n for all xO, and deduce from this relation that
00
1 e- x2 dx==!V n .
o
21.12) Show that if -1<al, then
00
f eX'+e-x' dx = tvn(l - ;3 + 5 - - 7 + ...).
o
21.13) Show that, for a>O, b>O,
n/2
f a+b
F(a, b) == log (a 2 cos 2 x+b 2 sin 2 x) dx = n log - 2 .
o
21.14) Show that F(r) ==/0 log(I-2r cos x+r2)dx, defined for all
real values of r, satisfies F(r) ==F( -r) and F(r2) =2F(r) for all r. De-
duce from these relations that
{ nlO g r 2 for r 2 >1,
F(r) -
o for r2 1.
21.15) Show, similarly, that if P(r, x) is the Poisson kernel
l-r 2
P(r, x) == ,
1-2rcosx+r 2

148
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 21
then
n { n
1 P(r, x) dx ==
o n
for r 2 > 1 ,
for r 2 <1.
21.16) Let a>O be fixed, and let b be real. Show that the integral
of LaPlace
00
F(b)== I e- ax2 cos 2bxdx
o
satisfies dF/db==-2a- 1 bF(b), and deduce from this equation that
F(b) = ! ( : J e- b2 / a .
Show also that
00
G(b)== I e- ax2 sin 2bxdx
o
satisfies dG/db+2a- 1 bG(b)==a- 1 , and deduce from this equation that "-
b
G(b) = + e- b2 / a f e t2 / a dt.
o
Note that lim G(b) ==0 and lim bG(b)= 1/2a as boo.
Show, finally, that
00 b
f e- ax2 sin :bX_ _ dx = ( : J f e- t2 / a dt.
o 0
21.17) Show that if q> -1 and rO, then the integral
I xq(log x)r (l-x/n)ndx tends to the finite number 100 xq(log x)r e-xdx
as noo.
21.18) Show, by means of the preceding exercise, that
00
f log x'e-Xdx == lim { log n - ( 1 +  +  + ... +  )} .
noo 2 3 n
o
21.19) Show that if f(x) is continuous on -oo<x< 00, and
gn(x) ==n{f(x+ l/n) -f(x)}
for n == 1, 2, . . .,

Ch. 4,  21J RIEMANN INTEGRAL AND LEBESGUE INTEGRAL 149
then
b
lim (R) I gn(x) ==f(b) -f(a)
a
as n-+oo
for any bounded interval [a, bJ. Show also that if the derivative f'(x)
exists and is bounded on [a, bJ, then
b
(L)J f'(x) dx==f(b) -f(a).
a
EXAMPLES RELATED TO THE RIESZ REPRESENTATION THEOREM
21.20) Let L be the collection of all real continuous functions on Rk
having a bounded carrier, and let Xo be a point in Rk. Show that
f(f)==f(xo) is a non-negative linear functional on L, and show also
that the corresponding measure v in Rk is such that every set E c Rk
is v-measurable, and v(E) is either one or zero according as E does or
does not contain the point Xo.
21.21) We use he same notations as in the preceding exercise. Let
v be a measure in Rk corresponding to some non-negative linear
functional on L, and let foO be v-summable over every bounded
closed interval. Show that v1(E) == JEfodv, defined at least on the a-
field of all v-measurable sets, is a measure of the same kind.
21.22) We use the same notations as in the preceding exercises.
Show that, for k== 1, the collection of all measures v, corresponding to
non-negative linear functionals on L, is the collection of all Stieltjes-
Lebesgue measures defined in sec. 10, Example 8.
21.23) Let C be the collection of all real continuous functions on Rk,
and let f(f) be a non-negative linear functional on C. Show that f(f)
is an elementary integral on C (communication by W. A. J. Luxem-
burg) .
21.24) Let f(f) be the integral of the preceding exercise, and # the
measure in Rk induced by f(f). Show the existence of a bounded and
closed set Eo c Rk such that #(F) ==0 for every F such that Eo and F
are disjoint.

150
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 22
22. The Exterior Measure of Ordinate Sets
This section is more or less an intruder in the present chapter, since
it might have appeared as well either in the preceding or in the next
chapter. Following M. H. STONE [lJ, we shall consider the exterior
measure il* in X X Rt (introduced in connection with the general Da-
niell integral) from a somewhat different angle, and this will pay its
dividends in the next chapter, where Stone's version of the Fubini
theorem on repeated integration is the main topic. The reason for in-
serting the section here is that the new look of il* will enable us to
round off the discussion on integration of complex functions (cf. sec. 19)
in a satisfactory manner.
Let, as before, L be the linear collection of real finite valued functions
I(x) on X, which serves as the initial domain of definition of the inte-
gral f(/). We assume, of course, that I, gEL implies max(/, g)EL and
min (I, g) EL, but we do not assume that IEL implies that min (I, 1) is
measurable. In order to extend the elementary integral f(/) we have
introduced in the preceding chapter the measure il on the semi-ring T
of all sets F -G (F and G exterior ordinate sets of I, gEL +, Ig) by
il(F -G)==f(/-g), and the extension procedure for measures was ap-
plied to (X X Rt, T, il). It will be useful to consider more closely the
exterior measure il*(P) of the exterior ordinate set P of an arbitrary
(not necessarily f-measurable) function P(x)O. If P is sequentially
covered by T, then (by definition) il*(P) ==inf  il(An) ==inf  il(F n -G n )
==inf  f(ln-gn) for all possible coverings  An==>P, where An==
Fn-GnE T. Now,  An== (Fn-Gn)==>P means that, for every point
XEX, we have  {/n(x)-gn(x)}p(x). Hence, if P is sequentially cover-
ed by T, then il*(P)==inf  f(hn) for all possible sums  hn(x)p(x),
where hnEL+. Furthermore, if there exists no sum  hn(x), hnEL+,
which is greater than or equal to P(x) on X, we have (by definition)
that il*(P) = +cx>. In order to get a shorter notation, we shall denote,
for an arbitrary real function P(x) on X (it is no longer assumed,
therefore, that pO on X), by Alp the exterior measure il*(IPI) of the
exterior ordinate set IPI of IP(x) I. Hence
Alp==inf  f(lhnl) lor all sums  Ihn(x)IIP(x)l, hnEL,
il such sums exist,
AlP=+cx>,
il there exists no sum
 Ihn(x)IIP(x)l, hnEL.

Ch. 4, 922J
EXTERIOR MEASURE OF ORDINATE SETS
151
Evidently, if IP(x) I is f-measurable, we have JVP==f(IPI), and if
P(x) is f-summable (i.e., pELi r »), then JVP==f(IPI)==IIPII<oo.
THEOREM 1. The functional JV P has the following properties:
(a) OJVpoo, and JVp===O if and only if P is a null function,
(b) JV(ap)==laIJVP for any finite real constant a,
(c) IP11IP21 imPlies JVP1JVP2,
(d) JVP===JV(IPI) ,
(e) IPIr IPnl imPlies JVpr JVPn. In particular, JV(P1+P2)
JV PI +JV P2.
PROOF. The properties (a), (c) and (d) are immediate consequences
of the definition. The proof of (b) is derived from sec. 13, Theorem 4,
or independently by observing that lapl Ihnl if and only if IPI
 lal-1lhnt for a=FO. Property (e), upon closer inspection, is a particu-
lar case of the property that exterior measure is a-subadditive; we
present, however, an independent proof. It may be assumed that
 JVPn<oo. Then, given E>O, there exists a sequence hniEL (i==
1,2, ...) such that IPnll Ihnil and JVPn> 1 f(lh ni l)-E/2 n for
n== 1,2, . . . . It follows that IPIni Ihnil, so JVPni f(lhnil)<
n JVPn+ E , i.e., JVp JVPn.
The next theorem, trivial though it may seem, is sometimes very
useful.
THEOREM 2. If fn is a sequence of real f-summable functions, and if
there exists a real function f (not yet assumed to be f-measurable) such
that JV(f-fn) tends to zero as noo, then f is f-summable (so, conse-
quently,
lim IIf- f nil ==lim f(lf- f n I) ==lim JV (f- f n) ==0,
and f(f)==lim f(fn)).
PROOF. It follows from lim JV(f-fn) ==0 that lim JV(fn-fm) ==0 as
m, noo, so fn is a fundamental sequence in the metric space Li r ).
Hence, by the completeness of Li r ), there exists an f-summable
function g(x) such that limlig-fnll===O, i.e., lim JV(g-fn) ===0. But then
JV(f-g) ===0, since JV(f-g)JV(f-fn)+JV(g-fn), so f-g is a null
function. Hence, f===g except on a null set, so f is f-summable.

152
STIELTJES-LEBESGUE INTEGRAL
[Ch. 4, 9 22
We shall make use of Theorem 2 in the proof of the next theorern.
THEOREM 3. If f and g are real f-summable functions, then (f2+g2)l
is f -summable.
PROOF. (a) We note first that if F(x)==(I+x2)i on Oxl, and
e>O, there exists a continuous function p(x) on Ox 1 such that
IF(x) -p(x) I <8, and the graph of p(x) consists of a finite number of
line segments which are either parallel to the x-axis, or pass (if ex-
tended) through the origin. Evidently, we may assume that p(O)==
F(O)== 1 and p(I)==F(I)==2 l . It follows that if H is any linear col-
lection of functions on [0, IJ, containing the functions po(x) == 1 and
PI (x) ==x, and with the property that 1p1, 1p2 EH implies max (1p1, 1p2) EH
and min (1p1, 1p2) EH, then p(x) EH. Indeed, if the graph of p consists
of p segments, then p is exactly the function pp of the sequence P2==
max (po, a2P1), p3==min (p2, a3PO), P4==max (p3, a4P1), p5==min (p4, a5PO),
. . " where a2, a3, ... are suitable positive constants; hence, si1).ce
po, P1 EH by hypothesis, we have pnEH for all n, so p==ppEH.
Defining now p(x) on [-1, OJ by p(x)==p(-x), it is easily seen that
IF -pi <e on [ -1, IJ, and pEH for any linear collection H of functions
on [-1, IJ, containing po(x)== 1 and p1(X)==X, and having the property
that 1p1, 1p2 EH implies max (1p1, 1p2) EH and min (1p1, 1p2) EH.
(b) Next, we consider the function F(, 'Y))=(2+'Y)2)! on R 2 . Let S
be the circumference of the square, having the origin as centre and
{(, 'Y)): -11, 'Y)=I} as one of its sides, and let O<e<l. It is not
difficult to derive now from the result in (a) that there exists a con-
tinuous function p(, 'Y)) on R 2 such that
(1) p is positive homogeneous, i.e. p(a, a'Y))==ap(, 'Y)) for any con-
stant aO,
(2) IF -pl<8 on the circumference 5,
(3) p is contained in any linear collection H of functions on R2
which contains the functions P1(, 'Y)) == and P2(, 'Y)) =='Y), and which
satisfies the condition that 1p1,1p2EH implies max(1p1,1p2)EH and
min (1p1, 1p2) EH.
(c) Let now f and g be f-summable over X, and let again F(, 'Y))==
(2+'Y)2)1. Then, by (a) and (b), IF(, 'Y)) -p(, 'Y)) I <e(I1 + I'Y)I) for any
point (, 'Y)) E R2, so
IF{f(x), g(x)}-p{f(x), g(x)}1 <8(lf(x) 1+ Ig(x) I)

Ch. 4, 9 22J
EXTERIOR MEASURE OF ORDINATE SETS
153
on X, except possibly on a null set. It follows that
%{F(j, g)-p(j, g)}E(%j+%g).
Furthermore, p(j, g) is f-summable. Indeed, p(j, g) is contained in any
linear collection H of functions on X, containing the functions j and g,
and satisfying the condition that hI, h 2 E H implies max (hI, h 2 ) E Hand
min (hI, h 2 ) EH; the collection of all real f-summable functions is such
a collection. The device of selecting a sequence En! 0, and determining
a corresponding sequence of f-summable Pn(j, g) shows now that
lim %{F(j, g)-Pn(j, g)}==O, so F(j, g)==(j2+g2)! is f-summable by
Theorem 2.
It is an immediate consequence of the last theorem that if g and h
are real f-summable functions, and j==g+ih, then Ijl===(g2+h 2 )lisf-
summable. Hence, the theorems on integration of complex functions
in sec. 19 are also valid in the case that f(j) is a general Daniell inte-
gral, which is not necessarily a Stieltjes-Lebesgue integral.
Exercises
FURTHER EXAMPLES
22.1) Show that if j and g are f -summable functions, then
(ljIP+lg/ p )l/ P is f-summable for all pl. Show also that Ijgl! is f-
summable.

CHAPTER 5
FUBINI'S THEOREM
Fubini's theorem in sec. 23 states a sufficient condition in order that the
evaluation of a multiple integral may be reduced to the evaluation of a re-
pea ted integral. The sections 24 and 25 are devoted to some applications of
Fubini's theorem to partial integration, the Gamma function and fractional
integration, and the exercises in sec. 23 show how to extend the theory of
multiple integrals to infinite product sets.
23. Fubini's Theorem
Let Xl and X 2 be (non-empty) point sets; the points of Xl and X 2
will be denoted by x and y respectively, and the points of X 3 ==X 1 XX 2
will be denoted by (x, y). Let #1 and #2 be measures in Xl and X 2
respectively, and let L(l) and L(2) be the corresponding collections of
real step functions. In view of sec. 17, Theorem 9 the collections
L(l) and L(2) may serve as the initial domains of definition for the
integrals Jx1f(x) d#l and Jx 2 f(y) d#2 respectively, although the same
integrals may be obtained also, perhaps, by extension of elementary
integrals having different initial domains of definition. We shall use
the notation JVf, introduced in the preceding section, and since we
have to distinguish between Xl and X 2 , we shall write JV 1 f for
functions f(x) on Xl, and JV 2 f for functions f(y) on X 2 . Furthermore
we denote, if the integrals exist, Jx1f(x) d#l and Jx 2 f(Y) d#2 by f 1 f and
f 2 ; respectively, in accordance with what we have already done
previously.
"--
LEMMA ct. If r is the collection of all subsets C == A X B of X 3==X 1 X X 2,
such that A is #l-measurable, B is #2-measurable, and #1(A)#2(B) <00,
then r is a semi-ring with the special property that the difference of
any two sets of r is a finite union of sets of r.

Ch. 5,  23J
FUBINI'S THEOREM
155
PROOF. The empty set 0 is in T on account of 0x 0==0. Further-
more, if A 1 xB 1 ET and A 2 xB 2 ET, we have
(AI X B 1 )n(A 2 X B 2 ) ==A1A2 X B 1 B 2 ET.
Finally, if A1xB1ET, A 2 xB 2 ET and A 2 xB 2 cA 1 xB 1 (so A 2 cA 1
and B 2 cB 1 ), then
A1XB1-A2XB2==(A1-A2) xB 1 +A 2 x (B 1 -B 2 )
is the union of two sets of T. It follows that T is a semi-ring with the
indicated special property.
Let L be the collection of all real functions
p p
f(x, y) ==  cnXCn(x, y) =  CnXAn(X)XBn(Y),
1 1
where all coefficients C n are finite real numbers, and all Cn==An X
BnET. Evidently, L is a linear collection such that f,gEL implies
max(f, g)EL and min(f, g)EL (cf. the remarks in sec. 12 concerning
this point; note that the special property of T, mentioned in Lemma cx,
plays a decisive role). In addition, fEL implies min (f, 1) EL.
THEOREM 1. If f 3 f is defined for any
p
f==  cnXAn(x)XBn(Y) EL
1
by
p
f 3 f==  c n #1(An)#2(Bn),
1
then f 3 f is an elementary integral on L (i.e., f 3 f is finite and linear,
f3f0 for fO, and f 3 fn!0 for fn!O).
PROOF. The uniqueness of f 3 f follows by writing f in its standard
representation as a linear combination of characteristic functions of
disjoint sets of T (note that any fEL assumes each of its values on a
finite union of sets of T). It is sufficient to prove that f n! 0 implies
J"3!n!0, the other assertions being evident. For this purpose we ob-

156
FUBINI'S THEOREM
[Ch. 5, 9 23
serve first that, given
p
f:::=  CnXAn(X)XBn(Y) EL
1
with all Cn=FO,
the integral
p p
f 2 f== J cnXAn(X)XBn(Y) d#2==  c n #2(B n )XAn(X)
X2 1 1
exists for almost every XEX 1 as a finite number (for a fixed XEX 1 the
integral does not exist, or fails to be finite, if for one or several indices
n it is true that xEA n and #2(Bn) ==00 hold sinlultaneously; in that
case, however, #l(A n) ==0). It follows that
p
f 1 f 2 f:::=f 1 (f 2 f) ==  c n #l(A n)#2(Bn) :::=f 3 f;
1
the fact that f 2 f either does not exist or is infinite on a null set in Xl
is of no importance for the existence. of f 1 f 2 f. Let now fnEL and
fn!O on X 3 . The subset of all XEX1, for which not all the f 2 fn are
finite, is then of #l-measure zero, and for any point x in its complement
we have f 2 fn!0 (by the theorem on dominated convergence), so
lf2fn!0 (once more by the theorem on dominated convergence).
Hence f 3 fn==f 1 f 2 fn!0.
The elementary integral f 3 f on L may now be extended (by applying
the extension procedure for integrals), and the so obtained Daniell inte-
gral is (due to the fact that fEL implies min(f, l)EL) the Stieltjes-
Lebesgue integral with respect to a measure v in X 3 . We recall that,
for any f 3 -measurable fO, the integral f 3 f is the #3-measure of the
ordinate set of f in X 3 X Rt. Furthermore, in accordance with the no-
tations in Xl and X 2, we shall denote by JV3f the exterior measure
#;(IFI) of the exterior ordinate set IFI of If(x, y) I, where f(x, y) is any
real function on X 3 (either f 3 -measurable or not). Hence, JV3f==
inf r f 3 (lfnl) for all sums  Ifnllfl, fnEL, if such sums exist, and
JV3f:::=oo if such sums fail to exist.
LEMMA fJ. For any real f(x, y) we have JViJV2fJV3f.
PROOF. We may assume that JV3f<oo. Then, given 8>0, there exist

Ch. 5, 923J
FUBINI'S THEOREM
157
functions fn EL such that Ifl ifni and
 f12Ifnl   3Ifnl<JV3f+E.
On account of Ifl Ifni we have also
A/2f A/2fn== f 2 1fnL
so
.AlA/2f .Alf2Ifnl= f 1 f 2 Ifnl<JV3f+e.
Hence .AlA/2fJV3f.
Note that it follows in particular from the last lemma that if f(x, y)
is an f 3 -null function, then, for almost every XEX 1 , we have f(x, y)==O
for almost every YEX 2 .
We shall denote by 51*52 the collection of all complex functions
f(x, y) having the property that for almost every XEX 1 the integral
J 2 f== Jx 2 f(x, y) d#2 exists as a finite number, and in addition f 2 f is #1-
summable over Xl. Shortly,
51 * 5 2 =={f(x, y) : lf2f exists and is finite}.
It has been shown already in the proof of Theorem 1 that L e 51 *5 2 .
THEOREM 2 (FUBINI'S THEOREM IN STONE'S VERSION; H. LEBESGUE,
1902 [2J, FOR LEBESGUE INTEGRATION OF BOUNDED FUNCTIONS; G.
FUBINI, 1907 [IJ, FOR LEBESGUE INTEGRATION OF SUMMABLE FUNC-
TIONS). If f(x, y) is f 3 -summable, then fE51*52, and f 3 f==f 1 f 2 f.
PROOF. It is, evidently, sufficient to give the proof for real f. Let
f(x, y) be real and f 3 -summable. Since L is dense in the metric space
of all real f 3 -summable functions, there exists a sequence fnEL
satisfying JV3(f-fn) <2- n . Letg(x) =  A/2(f-fn). ThenOg(x)oo, and
.Alg .AlA/2(f-fn) JV3(f-fn)< 2- n = 1.
It follows that the set E eX 1 on which g(x)==oo is an f 1 -null set
(indeed, if .AlXE>O, we should have .Alg.Al(kXE)==k.AlXE for k==
1, 2, 3, . . . on account of gkXE, i.e., we should have .Alg===oo). Hence,
Og(x)<oo for almost every XEX 1 , so that, for these values of x, we
have lim A/2(f-fn) ==0 by the definition of g(x). But then, by Theo-

158
FUBINI'S THEOREM
[Ch. 5, 9 23
rem 2 in the preceding section, the function f(x, y) is #2-summa ble
over X 2 for almost every XEX1 (since all fn(x, y) are #2-summable over
X2 for almost every XEX1). In other words, h(x)=f 2 f exists as a finite
number for almost every XEX 1 . On account of Ih-f 2 fnl == If 2 (f-fn) I
JV2(f-fn), we have then
AIi(h-f2fn)AliJV2(f-fn)AI3(f-fn) < 1/2 n ,
and this shows, once more by Theorem 2 in the preceding section, that
h(x)=f2f is f 1 -summable, since all f 2 fn are f 1 -summable. Finally,
f 1 f 2 f=f 1 h=lim f 1 (f 2 fn)==lim f 3 fn==f 3 f.
COROLLARY. Let 52 *5 1 = {f(x, y) : f 2 f 1 f exists and is finite}. Then, if
f(x, y) is f 3 -summable, we have fES2*Sl, and f 3 f==f 2 f 1 f. Hence
f3f=f1f2f==f2f1f
for any f 3 -summable f(x, y)}' the double integral f 3 f may, therefore, be
expressed in two different ways as a repeated integral.
As observed above, f 3 f is the Stieltjes-Lebesgue integral with respect
to a measure v in X 3 , i.e. f 3 f=/x 3 f(x, y)dv. Given the set EcX 3 and
the point XEX 1 , the subset Ex=={Y: YEX 2 , (x, Y)EE} of X 2 is called the
section of E corresponding to the given point x. Similarly, if EcX a
and YEX 2 are given, the subset Ey=={x: XEX 1 , (x, y) EE} of Xl is also
called a section of E. Concerning the measures of such sections, the
following theorem holds.
THEOREM 3. If E cX 3 is a v-summable set (i.e., if v(E)==f3XE<OO),
then /X 2 XE(X, y)d#2==#2(Ex) exists as a finite number for almost every
XEX 1 , and /xl#2(Ex)d#1 exists as a finite number. Similarly,
I XE(X, y) d#l==#l(E y )
Xl
exists as a finite number for almost every YEX 2 , and /x 2 #1(E y ) d#2 exists
as a finite number. In addition,
v(E) == I #2(Ex) d#l == I #l(Ey) d#2'
Xl X2
PROOF. Follows immediately from the preceding theorem.

Ch. 5, 923J
FUBINI'S THEOREM
159
It is possible to give a small extension to Fubini's theorem as stated
in Theorem 2; the extended version may be applied also to some not
necessarily f 3 -summable functions.
THEOREM 4. If f(x, y)O, and if there exists a sequence of non-nega-
tive f 3 -summable functions fn(x, y) such that fn if on X 3 , then f 1 f 2 f and
f 2 f 1 f exist (the value +00 is admitted now), and
f3f==f1f2f==f2flf.
In particular, if f=XE(X, y), where E==r En and all En are v-
summable, then f3f==f1f2f==f2f1f.
PROOF.
f 3 f==lim f 3 fn==lim f12fn==lim f 1 (f 2 fn)=f 1 (lim f 2 fn)==f 1 f 2 f
by repeated application of the theorem on integration of increasing
sequences. The last assertion of the theorem follows by observing that
the characteristic functions fn of the sets f=l E i are v-summable, and
fnif onX 3.
COROLLARY. If Xl is of a-finite # I-measure , and X 2 is of a-finite #2-
measure, then every v-measurable f(x, y)O may be approximated in the
indicated manner, so f3f==f1f2f==2flf holds for all v-measurable non-
negative f(x, y).
PROOF. Since Xl and X 2 are of a-finite measure, there exist sequences
An i Xl and Bn i X 2 respectively, such that An is #l-measurable, Bn is
#2-measurable, #l(A n )<oo and #2(Bn) <00 for all n. Then Cn==An X Bn
is an element of the semi-ring r (cf. Lemma a;) for all values of n,
C n i X 3 , and
v(C n) ==f 3 XCn =#l(A n)#2(Bn) <00
for all n.
Given now the v-measurable function f(x, y)O, the sequence of func-
tions fn(x, y) is defined by
{ min(f, n)
fn ==
o
on C n,
on X 3 -C n .

160
FUBINI'S THEOREM
[Ch. 5, 9 23
Evidently, the functions fn are v-measurable,
f 3 fn== Ifndvnv(Cn)<oo,
X3
and f n i f on X 3.
We present a counter example to show that, even though f(x, y)O,
the equality J3f==f1f2f==f2lf does not hold in all cases. Let Xl be
the linear interval [0, IJ with #l(A) equal to the number of points in A
(i.e., #1 is the discrete measure in Xl), and let X 2 be also the linear
interval [0, IJ with #2 the Lebesgue measure in X 2 . Consider now the
set E =={(x, y) : y==x} c XIX X 2 ; E is therefore the main diagonal of the
square [0, 1; 0, IJ. Then v(E) ==f 3 XE== +00, but f 1 f 2 XE==f 1 (0) ==0 and
f 2 f 1 XE==f 2 (1) = 1.
The next theorem is of frequent use in applications; it c on tains a
sufficient condition for applying Fubini's theorem.
THEOREM 5. Let the real or comPlex function f(x, y) be v-measurable,
and let there exist a sequence of f 3 -summable non-negative functions
fn(x, y) such that fnilfl. Then, if f 1 f 2 1fl<00 (the existence of f 1 f 2 1fl
follows by the preceding theorem), the function f is f 3 -summable}' hence,
3f==f1f2f==f2f1f.
PROOF. By the preceding theorenl we have 31fl ==f 1 f 2 Ifl, so f 3 1fl
<00, i.e. If I is f 3 -summable. Since f is 3-measurable by hypothesis, it
follows that f is f 3 -summable.
We add some remarks concerning the connection between the initial
measures #1 and #2 in Xl and X 2 respectively, and the measure v in X 3 .
In the first place we observe that the semi-ring r of all sets C ==A X B,
where A is # I-measurable, B is #2-measurable, and #1(A)#2(B) <00,
consists entirely of v-measurable sets, since 3XC==#1(A)#2(B). Hence,
any CEr is v-measurable, and v(C)==#1(A)#2(B). It follows that
#1(A)#2(B) is a measure on the semi-ring r, and this measure is the
restriction of the measure v to the sets of r. The question may be
raised if, conversely, given any v-summable set C of the particular
form C ==A X B, the sets A and B are necessarily #l-measurable and
#2-measurable, and if it is true then that v(C)==#1(A)#2(B). This is not

Ch. 5,  23J
FUBINI'S THEOREM
161
always so, as the following example shows. Let A eX1 satisfy fl1(A)=0.
Then A XX 2 Er, and v(A xX 2 )==0. It follows that A xB, where B is
any, not necessarily fl2-measurable, subset of X 2 , is v-measurable, and
v(A X B) =0. By a small modification ot the raised question, however,
we get a satisfactory result. Given any v-summable set C ==A X B, about
which it is known that either v(C) >0 or fli(A) >0 and fl'2(B) >0, we
may conclude that the sets A and Bare fl1-measurable and fl2-measur-
able respectively, and V(C)==fl1(A)fl2(B). The proof is immediately de-
rived from Theorem 3.
In the second place one may ask if, applying the extension procedure
for measures to (X 3 , r, fl1(A)fl2(B)), the result will be the measure v.
In order to see that the answer is affirmative, we observe first that the
collection A of all finite unions of sets of r is a ring (this is due to the
special property of r indicated in Lemma ex). Denoting, temporarily,
the measure fll(A)fl2(B) on r by VI, the measure VI may be extended
in an obvious manner to all sets of A, and the remaining question is
whether the extension procedure for measures, applied to (X3, A, VI),
yields the measure V in X 3 . It follows from sec. 17, Theorem 9 that
this is indeed the case. Since V is consequently the extension of the
initial measure fl1(A)fl2(B) on r, the measure v in X 3 is sometimes
called the product measure of fl1 and fl2, and denoted then by V=fl1 X fl2.
Accordingly, the integral J x 3 t(x, y) dv is written as J x 3 t(x, y) d(fll X fl2).
Thirdly, and this remark is related to the first remark, it seems a
reasonable conjecture that if P eXI and Q c X 2 are fl1-measurable and
fl2-measurable respectively, then 5 ==P X Q is v-measurable, and v(5) ==
fl1(P)fl2(Q). If P1(P)fl2(Q) < 00, this is true by what has already been
observed above, so it remains only to consider the case that fl1(P)fl2(Q)
==00. We wish to show that xs(x, Y)=XP(x)XQ(Y) is f 3 -measurable, and
for this purpose it will be sufficient to show that min (XS, xc) is f 3 -
measurable for any set C Er. Now, if C ==A X B Er, we have min (XS, xc)
=XPA(X)XQB(Y), and this function is f 3 -measurable since PA XQBEr
(indeed, PAis fl1-measurable, QB is fl2-measurable, and fl1(P A)fl2(QB)
<:fl1(A)fl2(B) <00). It remains to prove that fl1(P)fl2(Q)==00 implies
v(5) =00. If it were true that v(5)<00, then 5=PxQ would be v-
summable, so that we should have fl1(P)fl2(Q) ==v(5) <00 by Theorem 3.
This is a contradiction, hence v(5) ==00.

162
FUBINI'S THEOREM
[Ch. 5, 9 23
We use the last remark in the proof of the following theorem.
THEOREM 6. (a) If the non-negative functions f(x) and g(y) are #1-
measurable and #2-measurable respectively, then h(x, y) ==f(x)g(y) is v-
measurable, and f3h==(f1f)(f2g).
(b) If the real or comPlex functions f(x) and g(y) are #l-summable and
#2-summable respectively, then h(x, y) f(x)g(y) is v-summable, and f3h==
(f 1 f) (f 2 g).
PROOF. (a) By sec. 17, Theorem 4 there exists a sequence of #1-
measurable functions fn(x)O, each fn assuming only a finite number
of finite values, such that fn i f on Xl. Hence, each fn is of the form
fn==f=l CiXpt(X), where p depends on n, all Ci satisfy OCi<OO, and
the disjoint sets Pi are #l-measurable. Similarly, there exists"a sequence
of #2-measurable functions gn(y)O, each gn assuming only a finite
number of finite values, such that gn i g on X 2 . It follows readily from
the remark above that the non-negative functions hn(x, y) fn(x)gn(Y)
are v-measurable, and f3hn==(f1fn)(2gn). Hence, since fn i f on Xl,
gn i g on X 2 , and h n i h on X 3 , we obtain (by means of the theorem on
integration of increasing sequences) the desired result that h is v-
measurable, and f3 h == (f 1 f)(f 2 g).
(b) If f and g are real and each of the two is either non-negative or
non-positive, the result follows from (a). If f and g are real, then
hfg== (f+ + f-) (g+ +g-) ==f+g++ f+g-+ f-g++ f-g-
holds at each point (x, y) EX 3 , and the desired result holds for each
term in the last expression. Hence, since each term in f 3 (f+g+) +
f 3 (f+g-) +f3(f-g+) +f 3 (f-g-) is finite, the result holds for h. The ex-
tension to the case that f and g are complex, is similar.
Remark. It follows, of course, immediately from part (a) that if the
real or complex functions f{x) and g(y) are #l-measurable and #2-
measurable respectively, then h(x, y)==f(x)g(y) is v-measurable. It is
not true, hovvever, that if f and g are #l-integrable and #2-integrable
respectively, then h is necessarily v-integrable (cf. Exercise 23.7).
Finally, we direct our attention towards the particular case of Le-
besgue measure in real number space. If X 1 Rk with Lebesgue measure

Ch. 5, 9 23J
FUBINI'S THEOREM
163
#1, and X 2 ==Rl with Lebesgue measure #2, then the product measure
V==#lX#2 in X1XX2Rk+l is again Lebesgue measure. The (not en-
tirely trivial) proof is obtained by observing that v is the extension of
the measure #1(A)#2(B), initially defined on the semi-ring rvr of
all sets A X B such that A is #l-measurable, B is #2-measurable, and
#1(A)#2(B)<oo, whereas Lebesgue measure # in Rk+l is (by its defi-
nition) the extension of the measure #1(A)#2(B), initially defined on
the smaller semi-ring rJ.l of all sets A X B such that A and B are cells.
Evidently v# on rJ.l' and it is not difficult to prove that also #==v
on rv (if C==A XBEr v , assume to begin with that #l(A)<oo and
#2(B)<oo, and prove that #(C)v(C) first if A and B are a-sets with
respect to rJ.l' next if A and B are limits of descending sequences of
a-sets, and finally if either #l(A)==O or #2(B)0. Furthermore, if
#l(A)==O, show that #(A xX2)v(A xX 2 )==0; the case that #l(A)==O
and #2(B)oo follows then).
Concerning Lebesgue measure it is interesting to note that, even
in the simple case where f 1 f and f 2 f are one-dimensional Lebesgue
integrals, it may happen that f 1 f 2 f and f2lf exist as finite
numbers, without being equal. If LI is the interval [0, 1], f(x, y) ==
(X 2 _y2)f(X 2 +y2)2 on J xLi, and #1, #2 are Lebesgue measure, then
f[ff(X,Y)d fi2 ]d fi1 = f - 1;X 2 dfil=!n
  
and
f [f f(x, y) d fi1 ] dfi2 = f 1 2 dfi2 = -In.
  
Evidently, this shows that f{x, y) is not Lebesgue summable over J X J.
Exercises
In the following exercises the one-dimensional Lebesgue integral of
j(x) over [a, bJ will be denoted by J: f(x)dx.

164
FUBINI'S THEOREM
[Ch. 5, 9 23
A COUNTER EXAMPLE
23.1) Show that
1 00 00 1
1[1 (e-XY-2e-2XY) dx] dy#-/[I (e-XY-2e-2XY) dyJ dx,
o 1 1 0
although each of the repeated integrals exists as a finite number.
EXAMPLES IN LEBESGUE INTEGRATION
23.2) Show that if
f(x, t) == (1 + x 2 t 2 )-1( 1 +y 2 t 2 )-1,
"
where y>O, then
1 00 00 1
1[lfdt]dx= 1[lfdx]dt;
o 0 0 0
show also that g(y, t) = (arc tg t)j[t(1 +y 2 t 2 )] satisfies
1 00 00 1
1[1 gdtJ dy== 1[1 gdy] dt,
o 0 0 0
and combine the so obtained results to derive that
00
f (arc tg t)2
t 2 dt == n log 2.
o
23.3) Show that if
f (x, y) == y e - (1 + x 2 )y2 ,
then
00 00 00 00
1[1 fdx] dy= 1[1 fdy] dx,
o 0 0 0
and derive by means of this result that
00
1 e- x2 dx==!n!.
o
23.4) Show that if
f(x, y) ==e- Xy2 sin x,

Ch. 5, 923J
FUBINI'S THEOREM
165
then
b 00 00 b
f[ffdyJdx== f[ffdxJdy
o 0 0 0
for any b>O,
and derive from this equality that
b
f SIn x
A == lim 1- dx == (!n)i.
b-+oo X 2
o
Show, similarly, that
b
. f cos x
B == 11m dx == (!n)i.
b-+oo x!
o
The two improper Riemann integrals are called the integrals of Fresnel.
23.5) Show that, for a>O, bO,
00
00
f -a2x2-b2 / x2 d 1 -2ab f -t2 dt nl -2ab
e x==-e e ==-e .
2a 2a
o -00
For this purpose, introduce the new variable t==ax-bjx.
23.6) Show that, for aO, b>O,
00
00
00
f :: dx= f [cosax f e-<x'+b')Y'2 Y d Y ]dX.
o 0 0
Show also that the order of integration in the repeated integral may
be interchanged, and derive (by means of Exercise 21.16 and the pre-
ceding exercise) that
00
f cos ax d n - b
x==-e a.
x 2 +b 2 2b
o
Show also that
00
f SIn ax n
dx == - (1_e- ab ).
x(x 2 +b 2 ) 2b 2
o

166
FUBINI'S THEOREM
[Ch. 5, 9 23
A COUNTER EXAMPLE
23.7) Let f(x)==x- 1 -1 on J==[O, 2J. Show that, although !L1fdx ex-
ists and satisfies !L1fdx==+(X), the function h(x, y)==f(x)f(y) is not Le-
besgue integrable over J X J.
A UNIQUENESS PROPERTY OF THE RIEMANN INTEGRAL
23.8) Let Xl ==Rk and X 2==Rl, and let C(l) and C(2) be the col-
lections of all real continuous functions with bounded caTriers on Xl
and X 2 respectively. Furthermore, let f{f(x)} and f{g(y)} be non-
negative linear functionals on C(l) and C(2) respectively, and denote
the measures induced in Xl and X 2 by the extended integrals f(f) and
f(g) by VI and V2 respectively. Finally, let VI X V2 be the product
measure in Rk+Z' Show that any function h(x, y), defined and con-
tinuous on Rk+Z' is (VI X v2)-measurable.
23.9) Let C be the collection of all real continuous functions on Rk
with bounded carriers. Among all non-negative linear functionals on C
the Riemann integral f(f) has the property that f{f(x+a)}==f{f(x)}
for all a== (aI, . . " ak). Show that, except for a constant factor, f (f) is
the only non-negative linear functional on C having this property.
PRODUCT INTEGRAL FOR A FINITE NUMBER OF FACTORS
23.10) It is easy to extend Fubini's theorem to integrals over Carte-
sian products of more than two point sets. Show that if the sets Xi
(i== 1, . . " n) are non-empty point sets with measures fli and integrals
if==! xifdfli, then the collection T of all subsets C ==A 1 X . . . X A n of
Xl X . . . X X n , such that all Ai are fli-measurable and I1f= 1 fli(A i ) <(X),
is a semi-ring (proof by induction). Show also that if L is the collection
of all
p
f(X1, · · " x n ) ==  CkXCk
k=l
for Ck==A 1k X... XAnkET,
then
p n
ff==  Ck II fli(A ik )
k=l i=l
is an elementary integral on L. Finally, prove Fubini's theorem: If
f(X1, .. .,x n ) isf-summableove rX 1x... xX n , thenff==f 1 f 2 .. .fnl,

Ch. 5, 923J
FUBINI'S THEOREM
167
and also f f==f i1 f i2 . . . finf for any permutation (iI, "', in) of
(1, . . " n). If f' and f" are the product jntegrals of f 1 , . . . , f k and
fk+1, "', f n respectively, show that ff==f'f"f if f is f-summable,
but not necessarily if f is not f-summable (consider an example with
fl1 and fl2 Lebesgue measure, fl3 discrete measure, f' corresponding to
fl1 X fl2 and f" to fl3). Show also that f is not always the product inte-
gral of f' and f", but f is the product integral of f' and f" if all fli
are a-finite.
PRODUCT INTEGRAL FOR A COUNTABLE NUMBER OF FACTORS
23.11) Let Xi be a non-empty point set for i  1, 2, . . " and let fli
be a measure in Xi such that fli(X i )== 1 for all i. Let X m be the set of
all points (Xl, X2, . . .) such that XiEXi for i== 1, 2, . . ., and let xn) be
the set of all points (Xn+1, Xn+2, . . .) such that XiEXi for in+ 1,
n+2, ... . Note that EcX 1 x ... xX n implies Exxn)cXm. Show
that the collection r of all sets A 1 x... xAnxxn), where Ai (i==l,
. . " n) is fli-measurable, and where n may be different for different
sets, is a semi-ring. Note that X m itself is a set of r.
23.12) We use the notations of the preceding exercise. Let L be the
collection of all real functions
p
f==  CkXCk
k=l
for
C k ==A 1k X... XAnkkXxnk)Er.
Show that
p nk
ff==  Ck II fli(Aik)
k= 1 i= 1
is a uniquely determined non-negative linear functional on L. Note that
if C is a (real) constant, and fc on X m , then fEL and ffc. Show also
that if L(n) and f(n) are the analogs in xn) of Land f in X m , then
ff==f 1 f 2 . . .fnf(n)f for any fEL.
23.13) We use the notations of the preceding exercises. Show that ff
is an elementary integral on L.
23.14) The elementary integral ff of the preceding exercise, initially
defined on L, may be extended, and the measure v induced in X m by the
extended integral is called the product measure of the given measures
/-li (i == 1, 2, . . .). Product measures of this kind, for the particular case

168
FUBINI'S THEOREM
[Ch. 5, 9 23
that all Xi are equal to the cell (0, IJ and all fli are Lebesgue measure,
were introduced by P. J. DANIELL (1919, [2J) and B.JESSEN (1934,
[IJ). The product space X m js then sometimes called a torus'space. The
generalization to abstract measures is due to S. KAKUTANI (1943, [IJ)
and E. SPARRE ANDERSEN and B. JESSEN (1946, [IJ). Show that if, for
each i, E i is a fli-measurable subset of Xi, thenE==E 1 xE;2X.. y is v-
measurable, and v(E) == II 1 fli(E i ).
23.15) Show that if the set EcX 1 x...xX n is (fl1X".Xfln)-
measurable, then E X xn) is v-measurable, and v(E X xn») = (fl1 X . . .
X fln) (E). Somewhat more generally, show that if in addition the sets
Ai (i==n+l, "', n+k) are fli-measurable, then FExAn+1X... X
A n+k X xn+k) is v-measurable, and
n+k
v(F)=={(fl1X", Xfln)(E)} II fli(A i ).
i=n+ 1
Derive from this result that v is the product measure of fl1 X . . . X fln
and v(n), where v(n) is the product measure of fln+1, fln+2, . .. . Show,
finally, that fl==l" .f n f(n)/==f(n)f 1 . . .fnl for any v-summable I.
23.16) If I is v-summable, then fl. . .fnl and f(n)1 are functions of
Xn+1, Xn+2, . .. and Xl, . . " X n respectively. Hence, they may be re-
garded again as functions gn(x) and hn(x) of all variables Xl, X2, . . ..
Show that flgnlflfl and flhnlfl/l. Denoting fl. . .fnl/l by kn(x),
observe that Ign(x)lkn(x),
23.17) Let I be v-summable, and let gn(x)==f 1 .. .fnl and hn(x)==
f(n)/, just as in the preceding exercise. Furthermore, let g(x) be the
constant function whose value at each point XEX m is fl. Show that
flg-gnl and flf-hnl tend to zero as noo. Verify first that the as-
sertion is true for IEL, where L has the same meaning as in the Exer-
cises 23.13 and 23.14; it is then even so that f 1 ...f n l==fl and
f(n)/ I for n sufficiently large. Since L is dense in the metric space
of all real v-summable functions, we may, if the real v-summable
function I and 8>0 are given, select 1EL such that fl/-11<8. Now
apply the preceding exercise to 1-1.
23.18) Show that if I is non-negative and v-summable, gn(x)==
fl.. .fnl and Aa=={x: supn gn>a} for a>O, then av(Aa)f(IXA.J.
Show, similarly, that if hn(x)=f(n)1 and A=={x: supn hn>a} for a>O,
then av(A)f(IXA.a')' For the first part, set Bn=={x: sUPlkn gk>a}

Ch. 5, 923J
FUBINI'S THEOREM
169
and Pn=={x: gn>a}. Then Bn is the disjoint union of Qn==Pn, Qn-1==
Pn-1-Pn,Qn-2==Pn-2-Pn-1-Pn, .. .,Q1==P 1 -P2-.. .-Pn.Since
gk(X) does not depend on Xl, . . . , Xk, the same is true of Qk, so
f(fxQJ ==f(k)f 1 . . . fk(fxQJ ==f(k)(XQkf1. · · fkf) ==f(k) (gkXQJ.
The last integrand does not depend on Xl, . . " Xk, hence
f (fxQJ ==f 1 . . . fkf(k) (gkXQJ ==f (gkXQJ av( Q k)'
Add up these inequalities for k== 1, . . " n, and then let n-+oo. The
proof for A is similar; note that Bn is the disjoint union of R1==Pl,
R2==P 2 -P 1 , · . " Rn==Pn- P n-1-' · · -Pl.
JESSEN'S THEOREM
23.19) Show that if f is v-summable, gn(x) ==f 1 . . . fnf, hn(x) ==f(n)f,
and g(x) is the constant function whose value is ff, then gn(x) and
hn(x) converge pointwise v-almost everywhere on X m to g(x) and f(x)
respectively. For the proof, we may assume that f is real. Then, given
8>0, we have f==l+h with lEL and flhl<8. Since lEL, we have
fl.. .fnl=fl for n sufficiently large. It follows that
P(x)==lim sup If 1 .. .f n f-f 1 .. .fmfl
k-+oo m,n>k
=lim sup If 1 .. .f n h-f 1 . . .fmhl2 sup If 1 .. .fnhl.
k-+oo m,n> k n
By the preceding exercise,
av{x: P(x) >2a}flhl<8
for every a>O.
For any fixed a>O, this holds for every 8>0, so v{x: P(x) >2a}==0.
This holds for every a>O, so P(x)==O almost everywhere. In other,
words, gn converges pointwise almost everywhere on X m to a function
, g*(x). Apply Exercise 23.17 to conclude that g*==g almost everywhere.
The proof for h n is similar; observe that f(n)l==l for n sufficiently
large.
23.20) Let f be v-summable over X m , and let f have the property
that, for almost any point a== (aI, a2, . . .) EX m and any index n, we
,
have
f(X1, . · " x n , an+1, an+2, · · .) == f(a)
for all Xl,... , X n.

170
FUBINI'S THEORElVI
[Ch. 5, 9 23
Show, either by means of Jessen's theorem or by means of the more
elementary property in Exercise 23.17, that I is v-almost everywhere
equal to the constant value fl. Note that, by Exercise 23.17, the
sequence gn==f 1 .. .fnf satisfies lim flgn-ffl==O as n-+oo.",Hence
lim gnk==fl holds v-almost everywhere for some subsequence gnk' Actu-
ally, Jessen's theorem shows that it holds for the sequence gn itself,
but for the present purposes it is not necessary to know this. Then
lim gnk(a)==fl and
I(X1, · · . , X n , an+1, a n +2, . . .) == f(a)
hold at v-almost every point a. Derive from the last equality that gn==
fl. . . fnl has at such a point the value f(a) for all n; it follows then
that f(a) ==lim gnk(a) ==f I.
PRODUCT INTEGRAL FOR AN UNCOUNTABLE NUMBER OF FACTORS
23.21) Let {T} be an uncountable index set, let X T be a non-empty
point set for every index T, and let flT be a measure in X T such that
flT(X T )== 1. Furthermore, let Xo be the set of all functions X(T) such
that X(T) EXT for all T (axiom of choice), and let X&Tl,...,Tn) be the set of
all functions X(T), defined for all TE{T} except for T==T1, "', Tn, such
that X(T) EXT for all T for which X(T) is defined. Show that the col-
lection r of all sets AT! X . . . X A Tn X X&Tl,...,Tn), where ATi (i== 1, . . " n)
is fli-measurable, and where n and the indices T1, . . . , Tn may be differ-
ent for different sets, is a semi-ring. Show also that if the linear col-
lection L of functions and the functional f I are defined as in the
countable case, then fl is an elementary integral on L. The integral
fl may be extended, and the induced measure v in Xo is, also in this
case, called the product measure of the measures flT' Show that if, in
addition, each X T contains at least one measurable subset of measure
properly between zero and one, then the measure v is not separable,
although v(Xo)==I. Note that, in order to have non-separability, an
additional hypothesis such as the last one is necessary (consider the
case that each X T contains only the empty set and X T itself as measur-
able subsets).

Ch. 5, 9 24J
PARTIAL INTEGRATION
171
SEPARABILITY AND a-FINITENESS
23.22) By way of contrast to the example in the preceding exer-
cise, show that if # is a separable measure in X, having the property
that each #-measurable set of positive measure has a #-measurable
subset of finite positive measure, then # is a-finite.
24. Partial Integration
As an application of Fubini's theorem we shall derive an extension
of the well-known formula for partial integration. In its familiar
version for Riemann integrals the formula asserts that if F(x) and G(x)
have derivatives f(x)F'(x) and g(x)==G'(x) which are properly Rie-
mann integrable over the interval [a, bJ c R1, then
b b
I Fgdx+ I fGdx==F(b)G(b) -F(a)G(a).
a a
The proof is derived immediately from the observation that the ex-
pression on the left is the integral of the derivative of H(x) , where
H(x)==F(x)G(x), so
b b
I (Fg+fG) dx= I H' dx==H(b) -H(a).
a a
Note that the Riemann integrability of Fg+fG follows from the theo-
rem that the product of two Riemann integrable functions is Riemann
integrable. The extension to improper Riemann integrals is immedi-
ate, provided the appropriate limits exist as finite numbers (all this has
been used already in several of the exercises in the preceding sections).
The formula may be generalized in several respects. In the first
place Riemann integration (i.e., essentially Lebesgue integration) may
be replaced by integration with respect to an arbitrary Stieltjes-
Lebesgue measure # in R1; in the second place the Riemann inte-
grable f(x) ==F' (x) will be replaced by an arbitrary #-summable f(x) ,
and the natural substitute for the old F(x) is then the function F(x) ==
f: fd#+C, where C is a constant which may be taken equal to zero
without loss of generality. The functions g(x) and G(x) are generalized
similarly.

172
FUBINI'S THEOREM
[Ch. 5, 9 24
We shall begin by making some notational conventions. If, in k-
dimensional real number space Rk, we have an interval determined by
the vertices aI, b 1 ; . . . ; ak, b k , we shall also admit the case that ai==
-00 and (or) b i == +00 for one or more values of the index i, and we
shall say in that case that the interval is an unbounded interval. The
case of a degenerate interval (ai==b i for at least one i) was defined al-
ready previously.
Let now fl be an arbitrary Stieltjes-Lebesgue measure in R 1 . Then
all bounded cells are fl-measurable and of finite measure, and all inter-
vals (either closed, open or half open) are fl-measurable. If f(x) is fl-
integrable over [a, b], we shall denote the integral of f over this closed
interval by J: f(x) dfl, and, writing F(x) == r: fdfl for any x satisfying
axb, the integral of f over [a, x) will be denoted by F(x-), where
F(a-)==O by definition. The values of F(x) and F(x-) may be differ-
ent, since F(x)-F(x-)==f(x)fl({x}), where {x} is the set consisting of
the single point x, and f(x)fl({x}) may be different from zero.
THEOREM 1. Let v and T be measures in R 1 such that all bounded cells
(and hence all bounded intervals) are measurable and of finite measure
with respect to v and T. Let L1 == [a, b] be a given (bounded or unbounded)
closed interval such that v(L1) and T(LI) are finite. Finally, let
N(x)=v([a, x]),
T(x)==T([a, x]),
N(x-)==v([a, x)),
T(X-)==T([a, x))
for axb. Then
I N(x-) dT+ IT(x) dv==N(b)T(b).
L1 L1
(1)
By setting
Nm(x)=t{N(x) +N(x- )},
T m(x) ==!{T(x) +T(x-)},
the formula gets the more symmetric form
INm(x)dT+ IT m(x)dv==N(b)T(b).
L1 L1
(2)
PROOF. Since N(x-) is non-negative and non-decreasing on L1, the
set {x: N(x-»a} is an interval for any a>O, and therefore T-measur-

Ch. 5,  24J
PARTIAL INTEGRATION
173
able. It follows that N(x-) is T-measurable. Similarly, T(x) is v-
measurable. In addition, N(x-) and T(x) are bounded on J; hence,
the integrals on the left in (1) exist as finite numbers.
Let a==v X T be the product measure of v and T in J X LI. Then the
set E =={(x, y): (x, y) EJ X J, yx} is a-measurable, since E is the limit
of a descending sequence of sets En such that each En is a union of
rectangles. Hence, XE(X, y) is a a-measurable function, and by Fubini's
theorem
b x
J XE(X, y)da==J{JdT}dv==JT(x)dv,
x a a 
b b
J XE(X, y)da== J{Jdv}dT== J{N(b)-N(x-)}dT,
x a x 
so
J. T(x) dv== J {N(b) -N(x- )}dT=N(b)T(b) - J N(x-) dT,
  
which is formula (1). Similarly, of course,
J N(x) dT+ JT(x-) dv==N(b)T(b),
 
and by addition we obtain (2).
.
THEOREM 2 (THEOREM ON PARTIAL INTEGRATION). Let fl be a Stiel-
ties-Lebesgue measure in R1, and let J==[a, bJ be a given (bounded or
unbounded) closed interval. If f(x) and g(x) , real or comPlex, are fl-
summable over J, and if
x
F(x) = Jfdfl,
a
x
G(x) = Jgdfl
a
for axb, then
J F(x-)g(x) dfl+ JG(x)f(x) dfl==F(b)G(b).
 
(3)
By setting
Fm(x) ==i{F(x)+F(x- )},
Gm(x) == i{ G(x) + G(x- )},
the formula gets the more symmetric form
J F m(x)g(x) dfl+ J Gm(x)f(x) dfl==F(b)G(b).
 

174
FUBINI'S THEOREM
[Ch. 5, 9 24
/'
PROOF. We assume first that I(x) and g(x) are non-negative, and the
measures v and T are defined in LI by setting
v(E) == f Idfl,
E
T(E) == f gdfl
E
for any fl-measurable subset E of J. Then F(x) and G(x) take over the
roles of N(x) and T(x) in the preceding theorem, so
fF(x-)dT+ fG(x)dv==F(b)G(b).
L1 L1
Selecting a sequence In(x) of fl-measurable step functions such that
In(x)j F(x-) on J, we have
f F(x-) dT==lim f IndT==lim f Ingdfl== f F(x- )g(x) dfl.
Similarly f G(x) dv== f G(x)/(x) dfl, and formula (3) follows.
If 1 and g are real, we have 1==1++1- and g==g++g-, where 1+, -1-,
g+ and -g- are non-negative, and by applying the already obtained
result to all combinations of pairs out of these four functions, we get
the result for 1 and g. Similarly, if 1 and g are complex.
The extension to F m and G m is similar as in the preceding theorem.
We add some remarks.
- (a) In the particular case that the interval J is degenerate (i.e.,
b==a), the left side of (3) is G(a)/(a)fl({a}) , and the right side is F(a)G(a);
these numbers are indeed equal in view of F(a)==/(a)fl({a}).
(b) If fl has the property that any set, consisting of one point, is of
measure zero, then the distinction between F(x) and F(x-) becomes
redundant, and it is now also indifferent whether J is closed, open .or
half open. Hence, if for example fl is Lebesgue measure,
- b b
f F(x)g(x) dfl+ f G(x)/(x) dfl==F(b)G(b).
a a
(c) Theorem 1, in the general form as presented here, is due to E.
HEWITT (1960, [1 J).

Ch. 5,  25J GAMMA FUNCTION AND FRACTIONAL INTEGRATION 175
Exercises
THE SECOND MEAN VALUE THEOREM
24.1) Show that if # is a measure in X, f is #-summable over X, and
E c X is #-measurable, then IE If I d# tends to zero as #(E) tends to zero.
More precisely, show that, given E>O, there exists a number l5>0 such
that #(E) <l5 implies IE Ifl d#<E.
24.2) Let # be a Stieltjes-Lebesgue measure in R1 such that the
measure of any set consisting of one point is zero, and let f(x) be #-
summable over the interval [a, b]. Show that F(x) ==1: fd# is con-
tinuous on [a, b].
24.3) Let # be a Stieltjes-Lebesgue measure in R1 such that the
measure of any set consisting of one point is zero, let Q(x) be non-
decreasing on the bounded closed interval [a, b], and let f(x) be real
and #-summable over [a, b]. Show that there exists a point  in [a, bJ
such that
b  b
I Qfd#==Q(a) I fd#+Q(b) I fd#.
a a 
This is an extension of the classical second mean value theorem.
25. The Gamma Function and Fractional Integration
In this section the measure will be Lebesgue measure; in the one-
dimensional case we shall denote the Lebesgue integral of f over [a, bJ
by I: f(x) dx, and in the two-dimensional case we write similarly
I: led f(x, y) d(x, y).
It will be shown first how some well-known theorems concerning
the Gamma function are easily derived by means of the methods of
Lebesgue integration.
LEMMA 1. The function e- x xp-1 is summable over [0,00) for p>o,
but not for pO.
PROOF. Since
1 1 1
I e-1xp-1dx1 e-Xxp-1dx1 x p - 1 dx,
o 0 0

176
FUBINI'S THEOREM
-[Ch. 5,  25
the integral in the middle is finite for P>O and infinite for pO.
Furthermore, given the real number p, there exists a positive constant
k such that e- 1x x p - 1 <k for xl, so e- x x p - 1 <ke-!x on [1,00), and
this shows that e- x x p - 1 is summable over [1, 00).
In view of the lemma, the following (familiar) definition is justified.
DEFINITION. For p>O, the Gamma function r(p) of p is defined by
00
r(p) == J e- X x p - 1 dx.
o
THEOREM 1. For P>O we have r(p+ 1)=pr(p)}' hence r(p+ 1)==pI
if P is a positive integer.
PROOF. Let P>O, and O<a<b<oo. Then, by partial integration,
b b
J e-xxpdx=-e-bbp+e-aaP+p J e- x x p - 1 dx,
a a
so (a! 0 and b i 00)
00
00
r(p+ 1)== J e-xxpdx==p J e- x x p - 1 dx==pr(P).
o 0
It is easy to verify that, for P>O, q>O, the function x P - 1 (I-x)q-l
is summable over [0, IJ, and the following (familiar) definition is there-
fore justified.
DEFINITION. For p>O, q>O, the Beta function B(P, q) is defined by
1
B(P, q)== J x p - 1 (I-x)q- 1 dx.
o
Note that B(P, q) is the limit for 15!0 of the Riemann integral
fl- tJ x P - 1 (I-x)Q- 1 dx, so that, by the rules for Riemann integration,
1 1
B(P, q) = J x p - 1 (I-x)Q- 1 dx== J x Q - 1 (I-x)p- 1 dx=B(q, P).
o 0

Ch. 5, 925J GAMMA FUNCTION AND FRACTIONAL INTEGRATION 177
THEOREl\12. For P>O, q>O, we have
T(P)T(q)==T(p+q)B(P, q).
PROOF. Using that J: f(x)dx==J::cc f(x-c)dx holds for any finite
real c, we obtain, for y>O,
00 00 00
e-YT(p)== I e- x - Y x p - 1 dx= I e- X (x-y)p- 1 dx== If(x, y)dx,
o y 0
where
{ e-X (x-y)p-1 for Oy<x,
f(x, y) ==
o for Oxy,
hence
00 00 00
Idyl yq-lf(x, y) dx== I T(P) e- Y yq- 1 dy=T(P)T(q).
000
It is easily verified that f(x, y) is (two-dimensionally) measurable on
[0,00; 0, 00), and yq-lf(x, y) is therefore non-negative and measurable
on [0,00; 0, 00). It follows by Fubini's theorem that
00 00
00 00
T(P)T(q) == I dy I yq-1f(x, y) dx== I dx I yq-1f(x, y) dYe
o 0 0 0
Now, in view of the ordinary rules for Riemann integration,
00 x
I yq-1f(x, y) dy=e- X I yq-1(x-y)p- 1 dy
o 0
1
=e- X X P + Q - 1 1 t Q - 1 ( I-t)p- 1 dt
o
==e- X x P + Q - 1 B(P, q),
so
00
T(P)T(q)=B(P, q)1 e- X x p + Q - 1 dx==T(p+q)B(P, q).
o
The next theorem prepares the way for the introduction of fractional
integration.
THEOREM 3. If Ci>O, and f(x) is summable over [0, b], where O<b<oo,
then
x
h(x) == I (x-t)a-1f(t) dt
o

178
FUBINI'S THEOREM
[Ch. 5, 9 25
exists as a finite number for almost every XE[O, b], and h(x) is summable
over [0, b].
PROOF. Setting g(u)==u cx - 1 for Oub, and g(u) ==0 for all other real
values of u, we have to investigate the existence on [0, b] and the
summability over [0, b] of h(x)==lt g(x-t)f(t)dt. It is not difficult to
see that F(x, t)==g(x-t)If(t) I is non-negative and measurable on
[-00, +00; 0, b], so
 b b 
J J F(x, t) d(x, t) = J dt If(t) I J g(x-t) dx
- 0 o-
b 
= Jlf(t)ldt J g(x)dx<oo.
o -
This shows that g(x-t)f(t) is summable over [0, b; 0, b], so (by Fu-
bini's theorem)
b b b b b x
J J g(x-t)f(t) d(x, t) == J dx J g(x-t)f(t) dt== J dx J (x-t)a- 1 f(t) dt.
o 0 0 0 0 0
It follows that h(x) exists as a finite number for almost every XE[O, b],
and h(x) is summable over [0, b].
DEFINITION. If ex>O, and if f(x) is summable over [0, b], where 0<
b<oo, then the fractional integral fa(x)==af(x) of order ex of f(x) is de-
fined on [0, bJ by
x
.FIX/(X) = r;cx) f (X-t)lX-l/(t) dt. .
o
The name is justified by the particular case ex== 1 (note that f 1 f(x) =
It f(t)dt) and by the next theorem.
THEOREM 4. If ex>O and fJ>O, then the fractional integral of order fJ
of the fractional integral of order ex of f(x) is the fractional integral of
order ex+ fJ of f(x), i.e.,
fp.frxf(x) ==fcx+pf(x).

Ch. 5,  25J GAMMA FUNCTION AND FRACTIONAL INTEGRATION 179
PROOF. The definition implies that
x t
T(Ci)T(fJ)fpfrxj(x) == 1 dt(X-t)P-1 1 (t-u)a-1j(u) duo
o 0
(1)
From
x x-u
1 (X-t)P-1(t-u)rx- 1 dt== 1 (x-u-t)P- 1 t rx - 1 dt
u 0
1
== (x-u)a+P-1 1 (l-t)P- 1 t a - 1 dt
o
== (x-u) rx+p-1B (Ci, fJ) == (x-u)rx+P-l T(Ci)T(fJ)jT(Ci+fJ)
we derive
x t x x
1 dt(x-t)P-l 1 (t-u)a- 1 Ij(u) I du= 1 du Ij(u) 11 (x-t)P-1(t-u)a- 1 dt
o 0 0 u
x
== T(Ci)T(fJ) J  (X-U)iX+P-llf(u) I du
T(Ci+fJ)
o
==T(Ci)T(fJ)frx+plj(x) I,
and this is finite for almost every XE[O, bJ. We may invert therefore,
for these values of x, the order of integration in (1), and we obtain
x x
T (Ci)T (fJ)fpfrxj(x) == 1 du j(u) 1 (x -t)P-1(t-U) a-I dt
o u
x
== T(Ci)T(fJ) f (X-U)rx+p-1j(U) du==T(Ci)T(fJ)frx+pj(x).
T(Ci+fJ)
o
This shows that fpfaj(x) ==frx+pj(x) almost everywhere on [0, b J.
Exercises
FURTHER PROPERTIES OF THE GAMMA FUNCTION
The following sequence of exercises is mainly devoted to a method
of developing the theory of the Gamma function, due to H. Bohr,

180
FUBINI'S THEOREM
[Ch. 5, 9 25
J. Mollerup and E. Artin; a detailed account may be found in E. ARTIN
[IJ. .
, 25.1) Show that, for n== 1, 2, . .. and all x>O,
00
dnT(x)jdxn== I e- t t x - 1 (log t)ndt.
o
25.2) The function I(x) of one variable x, defined on the open inter-
val (a, b), is said to be convex whenever l{exX1+(I-ex)X2}ex/(X1)+
(l-ex)/(x2) for any pair of points Xl, X2 in (a, b) and any number ex
satisfying Oex 1. Show that I is convex on (a, b) if and only if
I(X2) - I(X1) I(X3) - I(X2)

X2- X 1 X3- X 2
for any triple Xl, X2, X3 in (a, b), satisfying Xl <X2<X3. Show also that
I is convex on (a, b) if and only if
I(X2) - I(X1) I(X3) - I(X1)

X2- X 1 X3- X 1
for any triple Xl, X2, X3 in (a, b), satisfying Xl <X2<X3. Show, finally,
that if 11, . . ., In are convex on (a, b), then 11 +. . . +In is convex on
( a, b).
25.3) Show that if I(x), defined on (a, b), is bounded on every closed
subinterval of (a, b), and satisfies 1{!X1 +!x2}1/(x1) +1/(x2) for any
pair of points Xl, X2 in (a, b), then I is continuous and convex on (a, b).
25.4) Show that if I(x) is convex on (a, b), then I is bounded on any
closed subinterval. Hence, by the preceding exercise, I is continuous on
(a, b).
25.5) Show that if the non-negative measurable functions I and g
have the property that 1 2 and g2 are summable over the set E, then Ig
is summable over E, and (JEIgdfl)2rEI2dfl JEg 2 dfl. This is the Schwarz-
Buniakowski inequality.
25.6) The function I(x), defined and positive on (a, b), is called loga-
rithmically convex whenever log I(x) is convex on (a, b). Show that if I
is logarithmically convex on (a, b), then I is continuous on (a, b). Show
that T(x) is logarithmically convex on (0,00), and show also that, for
any fixed y>O, B(x, y) is logarithmically convex on O<x<oo.

Ch. 5, 925J GAMMA FUNCTION AND FRACTIONAL INTEGRATION 18 I
25.7) Let f(x), defined for x>O, satisfy
(a) f(x+ 1) ==xf(x) for all x>O,
(b) f( 1) == 1,
(c) f is logarithmically convex.
Observe that f(x) ==T(x) satisfies these conditions, and show that
any f, satisfying the conditions, satisfies also
nXn'
f(x) == lim . ,
noo x(x+ 1). . . (x+n)
so that, consequently, there exists only one function satisfying the
conditions, namely T(x). Hence, Gauss' product representation
nXn'
T(x) == lim .
noo x(x+ 1) . . . (x+n)
holds.
25.8) Show by partial integration that B(x+ 1, y)=xB(x, y)/(x+y)
for x>O, y>O, and that, for any fixed y>O,
f(x) = B(x, y)T(x+y)
B(I, y)T(1 +y)
satisfies the conditions (a), (b) and (c) of the preceding exercise. This
furnishes a second proof 'Of the relation B(x, y)T(x+y)==T(x)T(y).
25.9) Show, either by direct computation or by computing first the
integral B(t, l), that T(l) ==n t .
25.10) Show that Legendre's relation
2 X - 1 T(lx)T(t{x+ I}) ==ntT(x)
holds, by making use of Exercise 25.7.
25.11) Define T(x) for negative X=F-l, -2, ... inductively by
T(X)==X-1T(x+ 1). Show that Gauss' product representation and Le-
gendre's relation continue to hold for these values of x.
25.12) Let cp(x) , defined and positive for all real x, have a con-
tinuous second derivative, and let cp(x+ 1) ==cp(x) for all x. Further-
more, let
cp(tx)cp{t(x+ 1) }===ccp(x)
for all x,
where c is a (necessarily positive) constant. Show that cp(x) ==c for all x.

182
FUBINI'S THEOREM
[Ch. 5, 9 25
Observe, for this purpose, that the second derivative g(x) of log cp(x) is
continuous, and satisfies
g(x) ==!g(!x) +!g{!(x+ I)}.
25.13) Let
cp(x)==r(x)r(l-x) sin nx
if the real number x is not an integer, and cp(x) ==n if x is an integer.
Show successively that cp(x+ 1) ==cp(x) for all x, cp(x) >0 for all x, and
cp(x) has derivatives of all orders for all x. Next, show that cp(!x)+
cp{!(x+ 1) }=ncp(x). Hence, by the preceding exercise, cp(x) ==n for all x,
I.e.,
n
r(x)r(l-x) == .
SIn nx
for all non-integer x.
25.14) Show that, for all real x,
n ( x2 )
sin nx = nx lim II 1 - 2 .
noo k=l k
25.15) Show that, for O<ex<l,
00
f X<x-l n
dx == .
1 +x sin nex
o
UNIQUENESS OF FRACTIONAL INTEGRATION
25.16) Let /(x) be summable over [0, bJ with O<b<oo, let ex>O,
and let fIX/ex) ==g(x) , where fIX is the fractional integral of order ex.
Show that if /l(X) is any other function, summable over [0, bJ and
satisfying f<x/1{X)==g(x), then /1=/ almost everywhere.
ABEL'S INTEGRAL EQUATION
25.17) Let p> -1, so that x P is summable over [0, bJ for any posi-
tive b, and let ex>O. Show that
f<xxp=xp+<xr(p+ l)/r(p+ 1 +ex),
.
I.e.
xp+<x==f<x{xpr(p+ 1 +ex)/r(p+ I)}.

Ch. 5, 925J GAMMA FUNCTION AND FRACTIONAL INTEGRATION 183
Show that, for q>Ci-l, the equation frxf(x)==x q has the unique so-
lution
f(x) ==xq-rxT(q+ 1 )jT(q+ l-Ci).
Show in particular, by choosing Ci==!, q==O, that Abel's integral equation
x
f f(t) dt == n!
(x-t)i-
o
has the unique solution f(x) = 1 j(nx)!.

CHAPTER 6
BANACH SPACE AND HILBERT SPACE; Lp SPACES
In this chapter we introduce Banach space and Hilbert space which, apart
from their usefulness for some problems of integration theory, are of vital im-
portance in their own right. Sec. 27 is devoted to the famous Hahn-Banach
extension theorem, and sec. 28 contains an application of the extension theo-
rem to a problem in measure theory. In sec. 30 Banach spaces are introduced
the elements of which are measurable functions; the most important examples
are the Lp spaces. Finally, in sec. 31, the Bochner integral, Le., the Banach
space valued equivalent of the Stieltjes-Lebesgue integral, is defined and its
main properties are derived.
26. Normed Linear Spaces and Banach Spaces
Let V be a linear vectorspace; i.e., V is an Abelian group with respect
to addition, and the real or complex numbers act as operators on V
such that, for x, Y E V and ex, fJ real or complex, the distributive laws
ex(x+y) ==exx+exy and (ex+ fJ)x==exx+ fJx hold; in addition, ex(fJx) == (exfJ)x
and lx==x. If we have the complex numbers acting as operators on V,
the space V is called a linear vectorspace over the complex numbers or
a comPlex linear vectorspace; if we have merely the real numbers acting
as operators on V, then V is called a linear vectorspace over the real
numbers or a real linear vectorspace. We shall assume here that V is a
complex linear vectorspace; everything which follows holds with small
modifications also for real linear vectorspaces. If the maximal number
of linearly independent elements in V is finite, say n, the space V is
said to be n-dimensional; if such a finite maximal number of linearly
independent elements does not exist, V is said to be of infinite di-
mension. The null element of V (when V is regarded as an additive
group) will be denoted by 0 (it will always be clear from the text
whether 0 represents the null element of V or the number zero).

Ch. 6,  26J NORMED LINEAR SPACES AND BANACH SPACES 185
The linear vectorspace V is called a normed linear space if there
exists a non-negative function I[xil on V such that
(a) Ilxll==O if and only if x==O,
(b) Ilx+yl[llxll+IlYII for all x, YEV (the triangle inequality),
(c) Ilexxll== lexl'llxll for all XE V and all complex ex (homogeneity proper-
ty).
The non-negative number I[xll is called the norm of the element XE V .
Evidently, any normed linear space is a metric space with respect to
the distance function d(x, y)==l[x-YII.
The normed linear space V is called a Banach space if V is a complete
metric space with respect to the distance d(x, y) == Ilx-yll, i.e., if
lim IIxm -xnll==O
as m, noo
implies the existence of an element XEV such that limllx-xnll==O as
noo. Banach spaces were introduced in the years around 1920 by
F. Riesz, S. Banach and N. Wiener, and a comprehensive theory of
the properties of these spaces was developed by the Polish school of
mathematicians headed by S. Banach. An account of this research may
be found in S. BANACH'S treatise (1932, [lJ). It is almost needless to
say that, after 1939, the work was vigorously carried on, mainly by
American and Russian mathematicians, and an extensive account of
this work is presented in the treatise by N. DUNFORD and J. SCHWARTZ
( 1958, [1]).
ExamPles. Real k-dimensional Euclidean space Rk with I[x!I==(xi+
. . . +x)! for x== (Xl, . . " Xk) is a real Banach space (cf. also Exer-
cise 27.1 and sec. 39, Theorem 3). The complex k-dimensional Eucli-
dean space C k of all k-tuples of complex numbers X== (Xl, . . " Xk) with
l[xll==(lxlI2+. . . + IXkI2)! is a complex Banach space. For k== 1 we ob-
tain the space C 1 of all complex numbers with IIxl[ == Ixi.
The complete metric space L 1 of all functions I, which are fl-summable
over the point set X, is a Banach space with respect to the norm 11/11==
Ix III dfl.
If [a, bJ is a bounded closed interval, the collection of all functions
I, continuous on [a, bJ, is a Banach space with respect to 11/11==
maxaxb If(x) I. The norm 11111 in this case is called the unilorm norm

186 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 9 26
of f, since the assertion that limllf-fnll==O is equivalent to the as-
sertion that the sequence fn(x) converges uniformly on [a, bJ to f(x).
The completeness of the space follows by observing that the limit
function of a uniformly converging sequence of continuous functions
on [a, bJ is continuous on [a, bJ.
Let V and W be linear vectorspaces (it is permitted that V and W
are identical). If, for all XEV, there is defined (unambiguously) a corre-
sponding element y==TXEW, we shall say that T is a transformation
on V into W. In other words, T is then a function, whose domain of
definition is the whole space V, and with values in the space W. The
transformation T is called a linear transformation if T(exxl +fJX2) ==
aTx1+fJTx2 for all X1,X2EV and all complex ex, fJ. If V and Ware
normed linear spaces, and T is a transformation on V into W, then the
transformation T is said to be continuous at the element XOEV if, given
8>0, there exists a number £5>0 such that IITx-Txoll<e whenever
IIx-xoll<£5. If T is continuous at all elements XEV, then T is called a
continuous transformation. The transformation T (on V into W) is
called bounded if there exists a positive constant C such that IITxII
Clixil for all XE V.
THEOREM 1. If V and Ware normed linear spaces, and T is a linear
transformation on V into W, then the following assertions are equivalent.
(a) T is continuous,
(b) T is continuous at one element of V,
(c) T is bounded.
PROOF. (a) =>(b) is evident.
(b)=>(c). Let T be continuous at XOEV. Then there exists a number
B>O such that IIx-xollB implies .
IIT(x-xo) II == IITx- Txoll 1.
Hence IITvll1 for all VEV satisfying IIvllB. For any XEV we have
IIBxlllxII II =B, hence IIT(BxlllxlI) II 1, i.e.,
IITxl1 = I" IIT(Bx/llxIDII  I" .
This shows that IITxIICllxll for C== liB and all XEV.

Ch. 6,9 26J NORMED LINEAR SPACES AND BANACH SPACES 187
(c)=>(a). Let IITxllCllxll for all XEV. Then, given e>O and XIEV,
any XEV such that IIX-x111<eIC satisfies
IITx - TX111 == IIT(x -Xl) IICllx-X111 <E.
Hence, T is continuous at the arbitrary element Xl E V, i.e., T is con-
tinuous.
If V and Ware normed linear spaces, and the linear transformation
T (on V into W) is bounded, then the smallest non-negative number C,
satisfying I!TxllCllxll for all XEV, is called the norm of T, and denoted
by IITII. Note that such a smallest number exists. Hence IITxll/llxIIIITIl
for all X=FO; on the other hand, given E>O, there exists an element
Xl E V such that II Tx 111/11 x 111 > IITII -c. Observing also that IIT(exx) II/IIcxxil
== IITxll/llxll if exX=F0, we have, therefore,
IITxll
IITII == sup == sup II Txll.
X=F 0 IIxil IIxll= 1
Given the linear transformations T1 and T 2 (on V into W) and the
complex number ex, the linear transformations T 1 + T 2 and exT 1 are de-
fined by (T1+T2)(X)==T1X+T2X and (exT 1 )(x)==exT 1 x respectively. The
null transformation 0 (on V into W) is defined by Ox==O for all XE V
(where 0 is the null element of W). Evidently, in view of these defi-
nitions, the collection of all linear transformations on V into W is a
linear vectorspace, with the null transformation as null element.
THEOREM 2. Let V and W be normed linear spaces, and let Y be the
collection of all bounded linear transformations on V into W. If, for each
TEY, the norm IITI! is defined as above (hence, IITII==supllTxll for IIxll== 1),
then Y is a normed linear space with respect to the norm IITII. If, in ad-
dition, W is a Banach space, then Y is a Banach space.
PROOF. IITII==O if and only if Tx==O for all XE V, that is, if and only
if T is the null transformation. Furthermore,
liT 1 +T 211= sup liT 1x+T2xllsup liT 1 x ll+ su p IIT 2 xl/= liT 111+ liT 211,
IIxll = 1
and
IIexTII== sup IIexTxll== lexl sup IITxll== lexl' IITII.
IIxll= 1
This shows that Y is a normed linear space.

188 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,  26
Let us assume now that W is, in addition, a Banach space, let
TnEY for n==I,2, "', and limIITm-Tnll==O as m,n-+oo. Then
limllT m x - T nxll==O for any XEV, so the element Tx==lim T nXEW
exists on account of the completeness of W. The thus defined transfor-
mation T is clearly linear, and it follows from lilT mil-liT nlll IIT m - T nil
that the sequence of numbers liT nil is bounded, so liT nllC for all n
and some C'>O. Then
IITxll==lim liT nxllCllxll
for all XEV,
so T is a bounded linear transformation, i.e. TEY. Finally, given e>O,
it follows from liT m- T nll<e for m, nno(e) that any XE V such that
Ilxll== 1 satisfies liT m x - T nxll<e for m, nno. Hence
IITx-T nxll== lim liT m x - T nxlle
for Ilxll== 1 and nno.
moo
In other words, liT -T nlle for nno, i.e., lim liT -T nll==O. This shows
that Y is complete, i.e., Y is a Banach space.
Any transformation on the linear vectorspace V into the Banach
space C 1 of all complex numbers is called a functional; if the transfor-
mation is linear, it is called a linear functional (note that the present
definition is in agreement with the definition in sec. 12). If, in ad-
dition, V is a normed linear space, and the transformation is bounded,
then it is called a bounded functional. The collection of all bounded
linear functionals on the normed linear space V is denoted by V*,
and it is customary to use for bounded linear functionals the
special notation F(x), or sometimes x*(x), instead of Tx. Hence, IIFII==
sUPllxll=l IF(x) I is the norm of FEV*. It is an immediate consequence
of Theorem 2 that V* is a Banach space. The Banach space V* is called
the conjugate space (or adjoint space) of V.
If VIis a subcollection of the linear vectorspace V such that, for
any two elements x, YEV1 and any two complex numbers ex, {J, we have
exX+{JYEV 1 , then VI is called a linear subspace of V. Evidently, if V is
a normed linear space, any linear subspace is (with respect to the same
norm) also a normed linear space. If V is a Banach space, and the linear
su bspace VIis closed (regarded as a subset of the metric space V), then

Ch. 6,9 26J NORMED LINEAR SPACES AND BANACH SPACES 189
VI is also a Banach space (if limllxm-xnll==O and all XnEV1, thenxEV
exists such that x==lim X n , and consequently, since VIis closed, we
bave XEV1). ExamPle: If V is the Banach space of all functions I(x),
continuous on the bounded interval [a, b J and with norm IIIII == max If(x) I
for axb, then the collection of all polynomials P(x) is a linear sub-
space of V, but not a closed linear subspace (indeed, I(x) =sin XE V
and in view of the power series expansion of sin x there exists a
sequence of polynomials Pn(x) such that lim Ilf-Pnll==O, but I itself is
not a polynomial). This collection is, therefore, an example of a normed
linear space which is not a Banach space. If [c, d] is a proper sub-
interval of [a, b], then the collection of all lEV, vanishing on [c, dJ, is
a closed linear subspace of V.
Exercises
HAMEL BASIS
26.1) Let V be a linear vectorspace. The finite or infinite subset
{Xa} of elements XaE V (where {ex} is an index collection) is called
linearly independent whenever every finite subset of {xa} is linearly
independent in the usual sense. If, in addition, every XE V is a finite
linear combination of elements from {xa}, then {Xa} is called a Hamel
basis of V. Show that in this case the coefficients in the finite linear
combination which represents XEV are uniquely determined by x.
Show also, by means of Zorn's lemma, that every linear vectorspace
V, consisting not exclusively of the null element, has at least one
Hamel basis.
EXISTENCE OF UNBOUNDED LINEAR TRANSFORMATIONS
26.2) Let V and W be normed linear spaces, where V is of infinite
dimension and W contains at least one element Y=FO. Show that there
exists at least one unbounded linear transformation on V into W.
A COUNTER EXAMPLE
26.3) Let V be the collection of all functions I(x), continuous on
Ox<oo and satisfying lim f(x) ==0 as X-+oo. Show that V is a Banach

190 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,  27
space with respect to the norm Ilfll==maxlf(x)1 for Ox<oo. Let Tn
(n== 1, 2, . . .) be the linear transformation on V into V, defined by
(T nf)(x)==f(x+n). Show that Tn is bounded; more precisely, show that
IITnll==l. Show also that lim Tnf==Of for any fEV, where 0 is the null
transformation, i.e., show that
lim liT nfll==O
for any fE V.
This shows that limllT nf-Tfll==O, holding for each fEV separately,
does not always imply that lim liT n - TII ==0.
27. The Hahn-Banach Extension Theorem
In this section we shall discuss the problem of extending a linear
functional, defined on a linear subspace of a linear vectorspace V, to
the entire space V.
We first give some additional definitions. The realvalued functional
P(x), defined on the linear vectorspace V, is called subadditive if P(x+y)
P(x) +P(y) for all x, y E V. If, in addition, p(ax) == ap (x) for all constants
aO, the functional P(x) is sometimes called sublinear.
Next, if V 0 is a proper linear su bspace of the real linear vectorspace
V, and z is an element of V not in V o , we shall denote by Vo[z] the
linear subspace of all elements v+az (VEV O , a real). The represen-
tation x==v+az for an element XEVO[Z] is unique. Indeed, V1+a1z==
V2+a2z implies (a1 -a2)z==v2-V1 E V 0, so a1-a2==0 (since z is not in
V 0), and then also VI ==V2.
LEMMA CX. If V 0 is a proper linear subspace of the real linear vector-
space V, if P(x) is a real sublinear functional on V and f(x) a real linear
functional on V 0 such that f(x)P(x) for all XE V 0, and if z is an element
of V not in Vo, then there exists a real linear functional f1(X) , defined on
V o[zJ, such that f1(X) ==f(x) on V 0 and f1(X)P(x) on V o[zJ.
PROOF. The problem is to find a suitable value for t==f1(Z); once
this has been done, the definition f1(v+az)==f(v)+at for every VEVO
and every real a extends f linearly to Vo[zJ. The requirement con-
cerning t is that f(v)+at==f1(v+az)p(v+az) shall hold for every VEV O
and every real a=FO. For a>O we set v==aw, and we write the ine-
quality as f(w)+tP(w+z), which is required to hold for all WEVO.

Ch. 6, 927J
. HAHN-BANACH EXTENSION THEOREM
191
For a<O we also set v==aw, and we now divide by -a, obtaining
-f(w)-tP(-w-z), holding for all WEV O . Hence, t has to be chosen
such that
-P{ -W1- Z ) -f(W1)tP(W2+Z) -f(W2)
for all WI, W2 E V o . But
f(W2) -f(W1) ==f(W2-W1)P(W2-W1)
==P{(W2+ Z ) + (-W1-Z)}P{W2+Z) +p( -W1- Z ),
so
-P( -WI -z) - f(W1)P(W2+Z) - f(W2)
for all WI, W2 E V 0,
and this shows that
P==sup{ -P( -W1- Z ) -f(w1)}inf{p(w2+z) -f{W2)}==Q.
Wl
W2
It follows that t can be taken as any number satisfying PtQ.
THEOREM 1 (HAHN-BANACH EXTENSION THEOREM; H. HAHN, 1927,
[2J; S. BANACH, 1929, [4J). If VIis a linear subspace of the real linear
vectorspace V, if P(x) is a real sublinear functional on V, and if f(x) is
a real linear functional on VI such that f(x)P(x) for all XE V 1, then
there exists a real linear functional F(x) on V such that F(x)==f(x) on VI
and F(x)P(x) on V.
PROOF. If V1== V, there is nothing to prove; we may assume, there-
fore, that VIis a proper linear subspace of V. We consider now all
linear extensions g(x) of f(x) satisfying g(x)P(x), i.e., all linear function-
als g(x), defined on a linear subspace V(g)  VI, such that g(x)==f(x) on
VI and g(x)P(x) on V(g). The collection of all these g(x) is not empty
(it contains f(x) itself), and is partially ordered by inclusion (gl  g2 if
g2(X)==gl(X) on V(gl)). Let gp, where {fJ} is an index set, be a chain in
this collection. If Xo is any element in the union of all V (gp) , then all
gp, insofar as they are defined at Xo, have the same value at Xo; hence,
if go(xo) is defined as this comn10n value, then go(x) is also a linear
extension of f(x), defined on the union of all V (gp) , and satisfying
go(x)P(x). This shows that go is an upper bound of the chain gp. It
follows by Zorn's lemma that there exists a maximal linear extension

192 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 27
F(x) of I(x), satisfying F(x)P(x), and the domain of definition V o of
F is the entire space V, since otherwise F could be extended further
by Lemma ex (which would contradict the maximality of F). Hence,
F(x) is the required extension.
COROLLARY. II VIis a linear subspace 01 the normed linear space V,
and il I(x) is a bounded linear lunctional on VI, then I can be extended
to the whole 01 V without changing its norm.
PROOF. We assume first that V is a real normed linear space (hence,
I is then assumed to be a real linear functional). Let IIIII be the norm
of I on VI, so I/(x) III/II · IIxll for every XE VI. Setting now P(x) == 11/11 '/Ixll
on V, the functional P(x) is evidently sublinear, and l(x)If(x) IP(x)
on the subspace VI. Hence, there exists an extension F of I to the
whole of V such that F(x)P(x) == 11/11' IIxll for all XE V. Since, then,
F( -x)II/II '11-xll==IIfll' IIxll,
the inequality F(x)==-F(-x)-II/II'llxll is also true, hence IF(x)l
11/11' IIxil for all XE V. This shows that IIFIIII/II, and since the inverse
inequality holds obviously, we obtain IIFII==II/II.
We now deduce the proof for the case that V is a complex normed
linear space from the real case, following an argument due to H. F.
BOHNENBLUST and A. SOBCZYK (1938, [IJ), and G. A. SOUKHOMLINOV
( 1938, [1]).
Note first that a complex normed linear space V becomes a real
normed linear space v(r) if the collection of numbers, acting as scalar
multipliers, is restricted to the set of all real numbers (the spaces V
and v(r) contain, therefore, the same elements, but the dimension of
v(r) is, so to say, twice the dimension of V, since for any X=FO the ele-
ments x and ix are, in v(r), linearly independent). Also, the real and
imaginary parts g(x) and h(x) of the complex linear functional I(x) ==
g(x) +ih(x) on VI are each real linear functionals on vir) such that
IIgllII/II and IIhllII/II. In addition,
g(ix) +ih(ix) == I(ix) ==i/(x) ==i{g(x) +ih(x)} == -h(x) +ig(x) ,
so h(x)==-g(ix), and therefore I(x)==g(x)-ig(ix) on VI. By the result
for the real case, there exists a real linear extension G of g to the

Ch. 6, 927J
HAHN-BANACH EXTENSION THEOREM
193
entire space such that IIGII==llgllll/ll. We now define
F(x) ==G(x) -iG(ix)
on V'
,
for XEV1 this implies
F(x) ==g(x) -ig(ix) == I(x).
Evidently F is real linear (i.e. F(ax) ==aF(x) for real a), and in order
to show that F is complex linear, it is therefore sufficient to observe
that
F(ix) ==G(ix) -iG( -x) == i{G (x) -iG(ix)} ==iF(x).
Finally, given XEV, we write F(x)==re icp , where r==IF(x) I and qJ is real.
Then x1==xe- icp satisfies F(x1)==r, hence G(x1)==r, and
IF(x) I ==r==G(x1)IIGII'llx11111/11'llx111== If/ll'llxll.
It follows, as in the real case, that IIFI/==ll/lf. Note the similarity be-
tween the final part of the present proof and the proof that If Id#l
f f/l d# for any complex #-summable I.
THEOREM 2. I I V is a normed linear space, not consisting exclusively
01 the null element, then there exists a bounded linear lunctional on V
which is not identically zero, i.e., the conjugate space V* does not consist
exclusively 01 the null element.
PROOF. Let XOEV, and xo=¥=O. If VI is the linear subspace of all
complex multiples x==cxXo of Xo, then I(x) == I (cxxo) ==cxllxo/I is a bounded
linear functional on VI, satisfying 11/11 == 1. Hence, by the preceding
corollary, there exists an element FE V* such that IIFII == 1.
Exercises
FINITE DIMENSIONAL BANACH SPACE
27.1) If V is a normed linear space with norm "xii, and IIxl11 is an-
other norm on V, then the two norms are said to be equivalent norms
whenever there exist two positive constants C 1 , C 2 such that the
quotient of Ilxll and IIxl11lies between C 1 and C 2 for all XEV. Let V be
k-dimensional, and let e1, . . " ek be a fixed basis in V. Show that if

194 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 9 28
IlxlI* is defined for x==  jej by Ilxll* == f Ijl, then IIxll* is a norm in
V, and V is a Banach space with respect to this norm. Show that any
other norm in V is equivalent to IIxll*, and deduce from this that any
two norms in V are equivalent, and that V is a Banach space with
respect to any norm.
27.2) Let V be a k-dimensional normed linear space with norm IIxll
and basis e1, . . " ek, and let, for x=== 1 jej, the functionals Fj(x), i ==
1, . . " k, be defined by Fj(x) ==j. Show that the Fj(x) are bounded
linear functionals on V, linearly independent regarded as elements of
V*. Deduce from this that any linear functional F(x) on V is bounded,
by showing that F is a linear combination of F 1, . . " F k. Note also
the implication that V* has the same dimension as V.
DIMENSIONS OF A SPACE AND ITS CONJUGATE SPACE
27.3) Show that if the normed linear space V contains at least k
linearly independent elements, then its conjugate space contains at
least k linearly independent elements.
28. The Problem of a Measure, defined for all Bounded Subsets
of the Real Line
Let, in this section, J be the ring of all bounded subsets of the real
line R1. In the earlier sections we have encountered several examples
of measures # defined on LI :
(a) #(E) ===0 for all E EJ;
(b) #(0) ==0, and #(E) ==00 for all other E EJ;
(c) #(E) ==0 if E consists of a finite or countable number of points,
and #(E) ==00 if E consists of an uncountable number of points;
(d) discrete measure in R 1 , i.e., #(E) is equal to the number of
points in E ;
(e) if Xo is a given fixed point of R1, then #(E)==1 if xOEE, and
#(E) ==0 if Xo is not contained in E.
If we impose the condition that, for any EEJ, the measure #(E)
should be invariant under all translations of E, then example (e) drops
out, and if we impose the additional condition that the measure of the
cell (0, IJ should be finite, then (b), (c) and (d) drop out as well. Hence,
if we require finally that # should not be identically zero, then none of

Ch. 6,  28J
PROBLEM OF A MEASURE
195
the above examples satisfies all the conditions. AS' a matter of fact,
there exists no measure satisfying the conditions, as shown by the
following theorem.
THEOREM 1. There exists no measure #, defined on the ring LI of all
bounded subsets of R1, such that
( 1 ) # (E) is invariant under all translations,
(2) the measure of the cell (0, 1 J is finite,
(3) there exists a set 5 EJ such that #(5) >0.
PROOF. Assume that # satisfies the three conditions. Denoting the
cell (n, n+ IJ by An for n==O, =i 1, =i2, ..., we have  #(5An)==
#(5) >0 (where the sum on the left contains only a finite number of
non-zero terms, since 5 is bounded), hence #(5An) >0 for some value
of n, and by (1) this implies that the cell (0, 1 J contains a subset of
positive measure. It follows, in view of (2), that the cell (0, 1 J is of
finite positive measure. The subset E of (0, IJ is now defined in ex-
actly the same manner as in sec. 10, Theorem 2 (where it was proved
that the thus defined set E is not Lebesgue measurable), and it follows
exactly as in the proof of the cited theorem that neither #(E) >0 nor
#(E) ==0. Hence, the measure # cannot exist.
Given an arbitrary ring of subsets of a point set X, the set function
v(E), defined for all sets E in the ring, is called a charge (also a content,
or a finitely additive measure) if v is monotone, v(0) ==0 and v is finitely
additive, i.e., V(E1 +E 2 ) ==v(E 1 ) +V(E2) for all disjoint sets E1, E 2 in
the ring. The rest of the present section will be devoted mainly to the
proof that there exists a charge v on the ring J of all bounded subsets
of R1 such that v(E) coincides with the Lebesgue measure of E for all
Lebesgue measurable sets EELI. The discussion which follows is due to
S. BANACH (1923, [2J).
Let V be the collection of all bounded real functions f(x) on R1 of
period 1; i.e., f(x+ 1) ==f(x) for all xER 1 . With the usual definitions of
addition and scalar multiplication by real numbers, V is a real linear
vectorspace. In order to avoid confusion, it may be useful to remark
that there is no identification of functions in V; as soon as two functions
in V differ in at least one point, they are regarded as different elements

196 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 28
of V. Let, for any fEV,
1 n
M(f; Ci1, · · " Cin) == sup -  f(X+Cii),
-oo<x<oo n i=l
where Ci1, . . " Cin is an arbitrary finite sequence of real numbers, and
let P(f) be the greatest lower bound of M (f; Ci1, . . " Cin) for all such
sequences (the number n, therefore, is also variable).
We shall prove that P(f) is a sublinear functional on V. Evidently,
p(af) == ap (f) for any constant aO. In order to prove now that P(f) is
subadditive, let f, gE V and 8>0 be given. Then there exist sequences
Ci1, . · " Cip and fJ1, "', fJq such that
M(f; Ci1, · · " Cip) <P(f) +8 and M(g; fJ1, · · " fJq) <P(g) +8,
so that, arranging the pq numbers Cii+fJj into a single sequence Y1, . . "
Ypq, we have
P(f+g)M(f+g; Y1, .. ., ypq)
p q
==sup p-1 q -l   (f+g) (X+Cii+fJj)
i= 1 j= 1
q p
Sup{q-l  p-1  f(X+Cii+fJj)}+
j= 1 i= 1
p q
+sup{P-1  q-l  g(X+Cii+fJj)}. (1)
i= 1 j= 1
But
p
p-1  f (x + Cii + fJ j ) M (f; Ci 1, . . " Cip)
i=l
for all x and all fJj, so
q p
q-1  p-1  f(X+Cii+fJj)M(f; Ci1, . . " Cip)
j= 1 i= 1
for all x, and this implies that the last expression in (1) does not exceed
M(f; Ci1, "', Cip)+M(g; fJ1, ..., fJq). Hence,
P(f+g)M(f; Ci1, "', Cip)+M(g; fJ1, "', fJq) <P(f) +P(g) +28;
this holds for every 8>0, so P(f+g)P(f)+P(g).

Ch. 6, 928J
PROBLEM OF A MEASURE
197
We show next that, for any Lebesgue measurable f(x) in V, the Le-
besgue integral f(j) over (0, IJ does not exceed P(j). It will be suf-
ficient to show that f (j) M (j; aI, . . " an) for any sequence aI, . . .,
an, and this follows by observing that f{j(x)}==f{j(x+a)} for any ex
in view of the periodicity of j, so that
1
f(j)==n- 1 !{j(x+a1) + ... +j(x+an)}dxM(j; aI, '.', an).
o
Hence, since the collection of all Lebesgue measurable j(x) in V is a
linear subspace VI of V, and f(j) is a linear functional on VI satis-
fying f(j)P(j), there exists in view of the Hahn-Banach extension
theorem a linear extension F(j) of f(j) to V such that F(j)P(j) for
all jE V. This has several consequences. In the first place, F(j) satisfies
F{j(x+xo)}==F{j(x)} for any real Xo. In order to prove this, we set
g(x) ==j(x+xo) -j(x) and a1 ==0, a2==XO, a3==2xO, ..', an== (n-l )xo, so
that
P(g)M(g; aI, "', an)==n- 1 sup{j(x+nxo)-j(x)}
x
for all n, and so P(g)O (since j is bounded). Similarly, P( -g)O.
Hence F(g)P(g)O and F(g)==-F(-g)?:;-P(-g)?:;O, so F(g) ==0.
Also, the linear functional F(j) is non-negative on V (i.e., F(j)O for
j(x)?3:-0). This follows by noting that P(j)O for any j(x)O, so F(j)
P(j)O for j(x)O, i.e., F(j)O for j(x)?3:-0.
Secondly, writing
f(j)==![F{j(x)}+F{j(l-x)}J,
and noting that, for any jE VI, the Lebesgue integral f(j) satisfies
f{j(x)}==f{j(l-x)} (this is true for continuous functions, and there-
fore, by extension, for all jEV1), we obtain f(j)==f(j) for all JEV1,
and f{j(x)}==f{j(l-x)} for all JEV.
We have proved, therefore, the following theorem.
THEOREM 2. I j V is the linear vectorspace oj all bounded real junctions
j(x) oj period 1 on R1, then there exists a non-negative linear junctional
f (j) on V, such that
(a) f (j) is the Lebesgue integral oj j over (0, 1 J jor any Lebesgue
measurable j in V,

198 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 9 28
(b) f{j(x+xo)}==f{j(x)} jor any real xo,
(c) f{j(l-x)}=== f{j(x)}.
Specializing j(x) such that j(x) coincides on (0, 1J with the charac-
teristic function XE(X) of an arbitrary subset E c (0, IJ, and denoting
f(j) for this special j by v(E), we obtain the following theorem.
THEOREM 3. There exists a non-negative set junction v(E), dejined jor
all subsets E oj (0, 1], such that
(a) V(E1 +E 2 ) ===V(E1) +V(E2) ij E 1 and E 2 are disjoint, i.e., v(E) is
jinitelyadditive,
(b) v(E) is the Lebesgue measure oj E jor any Lebesgue measurable E,
(c) v(E) is invariant under translations oj E (it will be evident how
this is to be understood ij ajter the translation there is a part oj E lying
outside (0, 1 J),
(d) v(E) is invariant under rejlection oj E with respect to x==!.
We add some remarks. It is evident how to extend v(E) by addition
to any bounded subset E of R1, such that the extended v(E) has the
properties (a), (b), (c) and (d). The extended v(E) is, therefore, a charge
on the ring L1 of all bounded subsets of R1. The statements in (c) and
(d) may be condensed into the single statement that congruent sets
have the same v.
Exercises
EXTENSION TO HIGHER DIMENSIONS
28.1) Show that Theorem 2 is valid in Rk for k> 1, where (c) is to
be read now as f{j(X1, . . " 1-x p , . . " Xk)}=== f{j(x, . . " x p , . . " Xk)}
for p== 1, . . " k. Note, however, that if we specialize j to character-
istic functions, obtaining thereby a parallel of Theorem 3, we do not
obtain the result that all congruent sets have the same v.
BANACH-MAZUR LIl\1:IT
28.2) Let V be the linear vectorspace of all bounded real functions
on [0, 00). Show that there exists a real functional Binz j(x) on V such

Ch. 6, 9 29J
HILBERT SPACE
199
that
(a) -Bin{af(x)+bg(x)}==a -Bin f(x)+b -Bin g(x) for real a, b,
(b) -Bin f(x)O if f(x)O on [0, 00),
(c) Blm f(x+xo)==Bin f(x) for any non-negative xo,
(d) Etin f(x) ==limX_H)o f(x) if the (ordinary) limit on the right exists.
The introduction of this generalized limit is due to S. MAZUR (1929,
[IJ).
28.3) Let V be the linear vectorspace of all bounded real sequences
Xn (n== 1,2, . . .). Show that there exists a real functional-Bin X n on V
such that
(a) -Bin (axn+bYn)==a ain xn+b Blm Yn for real a, b,
(b) Btin xnO if xnO for all n,
(c) Btin X n +1==-B"m Xn,
(d) Btin xn==limnoo X n if the (ordinary) limit on the right exists.
EXISTENCE OF A PURE CHARGE
28.4) In the last paragraph of the present section we have obtained
a charge v(E) on the ring J of all bounded subsets of R1, such that v
is not identically zero. Show that v is a pure charge on J. We recall
that, by Exercise 9.17, the charge v on J is called a pure charge when-
ever any measure # on J satisfying O#(E)v(E) for all EELI is identi-
cally zero on J.
29. Hilbert Space
Let V be a complex linear vectorspace, and let there be defined on
the Cartesian product V X V an inner product, i.e., corresponding to
each pair XEV, YEV there is a complex number, called the inner product
of x and y, and denoted by (x, y), such that
(a) (Xl +X2, y) == (Xl, y) + (X2, y),
(b) (exx, y) == ex(x, y) for any complex ex,
( C ) (x, y) == (y, x),
(d) (x, x) >0 for X=FO. _
The dash above (y, x) in (c) means that (y, x) is the complex conjugate
of (y, x); generally, if ex is any complex number, we shall denote by &
its complex conjugate.

200 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 29
Any linear vectorspace V with an inner product is called an inner
product space (or a general Euclidean space). In the case that the inner
product space V is a real linear vectorspace, it is assumed then that
the inner product (x, y) is realvalued, so that (c) becomes now (x, y)==
(y, x). For abbreviation we shall write (x, x)!==llxll. Note that x==O im-
plies (x, y)==O for ally by (b); hence, on account of (d), we have (x, x)==O
if and only if x==O, i.e., Ilxll==O if and only if x==o.
THEOREM 1 (SCHWARZ'S INEQUALITY). We have
1 (x, y) 1lIxll'llyll
for all x, y E V, with equality if and only if x and yare proportional.
PROOF. If (x, y)==O, the theorem is true (note, in particular, that
the assertion concerning equality is true in this case). Assume therefore
that (x, y) *0, so x*O, i.e., (x, x) >0. Observing now that
O(x-exy, x-exy) == (x, x) -ex(y, x) -&(x, y) + lexI2(y, y)
for all ex, and choosing ex==(x, x)/(y, x), we obtain
O(x, x) - (x, x) - (x, x) + (x, X)2(y, y)/I (x, y) 1 2 ;
hence, dividing by (x, x),
1 (x, y) 1 2  (x, x) (y, y).
Equality occurs if and only if x-exy==O.
THEOREM 2 (TRIANGLE INEQUALITY). We have
IIx+yllllx"+ IIYII
for all x, y E V, and there is equality if and only if x and yare proportional
with a non-negative factor of proportionality.
PROOF. For Ilx+YII==O the theorem is true. Assume therefore that
Ilx+yl/*O. It follows then from
/lx+y/1 2 ==(x+y, x+Y)I(x, x+y)I+I(y, x+y)1
llxll'/lx+YII+ Ily/I'llx+y/l,
upon division by Ilx+YII, that "x+yll/lxll+/lYII.

Ch. 6, 9 29J
HILBERT SPACE
201
Equality occurs only if x and yare both multiples of x+y, and if,
in addition, the two numbers (x, x+y) and (y, x+y) are non-negative,
i.e., if x and yare non-negative multiples of x+y.
We may conclude, therefore, that any inner product space V is a
normed linear space with respect to the norm IIxll==(x, x)!. If, in ad-
dition, V is a complete normed linear space (i.e., if V is a Banach
space with respect to the norm Ilxll==(x, x)!), then the space V is called
a Hilbert space. Abstract Hilbert spaces were introduced by J. VON
NEUMANN (1929, [2J), although the theory for the important special
example l2 (which will be discussed in the next section) was already
developed by D. Hilbert in 1906 (reproduced as "Vierter Abschnitt"
in D. HILBERT' S treatise [1 J) .
The elements x, y E V are called orthogonal if (x, y) ==0. If x and yare
orthogonal, we shall sometimes write x l..Y. Orthogonal elements x, y
satisfy the Pythagorean theorem: IIx+YIl2== Ilxll 2 + lIyll2. Note also that in
any inner product space V the parallelogram law holds:
Ilx+YI12+ Ilx-YI12==21I x I1 2 + 211yl12
for any pair x, YEV. Conversely, if V is a normed linear space in which
the parallelogram law holds, then it is possible to introduce an inner
product (x, y) in V such that (x, x)! becomes the existing norm IIxll in
V (P. JORDAN and J. VON NEUMANN, 1935, [IJ). This is done by defining
4(x, y) == Ilx+yl12 -lIx-YI12+illx+iYI12- i lix-iYll2,
and then it is an elementary, although tedious, task to show that the
function (x, y) has all the properties of an inner product, and that
(x, x) ==IIxll2.
ExamPles. Any k-dimensional complex linear vectorspace may be
made into a Hilbert space by selecting a linearly independent basis,
and defining
(x, y) == X 1Y 1 + . . . + x kY k f or x == (x 1, . . " X k) and y == (y 1, · · " Y k) ,
where Xl, . . . , Xk and Y1, . . " Yk are the coordinates of x and y with
respect to the selected basis. The proof, in particular the proof that
the space is complete, is left to the reader (cf. also Exercise 27. 1) .

202 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 29
The collection of all complex functions I(x), continuous on [0, IJ, is
made into an inner product space by defining (I, g) ==/0 1 I(x) g(x) dx.
The Banach space of all functions I(x), Lebesgue summable over
[0, 1 J and with norm 11/11 == IlI/(x) I dx, is not a Hilbert space since the
parallelogram law fails to hold: if F==[O, tJ and G==[!, IJ, l==xF and
g==XG, then
11/+gll== 1,
II/-gil == 1,
I1III == Ilgll ==!,
so
II/+gI12+ 11/-gI1 2 ==2
and
211f11 2 +21IgI1 2 == 1.
The subset S of the linear vectorspace V is called convex if it con-
tains the segment between any pair of its points, i.e., if Xl, X2ES im-
plies (XXI + (l-ex)x2 ES for all ex such that Oex 1.
THEOREM 3. A closed convex subset S 01 the Hilbert space H contains
a unique element 01 smallest norm.
PROOF. Let d==infllxll for XES. Then there exists a sequence XnES
such that IIxnll!d. We have !(xm+x n ) ES since S is convex, so IIxm+xnll
2d. It follows therefore, in view of
IIx m -xnll 2 ==2I1 x mI1 2 + 211 x nll 2 -llx m + xn11 2 ,
that lim IIxm-xnll==O. This shows that the sequence X n is a fundamental
sequence; let xo==lim X n (the limit element Xo exists, since H is a Hil-
bert space). Then XOES (since S is closed), and I[xol!==lim Ilxnll==d. Hence
xo is an element in S of smallest norm. Finally, if there would exist
another element XOES, xo*xo, IIxoll==d, we should have !(XO+XO)ES
and
11!(xo+xo) 112==!ll x oIl2+ !ll x oll2 -li!(xo-xo) 11 2 <d 2 ,
and thi,s is impossible.
THEOREM 4. II M is a closed linear subspace 01 the Hilbert space H,
and X is an arbitrary element 01 H, then there is a unique decomposition
X==X1+X2 such that Xl EM and X2l..M (i.e., (X2, y)==O lor all YEM).
PROOF. If xEM, the decomposition x==x+O is the required one. If
x is not in M, we may call the set of all x-y, where y runs through M,
the linear variety parallel to M through x. This linear variety is closed

'Ch. 6, 9 29J
HILBERT SPACE
203
.and convex, and contains therefore a unique element X2==X-X1 (Xl EM)
-of smallest norm, i.e., Ilx2-ayIl2I[X2112 for all YEM and all complex a.
We select an arbitrary YEM, Y=FO, and set a==(x2, y)/(y, y). Then
IIx2112  Il x 2- a yl12 == (X2, X2) - a(y, X2) - &(X2, y) + laI 2 (y, y)
= (X2,X2) _2 1 (X2,Y)1 2 + I(X2,Y)1 2 = Il x 211 2 _ I (X2,Y)1 2 ,
(y , y) (y, y) (y, y)
hence (X2, y)==O for all YEM. This shows that X==X1+X2, where X1EM
and x21-M.
The decomposition is unique, since Xl +x2==xi +X2 (Xl, xi EM and
..%2, X21- M ) implies P==x1- x i==X2- X 2; hence pEM as well as P1-M.
This implies that p is orthogonal to itself, i.e., (P, P) ==0. But then p==O,
, d '
so Xl ==X1 an X2==X2.
THEOREM 5. If F(x) is a bounded linear functional on the Hilbert
.space H, there exists a uniquely determined element YEH such that F(x)==
(x, y) for all xEH; in addition, IIFII==lIyll.
PROOF. If F(x)==O for all xEH, the element y==O satisfies the re-
.quired conditions. If F(x) is not identically zero, the set of all X such
that F(x) ==0 is a proper closed linear subspace M of H, and it follows
that there exists an element Z=FO such that z 1-M (select an arbitrary
.x not in M, and write X==X1 +z, Xl EM, z 1-M; this is possible by the
preceding theorem). Multiplying z, if necessary, by a suitable constant,
we may make F(z)== 1. Then, given an arbitrary xEH, the element m==
x-F(x)z is in M (since F(m)==F(x)-F(x)F(z)==O); hence, x==m+
F(x)z is the decomposition of X into an element mEM and an element
F(x)z 1-M. Let now y==z/ll z I1 2 , so lIyll== 1/llzll and
F(y) ==F(z) /llzIl 2 == 1 /lIzl!2== IIYIl2== (y, y).
Then x==m+F(x)z may be rewritten as x==m+ay for a suitable com-
plex a, and
F(x) ==F(m+ay) ==aF(y)==a(y, y)==(ay, y)==(m+ay, y)==(x, y).
The uniqueness of the thus obtained element y is evident (if (x, Y1) ==
(x, Y2) for all x, the choice X==Y1-Y2 yields (Y1-Y2, Y1-Y2) ==0).
Finally, it follows from IF(x) I == I (x, y) 1IIYII'lIxll that IIFIIlIyll, and
from F(y)==(y, y)==I[YII'IIYII that IIFIIIIYII. Hence IIFII==IIYII.

204 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,  29
The present proofs of the Theorems 4 and 5 are due to F. RIESZ
(1934, [3J).
We conclude the section with a so-called separation theorem; a par-
ticular case of a much more general theorem.
THEOREM 6. If 5 is a closed convex subset of the Hilbert space H, and
Xo is an element of H not in 5, then there exists a bounded linear functional
F(x) on H and a real number c such that the real part G(x) of F(x) satis-
fies G(x) <c for all XES and G(xo) >c.
PROOF. Assume first that xo==O. Let Xl be the element of 5 of
smallest norm. The bounded linear functional F(x) on H is defined by
F(x)==-(x, Xl), and the real number c by c==F(tx1)==-!lIxlI12. Then
the real part G(x) of F(x) satisfies G(xo) ==G(O) ==F(O)==O>c. It remains
to prove that G(x)<c for all XES. Assume that G(X2)C for some X2ES.
Then x==ax2+(I-a)x1ES for Oal, and
IIx11 2 == (x, x) ==ex 2 I1x2112- 2a( l-ex)G(x2) + (l-ex)21IxlI12
== IIxl112 - 2a{lI x lI1 2 + G (x2)}+ Dex 2 .
Since IIxlI12+G(x2)==-2c+G(x2)-2c+c==-c>0, it follows that Ilxll
< IIx111 for sufficiently small ex, contradicting the fact that Xl is the
unique element of S of smallest norm. Hence, G(x)<c for all XES.
If xo*O, we consider the set 5' of all Y==X-Xo, where X runs througll
S. Let F(x) be a bounded linear functional and c' a real number such
that the real part G(x) of F(x) satisfies G(O) >c' and G(y) <c' for all
YES'. Then, if c==c'+G(xo), we have G(xo»c and G(x)<c for all XES.
Exercises
SEPARATION THEOREM
29.1) Let S be a closed convex subset of the Hilbert space H with
the property that if XES then xeiffJES for all real qJ. Show that if Xo is
an element of H not in 5, then there exists a bounded linear functional
F(x) on H and a real number c such that IF(x) I <c for all XES and
IF(xo) 1 >c.

Ch. 6, 9 30J
BANACH FUNCTION SPACES; Lp SPACES
205
CONVEXITY
29.2) Let 5 be a convex subset of the Banach space V, having the
null element of V as an interior point, and let xo=¥:O be on the boundary
of S. Show that all points aXo, Oa< 1, are interior points of S.
30. Banach Function Spaces; Lp Spaces
We return to measures and integrals. Given the measure # in the
non-empty point set X, we denote again by M+ the collection of all
non-negative #-measurable functions, and we assume that to each
IEM+ corresponds a number p(f) such that Op(/)oc> (the value +oc>
is, therefore, admitted), and such that
(a) p(/) ==0 if and only if 1==0 (almost everywhere on X),
p(/1 + 12) p(/1) + p(/2),
p(al) ==ap(/) for any constant aO;
(b) if InEM+ (n==l, 2, .. .), and Inil (almost everywhere on X),
then p(1 n) i p(/).
Note that it follows from (b) that p(f) is order preserving, i.e., if
111 almost everywhere on X, then p(/1)p(f). The proof is derived by
considering the sequence 11/f. · · i I.
Given any #-measurable complex function I(x) on X, we now define
the number 11/11 by 11/11==p(I/I), and we denote by L p the collection of all
I satisfying II/II<oc>. Then, with the now already familiar convention
that almost equal functions are identified, L p is a normed linear space
with respect to 11/11 as norm. Indeed, 1I/11==p(l/l) ==0 if and only if 1/1=0
almost everywhere, that is, if and only if 1==0 almost everywhere.
Furthermore,
11/1 + 1211==p(1/1 + 121)p(1/11 + 1/21)p(1/11) +p(1/21) == 11/111+ 11/211,
and
II all I ==p(la/l) == lalp(l/l) == lal'lI/lI
for any complex a.
THEOREM 1. (a) II 11 is #-measurable, f2ELp, and 1/1(X)I1/2(X)1 al-
most everywhere, then 11 ELp, and 1I/1111I/211.
(b) II IEL p , then f is linite almost everywhere.
(c) If InELp and InO on X (n== 1,2, . . .), and In i I almost every-
where, then either IEL p and Ilfnlli 1I/11<0c>, or IIlnlli II/II==oc>.

206 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 30
(d) II InELp (n== 1,2, . . .), and lim In(x)==/(x) almost everywhere,
then 11/11lim inflllnil (this is a parallel 01 Fatou's lemma lor integrals).
PROOF. Only (b) and (d) need a proof.
(b) Let E =={x: I/(x) I ==oo}. In view of I/(x) lnXE(x) for n== 1, 2, . . "
we have n IlxEIIII/II<oo for n== 1,2, . . " so IIXEII==O. This shows that
XE==O almost everywhere, i.e., #(E) ==0.
( d) Let
km(x) ==inf (1Im(x) I, I/m+1(x) I, . . .) for m== 1, 2, · · · .
Then Ilkmlllllnll for all nm, so Ilkmlllim inf Il/nll. Also Okm(x) i I/(x) I
almost everywhere on X, so
IIIII == lim Ilkmlllim inf III nil.
m-».oo
THEOREM 2. The normed linear space L p is a Banach space.
PROOF. The proof is similar to the proof that the collection of all
#-summable functions is a Banach space with respect to the norm
Jx III d# (cf.sec.16, Theorem 1). LetlnEL p (n== 1, 2, · · .), andlim 1[lm-Inll
==0 as m, noo. Then there is a subsequence gn(x) of In(x), i.e. gl==lnl'
g2==ln2 with n2>n1, and so on, such that r Ilgn+1-gnll<oo. Defining
g(x) by
00
g(x)== Ig1(X) I +  Ign+1(X) -gn(x) I,
1
we have
p 00
Ilgll==lim Illg11+ Ign+1-gnlllllg111+ IIgn+1-gnll<oo,
p-».oo 1 1
so gEL p . It follows by Theorem 1 (b) that the set E=={x: g(x)==+oo}
is of measure zero; hence, on the set X -E the series r Ign+1(X) -gn(x) I
converges in the classical sense, and the same holds then for the series
 {gn+1(X)-gn(X)}. Setting I(x) ==0 on E and
00
I(x) ==gl(X) +  {gn+1(X) -gn(x)}
1
on X - E,
we have then I/(x)lg(x), so 1I/11lIgl[<oo, i.e., IEL p . Furthermore,
00
I/(x) -gp(x) I  Ign+1(X) -gn(x) I
n=p
on X - E

Ch. 6,  30J
BANACH FUNCTION SPACES; Lp SPACES
207
(Le., almost everywhere); hence,
00
Ilf-gplI Ilgn+1-gnll0
p
as poo.
This shows that
Ilf - f nll Ilf- g pll + Ilg p- f nil O
as n (andp)oo.
DEFINITION. A ny Banach space L p of the type introduced here is called
a Banach function space. .
We shall postpone the investigation of further properties of general
Banach function spaces until Chapter 15; here we shall present some
important examples, occurring in many branches of analysis.
Given fEM+, we define, for any number p such that lp<oo, the
number pp(f) by pp(f)==(Jxf P d#)l/p. Note that, for p== 1, this is simply
the integral of f over X. We also introduce the number Poo(f) , as follows.
Let, for any a such that Oa<oo, the set Xa be defined by Xa==
{x: f(x) >a}. If #(X a ) >0 for all a, we define Poo(f) == +00; if, however,
#(Xa)O for some value of a (so #(Xb)==O for all b>a), we define
poo(f)==ao, where ao==inf a for all a such that #(Xa)O. Obviously,
#(X ao ) is still zero, but #(X b ) >0 for all b<ao. The number Poo(f) is
called the esselltial upper bound of f in view of the property that, for
Poo(f) <00, there clearly exists a function f1 EM +, almost equal to f,
such that f1 is bounded on X with its least upper bound equal to Poo(f).
In other words, Poo(f) is the least upper bound of f if sets of measure
zero are neglected.
For lpoo the function pp(f), defined on M+, obviously has the
properties that pp(f) ==0 if and only if f==O (almost everywhere), pp(af)
==app(f) for any constant aO, and Ofn j f (almost everywhere) im-
plies Pp(jn)jpp(f). The last property follows, for lp<oo, by means of
the theorem on integration of increasing sequences, and for p==oo the
assumption that lim Poo(f n) <Poo(f) yields an immediate contradiction.
Furthermore, for p== 1 and p==oo it is also evident that PP(j1 +f2)
pp(f1) + pp(f2). In order to show that this triangle inequality holds
for all values of p in 1 poo, we first prove:

208 BANACH SPACE AND HILBERT SPACE; L p SPACES [Ch. 6,9 30
THEOREM 3 (HOLDER'S INEQUALITY). Let 1 poo, and let the number
q be defined by 1 /p+ 1 /q== 1 (so 1 qooJ. furthermore, q==oo for p== 1,
and q== 1 for poo). Then, if f, gEM+, we have
J fg d#pp(f)pq(g).
x
PROOF. For p== 1 and p==oo the proof is trivial; we may assume,
therefore, that 1 <p<oo (so that, consequently, also 1 <q<oo). If one
at least of the numbers pp(f) and pq(g) is zero, the theorem is true; let,
therefore, pp(f) *0 and pq(g) *0. If, in that case, one at least of the
numbers pp(f) and pq(g) is infinite, the theorem is true as well; let,
therefore, O<pp(f) < 00 and O<pq(g) < 00, so that, in view of the defi-
nition of pp(f) and pq(g), the functions f and g are finite almost every-
where on X. Without loss of generality we may assume that f and g
are finite at each point XEX.
For t> 1 and O<m< 1 we have t m - 1 < 1; integration over lty
yields ym-l<m(y-l) fory>l. Settingy==a/b (a>b>O) andm==l/p,
we obtain
a 1 / p b- 1 /p-l <p-1(ab- 1 -1),
I.e.,
a1/pb1-1/p-b<p-l(a -b).
Since 1-1 /p== l/q, the last inequality is equivalent to
a 1 / p b 1 /q < b+p-1(a -b) ==a/p+ b/q.
Since the roles of p and q may be interchanged, the same inequality
holds for b>a>O, and for a==b>O it becomes an obvious equality.
Hence, setting a==fP(x)/p(f) and b==gq(x)/p(g) for any XEX, we find
f(x)g(x) fP(x) gq(x)
 + ,
pp(f)pq(g) Pp(f) qp(g)
so
1 I' 1 1
pp(f)pq(g) J fgdft  P + q = 1,
x
.
I.e.,
J fg d#pp(f)pq(g).
x

Ch. 6, 930J
BANACH FUNCTION SPACES; Lp SPACES
209
THEOREM 4 (TRIANGLE INEQUALITY). II Ipoo, and I, gEM+, then
pp(1 + g) pp(/) + pp(g).
PROOF. For p 1 and poo the proof is trivial; let therefore 1 <
P<oo, and let again I/P+l/q==l, so qP/(P-I). If pp(/+g) ==0, or if_
pp(/) +pp(g) oo, the theorem is true; we assume therefore that pp(/+g)'
*0 and pp(/) *00, pp(g) *00. It follows then from
(/+g)P{2 max(/, g)}p2P(lp+gp)
that pp(/+g)<oo, and, by means of Holder's inequality, we obtain
(writing J instead of Jx)
! (/+g)pdfl== ! 1(/+g)p- 1d fl+ ! g(/+g)p- 1 dfl
pp(/){! (/+g)Pdfl}l/q+pp(g){! (/+g) pd fl}l/ q ,
where we have used that q(P-I)p. Dividing by the finite number
{J (/+g) pd fl}l/ q , the desired result follows.
In view of what has been proved now, it is evident that, for lp(X),
the function pp(/) satisfies all conditions summed up at the beginning
of this section. We may conclude, therefore, that the collection Lp of
all fl-measurable complex I(x) , satisfying pp( III) < 00, is a Banach
function space with respect to the norm "/lIppp(I/I). For p== 1 this
is the Banach space L1 of all fl-summable functions, the properties of
which were already discussed in sec. 20. The collection L) of all fl-
measurable real functions I(x), satisfying 1I/IIppp(IfI)<oo, is a real
Banach space with respect to the norm 11/11p. If it is desirable to indi-
cate the set X on which the functions are defined, and the measure fl,
we shall write Lp(X, fl).
The space L 2 deserves special attention, since it may be regarded
not merely as a Banach space, but also as a Hilbert space. Note first
that, for p==2, the number q determined by I/P+ I/q== 1 satisfies q==2,
so that Holder's inequality for I, gEL 2 becomesJ I/gl dfl11/1121IgI12. Hence,
if I, gEL 2 , the integral J I(x)g(x) dfl exists' as a finite number, and it
follows then immediately that (I, g) == J Iff, dfl is an inner product in L2,
and that (/,/)==JIf12dflII/II. The Hilbert space norm (/,/)i and the
already existing norm 111112 are, therefore, identical, and Schwarz's ine-

210 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 930
quality I(f, g)Illfll'llgll is essentially the same as Holder's inequality
for the particular case p==2. The following theorem is now an immedi-
ate consequence of Theoren1 5 in the preceding section; for the case
that X is a bounded interval on the real line and fl is Lebesgue measure,
it is due to M. FRECHET (1907, [1 J).
THEOREM 5. If F(f) is a bounded linear functional on L2, then there
exists a unique function gEL 2 such that F(f) == J f(x)g(x) dfl for all fEL 2 ,
and the norm IIFII of F satisfies IIFII== IIg112. Conversely, it is evident that
any gEL 2 determines a bounded linear functional F(f) on L 2 by F(f) ==
J fg,dfl.
We assume now that the Stieltjes-Lebesgue integral J fdfl is the ex-
tensjon of the elementary integral f(f), initially defined only on the
collection L of realvalued functions f(x). It has been proved in sec. 16
that fEL implies fELi r ); more explicitly, we have shown that L is dense
in Li r ). This result may be extended if L consists of bounded functions
and if, in addition, fEL always implies min (f, 1) EL; a case which oc-
curs often in practical applications (e.g., when L consists of step func-
tions, or continuous functions possessing bounded carriers).
THEOREM 6. Let L consist of bounded functions, and let fEL imPly
min(f, I)EL. Then L is included in L) for lpoc>, and L is dense in
L) for lp<oc>. In general, L is not dense in L<:O).
PROOF. Evidently, L is included in L). Let, therefore, lp<oc>. If
fEL, then If I is summable; in addition, If I is bounded, so there exists a
positive constant a such that laf(x) I 1 on X. On account of laf(x) IP
laf(x) I and the summability of laf(x) I it follows then that laf(x) IP is
summable, hence Ifl P is summable, i.e., fEL).
In order to show that L is dense in L) for lp<oc>, we may either
give a proof similar to the proof for p== 1 (sec. 16, Theorem 2), or, using
the fact that the theorem holds for p== 1, we may reduce the case
1 <P< oc> to the case p== 1. Here we choose the second alternative.
If 1 <P<oc>, and fEL), then f==f+-( -f-), where f+ and -f- are
non-negative, and f+, -f-EL). It will be sufficient, therefore, to prove
that any non-negative f in L) may be approximated (with respect to
the Lp norm) with a prescribed degree of accuracy by functions of L.

Ch. 6, 930J
BANACH FUNCTION SPACES; Lp SPACES
211
Given OfEL), let An=={x: n- 1 <f(x)<n} for n==l, 2, "', and gn==
fXAn' Then #(An)<oo for all n (since fp is summable over An), and
f-gn!O almost everywhere on X, so Ilf-gnllp!O by the theorem on
dominated convergence. Hence, given e>O, we may fix n such that
IIf-gnllp<l e . For this fixed n, it follows from OgnnXAn and #(An)
<00 that gnELr), and we may choose, therefore, hEL+ such that
IIgn -hill < (e/2n)P; this is possible since L is dense in Lr). Replacing
h(x) by min(h(x), n)EL+, the number IIgn-hl11 is not enlarged; conse-
quently, we may immediately assume that Oh(x)n for all XEX.
Then J Ign-hlpd#nP-1 J Ign-hl d#, so IIgn-hl[p(nP-1I1gn-hI11)1/p
nllgn -hll/P < le. Hence
IIf-hllpllf-gnllp+ IIgn -hllp<e.
Finally, in order to show that, in general, L is not dense in L), we
consider the example that the integral J fd# is the Lebesgue integral
over the real line (or over a bounded subinterval J of the real line). If
L is the collection of all realvalued continuous functions possessing
bounded carriers (or the collection of all realvalued continuous functions
on J, respectively), then L is not dense in L), since the function
f(x) EL), defined by f(x) ==0 for Ixl > 1, and f(x) == 1 for Ixl 1 (or
f(x) == XL1 1 , where Lll is a proper subinterval of the above-mentioned
interval J), cannot be approximated in the Loo norm by functions
hEL (any hEL satisfies Ilf-hl[ool)' If L is the collection of all step
functions s(x)==l cnXAn(x), where the sets An are disjoint cells, and
fEL) is the function, zero outside O<x 1 and assuming the value
(-I)n on {x: 1/2n+1<xI/2n} for n==O, 1,2, "', then f cannot be ap-
proximated in the Loo norm by functions sEL.
Denoting the collection of all functions f+ig, \vhere f, gEL, sym-
bolically by L+iL, it follows now immediately (still under the as-
sumption that L consists of bounded functions, and fEL implies
min(f, I)EL) that L+iL is dense in Lp for Ip<oo. In particular,
the collection of all complex step functions is dense in Lp for lp<oo.
By making use of this fact for p== I, we have proved in sec. 20 that
the space L 1 is separable if and only if the measure # is separable.
Making a few small changes in the proof, it may be shown similarly
that Lp, for Ip<oo, is separable if and only if # is separable. For

212 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 30
Loo the situation is usually different. In fact, as soon as there exists a
decomposition X==r X n , where all X n are disjoint and of positive
measure, the space Loo(X, #) is not separable. It will be sufficient to
prove that the subcollection of Leo, consisting of all I(x) that are
constant on each X n and assume only the values zero and one, is not
separable. This follows by observing that the subcollection contains an
uncountable number of functions (in view of the well-known fact,
proved by means of Cantor's diagonal procedure, that the number of
sequences aI, a2, . . " where each an is either zero or one, is uncountable),
and II/-glloo== 1 for any pair I, g of different functions in the subcol-
lection.
If the point set X consists of the points {I, 2, 3, . . .}, and each point
has the measure one (i.e., if we have the discrete measure on X), then
each complex (or real) function on X is a sequence 1=={ln} of complex
(or real) numbers, and each function is measurable (since each subset
of X is measurable with respect to discrete measure). The function
space Lp (lp<oo) consists now of all 1=={ln} such that II/lIp==
(r IlnI P )l/p<oo, and Loo consists of all 1=={ln} such that 1111100==
sup Ilnl <00. In this particular case Lp is usually denoted by lp, and
lp is called a sequence space. In the Hilbert space l2, the oldest example
of a Hilbert space of infinite din1ension, the inner product (I, g) is now
(I, g)==r Ingn. For lp<oo, the space lp is separable, since the dis-
crete measure in X is separable (there is only a countable number of
sets of finite measure). As follows from the remarks in the preceding
paragraph, the space loo is not separable. Similar remarks may be made
if the point set X consists of the points {. . " -2, -1, 0, 1, 2, . . .}, all
summations being then extended over -oo<n<oo.
Exercises
NORM OF BOUNDED LINEAR FUNCTIONALS
30.1) Let 1 <poo, and 1 /p+ 1 /q== 1. For any fixed gELq, the inte-
gral f Igd# exists for all IELp, and If Igd#I! I/gl d,ullgllqll/lip by Hol-
der's inequality. It follows that F (I) == f Ig d# is a bounded linear
functional on Lp, and IIFIIllgllq. Show that IIFII==llgllq; more explicit-

Ch. 6,  30J
BANACH FUNCTION SPACES; Lp SPACES
213
ly, show that
Ilgllq==max Iffgd#1 ===max f Ifgl d#
for all fEL p satisfying Ilfllpl. Note that, for p===2, every bounded
linear functional on L 2 is of this type by Theorem 5.
30.2) For any fixed gELoo, the integral I fgd# exists for all fELl,
and II fgd#I1 Ifgl d#llglloollfI11 by Holder's inequality. It follows that
F(f) ==1 fgd# is a bounded linear functional on L1, and IIFIII[glloo' Let
now, in addition, the measure # in X have the property that each set
of infinite measure contains a subset of finite positive measure. Show
that, in this case, IIFII==IIglloo' Show also that if the additional condition
is not satisfied, then it may happen that IIFII==O and Ilglloo>O.
COMPARISON OF Lp SPACES
30.3) Show that if #(X)<oo and fELp(X, #), then fELr(X, #)
for any r such that 1 rp< 00. Show also that if M p(f) ==
{I Ifl P d#/ #(X) }l/ P , and M r(f) is similarly defined, then M r(f)M p(f).
30.4) For sequence spaces the situation is reversed. Show that if
lpr<oo, and IIfllp==( IfnI P )l/p<oo, then l[fllrllfllp,
30.5) Let the measure # in X be a-finite. Show that, for every r in
lr<oo, there exists a subset XrcX such that #(Xr)r for all
r, XrcX r1 for rr1, and Xrj X as rjoo. Show also that if f(x) is #-
measurable on X, then
Ilflloo== lim {f Ifl r d#/#(X r )}l/r,
roo X r
where both sides may be +00. Hence, with the notations of Exercise 3,
Ilflloo== lim M r(f) === lim IIflir
if #(X) is finite.
roo roo
30.6) Show that if # is Lebesgue measure on X==[O, IJ, and l
P<oo, then the function x- 1 / p (1 +log X- 1 )-2/p is in Lp(X, #), but not
in Lr(X, #) for any r>p. Show also that x- 1 /p is in Lr(X, #) for all r
satisfying lr<p, but not in Lp(X, #). Show, finally, that log x-IE
Lp(X, #) for each p such that lp<oo, but log X-1ELoo(X, #) is false.
30.7) Show that if # is Lebesgue measure on X == [0,00), then the
function x-!(l + Ilog xl)-l is in L 2 (X, #), but not in Lp(X, #) for any
other value of p in lp<oo.

214 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 30
A CONVERGENCE THEOREM
30.8) Let 1 Poo and 1 /P+ 1 /q== 1. Show that if lim 11/- Inllp==O
and lim IIg-gnllq==O, then J Igdfl===lim J Ingndfl.
BANACH FUNCTION SPACES
30.9) If, in the Banach function space L p , the norm III1I satisfies
IIlnll! 0 whenever InELp (n=== 1, 2, . . .) and Iln(x) I! 0 on X, then L p is
said to possess an absolutely continuous norm. Show that there exists
a Banach function space such that its norm is not absolutely continu-
ous. Show also that in a Banach function space L p with an absolutely
continuous norm the dominated convergence theorem holds, i.e., if In
is fl-measurable for n== 1,2, . . " lim In==1 on X, and if Ilnlg on X for
some gEL p , then InELp, IEL p and lim 1I/-lnll==O.
30.10) Let, on the collection M+ of all non-negative fl-measurable
functions I(x), the functions Pn(/) (n== 1, 2, . . .) be defined such that
each Pn(/) satisfies the conditions summed up in the beginning of this
section, and let Kn==Lpn be the corresponding Banach function spaces
with norm 11/1!n==pn(l/l). Show that p*(/)==sup Pn(/) satisfies on M+ the
same conditions, and the unit sphere {I: II/II* I} of the corresponding
Banach function space K* is the intersection of the unit spheres of all
the Kn. It follows that the space K* is included in the intersection of
the spaces Kn.
30.11) Let Pn(/), for n== 1, 2, . . ., be the same as in the preceding
exercise, and let an (n== 1, 2, . . .) be positive constants. Show that
p**(/) == r anPn(/) generates a Banach function space K** included in
the intersection of the spaces Kn. Show also that if r an<oo, then
the space K* of the preceding exercise is included in K**.
30. 12) Show that the inclusion K * c K ** in the preceding exercise
may be proper by considering the following example. Let Po be a fixed
number such that 1 <Po<oo, and let Pn==max(l, po-n- 1 ) for n=
1, 2, . . . . Furthermore, let fl be Lebesgue measure in X == [0, 1 J. The
norm in Lp(X, fl), corresponding to the values P==Po and P==Pn re-
spectively, will be denoted by If/lio and 1I/IIn respectively. Show that the
norm 11/11*==sup 1I/IIn satisfies 11111*==11/110. Let a n ==n-(2+1/ p o), and 11/11**=
r anl!/lln. Show that if l(x)==x- 1 / po , then 11/11**<00 and 11/11*==00. Show
also that if g(x) ==X- 1 / P1 on (Xl, xoJ, g(x) ==x- 1 /P2 on (X2, X1J, and so on,

Ch. 6, 9 30J
BANACH FUNCTION SPACES; Lp SPACES
215
where the numbers Xn (n===O, 1, 2, . . .) are inductively chosen such
that. . .X2<X1 <xo== 1 and (J:nn-l gPnd#)1/Pn>n2+1/po, then Ilglln<oo for
all n, but Ilg!I**===oo. The space K**, corresponding to 11/11**, is therefore
properly included in the intersection of all spaces Lr(X, #) with r<po,
and K** properly includes Lpo(X, #).
A SPACE INCLUDING ALL Lp SPACES
30.13) Let # be a measure in X such that #(X) ==00, and such that,
given any #-measurable set E of positive measure and the number a
satisfying O<a<#(E), there exists a subset F of E such that # (F) ===a.
Note that Lebesgue measure in Rk satisfies this condition. Show that
if the complex #-measurable function I(x) is summable over each set
of finite measure, then there exists a set A of finite measure such that
IXX-A ELoo. Show also that I is summable over each set of finite measure
if and only if 1===/1 +12, where I1 EL 1(X, #) and 12ELoo(X, #).
30.14) Let X and # satisfy the same hypotheses as in the preceding
exercise, and let the function p(/) be defined on M+ by p(/) ==sup JE III d#,
where this least upper bound is taken over all sets E satisfying #(E) == 1.
Show that p(/) satisfies the conditions summed up in the beginning of
this section, and show also that p(XF) 1 and
Ild#{#(F)+ l}p(/)
F
for any #-measurable set F and any IEM+. Let L p be the Banach
function space with norm 1!/II===p(I/I). Show that IEL p if and only if
1==/1+/2 with 11 ELI and 12 EL oo' The introduction of the space L p is
due to G. G. GOULD (1959, [IJ) and J. J. SCHAFFER (1959, [IJ). Note
that L p includes all spaces Lp for 1 poo. Note also that if 1==11 + 12,
where 110 is summable and essentially unbounded, and 12 (X) = 1 for
all x, then I is in L p , but not in any Lp (IPoo).
30.15) Let L p be the same Banach function space as in the pre-
ceding exercise. Show that, given I(x) EL p , there exists a (not necessa-
rily uniquely determined) set EI such that #(Ef)== 1 and 11/11== JEt III d#.
Note the existence of a number Ao such that 1/1Ao on EI and 1/1Ao
on X-Ef.
30.16) Let L p be the same Banach function space as in the pre-
ceding exercises. Show that if IEL p , then 11/11==min (11/1111 + 11/21100), where

216 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 930
the minimum is taken over all decompositions f== 11 + f2 with 11 ELI
and f2ELoo.
THE CONVOLUTION
30.17) Let # be Lebesgue measure in k-dimensional space Rk, and
let IP<oo. Show that if fELp(Rk, #), then limllf(x+h)-f(x)llp==O as
hO. Note for this purpose that, given e>O, there exists a continuous
function g(x), possessing a bounded carrier, such that Ilf-gllp<le.
30.18) Let # be Lebesgue measure in Rk, let IPoo and I/P+ l/q
== 1. Show that if IELp(Rk, #) and gELq(Rk, #), then the convolution
F(X)==!Rk f(x-t)g(t)dt is a continuous function of x on Rk (in order to
make it unambiguously clear what is meant, we have written dt in-
stead of d#).
30.19) Let # be Lebesgue measure in Rk, let IPoo, and gE
L1(Rk, #). Show that, for any fELp(R k , #), the convolution
F(x) == I g(x-t)f(t) dt== I f(x-t)g(t) dt
is also in Lp(Rk, #), and IIFllpl!gl!ll[fllp, Setting (/) ==! g(x-t)f(t) dt
for all IELp, the transformation  is therefore a bounded linear
transformation on Lp into Lp such that 1I1IllgIl1.
THE FRACTIONAL INTEGRAL
30.20) Let # be Lebesgue measure in R1, and let X == [0, b], where
O<b<oo. Show that if fELp(X, #) for a value of P satisfying 1 <poo,
then the fractional integral od(x) is again in Lp for all ex>O, and
()(,f(x) is continuous on X for ex> lip. Show also that if fEL 1 (X, #),
then faf(x) is continuous on X for ex 1.
A CONVERGENCE THEOREM FOR Lp SPACES
30.21) Let IP<oo, fnELp (n==l, 2, ...) and IELp, and let fnf
pointwise #-almost everywhere on X. There are many simple examples
showing that these hypotheses do not yet imply that Ilf-fnllpO as
noo. Show, however, that if in addition limlllnllp==ll/lIp, then
lim Ilf-fnllp==O. For this purpose, observe first that X may be assumed
to be of a-finite #-measure. Next, given e>O, take Xl C X such that Xl
is of finite measure and!x-x1IfIPd#<!e. Observe now that, in view of

Ch. 6,  31J
BOCHNER INTEGRAL
217
Exercise 24.1, there exists c5 >0 such that fl(H) <c5 implies IH I/lpdfl< Ie.
By Egoroff's theorem (Exercise 18.12) we have X1E+H, where
fl(H)<c5 and In converges to I uniformly on E. It follows already that
IE I/IPdfl>lx I/IPdfl-e and IE 1/-lnlpdflO as noo. Hence IE Ilnlpdfl
IE I/IPdfl as noo, and so
!llnIPdfl> !I/IPdfl-e
E x
for nN1.
On the other hand, since limll/nllll/ll, we have
!llnIPdfl<!l/lpdfl+e
x x
for nN 2,
and it follows easily that IX-Ellnlpdfl<2e for nN3==max(Nl, N 2 ).
Then
II (/- I n) XX-EIII[/xx-EIi + III nXx-EII2(2e) lip,
so
! 1/-lnIPdfl2p+1e
X-E
for nN 3.
It follows that Ix 1/-lnlpdflO as noo.
Show that the theorem is not valid for P==oo. Show finally that, for
lp<oo, the theorem continues to hold if the pointwise convergence
of In to I is replaced by convergence in measure on every subset of
finite measure. Use Exercise 18.14.
31. Bochner Integral
If X is a non-empty point set and V is a complex (or real) Banach
space, and if I(x) is a function on X with values in V, then I is called
a vectorvalued lunction (on X into V). In the present section we assume
that fl is a measure in X, and we shall assign an integral to certain
vectorvalued functions I (on X into V); for each I the value of the
integral will be an element of V, so that we obtain a vectorvalued inte-
gral. In the particular case that V == C 1 (the space of complex numbers),
the thus extended notion will be again the ordinary numerical Stieltjes-
Lebesgue integrallxldfl.
Each vectorvalued function (on X into V) assuming only a finite
number of different values, each value *0 on a fl-measurable set of

218 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,931
finite measure, is called a step function. Exactly as in the numerical
case, it follows that if t(X)==f=l ViXEi(X), where the elements ViEV
and the sets E i are of finite measure (but not necessarily disjoint),
then t(x) is a step function. Assigning to each such t(x) the element
(t)==f=l vifl(E i ) of V, it follows then that (t) is uniquely de-
termined (as in the numerical case), and (t) is linear on the linear
collection of all step functions. Finally, IIf(t) 1I!x lit (x) II dfl for any step
function t(x), where the integral on the right is an ordinary Stieltjes-
Lebesgue integral. For the proof we may assume that t(x) == l ViXEi(X)
with disjoint sets E i . Then IIt(x) 11== lllvillxEi(x) for each XEX (since at
each XEX one term at most is different from the null element). Hence
p p
Ilf(t) II == II vifl(E i ) II IIvillfl(E i ) == jllt(x) II dfl.
11](
THEOREM 1. Let the sequence of vectorvalued step functions tn(x) satisfy
limjlltm(x)-tn(x)lIdfl==O as m, n-+oo.
](
Then there exists a unique vectorvalued f(x) such that Ilf(x)11 and all
functions I[tn(x) -f(x) II are fl-measurable and lim Ix !It n -fll dfl==O. Unique-
ness is to be understood here in the sense that functions differing only on
a null set are identified.
PROOF. The proof is similar to the proof that the space L 1 of all
numerical fl-summable functions is a complete metric space (cf. sec. 16).
We shall need the Banach space parallel of the theorem that an abso-
lutely convergent series is convergent: If VnEV, and r IIvnll<oo,
then r V n converges to a limit VEV, i.e., the partial sums sP==l Vn
satisfy limllv-spll==O. The proof is immediate. If p<q, then
q q
Ilsp-sql!==1I  vnll  IIvnll-+O
p+l p+l
as p, q-+oo.
Hence, sp is a fundamental sequence in V, and since V is a Banach
space, sp has a unique limit VEV.
Given now the sequence tn(x) of step functions, satisfying the hy-
potheses of the theorem, we select a subsequence t nk such that
00
 !lItnk+l-tnklldfl<OO.
k=l

Ch. 6,9 31J
BOCHNER INTEGRAL
219
Then Ilt n1 (x)ll+ r Ilt nk +1(x)-t nk (x)II is fl-summable, and hence finite for
almost every XEX. It follows by the remark in the preceding para-
graph that the vectorvalued series t n1 (X) + r {t nk +1(x)-t nk (x)} con-
verges for the same values of x; let f(x) be its sum where it converges
(i.e., almost everywhere), and let f(x)==O elsewhere. Since, therefore,
f(x) ===lim tnk(x) almost everywhere, we have Ilf(x) 11===lim IItnk(x) II and
I[tn(x) - I(x) II ==limk Iltn(x) -tnk(x) II almost everywhere, so Ilf(x) II and
IItn(x)-f(x)11 are fl-measurable. It follows from
00
f(x) -tnp(x) ==  {t nk +1 (x) -tnk(x)}
k=p
that lim J IIf-tnpll dfl==O as poc>, and hence
IlItn-flldfl/lltn-tnplldfl+ Illt np -flldflO
as noc>.
Assume that J Iltn-glldflO for another vectorvalued g(x). Then, in
particular, J Ilt nk -glldflO, where nk is the subsequence introduced
above. But the L1 convergence of Iltnk(x) -g(x) II to zero implies the
pointwise convergence almost everywhere to zero for a subsequence.
This subsequence of tnk(x) converges then pointwise to f(x) and g(x)
simultaneously, so f(x) ==g(x) almost everywhere. This shows that f(x)
is uniquely determined.
Let, again, the sequence tn(x) of step functions satisfy
limJxlltn-tmlldfl==O as m, noc>, and let f(x) be the unique function
such that lim Jxlltn-flldfl==O. Then the sequence of elements f(t n )
converges. Indeed,
IIf(t n ) -f(t m ) 11== Ilf(t n -t m ) II/I[tn -tmll dflO
as m, noc>.
The limit element of f(t n ) in V is denoted by f(f), and is called the
Bochner integral of f(x) (S. BOCHNER, 1935 [2J; N. DUNFORD, 1935 [1 J).
In order to show that the definition is justified, we have to prove that
the element f(f) is uniquely determined by I in the sense that if tn(x)
and t(x) are two sequences of step functions such that limJ Iltn-/lldfl
==limJllt-flldfl==O, then limf(tn)==limf(t). This follows from
I!f(t n ) -f(t) 11/lltn -t11 dfl/lltn -II! dfl+ Illt -fll dfl.

220 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,9 31
We shall denote the collection of all/(x) for which the Bochner inte-
gral is thus defined by B1==B1(X, V; fl). It is evident that B 1 is a
complex linear collection, and f(/) is linear on Bl' Any fEB 1 is called
a Bochner integrable lunction.
THEOREM 2. If f is Bochner integrable, then Ilf(f)II! Ilflldfl<oo.
PROOF. If fEB 1 and e>O, then lim J IIf-tnlldfl==O for some sequence
of step functions tn(x) , so J Ilf-tnil dfl<e for nN. Since IIf(x) II is measur-
able, it follows that
111/11 dflJ IltN11 dfl+e<oo,
and also
IJllflidfl- Jlltnlldfll/llf-tnlldfl<E
for nN.
Hence Jllflldfl==limJlltnlldfl, so that we may conclude from Ilf(tn)ll
J IItnll dfl, by making n-+oo, that IIf(f) II! Ilfll dfl.
Evidently, the collection B1 of all Bochner integrable functions f(x)
is a normed linear space with respect to J Ilflldfl as norm.
THEOREM 3. The collection B1 of all Bochner integrable functions f(x)
is a Banach space with respect to the norm J IIfl!dfl.
PROOF. Let fnEB1 for n==l, 2, "', and limJllfn-fmlldfl==O as
m, n-+oo. By the definition of Bochner integrability there exists for
each fn a step function t n such that J Ilfn-tnlldfl<n-1. It follows by
means of the triangle inequality that lim J IIt n -tmll dfl==O as m, n-+oo;
hence, by Theorem 1, there exists a function fEB 1 such that
limJlltn-flldfl==O. But then, once more by means ofthetriangleinequali-
ty, we have also limJllfn-flldfl==O. Evidently, f is uniquely determined.
THEOREM 4 (DOMINATED CONVERGENCE THEOREM FOR THE BOCHNER
INTEGRAL). If fnEB1 and Ilfn(x)llg(x) forn== 1,2, "', where g(x) is
fl-summable over X, and if fn(x) converges to f(x) almost everywhere on
X, then fEBl and lim J IIfn -fll dfl==O. I n particular, f(/) ==lim f(fn).
PROOF. Since fn-f==limrn(fn-fm) almost everywhere, we have
IIfn-fll==limmllfn-fmll almost everywhere, so Ilfn-fll is fl-measurable.
Furthermore, Ilfn(x) - f(x) 1I2g(x) and lim Ilfn(x) - f(x) II ==0 almost every-

Ch. 6,  31J
BOCHNER INTEGRAL
221
where on X. Hence, by the dominated convergence theorem for nu-
merical functions, lim J Ilj n - jll dfl == o. Then j E B 1 by the preceding
theorem, and f(j) ==lim(jn) on account of I[f(jn) -f(j)II!lljn-jll dfl.
If W is a second Banach space, it will not be necessary to have a
different symbol for the Bochner integral of any function g(x) E
B1(X, W; fl) ; the use of the same notation f(g) will cause no confusion.
THEOREM 5. Let T be a bounded linear transjormation on V into a
second Banach space W. Ij j(x) EB 1 (X, V; fl), then Tj(x) EB 1 (X, W; fl)
and f(Tj)==T(fj). In particular, ij F is a bounded linear junctional on
V, then J F{j(x)}dfl==F{f(j)}, where the integral on the lejt is now an
ordinary Stieltjes-Lebesgue integral.
PROOF. If t(x) is a step function with values in V, then Tt(x) is a
step function with values in W, and the equality f(Tt)==T(ft) is im-
mediately verified. Let jEB 1 (X, V;fl), and let limJlltn-jlldfl==O for
the sequence of step functions tn. Then Ilt nk - jll O pointwise almost
everywhere for son1e subsequence nk, and this implies that II Tt nk -Tjll
O pointwise almost everywhere. Hence
IITt n - Tjll ==limk IITt n - Ttnkll
almost everywhere,
so IITtn-Tjll is fl-measurable. It follows that
I IITt n - Tjll dflIITll/lltn - jll dfl O,
so TjEB1(X, W; fl) and
f(Tj)==lim f(Ttn)==lim T(ftn)==T(fj).
The collection B 1 of Bochner integrable functions is, obviously, the
.equivalent of the class L 1 of summable functions in the numerical case.
The question may be raised what is the equivalent of the class of
measurable functions. Evidently, in order that the vectorvalued
function j(x) be called fl-measurable, we shall require that IIj(x) II is
}t-measurable in the ordinary sense, but this alone is not sufficient.
Indeed, if E is a set of finite measure and the measurable function j(x)
is bounded on E, we shall require that jXE is Bochner integrable, and
this implies the existence of a sequence of step tunctions tn(x) satisfying

222 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 931
lim JE Ilf-tnil d#O, and hence the existence of a subsequence tnk(x)
converging pointwise almost everywhere to f(X)XE(X). This suggests the
following definition.
DEFINITION. The vectorvalued function f(x), on X into V, is called
#-measurable if the following condition is satisfied: Corresponding to any
#-measurable set E of finite measure there exists a sequence of step functions
tn(x) converging pointwise almost everywhere to f(x) XE(X).
Immediate consequences are:
(a) The collection of all measurable functions is a complex linear
collection.
(b) Any Bochner integrable function is measurable. Indeed, if fEBl,
there exists by the proof of Theorem 1 a sequence of step functions
converging pointwise almost everywhere on X to f(x).
(c) If f(x) is measurable, then IIf(x) II is measurable. Indeed, if lim tn(x)
==f(x) on E, then limlltn(x)IIllf(x)11 on E, so IIfxEIi is measurable for
any set E of finjte measure. This implies that min (lIf(x) II, s(x)) is measur-
able for any non-negative step function s(x), whence it follows by sec. 13,
Theorem 3 (2) that Ilf(x) II is measurable.
(d) If f is measurable and E is of a-finite measure, then there exists
a sequence of step functions sn(x) converging pointwise almost every-
where on E to f(x).
PROOF. Let E==r En, where the sets En are disjoint and of finite
measure. Furthermore, let tnk(X), k== 1, 2, . . " be a sequence of step
functions converging pointwise almost everywhere on En to f(x) as
k(X); we may assume without loss of generality that tnk(X) ==0 out-
side En. Then sn(x) ==t1n(X) +t2n(X) + . . . +tnn(x) is the required se-
quence.
(e) If f(x) is measurable and Ilf(x)lIg(x), where g(x) is a numerical
summable function, then f is Bochner integrable. In particular, if f is
measurable and Ilfil is summable, then f is Bochner integrable.
PROOF. Let t(x) be a sequence of step functions converging point-
wise to f(x) almost everywhere on the set E of a-finite measure where
g(x) >0; we may assume that each t;,,(x) vanishes outside the set E. The
step functions tn(x) are now defined by tn(x) t(x) at all x where
I It;" (x) 112g(x), and tn(x) o elsewhere. Then each tn(x) is a step function

Ch. 6,9 31J
BOCHNER INTEGRAL
223
satisfying IItn(x)II2g(x) on X. Furthermore, at each point XEE where
t(x) converges to f(x), we have IIt(x)II<2g(x) for nnx (we use here
that IIf(x) II is finite at each XEX, since f(x) is an element of the Banach
space V). It follows that tn(x) ==t(x) for nnx; hence, lim tn(x) == f(x)
almost everywhere on X. An application of the dominated convergence
theorem for vectorvalued functions shows then that f is Bochner inte-
grable.
(f) If f(x) is measurable and bounded on the set E of finite measure,
then f is Bochner integrable over E. This follows immediately from (e).
(g) In the numerical case the present definition of #-measurability
is consistent with our earlier definitions. For the details of the simple
proof, we refer to Exercise 20.1.
(h) If f(x) is a vectorvalued measurable function and g(x) a finite
numerical measurable function, then g(x)f(x) is measurable. Indeed, if
E is a set of finite measure, and the sequences of step functions tn(x)
and sn(x) converge pointwise almost everywhere on E to f(x) and g(x)
respectively, then sn(x)tn(x) is a sequence of step functions converging
almost everywhere on E to g(x)f(x).
(i) If the sequence of vectorvalued measurable functions fn(x) con-
verges pointwise to f(x), then f(x) is measurable.
PROOF. Since all IIfn(x) II are measurable, the same holds for Ilf(x)ll==
limllfn(x)ll. On account of the result in (h) the functions gn==
fn/( 1 + IIfnll) are measurable, and IIgn(x) II 1 for all n and all XEX. By the
dominated convergence theorem the limit function g== f / (1 + IIfll) is then
Bochner integrable over any set E of finite measure; hence gXE is
measurable, which implies (by the definition of measurability) that g
is measurable. It follows that f==g+lIfllg is measurable, once more by
means of the result in (h).
It is evident now how to define the vectorvalued equivalent Boo of
the space Loo. By definition, Boo ==Boo (X, V; #) is the collection of all
vectorvalued measurable f(x) such that IIf(x) " E Loo. Denoting the Loo
norm of IIf(x) II by Poo(llfll), it is obvious that the collection Boo is a
normed linear space with respect to Poo(llfll) as norm. This normed linear
space is a Banach space. Indeed, if fn(x) EBoo for n== 1,2, . . " and
limpoo(llfn-fmll)==O as m, n-+oo, then lim Ilfn(x)-fm(x)II==O for almost
every XEX, so f(x)==lim fn(x) exists for these points XEX. The function

224 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,  31
f(x) is measurable since all fn(x) are measurable, and it follows easily
that Ilf(x)IIELoo and lim Poo(llfn-fll) ==0 as n-+oo.
It remains to introduce the vectorvalued equivalent Bp of the space
Lp for 1 <P<oo.
THEOREM 6. Let 1 <P<oo, and let the sequence of vectorvalued step
functions tn(x) satisfy
lim Jlltn-tmIIPdfl==O
x
as m, n-+oo.
Then there exists a unique vectorvalued measurable f(x) such that
lim Ix Iltn-fllpdfl==O. In addition, Ilf(x)IIP is fl-summable, and
J Ilfil P dfl==lim J Iltnll P dfl.
x x
PROOF. The proof is along the same lines as in Theorem 1. We shall
denote the Lp norm (Ix Ig(x) I P dfl)l/p of a numerical function g(x) by
pp(g). The first step is now to select a subsequence tnk(x) such that
=1 pp(lltnk+1-tnkll) <00. Then Iltn1(x)ll+ r=l Iltnk+l(x)-tnk(x)11 is in Lp
(cf. sec. 30, Theorem 1 (c)), and hence finite for almost every XEX. It
follows that the vectorvalued series t n1 (x) + = 1 {t nk + 1 (x) -tnk(x)} con-
verges for these values of x; let f(x) be its sum where it converges (i.e.,
almost everywhere), and f(x) ==0 elsewhere. Since f(x) ==lim tnk(x) almost
everywhere, the function f(x) is measurable. It follows then from
f-t nj == =j (tnk+l -tnJ that lim pp(llf-tnjll) ==0 as i -+00, and hence
pp(llf-tnll)Pp(llf-tnjll) +pp(lltn.1-tnll)-+O
as n-+oo.
Finally, since
Ipp(lltnll) -pp(lltmll) IPp(lltn -tmll),
the sequence pp(lltnll) converges to a finite limit, and the inequality
Ipp(llfll) -pp(lltnll) lpp(l!f-tnll)
shows that I IlfllPdfl==lim I IltnllPdfl.
The uniqueness of f(x) is derived in the same manner as in Theorem 1.
The collection of all vectorvalued f(x) such that lim I Iltn-fllpdfl==O
holds for some sequence of step functions tn(x), will be denoted by

Ch. 6, 9 31 ]
BOCHNER INTEGRAL
225
Bp=Bp(X, V; fl). Evidently, B p is a normed linear space with respect
to pp(llfll)==(!xllf(x)IIPdfl)l/p as norm. The proof that Bp is a Banach
space with respect to this norm, is similar to the proof for p== 1 1n
Theorem 3.
As shown above, any fEBp (Ip<oo) is measurable, and IlfliELp.
The converse holds as well. If the vectorvalued function f(x) is measur-
able, and if Ilf(x)IIELp, then f(X)EBp. For p=1 this was proved in the
remark (e) above. The proof for 1 <P<oo is similar. We show the
existence of a sequence of step functions tn(x), converging to f(x)
almost everywhere, and satisfying Iltn(x) 1121If(x) II. Then IItn(x) - f(x) liP
converges to zero almost everywhere, and I!tn(x) - f(x) "P3Pllf(x) liP;
hence, lim! IIt n - fliP dfl==O by the dominated convergence theorem. It
follows that fEBp.
We recall that the conjugate space of the Banach space V (i.e., the
Banach space of all bounded linear functionals on V) is denoted by
V*. If f(x) and g(x) are vectorvalued functions, f(x) on X into V and
g(x) on X into V*, then we shall denote the value of the bounded linear
functional g(X)EV* applied to the element f(X)EV by <g(x), f(x). At
each point XEX the value <g(x), f(x) is a complex number, and
I<g(x), f(x)IlIg(x)II'lIf(x)lI. If f(x) and g(x) are measurable, then h(x)==
<g(x), f(x) is measurable. In order to prove this assertion, it will be
sufficient to show that h(X)XE(X) is measurable for any set E of finite
measure. We have hXE==<gXE, fXE), and there exist sequences of step
functions tn(x) and t(x), on X into V and V* respectively, converging
pointwise almost everywhere to fXE and gXE respectively. Then the
sequence of numerical step functions <t(x), tn(x) converges almost
everywhere to <gXE, fXE), and since all functions <t, t n ) are measur-
able, the same holds tor <gXE, fXE).
The following extension of Holder's inequality is now an immediate
consequence.
THEOREM 7. If lpoo, I/P+ l/q== I, f(x) EBp(X, V; fl) and g(x) E
Bq(X, V*; fl), then
If <g(x), f(x)dfllfllg(x)II'lIf(x)lldflpp(llfll)pq(llgll).
x x

226 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6,  31
Finally, we show that Fubini's theorem holds for Bochner inte-
gra tion.
THEOREM 8. Let #1 and #2 be measures in the non-empty point sets Xl
and X 2 respectively, and let v be the product measure #lX#2 in X 3 ==
X 1 XX 2 . The Bochner integrals of f(X)EB 1 (X 1 , V; #1), f(Y)EB1(X2, V;
#2) and f(x, Y)EB1(X3, V; v) will be denoted by f 1 f, f 2 f and 3f re-
spectively. Then f3f==f1f2f==f2f1f for any f(x, y) EB 1 (X 3 , V; v).
PROOF. The notations JVif==JVi(lIfll) , JV2f==JV2(llfll) and JV3f==JV3(IIfll)
will have the same meaning as in the proof of Fubini's theorem in
sec. 23. For any v-summable set E c X 3 we have
I XEdv== I {J X E d#2}d#1,
X3 Xl X2
and this implies that f 3 t==f 1 f 2 t for any v-step function t(x, y). The
proof that f 3 f==f 1 f 2 f holds for any fEB 1 (X 3 , V; v) is now almost the
same as for a numerical function, if one starts by observing the ex-
istence of a sequence of step functions tn(x, y) satisfying JV3(llf-t n ll)
< 1 /2 n .
Exercises
VECTORVALUED BANACH FUNCTION SPACES
31.1) Let, on the collection M + of all non-negative #-measurable
functions, p(f) be defined such that Op(f)oc> for any fEM+, and
such that p(f) ==0 if and only if f==O almost everywhere, p(f1 +f2)
p(f1) +p(f2), p(af) == ap (f) for any constant aO, and p(fn) i p(f) whenever
fn i f almost everywhere. Let, for any vectorvalued measurable function
f(x) on X into V, the number p(f) be defined by p(f) ==p(llfll), and let
L p be the collection of all such f for which p(f)<oc>. Show that L p is a
Banach space with respect to p(f) as norm. Show also that if fnELp
for n==l, 2, ... and limfn(x)==f(x) almost everywhere, then p(f)
lim inf p(fn). Finally, show that if the norm in L p is absolutely con-
tinuous (cf. Exercise 30.9), then the dominated convergence theorem
holds in L p .

Ch. 6,9 31J
BOCHNER INTEGRAL
227
NORM OF BOUNDED LINEAR FUNCTIONALS
31.2) Let 1 <poo, and 1 /p+ 1 /q== 1. For any fixed g(x) EBq(X, V*;
fl), the integral! <g(x), f(x»dfl exists for all f(X)EBp(X, V; fl), and
II <g, f>dfllpq(llgll)' pp(lIfll)
by Theorem 7. It follows that F(f)==! <g, f>dfl is a bounded linear
functional on Bp(X, V; fl), and IIFIIpq(llgll). Show that IIFII==pq(llgID;
i.e., show that
pq(llglI) ==sup II <g, f>dfll
for all fEBp(X, V; fl) satisfying pp(lIfll) 1.
31.3) For any fixed g(x) EBoo(X, V*; fl), the integral! <g, f> dfl exists
for all f(x) EB 1 (X, V; fl), and
II <g, f> dfllPoo(lIgll)p1(lIfll)
by Theorem 7. It follows that F(f)==! <g, f>dfl is a bounded linear
functional on B 1 (X, V; fl), and IIFIIPoo(lIgll). Let now, in addition, the
measure fl in X have the property that each set of infinite measure
contains a subset of finite positive measure. Show that, in this case,
IIFI[==Poo(lIgll). Show also that if the additional condition is not satis-
fied, then it may happen that IIFII==O and Poo(lIgll) >0.
31.4) Show that the results in the preceding two exercises remain
valid if V and V* are interchanged in the hypotheses as well as in the
conclusions.
A BOUNDED LINEAR TRANSFORMATION
31.5) Let T be a bounded linear transformation on the Banach
space V into the Banach space W, and let lpoo. Show that if
f(X)EBp(X, V;fl), then Tf(x)EBp(X, W;fl) andpp(IITf(x)II)IITllpp(lIf(x)II).
, Hence, T may also be regarded as a bounded linear transformation on
Bp(X, V; fl) into Bp(X, W; fl). Show that the norm of this transfor-
mation is again IITII.
ANOTHER EXTENSION OF HOLDER'S INEQUALITY
31.6) Let lpooand I/P+l/q==l. Show that if the finite numeri-
cal function f(x) satisfies f(x) ELp(X, fl) and the vectorvalued function

228 BANACH SPACE AND HILBERT SPACE; Lp SPACES [Ch. 6, 931
g(x) satisfies g(x)EBq(X, V; #), then j(x)g(x)EB 1 (X, V; #), and the
Bochner integral f(fg) satisfies
I If (jg) IIJ If(x) l'l!g(x) II d#pp(ljl)pq(llgll).
CONTINUITY
31.7) Let # be Lebesgue measure in k-dimensional space R k ; the
Euclidean distance between the points x, x' ERk will be denoted by
d(x, x'). The vectorvalued function g(x), on Rk into the Banach space
V, is continuous at Xo if, given E>O, there exists £5>0 (depending on
E and xo) such that d(x, xo)<£5 implies Ilg(x)-g(xo)ll<e. Show that any
g(x), continuous on Rk and possessing a bounded carrier, is uniformly
continuous. Show also that in this case g(x) is Bochner integrable.
Next, show that if f(X)EBp(Rk, V; #) and e>O are given, where 1
P<cx>, then there exists a continuous g(x), possessing a bounded carrier,
such that pp(IIf-gll) <E. Finally, show that if j(x) EBp(Rk, V; #) for
some p satisfying lp<cx>, then lim pp(IIj(x+h)-j(x)II)==O as h-+O.
THE CONVOLUTION IN TWO VARIANTS
31.8) Let # be Lebesgue measure in Rk, let lpcx> and I/P+
1 /q== 1. Show that if jEBp(Rk, V; #) and gEBq(Rk, V*; #), then
F(x) J <g(t) , j(x-t)dt
Rk
is a numerical continuous function of x on Rk.
31.9) Let again, # be Lebesgue measure in Rk, let 1 pcx> and
1 /p+ 1 /q== 1. Show that if the finite numerical function f(x) is in
Lp(Rk, #) and the vectorvalued function g(x) is in Lq(Rk, V; #), then
the Bochner integral
F(x) ==f{j(x-t)g(t)}
(t is the integration variable)
is a continuous vectorvalued function on Rk.
31.10) Let # be Lebesgue measure in Rk, let 1 pcx> and g(x) E
B1(Rk, V*; #). Show that, for any j(x)EBp(R k , V; #), the numerical
function
F(x) == J <g(t), f(x-t) dt== J <g(x-t), j(t) dt

Ch. 6,  31J
BOCHNER INTEGRAL
229
is in Lp(Rk, #), and
pp(IFI)P1 (1Igll)pp(llfll).
Formulate and prove a similar result if V and V* are interchanged.
31.11) Let # be Lebesgue measure in Rk, 1 p(X), and let the finite
numerical function g(x) be in L1(Rk, #). Show that, for any f(x) E
B p(Rk, V; #), the Bochner integral
F(x) ==f{g(x -t)f(t) }==f{g(t)f(x -t)}
is also in Bp(Rk, V; #), and
p p (IIF II) P1 ( I g I) p p (1Ifll) .
FRACTIONAL INTEGRAL FOR VECTORVALUED FUNCTIONS
31.12) Let # be Lebesgue measure in R1, and let X==[O, bJ, where
O<b<(X). If fEBp(X, V; #) for a value of p satisfying lp(X), then
the fractional Bochner integral faf(x) is defined for ex>O as I/T(ex)
times the Bochner integral over [0, x) of the function (x-t)rx- 1 f(t) ; the
integration is with respect to t. Show that, for all ex>O, the function
frxf(x) is again in Bp(X, V; #); show also that for 1 <p(X) the function
faf(x) is continuous on X for ex> I/P, and for p== 1 the function faf(x)
is continuous on X for exl. Finally, show that, for ex>O, fJ>O, we have
fpfaf(x) ==fa+pf(x).

CHAPTER 7
THE RADON-NIKODYM THEOREM
In the Radon-Nikodym theorem two integrals Jf and ff are compared. The
theorem in its simplest form states, roughly, that if J and f generate a-finite
measures, and every J-null set is a f-null set, then f may be expressed in
terms of J; more precisely, there exists an J-measurable function fo such that
f (I) = J (ffo) for all f -summable f. This is proved in sec. 32, and the exercises
contain supplementary results such as the Lebesgue decomposition of an inte-
gral f with respect to an integral J and Kakutani's theorem on product inte-
grals. The remaining sections in this chapter are devoted to extending the
Radon-Nikodym theorem to the non-a-finite case, culminating in Segal's theo-
rem that localizability of the measure generated by J is a necessary and suf-
ficien t condition for the theorem to hold.
32. The Radon-Nikodym Theorem
In the preceding chapters we have concentrated our attention upon
the various aspects of the integral ff over the point set X, but we have
not yet compared two different integrals ff and ff, taken over the
same point set X. This is what we shall do now. Let, therefore, ff and
ff be Daniell integrals over X, extensions of elementary integrals with
the same initial domain of definition L. The integral ff is called f-
absolutely continuous if every f-null set is also a f-null set, in other
words, if for every set E c X, satisfying f XE==O, the number f XE ex-
ists, and fXE==O. It follows that every f-null function is a f-null
function. Indeed, if flfl ==0, then E =={x: If(x) I >O} is an f -null set, i.e.,
fXE==O. But then fXE==O by hypothesis, so flfl==O (since Ifloo'xE
and f(OO'XE)==O).
LEMMA cx. If ff is f-absolutely continuous, then any (real or com-
Plex) f -measurable function is also f -measurable.
PROOF. Assume first that f(x) is non-negative and f-summable.

Ch. 7, 932J
THE RADON-NIKODYM THEOREM
231
Then ff==inf r fh n for all  hn(x)f(x), hnEL+ (cf. sec. 22). Hence,
given e>O, there exists (on account of ff<oc» a function he(x)==
 hn(x)f(x) such that fhe==  fhn<ff+e, and the function he(x),
as a sum of functions hnEL+, is f-measurable as well as f-measur-
able. Let en! 0, and consider the corresponding sequence of functions
hen(x); we may assume that (for each XEX) the sequence hen(x) is non-
increasing (replace, if necessary, h e2 by min (h el , h e2 ), and so on; the f-
measurability and f-measurability are preserved by this procedure).
Then hen!f11 on X, where the function 11 is f-measurable as well as
f-measurable, and fh en !f/1 by the theorem on dominated conver-
gence. But we have also fhen!ff (since fh en <ff+en), so f(f1-f) ==0.
It follows that f(f1-f) ==0, so that, in view of the f-measurability of
f1, the function I is seen to be also f -measurable.
Next, let f(x) be non-negative and f-measurable. Then min (f, h) is
non-negative and f-summable for any h(x) EL +. Hence, by what has
already been proved, min (I, h) is f-measurable for any h(x) EL +, and
this shows that I is f-measurable. It follows now immediately that
any (real or complex) f-measurable function is f-measurable.
LEMMA fJ. Let f f be f -absolutely continuous, and let the elementary
integral .%f be delined on L by .%f==fl+ ff. ApPlying the extension
procedure, the Daniell integral .%f is obtained. Then the (real or comPlex)
function f(x) on X is .%-measurable if and only if f(x) is f-measurable.
Furthermore, .%I==ff+ ff holds for any f-measurable f(x)O, and also
for any (real or comPlex) .% -summable function I(x).
PROOF. We recall sec. 9, Theorem 3, asserting that if #1 and #2 are
measures, initially defined on the same semi-ring T, then #==#1 +#2 is
a measure on T, and the corresponding exterior measures satisfy #*==
#i +#2. Furthermore, if the set E is #l-measurable as well as #2-measur-
able, then E is #-measurable, while, if in addition #l(A) and #2(A) are
finite for all sets A ET, the #-measurability of E implies, conversely,
the #l-measurability and #2-measurability of E. If, for any f(x) on X,
we use the notations f*l/l, f*1/1 and .%*Ifl for the corresponding ex-
terior measures of the ordinate set of If(x) I, and we let f, f and .%
correspond to #1, #2 and # in the theorem referred to, it follows now
immediately that .%*1/1 ==f*lfl + f*1/1 for all I. Furthermore, since f-
measurability implies f -measurability by the preceding lemma, and

232
THE RADON-NIKODYM THEOREM
[Ch. 7,  32
since ff and ff are finite for all fEL+, the second part of the theorem
referred to shows that any f?:-O is g-measurable if and only if f is f-
measurable. In addition, gf==ff+ ff for any such f?:-O. .It follows
easily that g-measurability and f-measurability are equivalent no-
tions for any real or complex f(x) , and gf==ff+ ff holds for any:/t-
summable f(x) (this is first proved for real f; note that g -summability
implies f-summability and f-summability, since ffgf and ff
gf for any g-summable f?:-O).
From this point on, we shall assume that ff is f-absolutely con-
tinuous, and that ff is a Stieltjes-Lebesgue integral. The second as-
sumption, by the representation theorem in sec. 17, Theorem 7, is
equivalent to the assumption that the characteristic function xx(x) of
the entire set X is f-measurable. Then, by Lemma cx, xx(x) is also f-
measurable, which shows, once more by the representation theorem,
that ff is also a Stieltjes-Lebesgue integral. For lp<oc>, the Banach
space of all complex f(x) , for which If(x)IP is f-summable over X, will
be denoted by Lp(f). The space Lp(f) is defined similarly. The corre-
sponding real Banach spaces will be denoted by L)(f) and L)(f).
The next lemma may be regarded as the basic core of the Radon-
Nikodym theorem, since the different existing versions of the theorem
are more or less immediate extensions of the assertion in the lemma.
LEMMA y. Let A be an f-summable (and hence f-measurable) subset
of X, and let A be also f-summable, i.e., we assume that fXA and fXA
are finite. Then there exists an f-summable function fo(x)?:-O, vanishing
outside A, such that any f(x) vanishing outside A is f -summable if and
only if f(x)fo(x) is f-summable. Furthermore, ff==f(ffo) for any such f.
Finally, the function fo(x) is f-uniquely determined by these conditions
(i.e., any other function, vanishing outside A and satisfying the same con-
ditions as fo, is f -almost equal to fo).
PROOF. The proof is divided into several parts.
(a) Let the integral gf be defined as in Lemma (J, that is to say,
:/tf is the extension of the elementary integral :/tf==ff+ ff on L. It
follows then immediately that :/t XA ==f XA + f XA; hence, the set A is
:/t -summable. Evidently, the collection of all real f(x), vanishing out-
side A and satisfying :/t(f2) <OC>, is a real Hilbert space H with respect

Ch. 7, 932J
THE RADON-NIKODYM THEOREM
233
to the norD;111/112=={:/t(f 2 )}!; this norm js derived from the inner product
(f, g)==.%(fg). Note that H is the closed linear subspace of Lr)(:/t),
consisting of all fELr)(.%) vanishing outside A ; note also that XA EH.
Hence, if I(x) is an arbitrary element of H, it follows by Schwarz's
inequality that l==fxAEL 1 (:/t), i.e., f is then .%-summable, and
I f IIflfl .%Ifl IlxA1I211f112.
This shows that f I is a bounded linear functional on the Hilbert
space H. Then, by sec. 29, Theorem 5, there exists a .%-unique real
function g(x) EH such that f (f) == (f, g) ==.%(fg) holds for every I(x) EH.
Evidently, the thus obtained function g(x) EH is .%-almost everywhere
non-negative (substitute in ff==:/t(fg) for I the characteristic function
of the set where g<O), so g(x) is also f-almost everywhere non-negative.
We show next that ff==:/t(fg) holds for any .%-measurable fO
which vanishes outside A. For this purpose, let fn==min(f, n) for n==
1, 2, . . . . Then all In are elements of the Hilbert space H, so it follows
from In j I, Ingj fg and fln==.%(fng) that ff==.%(lg). In particular,
f(fgP)==.%(fgP+1) for any .%-measurable IO and p==O, 1,2, . .. .
Let now l(x)O be an element of the Hilbert space H. The expansion
00> ff==.%(fg) ==f(/g) + f(fg) ==f(fg) +.%(fg2)
n
==f(/g)+f(/ g 2)+f(/g2)==.. .==f(1  gp)+f(fgn)
p=l
shows that the set on which g(x) 1 is an f-null set (set I equal to the
characteristic function of this set), and therefore also a f-null set.
Then Ign! 0 holds f -almost everywhere as well as f -almost every-
where, hence f (fgn) ! 0 on account of f 1< 00. Writing 10== r gP on
{x: g(x) < I} and 10==0 on {x: g(x)I}, we have therefore I =1 gPjlfo
almost everywhere, so f(f  gp) j f(ffo) , and this shows that ff==
.f(flo) for all non-negative IEH. It follows immediately that fl==
f(lfo) holds for all IEH. Note that, by its definition, 10 vanishes out-
side A. Furthermore, since XAEH and fXA<OO, we obtain ffo==
f(XAfo)==fxA<OO, i.e., 10 is f-summable.
(b) We have the task of supplying further details, and we investi-
gate first the behaviour of the obtained function 10 with respect to the
integral ff. It will be shown that the intersection G of A and the set

234
THE RADON-NIKODYl'fI THEOREM
[Ch. 7, 9 32
{x: fo(x)==O} is a f-null set, and that conversely, if EcA is a f-null
set, then fo==O holds f-almost everywhere on E. The first part is easy:
since fo is g-measurable, the set G is g-measurable, so g(X)==
:/tXGgXA<(X), i.e., XG is an element of the Hilbert space H. It follows
then by the final result in part (a) that f XG==f(xGfo) ==0.
Let, conversely, EcA be a f-null set, i.e., fXE==O (it does not
follow that E is an f-null set, it is possible that E is not even f-
measurable). Starting from the formula fXE==inf  fh n for all
r hn(X)XE(X), hnEL+, the same argument as in Lemma ex yields a
function f*(x), f-measurable as well as f-measurable, such that f*(x)
XE(X) and f(f*-XE) ==0, so ff*==O. We may assume that f*==O
outside A, and Of* 1 on A (if necessary, we replace f* by min (f*, XA)),
hence f* is an element of the Hilbert space H. It follows then that 0==
ff*==f(f*fo) , so that also f(XEfo) ==0 on account of f*foXEfoO. Hence
10==0 holds f-almost everywhere on E.
(c) In part (a) we have obtained the result that f f==f(ffo) holds
for all f in the Hilbert space H. It does not follow that ff==f(ffo)
holds for all fEL, since in general the functions fEL do not vanish
outside of A. Setting g==fXA for a given fEL, it is not even immedi-
ately evident that f g==f(gfo) holds for this g, since g is not neces-
sarily an element of the Hilbert space H Of L consists exclusively of
bounded functions, then of course g==fXAEH for all fEL; this, how-
ever, is not necessarily the case). We shall prove that, nevertheless,
f g==f(gfo) holds for all g==fXA, fEL. For this purpose, let first gO
be f-measurable and vanish outside A. Then gn==min(g, n)EH for
n== 1, 2, . . ., so f gn==f(gnfo) , and since gn i g, gnfo i gfo, "ve obtain
fg==f(gfo). In particular, if g==fXA where fEL+, we have f(fxA)==
J(fXAfo)==f(ffo); in addition, Of(fxA)<(X) on account of f(fxA)
ff<(X). Hence, by addition, f(fxA)==f(ffo) for all fEL. Note, for the
following, that in particular f(lflxA)==f(lflfo) for all fEL (since fEL
implies If I EL).
(d) We shall prove in this part that any f-summable f, vanishing
outside A, satisfies ffoEL1(f) and ff==f(ffo). Assuming first that
fELir)(f), there exists a sequence fnEL such that lim flf-fnl==O.
Since If-fnXAIlf-fnl at all XEX, this implies lim flf-fnXAI==O, so
fnXA is a f-fundamental sequence. It follows then from flfmXA-fnXAI
==f(lfm-fnlfo) that fnfo is an f-fundamental sequence, and passing

Ch. 7,  32J
THE RADON-NIKODYM THEOREM
235
if necessary to a subsequence, we may assume that InXA converges
pointwise to its limit IELir)(f) f-almost everywhere on A, and that
simultaneously Inlo converges pointwise to its limit, say gELir)(f),
J-almost everywhere on A. Since, therefore, lim InXA==I, except on a
f-null set E, and since by part (b) 10==0 holds f-almost everywhere
on E, we have lim Inlo==llo holding f-almost everywhere on A as
well as lim Inlo==g holding f-almost everywhere on A, so g and 110
are f-almost equal. It follows that fl/-InXAI and filio-iniol tend
to zero, hence
fl==lim f(lnxA)==lim f(lnlo)==f(llo).
The same equality fl==f(llo) is then true for any complex IEL1(f)
vanishing outside A. It is now also easy to verify that f l==f(llo)
holds for any non-negative f-measurable I which vanishes outside A.
(e) We have to show next that if I(x) vanishes outside A and IloE
L 1 (f), then IEL1(f). Let first 110);0 and IloELir)(f). Since the set
.G==An{x: 10(x)==0} is a f-null set by part (b), it is immaterial for
J(llo) as well as for the number fl whose existence we still have to
prove if we assume that 1==0 on G. It follows then from 1==0 on G and
the f-measurability of 110 that I itself is f-measurable, and therefore
f-measurable. Furthermore, since 1==0 on G and IloO on A, we have
1);0 on A. Hence, since I is now non-negative as well as f-measurable,
we may conclude by the final result in part (d) that fl==f(llo) , so
that IELir)(f) on account of f(llo)<oc>. The extension to real IloE
Lir)(f), and then to complex IloEL 1 (f), is evident.
(f) It remains to prove that 10 is f-uniquely determined. Assume,
therefore, that for some 10, vanishing outside A, the equality fl==
J(llo)==f(llo) holds for all IEL1(f) which vanish outside A. If EcA
is an arbitrary f-measurable set, then E is f-measurable, and hence
f-summable, so f{XE(/o-/o)}==O. Obviously, 10 is f-measurable (set
l==xA in fl==f(llo)); hence, if lo==go+iho (go and ho real), we may
first choose for E the sets {x: 10-go>0} and {x: 10-go<0} respectively.
This shows already that 10 and go are f-almost equal. Similarly, ho===O
holds f-almost everywhere. Hence, 10==/0 holds f-almost everywhere.
There is enough material available now to prove the first version of
the Radon-Nikodym theorem.

236
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 32
THEOREM 1 (RADON-NIKODYM THEOREM; INTEGRAL VERSION). Let
ff and ff be Stieltfes-Lebesgue integrals over X, extensions of elementary
integrals on the same initial domain of definition L. Let ff be f-abso-
lutely continuous, and f xx< 00 (the last assumption says, therefore, that
the measure induced in X by the integral f is a finite measure). Then
there exists an f-unique -summable fo(x)?:::O on X such that fEL 1 (f)
if and only if ffo EL 1(f), and ff==(ffo) for any such f. In addition,
the measure induced in X by the integral f is a-finite (i.e., there exists a
decomposition X ==  X n, where the sets X n are disfoint and f XXn < 00
for all n).
PROOF. Let # be the measure induced in X by the integral , and
observe first that any l(x) EL + is -summable, so that the set {x: l(x) >O}
is of a-finite #-measure. Hence, if h(x)==r hn(x), hnEL+, the set
{x: h(x) >O} is also of a-finite #-measure. Indeed, the partial sums
In(x)==f=l hi(x) are functions in L+, and
00
{x: h(x) >O}==  {x: In(x) >O}.
n=l
Since fxx<oo, we have fxx==inf  fh n for all r hn(x)XX(x),
hnEL+, so there exists a function h(x)==r hn(x), hnEL+, satisfying
h(x)xx(x). This implies that X=={x: h(x»O}, so X is of a-finite #-
measure by the above remark.
Let, then, X == r X n , where the sets X n are disjoint and XXn <00
for n == 1, 2, . . . . Note that each X n is f -measurable, and f XXn 
f xx<oo for all n. By Lemma y there exists, for each n== 1, 2, . . ., a
non-negative function fon, vanishing outside X n , such that ff==
f(ffon) for f vanishing outside Xn. Let fo(X)===l fon(x) on X. Each
f on X may be expressed in the form f== r f n, where f n== fXxn; hence,
if OfELir)(f) on X, then OfnELir)(f) on X n , so ff== ffn==
 f(fnfo)==(ffo). The extension to complex fEL 1 (f) is evident. Con-
versely, if OffoELir)(f) on X, then OfnfoELir)(f) on X n , so f(ffo)==
 f(fnfo)== ffn==ff. The extension to complex ffoEL 1 (f) is again
evident. Finally, it follows from f xx<oo that ffo<oo, so fo is f-
summable over X. The -uniqueness of fo is proved exactly as in the
last part of the proof of Lemma y.

Ch. 7,  32J
THE RADON-NIKODYM THEOREM
237
The Radon-Nikodym theorem is often stated not in terms of the
Stieltjes-Lebesgue integrals ff and ff, but in terms of the measures
# and v induced in X by ff and ff respectively. The formulation is
then chosen such that the integrals f and f themselves have disap-
peared altogether from the hypotheses. In order to state this measure
version of the theorem, let # and v be measures in X, extensions of
measures initially defined on the same ring A of subsets of X. The
measure v is then called #-absolutely continuous if #(E) ==0 implies
v(E) ==0, i.e., if, after extension of the measures, any #-measurable set
of #-measure zero is v-measurable and of v-measure zero.
THEOREM 2 (RADON-NIKODYM THEOREM; MEASURE VERSION). Let
# and v be measures in X, extensions of measures initially defined on the
same ring A of subsets of X, such that #(E) and v(E) are finite for all
E EA. Let v be #-absolutely continuous, and v(X) finite. Then there exists
a #-unique #-summable fo(x)O on X such that f is v-summable if and
only if ffo is #-summable, and Ixfdv==lxffod# for any such f. In par-
ticular, v(E) == IEfod# for all #-measurable sets E eX. Furthermore, the
measure # is a-finite (i.e., X is the union of an at most countable number
of sets of finite #-measure).
PROOF. Let L be the collection of all step functions f(x) ==
=1 cnXEn(x), with En EA for n==l, "', p, and let ff==l cn#(En)
and ff==l cnv(En) on L. Note that, by the hypotheses, ff and ff
are finite on L. By sec. 17, Theorem 9 the measures induced in X by
the extended integrals ff and ff are the same as the measures ob-
tained by immediately applying the extension procedure for measures
to (X, A, #) and (X, A, v) respectively, so the #-absolute continuity of
v implies the -absolute continuity of f. Hence, all hypotheses for
applying the integral version of the Radon-Nikodym theorem are satis-
fied. It follows that there exists a #-unique #-summable foO such that
f is v-summable (that is, f-summable) if and only if ffo is #-summable
(that is, f-summable), and ff==f(ffo) for any such f, i.e., Ixfdv==
Ixffod# for any such f. Furthermore, the measure induced by ff is a-
finite, i.e., # is a-finite.
We conclude the discussion with some remarks.
(1) We have derived the measure version of the theorem from the

238
THE RADON-NIKODYM THEOREM
lCh. 7,9 32
integral version, but it is not difficult to see that the two versions are
in fact equivalent. Indeed, assuming the measure version to hold, and
given the integrals f I and f I satisfying the hypotheses of the integral
version, we let # and v be the measures induced in X by f I and f I
respectively. Let A be the ring of all #-measurable sets of finite #-
measure. It follows from the representation theorem that by applying
the extension procedure for integrals to the elementary step function
integral with respect to # on A the given integral fl is reobtained.
Similarly, by extending the elementary step function integral with re-
spect to v on the ring Al of all sets of finite v-measure, we obtain fl.
By the f-absolute continuity of f any set EEA is v-measurable, so
AcA 1 (the finiteness of v(E) for any EEA follows from fxx<(X)). It
may happen, however, that Al is properly larger than A, and for that
reason it will simplify the situation if we prove first that f I is already
obtained by extending the elementary step function integral with re-
spect to v on A. Evidently, this elementary integral is the restriction
of f I to the collection of the #-step functions. Since any IE L + is f-
summable, there exists a sequence In of #-step functions such that
O/ni/; hence fin if I, and this shows that fl on L, and hence the
entire extended integral fl, is reobtained by starting from the #-step
functions. It follows then by sec. 17, Theorem 9 that the extended
measure v may be obtained by applying the extension procedure for
measures to (X, A, v). In other words, the measures # and v in X may
both be regarded as extensions of their own restrictions to the sets of
A. All hypotheses of the measure version are satisfied therefore for #
and v, and the desired integral version for f and f follows.
(2) The condition (in the integral version) that fxx<(X) may be
relaxed somewhat. If X == r X n, where the sets X n are disjoint and
f-measurable (and hence f-measurable), and if fXxn <(X) for each n,
the theorem may be applied for each X n separately, and by addition
we obtain an f-measurable 100 on X such that IEL1(f) if and only
if IloEL 1 (f), and fl==f(llo) for any such I. Furthermore, the measure
# induced by fl is a-finite in each X n , and hence a-finite in the whole
of X. In the present case, however, it is no longer necessary that 10 is
f-summable over X, although 10 is f-summable over each Xn.
{3) In some practical cases one meets the situation that it is desira-
ble to determine explicitly the function 10 in fl==f(llo) , and there

Ch. 7, 932J
THE RADON-NIKODYM THEOREM
239
sometimes appears in the course of the argument a function 1'0 which
is conjectured to be the required 10. It is not necessary then to verify
that fl==f(II'O) holds for each IEL 1 (f), but it is already sufficient to
prove this for each I==XE, where E e X is f-measurable, since f(XEI'O)
==fXEf(XElo), holding for all f-measurable EeX, already implies
that 1'0 is f-almost equal to the function 10 whose existence is guaran-
teed by the Radon-Nikodym theorem. In the case that fl==Jxld#,
and the measure # is the extension of the measure # initially defined
on the semi-ring r, it is even sufficient to show that fXEz:=JEI'Odfl
holds for all EEr (cf. sec. 18, Theorem 4).
(4) The first assumption in the measure version is that the measures
# and v may be regarded as extensions of measures initially defined on
the same ring A of subsets of X, such that #(E) and v(E) are finite for
every E EA. The question may be raised to which extent the finiteness
assumption is essential for the truth of the theorem. The assumption
that v(E) is finite for all E EA is subsumed under the further and
stronger assumption that v(X) is finite; the question remains, there-
fore, whether the theorem is still true if it is no longer assumed
that #(E)<oc> for all EEA. We shall investigate first where, in
this case, the proof may possibly go wrong, and then we present
an example where it happens indeed that the proof goes wrong,
simply because the Radon-Nikodym theorem fails to hold for the
example.
Let us assume, therefore, that # and v are measures in X, extensions
of measures initially defined on the same ring A of subsets of X. Let 'V
be #-absolutely continuous and v(X) < oc>. Similarly as in the proof of
Theorem 2, we let L be the collection of all step functions I(x) 
) cnXEn(x) with EnEA, but now with the additional condition that
#(En)<oc> for nl, "', p, and we set fl==) cn#(E n ) on L. Then it
remains true, by sec. 17, Theorem 9, that the measure induced in X
by the extended integral fl is the same as the extended measure #. It
is not certain, however, that if fl is now defined on L by fl==
) cnv(En) , then the measure induced in X by the extended fl is the
extended measure v. The theorem referred to (sec. 17, Theorem 9)
guarantees only that if ff==) cnv(En) on the collection L of all
functions I(x)==) cnXEn(X) with EnEA, then the measure induced in
X by the extended fl is the extended v. On account of the additional

240
THE RADON-NIKODYl\f THEOREl\!
[Ch. 7, 9 32
condition that #(E) <00 for the sets E occurring in the step functions
of L, the collection L may be considerably smaller than L.
This is illustrated by the following example. Let X==[O, IJ, and A
the a-field of all Lebesgue measurable subsets of X. We let # be the
discrete measure on A and v the Lebesgue measure on A. Extension
yields for the extended # the discrete measure in X (each subset of X
is #-measurable), and the given v is already extended as much as possi-
ble. Obviously, since #(E) ==0 only for the empty set E == 0, the measure
v is #-absolutely continuous. Furthermore, v(X) == 1. Assuming now
that the Radon-Nikodym theorem holds for this example, there exists
a #-summable foO such that V(E)==/Efod# for all EEA. The set P==
{x: fo(x) >O} is then of a-finite #-measure, i.e., it contains an at most
countable number of points, so v(P) ==0 (since v is Lebesgue measure).
Then l==v(X-P)==/x-pf o d#==O, a contradiction. Hence, the Radon-
Nikodym theorem fails to hold. Note that the collection L of the pre-
ceding paragraph consists here of all (finite-valued) functions differing
from zero only at a finite number of points, whereas L consists of all
Lebesgue measurable step functions. The elementary Lebesgue inte-
gral on L yields upon extension the Lebesgue integral in X, but the
elementary Lebesgue integral ff on L is identically zero, and so re-
mains zero for all f -summable functions when extended.
In the special case that # is Lebesgue measure in R1, and v is a
Stieltjes-Lebesgue measure in R1, the measures # and v may be re-
garded as extensions of measures initially defined on the ring A of all
finite unions of cells. Hence, if v is #-absolutely continuous and v(R 1 )
is finite, there exists a Lebesgue summable foO such that v(E) ==
/Efod# for all Lebesgue measurable sets E; in particular v{(xo,xJ}==
/x lo(t) dt for all cells (xo, x] c R 1 . This implies that g(x) ==g(xo) +
/x fo(t)dt for all xo, xER 1 , where g(x) is one of the functions generating
the Stieltjes-Lebesgue measure v (cf. sec. 10, Example 8). Since v and
g(x) -g(xo) determine each other uniquely, it must be possible to ex-
press in terms of g(x) the conditions that v is #-absolutely continuous
and v(R 1 )<00; this was done already by H. LEBESGUE (1904, [IJ,
p. 129). We shall return to this subject in sec. 41. The Radon-Nikodym
theorem was proved by J. RADON (1913, [lJ) for measures v in finite-
dimensional space Rk with Lebesgue measure #, and by O. NIKODYM

Ch. 7, 932J
THE RADON-NIKODYM THEORElVI
241
(1930, [lJ) for measures in abstract sets. The proof of the theorem, as
we have presented it here (with the conclusion that IEL1(f) if and
only if IloEL 1 (f); older proofs were restricted to the assertion that if
I is f-measurable and f-summable, then IloEL 1 (f)), is essentially
M. H. STONE'S version (1949, [IJ, Note IV); the device to base the
proof on the representation theorem for bounded linear functionals in
Hilbert space is due to J. VON NEUMANN (1940, [4J, p. 127-129). It
should be observed that a theorem which comes very near to the inte-
gral version was already proved by P. J. DANIELL (1923-24, [4J).
Exercises
CONVERSE OF THE RADON-NIKODYM THEOREM
32.1) There js a partial converse to the Radon-Nikodym theorem
as follows. Let the Stieltjes-Lebesgue integral fl, the extension of an
elementary integral on the function collection L, induce a a-finite
measure in X, and let the f-summable function 10(x)0 have the
property that f(llo) is finite for all IEL. This will happen, e.g., if
either 10 is bounded f-almost everywhere or if all functions in L are
bounded. Show that f l==f(llo) is an elementary integral on L. Show
also that (after extension) the equality fl==f(llo) holds for all f-
summable I. Next, prove that f is f-absolutely continuous, and
f xx< 00. Hence, since f I and f I are extensions of elementary inte-
grals on L, the Radon-Nikodym theorem may be applied; there exists
an f-summable to(x)O satisfying fl=f(llo) for all IEL1(f). Show
that 10 is f-almost equal to the given function 10.
32.2) Show that if in th preceding exercise we do not assume that
10 is f-summable over X, but we assume instead the existence of a
decomposition X ==  r X n into disj oint f -measurable sets X n such
that 10 is f-summable over each X n and f(IXxnl o ) is finite for all/EL
and all n, then the same conclusions (except fxx<oo) hold for the
 )I
integral fl, the extension of the elementary integral ftf(tlo) on
A
L, where L is now the collection of all finite linear combinations t of
functions IXxn with IEL.
32.3) Let fl be a Stieltjes-Lebesgue integral inducing a a-finite
measure in X, and let 10(x)0 be f-summable over X. Let Land L

242
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 32
be function collections both of which may be regarded as initial do-
A
mains of definition of fl, and such that f(llo) and f(tlo) are finite for
A A A
all IEL and tEL respectively. Hence, fl==f(llo) and fl==f(tlo) are
elementary integrals on Land L respectively. Show that the extended
A
integrals f I and f I are identical.
32.4) Let fl be a Stieltjes-Lebesgue integral inducing the a-finite
measure # in X, and let 10(x)0 be f-summable over X. Then fl==
f(llo) is an elementary integral on the collection Ls of all #-step
functions, and the results in Exercises 32. 1 and 32.3 hold. Show that
if 10 is not bounded f-almost everywhere, then it is not always possi-
ble to select for the initial domain of definition of fl every function
collection L which may be regarded as the initial domain of definition
of fl.
LEBESGUE DECOMPOSITION
32.5) The Stieltjes-Lebesgue integrals fl and fl, extensions of ele-
mentary integrals on the same initial domain of definition L, are called
orthogonal (or singular with respect to each other) whenever there exists
a decomposition X ==F +G of X into disjoint sets F and G such that
f XG==O and f Xp==O (all the power of f is concentrated, therefore, in
the set F, and all the power of f in the set G). We shall write f 1..f
whenever f and f are orthogonal.
Let the Stieltjes-Lebesgue integrals fl and fl, with L as initial
domain of definition, induce a-finite measures in X, and let .%1 be the
extension of the elementary integral .%1 == fl+ fl on L. Show the
existence of .% -n1easurable functions 11 and 12 such that 0/1  1, O
121, fl==.%(111) for all IEL 1 (f) and fl==.%(112) for aII/EL1(f).
LetF=={x:/1(x»0}and G==X-F=={x: 11(X)==0}, and set 10==xp/2 and
10 == XGI2' Let f11 and f21 be the. extensions of the elementary integrals
fl/:=:::.%(llo) and f2/==.%(llo) on L. Show that fl==f1/+f21 for all
IEL 1 (f), and show finally that f11 is f-absolutely continuous and
f21..f. Note that it follows then that f11..f2. The decomposition f ==
f1 + f2 is known as the Lebesgue decomposition of f with respect to f.
32.6) Show that the decomposition of f into an f-absolutely con-
tinuous integral f1 and an f -orthogonal integral f2 in the preceding
. . .
exerCIse IS unIque.

Ch. 7, 932J
THE RADON-NIKODYM THEOREM
243
PARTIAL ORDERING WITH RESPECT TO ABSOLUTE CONTINUITY
32.7) We consider the collection of all integrals, having the same
initial domain of definition L, and inducing a-finite measures in X.
Two such integrals ff and ff, such that f is f-absolutely continuous
and f is f-absolutely continuous, are called equivalent for the present
purpose, and we denote this by f - f. Show that f - f, f - g im-
plies f - g, so that, consequently, the collection of all integrals may
be divided into classes of mutually equivalent integrals. Denoting
these classes by [f], we shall write [f] -<[f] if there exist integrals
f E [f] and f E [f] such that f is f -absolutely continuous. Show that
if [f] -<[f], and f 1 and f1 are arbitrary integrals from [f] and [fJ
respectively, then f 1 is fl-absolutely continuous. Show that, conse-
quently, the relation [f] -<[f] defines a partial ordering in the col-
lection of all classes [f].
32.8) We use the same notations as in the preceding exercise. Let
[f] -<[g], and let f 1 and f 2 be two integrals from [fJ. Hence, if
gE[g], there exist non-negative g-measurable functions gl and g2
such that f 1 f==g(fg1) and f 2 f==g(fg2). Show that the sets G 1 ==
{x: gl{X»O} and G 2 =={x: g2(X»0} are g-almost equal. Next, if g' is
another integral from [g], and f 1 f==g'{fg'), show that G 1 and G'==
{x: g'(x) >O} are g-almost equal. Note that sets are g-almost equal
if and only if they are g'-almost equal. Finally, show that the integral
ff==g(fxG) is in [fJ for G==G 1 (or G==G 2 , or G==G').
32.9) We use the same notations as in the preceding exercises. Let
{T} be a non-empty index collection (finite, countable or uncountable),
and let the collection [fJT be bounded with respect to the partial
ordering, i.e., there exists a fixed [g] such that [fJT-<[gJ for all T.
Show that the collection [fJT has a least upper bound.
32.10) Show that any non-empty collection [fJT' not necessarily
having an upper bound, has a greatest lower bound.
AN IMBEDDING IN L 2 SPACE
32.11) Let the Stieltjes-Lebesgue integrals f, f, g induce in X
the a-finite measures flI, flJ and flK respectively. If f is f-absolutely
continuous (notation: f -<f), the Radon-Nikodym theorem shows the
existence of a non-negative f-measurable fo(x) such that f f==f(ffo) ,

244
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 32
i.e., ! /dflJ==! //odflI' It is sometimes convenient to denote /o(x) by
(dfldf)(x), as if /o(x) were a differential quotient; hence
f fdftJ= f f  dftl.
Show that if f -<f -<.%, then dfld.:Yt and (dfldf) (dfld.%) are .%-
almost equal.
32.12) We consider the collection of all Stieltjes-Lebesgue integrals,
having the same initial domain of definition, and inducing finite
measures in X. Let f, f, .% be three such integrals such that f-<.%
and f -<.%. Then the functions (dfld.%)t and (dfld.%)t are evident-
ly elements of the Hilbert space L 2 (.%), and their inner product, de-
noted by p(f, f), is
p(f, f)== j{(dfld.%) (dfld.%)}tdflK'
Show that if flJ is another integral in the collection such that f -<flJ,
f -<flJ, then
j {(df IdflJ) (d f IdflJ)}! dflP== j {(df Id.%) (d f Id.%)}i dflK' (1)
The number p(f, f) is independent, therefore, of the choice of the
majorant .%. Show that, given the integrals f and f in the collection,
there exists an integral .% (in the collection) such that f -<.% and
f -<.%. Hence, the number p(f, f) is defined for any pair f, f in
the collection. Finally, show that p(f, f){fxx' fxx}t, and p(f, f)
==0 if and only if the integrals f and f are orthogonal.
BOREL SETS
32.13) Let fl be a Stieltjes-Lebesgue integral over X, having the
function collection L as initial domain of definition, and let A be the
smallest a-field of subsets of X such that A contains all sets of the
form E =={x: /(x) >a}, where / varies through L + and a varies through
the positive numbers. The existence of A follows by Exercise 4.1. We
shall say that the sets of A are the Borel sets with respect to L. Note
that A depends only on L, and not on the integral fl. Show that any
Borel set is f-measurable. If any finite linear combination of f-
summable characteristic functions of Borel sets is called a Borel step

Ch. 7, 932J
THE RADON-NIKODYM THEOREM
245
function (with respect to f), show then that ff is an elementary inte-
gral on the Borel step functions, such that after applying the extension
procedure the given integral ff is reobtained. This extends the result
in the representation theorem of sec. 17, Theorem 7. Derive from this
extended result that any f-summable subset of X is included in and
f-almost equal to a Borel set. Finally, assume that the measure fl
induced in X by fl is a-finite, and fo(x)O is f-summable over X.
Show that ff==f(lfo) is an elementary integral on the collection ot
the Borel step functions (with respect to f), and the extended inte-
gral f I is identical with the extended integral f I in Exercise 32.4, where
we started from the elementary integral on the larger collection con-
sisting of all fl-step functions.
PRODUCT INTEGRAL
32.14) Let Xi (i== 1, . . " n) be non-empty point sets and, for each i,
let fif be a Stieltjes-Lebesgue integral over Xi, having the function
collection L(i) as initial domain of definition and inducing the finite
measure fli in Xi. According to the method of Exercise 23.10the product
integral f f==f 1 . . . f nf over the Cartesian product XIX . . . X X n is first
defined on the collection of all step functions of the form f(x) ==
k=l CkXCk with C k ==A 1k X ... XAnk such that Aik is fli-measurable,
and for this f(x) the definition is then that ff===l Ck IIi=l fli(A ik ).
By applying the extension procedure the integral ff is obtained. Show
that the same product integral is obtained if the sets Aik C Xi are re-
stricted to the Borel sets (with respect to L(i») in Xi.
32.15) Let Xi (i== 1, . . ., n) be non-empty point sets and, for each
i, let fil, AI, . .. be Stieltjes-Lebesgue integrals over Xi, having the
same function collection L(i) as initial domain of definition, and in-
ducing finite measures in Xi, such that the measures of the whole set
Xi are strictly positive. Note that if f i and A are such integrals over
Xi for i==l, "', n, and f==f 1 .. .n, f==f1.' .fn are the product
in tegrals over XIX . . . X X n, then f f and f f may be regarded as ex-
tensions of elementary integrals with the same initial domain of defi-
nition. Show now that p(, f)== IIf=l p(fi' A), where p(f, f) has
the same meaning as in Exercise 32.12. Observe, for the proof, that if
 (i == 1, "', n) is an auxiliary integral satisfying fi--<, A--<,

246
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 32
and we set g==gl... g n, then f«g, f«g, and df/dg==
Ilf=l dfi/dXi. Show also that f l..f if and only if f i l..A for at least
one value of i. Show, finally, that f «f if and only if A«f i for i==
1, ..., n, and df/df==Ilf=l dA/df i in this case.
,
INFINITE PRODUCT INTEGRAL; KAKUTA.NI S THEOREM
32.16) Let Xi (i== 1, 2, . . .) be non-empty point sets and, for each i,
let fil and AI be Stieltjes-Lebesgue integrals over Xi, having the same
function collection L(i) as initial domain of definition, and satisfying
fiXXt==AXXt== 1. Show that p(fi, f i ) ==p(A, A) == 1 and Op(fi, A)
 1. Let f I and f I be the product integrals over X ro==X 1 xX 2 X . . .
of the integrals f i (i== 1, 2, . . .) and A (i== 1, 2, . . .) respectively (cf.
Exercises 23.12-23.14). Note, as in the preceding exercise, that f I and
fl may be regarded as extensions of elementary integrals with the
same initial domain of definition.
Let now A «f i for all i. Show that f «f if II r p(f i , A) >0, and
f l..f if Ilr p(fi, A) ==0. This result is essentially due to s. KAKU-
TANI (1948, [2J). Observe for the proof of the first assertion that !2 n ==
fl. · . fn «fl. · · fn==&P n for all n, and the functions1pn(x) == (d!2 n /d&P n)t,
regarded as functions of XEX ro , form a fundamental sequence in the
Hilbert space L 2 (f) ; the limit function 1p(x) will be (d f /df) ( For the
proof of the second assertion, show that if Il p(fi' A)<e, and B==
{x: II dA/df i > I}, then fXB<e and fXXw-B<e. Note that p(f, f)
== Ilr p(fi, A) holds in either case.
EXTENSION OF KAKUTANI'S THEOREM
32.17) Let f and f be Stieltjes-Lebesgue integrals over X, having
the same function collection as initial domain of definition, and satis-
fying fxx==fxx==l. Let O<cx<l. Set g==(l-cx)f+cxf, and show
that gxx== 1, f «g, f «g, and (df/dg)(x)(I-cx)-l for g-
almost every XEX. Show also that p(f, f)(I-cx)-!p(f, g) and
p(f, g)I-cx.
32.18) Let the same hypotheses as in Exercise 32.16 be satisfied,
except that we do not assume that A«f i holds for all i. Show that,
nevertheless, p(f, f) == II r p(fi, A) holds. Select, for the proof, a
sequence of numbers CXi (i== 1,2, ...) such thatO<cxi< 1 and r CXi<oo.

Ch. 7,9 32J
THE RADON-NIKODYM THEOREM
247
Set ==(I-ai)i+ai and show that Ilr p(fi, ) >0, so f «g
by Exercise 32.16. Since « for all i, there are two possibilities by
Exercise 32.16: either f «g or f 1-g. In the first case, observe that
(dfjdg)t is the L2(') limit of {d(f1.. .fn)jd(gl" .gn)}t, and in the
second case use that p(fi, )(I-Cii)-tp(, ).
Show that f «f implies «fi for all values of i, and f --Lf if
and only if Ilr p(fi, )O. Show by an example that it may occur
that neither f 1-f, nor f «f or f «f holds.
EXAMPLES
32.19) In Exercise 11.4, it was asked to prove that if the series
r an of non-negative numbers an converges, then the sequence Pn
IIk=l (1 +ak) converges. Show that the converse assertion holds as
well, i.e., if anO and Pn converges, then 1 an converges. In the same
exercise it was also asked to prove that if Oan< 1 for all n, and r an
converges, then qn== II (l- a k) converges to a number q>O. Show
that the converse holds as well, i.e., if Oan< 1 for all n, and q==
lim qn==lim Il (l- a k), then q>O or qO according as r an con-
verges or diverges.
32.20) With the same notations as in Exercise 32.16, let each Xi
consist of two points Xi and Yi, let O<Cii< 1 and O<fJi< 1, fiX{Xi}
==Cii and fiX{Yi}== l- Ci i, whereas X{Xi}fJi and X{Yi}== I-fJi. Show
that the integrals i and  over Xi are equivalent (i.e., «fi as well
as fi«), and
p(fi, )==(CiifJi)t+{(I-Cii)(I-fJi)}!.
Denoting the product integrals over XIX X 2 X . . . by f and f, show
that f and f are equivalent or orthogonal according as the series
00 00
 [(Ci-fJ)2+{ (I- Ci i)l- (l-fJi)!}2J =  Yi
1 1
converges or diverges.
Show that if aCiil-a and afJil-a for a fixed number a>O,
then f and f are equivalent or orthogonal according as the series
r (ai-fJi)2 converges or diverges. In particular, note that if aia,
PifJ for all values of i, and Ci=l=-fJ, then f 1-f.

248
THE RADON-NIKODYM THEOREM
[Ch. 7,  33
POLAR MEASURE
32.21) In R2, let n be polar measure (cf. Exercise 10.6) and fl Le-
besgue measure. Set ff===Jfdfl and ff==Jfdn. Show that each of f
and f is absolutely continuous with respect to the other, and de-
termine the function fo satisfying ff===f(ffo) for all f-summable f.
INCREASING SEQUENCES OF MEASURES
32.22) Let fln, fl (n== 1, 2, . . .) be a-finite measures, and Vn, v (n==
1, 2, . . .) finite measures, in the point set X, where all these measures
are extensions of finite valued measures, initially defined on the ring A
of subsets of X. Let fln j fl and Vn jv on A, and let, finally, each Vn be
fln-absolutely continuous, i.e., vn(E) == J XEfndfln for some non-negative
fln-summable fn and all vn-measurable E. Sho"v that v is fl-absolutely
continuous, i.e., v(E) == J XEfdfl for some fl-summable f and all v-measur-
able E, and show also that f===lim fn holds fl-almost everywhere on X.
Use the result in Exercise 9.21.
33. The Radon-Nikodym Theorem for the Non-a-finite Case
Let ff===Jxfdfl be a Stieltjes-Lebesgue integral over X. The function
f(x), defined on X, is called a local f-null function (local fl-null function)
whenever fXE is an f -null function for every f -summable set E eX.
Accordingly, the set FeX is called a local f-null set (local fl-null set)
whenever the characteristic function Xp(x) is a local f-null function.
THEOREM 1. (a) Iff is a local null function, then f is measurable, and
If I is also a local null function.
(b) If f is a local null function, and EeX is of a-finite fl-measure,
then fXE is a null function.
(c) If f is a local null function, and Ig(x) I If(x) I on X, then g is also
a local null function.
(d) If f is a null function, then f is a local null function. If the measure
fl is a-finite (i.e., if X is of a-finite fl-measure) , then any local null
function is a null function.
(e) If the sets F n (n== 1, 2, . . .) are local null sets, then r F n is a
local null set.

Ch. 7, 933J
THE NON-a-FINITE CASE
249
PROOF. (a) If f is a local null function, and #(E) < 00, then fXE is a
null function, so fXE is measurable. This holds for all E satisfying
#(E) <00; hence, f is measurable. Since fXE is a null function if and
only if IfxEI == IfixE is a null function, it follows that If I is a local null
function.
The assertions in (b), (c), (d) and (e) are immediate consequences of
the definitions.
The functions f(x) and g(x), defined on X, are called locally f-almost
equal functions whenever f-g is a local f-null function.
THEOREM 2. (a) The property for functions to be locally f-almost
equal is an equivalence relation, i.e., if f1 - f2 and f2 - f3 are local null
functions, then f1-f3 is a local null function.
(b) The functions f and g are locally almost equal if and only if the set
{x: f(x) =Fg(x)} is a local null set.
PROOF. The assertion in (a) is evident, and (b) follows by observing
that fXE and gXE are almost equal if and only if the set {x: fXE-gXE
=FO}is a null set (cf. sec. 15, Theorem 1).
It follows now that the collection of all f-measurable functions, de-
fined on X, may be divided into equivalence classes of mutually locally
f -almost equal functions. In order to describe this situation, we intro-
duce the following notations.
Given the measurable function f(x) , we denote by f* the equivalence
class of all functions g(x) which are almost equal to f(x) , and by f**
the equivalence class of all functions g(x) which are locally almost
equal to f(x). Then:
(A) f** consists of all g(x) such that (gXE)*==(fxE)* for every summable
set EcX.
It follows that f** may be identified with the collection of all classes
(fXE) * ; where E runs through the summable subsets of X, in the
following sense: We have g(x) Ef** if and only if gXEE(fxE)* for every
summable E.
The situation described in (A) is a particular example of the following
more general situation:

250
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 33
(B) To each summable set E c X corresponds a measurable function
fE(X) vanishing outside E, such that for E and F summable the
corresponding functions fE and fF are almost equal on the inter-
section EF, i.e., (fEF) * == (fEXEF) * == (fFXEF) *.
Given (A), one obtains (B) by defining fE==fxE. Given (B), one does
not necessarily have (A), since there does not always exist a function
f(x) on X such that (fE)*==(fxE)* holds for all summable E eX simul-
taneously. Evidently, if X is of a-finite measure, there always exists
such a function f. The necessity to distinguish between null functions
and local null functions, as well as between the situations (A) and (B),
disappears as soon as X is of a-finite measure.
Consider the collection of all measurable functions vanishing outside
the given summable set E, identify almost equal functions in the col-
lection, and let M; be the collection of the thus obtained equivalence
classes. The collection {M;} of all such M;, for all summable E e X, is
called a sheaf (in accordance with the usage of the same term for simi-
lar concepts in several other branches of mathematics). Selecting now
an element f; from each M; such that (fEF)*==(fEXEF)*==(fFXEF)* for
all summable E and F, the collection of these f; is called a cross
section in the sheaf. We shall use the notation <f;> , or shortly <f*>,
to denote such a cross section. What we have in the situation (B) is,
therefore, simply a cross section <f;> in the sheaf {M;}. If fE(X)O
almost everywhere for each E, the cross section <f;> will be called a
non-negative cross section. Finally, we shall say that the cross section
<f;> is determined by the function f(x) , defined on the whole of X,
whenever f;== (fXE) * holds for each f; in the cross section.
THEOREM 3. If X is the direct sum of the summable subsets X ex , i.e.,
if there exists a collection {Xex} , finite, countable or uncountable, of dis-
foint summable subsets X ex of X such that X == U ex X ex and such that each
summable E e X is contained, except for a null set, in an at most count-
able union of the sets X a, then each cross section <f;> is determined by a
function f.
PROOF. Any x E X is a point of exactly one X a; define f(x) == f x rx (x)
for this particular point x and this particular X a (where f X rx is one of
the functions in the equivalence class f*xrx)' Then f(x) is defined on the

Ch. 7, 933J
THE NON-a-FINITE CASE
251
whole of X, and since any summable E is included, except for a null
set, in a set :'1 X(Xt' it follows easily that f;=(fxE)* for the thus
defined function f.
LEMMA cx. (a) Let f f== J f d# and J f= J fdv be Stieltfes-Lebesgue inte-
grals over X, extensions of elementary integrals initially defined on the
same function collection. Then any v-measurable set of a-finite v-measure
is included in a #-measurable set of a-finite #-measure (and the same
with # and v interchanged).
(b) Let ff and Jf be the same as above. Then the following assertions
are equivalent:
(1 ) J is f -absolutely continuous (that is, every f -null set is a J -null
set) ,
(2) every f -null set is a local J -null set,
(3) every local f-null set is a local J-null set (that is, J is locally
f -absolutely continuous).
PROOF. (a) It is evidently sufficient to prove that any v-summable
set E c X is included in a #-measurable set of a-finite #-measure. Given
such a set E, there exists a a-function S(X)XE(X) such that Js<(X).
In addition, E1=={X: s(x»O}=>E, and E1 is a #-measurable set of a-
finite #-measure (cf. the first paragraph of the proof of Theorem 1 in
the preceding section). Hence, E 1 is a set of the desired kind.
(b) Evidently, we have (1)=>(2). In order to prove (2)=>(1), let E be
an f-null set. Then E is a local J-null set by (2) and, by (a), the set
E is included in a J-measurable set E 1 of a-finite v-measure. It follows
that E=EE1, and since EEl is now a J-null set, the same holds for E.
Evidently, we have (3)=>(2). In order to prove (1)=>(3), let E be a
local f-null set. We have to show that E is a local J-null set, i.e.,
that EEl is a J-null set for any J-summable set E 1 . By (a), the set
E1 is included in an f-measurable set E2 of a-finite #-measure, and
since EE 2 is, therefore, an f-null set, the set EEl cEE 2 is also an f-
null set. It follows by (1) that EEl is a J -null set.
Having shown thus that it is not necessary to distinguish between
f-absolute continuity and local f-absolute continuity, we make an-
other remark before stating the main theorem in the section. Given the
cross section <f,E>' there is in general no function fo(x) on X which

252
THE RADON-NIKODYM THEOREM
[Ch. 7, 933
determines the cross section. However, if f(x) is a #-measurable function
vanishing outside some #-measurable set S of a-finite #-measure, then
it makes sense to speak about the function f(x)fo(x) , defined f-almost
unambiguously on the whole of X. Indeed, since S ==  r En, where the
sets En are disjoint and #-summable, we set f(x)fo(x)f(x)fo,En(X) for
xEE n (nl, 2, .. .), and f(x)fo(x)O outside S. We shall use this no-
tation whenever it is convenient.
THEOREM 4 (RADON-NIKODYM THEOREM; NON-a-FINITE VERSION).
Let ff J fd# and ff J fdv be Stieltfes-Lebesgue integrals over X, ex-
tensions of elementary integrals on the same initial domain of definition
L, and let ff be f-absolutely continuous. Then there exists a unique
non-negative cross section <f'O) == <f)E)' where E runs through the f-
summable subsets of X, such that for any f-summable function f the
function ffo is f-summable, and fff(ffo) for any such f. Conversely,
if f(x) vanishes outside a #-measurable set of a-finite #-measure, and ffo
is f-summable, then f is f-summable, and f(ffo)==ff.
PROOF. Observe first that any f-measurable set is f-measurable
by Lemma cx in the preceding section. For any set EeX, satisfying
#(E)==fXE<OO and v(E)fxE<OO, the non-negative function fO,E(X)
is defined now as in Lemma y of the preceding section. Hence, fO,E(X)
vanishes outside E, and any f vanishing outside E is f-summable if
and only if ffo,E is f-summable, with fff(ffo,E) for any such f. The
thus defined fO,E(X) is f-uniquely determined, i.e., the equivalence
class (fO,E)* is uniquely determined. Next, let fXE<OO but fXE==OO.
Then, by Lemma cx, E is contained in a f-measurable set of a-finite
v-measure. But E itself is f -measurable; hence, E is of a-finite v-
measure; E==r En with #(En) as well as v(En) finite for all n, and
all En disjoint. Setting fO,E(X)  r fO,En(X), the thus defined fOE
vanishes outside E, and it has the property that any f vanishing
outside E is f-summable if and only if ffo,E is f-summable, with
fff(ffo,E) for any such f. In addition, fO,E is f-uniquely determined
by this condition. Evidently, by the uniqueness, fO,E and fO,F are f-
almost equal on the intersection of the f-summable sets E and F.
Hence, </,E) is a non-negative cross section.
Let now f be non-negative and f -summable over X. Then there
exists a f-measurable set 5 of a-finite v-measure such that f vanishes

Ch. 7, 933J
THE NON-(]-FINITE CASE
253
outside S. By Lemma ex the set S is included in an J-measurable set of
a-finite fl-measure; we may assume therefore that S is this f-measur-
able set (which is then also f-measurable). Then S==r En with dis-
joint fl-summable En; hence, lo,s==r 10,En makes sense, and ff
/sfdv==  /Enfdv  JEnffodfl== /slfodfl /xflodfl==f(ffo) by the no-
tational conventions immediately preceding the present theorem. The
extension to the case that I is complex is evident.
Conversely, let the non-negative function I vanish outside the f-
measurable set S of a-finite fl-measure, and let Ilo Ilo,s be f -summable.
Since this implies that f(lfoXE)  f(fxE) holds for any fl-summable set
E, it follows by a similar argument as above that f(llo)ff. The
extension to complexvalued functions I is evident.
Finally, the cross section </,E> is uniquely determined by the con-
dition that fl==f(flo) holds for any f-summable f, since for any fl-
summable E the function fO,E is f-uniquely determined.
This general version of the Radon-Nikodym theorem is not very
satisfactory as long as we do not possess more detailed information
concerning some conditions which are sufficient, or necessary as well
as sufficient, in order that <f,E> be determined by a function fo, de-
fined on the whole of X. As shown above in Theorem 3, the condition
that X be the direct sum of the f-summable sets Xex (in particular,
the condition that fl be a-finite) is sufficient. A first necessary and
sufficient condition is presented in the next theorem. In order to state
the theorem more concisely we shall say that the measure fl, induced
by fl== / fdfl, has the Radon-Nikodym property whenever, given any
f-absolutely continuous integral fl, there exists an f-measurable
fo(x)O on X (fo depending on f) such that ff==f(flo) for any f-
summable f. We recall that if f is f-absolutely continuous, then (by
definition) f and f are extensions of elementary integrals with the
same initial domain of definition.
THEORElVI5. The measure fl, induced by fl==/ fdfl, has the Radon-
N ikodym property il and only if any cross section <I;> is determined by
a lunction f defined on the whole 01 X.
PROOF. Assume first that /1 has the property that any cross section
is determined by a function, and let f be an f-absolutely continuous

254
THE RADON-NIKODYM THEOREM
[Ch. 7, 933
integral. By the preceding theorem there exists a non-negative cross
section <f,E> such that ff==f(ffo) for any f-summable f; hence, by
hypothesis, this cross section is determined by a function fo(x) , defined
on the whole of X. Then fo(x)O holds almost everywhere on any #-
summable set, and this shows that the #-measurable set on which
fo(x) <0 is a local #-null set. Making fo(x) ==0 on this set does not
affect the relation ff==f(ffo). Hence, # has the Radon-Nikodym
property.
Assume now, conversely, that # has the Radon-Nikodym property,
and let, first, <g;;> be a cross section such that OgE(X) 1 holds almost
everywhere for all g;; in the cross section. As observed before, for any
measurable f(x) vanishing outside a set of a-finite measure it makes
sense now to use the notation f(x)g(x). This applies in particular to any
f E L (where L is the initial domain of definition of f), and fg is then
f-summable over X on account of OgE(x)I. Hence, ff==f(fg) ex-
ists as a finite number for any fEL. In addition, fnEL, fn!O implies
fng!O, so ffn==f(fng)!O by dominated convergence. This shows that
ff is an elementary integral on L.
We apply the extension procedure to ff. Since ff==f(fg)ff for
any non-negative fEL, we have f*f:::(f*f for the exterior measures
f*f and f*f of the ordinate set of any f(x)O. Hence, if fXE==O for
the set EeX, then fXE==O, so f is f-absolutely continuous. In view
of the Radon-Nikodym property of # there exists now an f-measur-
able function fo(x)O on X such that ff==f(ffo) for all f-summable
f; in particular f(fg)==ff==f(ffo) for all fEL.
We wish to show next that f(fg)==f(ffo) holds for all f-summable
f, and for this purpose we prove first that fo(x) 1 holds almost every-
where on any #-summable set E (and hence almost everywhere on any
set of a-finite #-measure). Assume, on the contrary, that fo(x) > 1 on
the subset F of E of positive measure. Then XF is f-measurable (by
the f-absolute continuity of f), and on account of fXFfXFfXE
<00 it follows that XF is f-summable. Hence fXF==f(XFfo»fXF
since fo(x) > 1 on F and #(F) >0. This contradicts f XFf XF, and so
fo(x) 1 almost everywhere on any #-summable set. Given now any
f-summable f(x) , there is a sequence fnEL such that lim f(lf-fnl)
==0; hence, lim f(lf-fnlfo) ==0 and lim f(lf-fnlg) ==0 on account of
OgE(x)1 almost everywhere and Ofo(x)1 almost everywhere on

Ch. 7, 934J
THE FINITE SUBSET PROPERTY
255
any set of a-finite measure. But then
f(/g)==lim f(lng)==lim f(lnlo)==f(llo).
The last result holds in particular for I==XE, where E is any /1-
summable set, so fgE==f(XElo), and this implies that gE and xElo are
f -almost equal. In other words, the cross section <g> is determined
by the function 10.
Finally, let <I;> be an arbitrary non-negative cross section and, for
n==I,2, ..., let gn,E(x)==min{/E(x),n}. Then, by what has already
been proved, <g:,E> is determined by a measurable function gn(x)O
on X. The measurable set on which g2(X)<gl(X) is evidently a local
null set, so that, replacing g2 by g2==max (gl, g2), the cross section
<g;,E> is still determined by g2. Similarly, g3 may be replaced by g3==
max (g2, g3), and so on. We may assume, therefore, that glg2g3
. . . on the whole of X. Then the given cross section <I;> is evidently
determined by I(x)==lim gn(x). This completes the proof.
We make one final remark. If the measure fl, induced by fl== J Idfl,
has the Radon-Nikodym property, then (by definition) there corre-
sponds to any f-absolutely continuous integral f an f-measurable
10(x));0, defined on X, such that fl==f(llo) for all f-summable I.
Since f-absolute continuity of f implies (by definition) that f and
f are extensions of elementary integrals on the same initial domain of
definition L, it appears at first as if the Radon-Nikodym property for
fl depends not merely on the integral f itself, but also on the initial
domain of definition L. The last theorem shows that this is only ap-
parent; it shows, in fact, that if fl has the Radon-Nikodym property
with respect to L, then fl has the Radon-Nikodym property with re-
spect to any other function collection which may be regarded as initial
domain of definition of f.
We shall return to the Radon-Nikodym property in sec. 35.
34. Partial Order in the Measurable Functions, and the Finite
Subset Property
As in the preceding section, let fl== J Idfl be a Stieltjes-Lebesgue
integral over X and, for any f -measurable function I(x), let 1* denote

256
THE RADON-N"IKODYM THEOREM
l Ch. 7,  34
the equivalence class of all functions which are f-almost equal to f.
The collection of all such equivalence classes f*, corresponding to real
functions f, is partially ordered in the natural manner, i.e., f*g* if
and only if, for any real fEf* and real gEg*, the inequality f(x)g(x)
holds f-almost everywhere on X. This partial ordering induces a
partial ordering in the sub collection of all X;, where E runs through
the f-measurable subsets of X. Hence, denoting by E* the equiva-
lence class of all sets which are almost equal to E, we obtain a partial
ordering of the equivalence classes E*. Note that E*F* means that
any EEE* is almost included in any FEF*. The equivalence classes
E* and F* will be called disjoint whenever, for any E EE* and any
FEF*, the intersection EF is a null set.
According to the terminology for partially ordered sets, f* is an
upper bound of the collection {f;} when f*f; for all a, and the upper
bound f* is the least upper bound of {f} (notation: f* ===sup f) when-
ever f*g* for any other upper bound g*. Note that we do not assert
that every collection {f} has a least upper bound. If the index set {ex}
is countable, the least upper bound f* ==sup f exists, and f(x) is then
the ordinary pointwise least upper bound of the functions frx(x). If,
however, the index set {a} is uncountable, this need not necessarily
remain true. Similarly as for functions, E* is an upper bound of the
collection {E;} when E*E for all ex, and the upper bound E* is the
least upper bound of {E} (notation: E*===sup E) whenever E*F* for
any other upper bound F*.
The following simple remark concerning partially ordered sets will
be useful. If {vrx} is a subset of the partially ordered set V, and corre-
sponding to each Va there exists a subset {wa} of V such that V rx ===
sup Wa, then {va}, for variable a, has the same upper bounds in V as
{Wa}, for variable a and . In particular, if one of these sets has a least
upper bound then so has the other, and the two least upper bounds
are then the same.
THEOREM 1. If #(X) < 00, and {fa} is a collection of non-negative #-
measurable functions fa(x) on X, then f*===sup f exists. In particular, if
{Ea} is a collection of #-measurable subsets of X, then E* ===sup E exists.
PROOF. Assume first that the functions fa are uniformly bounded;

Ch. 7, 934J
THE FINITE SUBSET PROPERTY
257
Ofa(x)M on X. We replace the collection {fa} by a larger collection
{g}, where each g is of the form g==max (Ial' . . ., fap) for some finite
but variable p, and fCXiE{/a}. Evidently, {g} has now the property that
Og(x)M on X for every , max(gl' "', g;p)E{g} for giE{g}, and
any upper bound of {/} is still an upper bound of the larger collection
{g}. It will be sufficient, therefore, to prove that sup g exists. The
number a==sup fg is finite (aM#(X)), so there exists a sequence
gnE{g} such that lim fgn==a. Obviously, we may assume that the
sequence gn(x) is non-decreasing at each point XEX (if necessary, we
replace g2 by g2==max (gl, g2), g3 by g3==max (g2, g3), and so on).
Hence, f(x)==lim gn(x)==sup gn(x) is #-measurable, and fl==a. Then
f* is an upper bound of {g}. Indeed, if this were false, there would
exist a function gOE{g} such that go-I>O on a set P of positive
measure, so Jp (go-f) d#==c>O. It follows that the functions g==
max(gn, go)E{g} would satisfy fgfgn+c, hence fg>a for suf-
ficiently large n. This contradicts the definition of a as a==sup f g.
Having shown thus that f* is an upper bound of {g}, it follows im-
mediately that f* ==sup g, since f* is already the least upper bound of
the sequence g.
In the general case the functions I ex are not necessarily uniformly
bounded on X. For any integer nl, set Ina==min(fa, n), and let f:==
supa fa' Obviously, I;f..., i.e., f1(X)f2(X)... holds almost
everywhere. Let f==lim fn, so f*==sup f. Then, by the remark which
precedes the present theorem, f*==suPn,ex fex==sup f;.
Finally, for the statement concerning E*==sup E;, note that if all fa
are characteristic functions of #-measurable sets, then the proof shows
that f* is the equivalence class of a characteristic function of a #-
measurable set.
DEFINITION . We shall say that the measure # has the finite subset
property whenever any #-measurable set of positive measure contains a
subset ollinite positive measure.
THEOREM 2. The following assertions are equivalent.
(a) The measure # has the finite subset property.
(b) For any #-measurable set E, we have #(E)==suP#(F) lor all #-
measurable sets FeE ollinite measure.

258
THE RADON-NIKODYM THEOREM
[Ch.7,9 34
(c) Any local #-null set is a #-null set.
(d) For any #-measurable set E, we have E*==sup F* for all #-
measurable sets FeE of finite measure.
PROOF. We prove first (a)=>(c)=>(b)=>(a), and then (a) =>(d) =>(c).
(a)=>(c). Assume that E is a local null set; hence, #(F)==O for all
measurable FeE of finite measure. If we should have #(E) >0, then
O<#(F)<oo for some measurable subset F of E by (a), and this
contradicts the assumption. Hence, #(E) ==0, i.e., E is a null set.
(c)=>(b). Let E be measurable, and set ex==sup #(F) for all measur-
able FeE of finite measure. We have to prove that #(E)==ex, and we
may assume that ex<oo (otherwise there is nothing to prove). Since
there exists an ascending sequence F n of measurable subsets of E
satisfying #(Fn) <00 and #(Fn)jex, the set F==r Fn is measurable,
and # (F) ==ex. It follows then from the definition of ex that any measur-
able subset F 1 of E-F for which #(F1)<00 satisfies #(F 1 )==0. This
shows that E-F is a local null set, and therefore, by (c), a null set.
Hence #(E)==#(F)==ex.
(b)=>(a). It follows immediately from (b) that the existence of a
measurable set of positive measure without a subset of finite positive
measure is an impossibility.
(a)=>(d). Given the measurable set E, it is obvious that E* is an
upper bound for the collection {F*} of all F*E* with #(F) <00. As-
suming that Ei is another upper bound, we consider the set E-E1.
In order to show that E*==sup F* for the collection {F*}, it will be
sufficient to prove that #(E -E1) ==0. Assuming now that #(E -E 1 ) >0,
the set E-E 1 contains, in view of (a), a measurable subset F of finite
positive measure. Then F and E1 are disjoint, but on the other hand F
is almost included in E1 since Ei is an upper bound of the collection
{F*}. This is impossible in view of # (F) >0. Hence #(E-E 1 ) ==0.
(d)=>(c). Assume that (d) holds, but (c) fails to hold. Then there ex-
ists a local null set E which fails to be a null set. This implies obvi-
ously that #(E)==oo, and #(F)==O for all measurable FeE of finite
measure. It follows that sup F*==0* for the collection {F*} correspond-
ing to these sets F, and this contradicts (d).
It has been shown thus that the finite subset property (and not the
a-finiteness of the measure) is a necessary and sufficient condition for

Ch. 7,  34J
THE FINITE SUBSET PROPERTY
259
identity of null sets and local null sets. The discrete measure in an
uncountable point set X is an example of a non-a-finite measure pos-
sessing the finite subset property.
In the following we assume that # has the finite subset property,
and we consider a collection {X} such that the corresponding subsets
Xcx eX are of finite positive measure, and all X in the collection are
disjoint (that is, XaXa' is a null set for a=Fex'). Since # has the finite
subset property there exists at least one such collection consisting of
one single X (except in the trivial case that #(X) ==0 which we shall
exclude). The family {} of all such collections  =={X} is, therefore,
not empty, and {} is partially ordered by inclusion; {X} c {Y;} when-
ever each X is one of the Y;. Evidently, each chain in this partially
ordered set has a least upper bound which is simply the collection of
all X occurring in the elements of the chain. Hence, by Zorn's lemma,
there exists at least one maximal collection m=={X}, We shall in-
vestigate some of the properties of such a maximal collection m,
which will be kept fixed. The maximality of m=={X;} means that if
E is a #-summable subset of X such that EX a is a null set for all ex,
then #(E) ==0. Observe that if X is the direct sum of the summable
sets Xcx of positive measure, then {X} is such a maximal collection.
In the general case, however, if m=={X} is the maximal collection
and the index set {ex} is uncountable, it may not be possible to select
sets XaEX such that the intersections XaXa' are empty for all pairs
ex, ex' simultaneously.
LEMMA ex. If E is #-summable and m=={X} is the maximal col-
lection, then # (EXa) >0 for an at most countable number of indices ex,
say ex==ex1, ex2, . . " and then E is almost equal to  1 EX cxt . Hence,
E* ==SUPi (EXaJ * ==supa (EX a) *.
PROOF. Since #(E)<cx>, and the sets Xcx are mutually almost dis-
joint, there is only a finite number of Xcx such that #(EX a ) > 1 ; simi-
larly, #(EX a »! for only finitely many ex, #(EX a »! for only finitely
many ex, and so on. Hence, #(EX a) >0 for an at most countable number
of indices a, saya==ex1, ex2, ... . ThesetE1==i:lEXaiisthenmeasur-
able, and E1 cEo Furthermore, (E-E1)X a is a null set for all ex, and

260
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 34
by the maximality of {X} this implies that #(E-E 1 ) ==0. Hence, E is
almost equal to E 1 ==  EX at .
LEMMA fJ. If {E} is a collection of summable sets such that 5*==
sup E exists, then 5*==suP,a E;a, where Ea==EXa.
PROOF. By the preceding lemma, E==supa E;a for each . Hence
5* ==suP,a E;a, by the remark which immediately precedes Theorem 1.
THEOREM 3. For each measurable set 5 c X we have 5*==sup (5X a )*.
In particular, X*==sup X;.
PROOF. By the equivalence of (a) and (d) in Theorem 2, we have
5* ==sup E* for all measurable E c 5 of finite measure, and by Lemma a
we have E*==sup(EX a )* for each such E. Hence, 5*==suPE,a(EX a )*.
On the other hand, (5X a )*==SUPE(EX a )* for each cx, so that in view
of the existence of sUPE,a (EX a ) * ==5* the least upper bound supa (5X a ) *
exists as well and equals 5*.
One might conjecture that the following converse theorem holds.
Given the measurable sets PacX a , the least upper bound P*==sup P
exists. It would also, at first sight, seem reasonable that the union
U a Pa is then in the equivalence class P*. This converse theorem is not
necessarily true, however, the reason being that for a fixed cxo the inter-
section of Xa o and the union Ua¥=a o P a may very well fail to be a null
set (it is a union Ua¥=a o P aXao of null sets, but the union of an un-
countable number of null sets is not necessarily a null set). In the
particular case that X is the direct sum of the summable sets X a
(always assuming that # has the finite subset property), the converse
theorem holds.
THEOREM 4. If there exist summable sets X a such that X is the direct
sum of the sets X a, and the sets Pac X a are measurable, then P== U a P a
is measurable, and P*==sup P;.
PROOF. Obviously, we have now PXa==P a , which shows that PX a
is measurable for all cx. It follows then by the definition of a direct sum
that PE is measurable for any summable E, and this implies that P
itself is measurable. The preceding theorem shows finally that P* ==
sup (PXa)*==sup P.

Ch. 7, 934J
THE FINITE SUBSET PROPERTY
261
Exercises
EQUIVALENT CONDITIONS FOR THE FINITE SUBSET PROPERTY
34.1) Show that the measure fl, induced by ff== J fdfl, has the finite
subset property if and only if the following holds. Given any f-measur-
able f(x)O such that ff>O, there exists an f-measurable g(x) such
that Og(x)f(x) and O<fg<oo.
34.2) Let the measure fl be induced by ffJ fdfl. Show that the
following assertions are equivalent.
(a) fl has the finite subset property.
(b) Given the measurable fO such that ff>O, there exists a
measurable g(x) such that Ogf and O<fg<oo.
(c) For any measurable f(x)O, we have ffsup fg for all measur-
able g(x) satisfying Ogf and fg<oo.
(d) Any local null function is a null function.
(e) For any measurable f(x)O, we have f*sup g* for all measur-
able g(x) satisfying Ogf and fg<oo.
THE CONTRACTED INTEGRAL
34.3) Let the measure fl be induced by ff== J fdfl. Then the con-
tracted integral fef induces the contracted measure fle, i.e., fef==
J fdfle (by Exercise 17.8). Show that fle has the finite subset property.
34.4) Let ff== J fdfl and ff== J fdv be extensions of elementary inte-
grals initially defined on the same function collection, and let fle and Ve
be the contracted measures corresponding to fl and v respectively. Show
that f is f-absolutely continuous if and only if any set E, satisfying
fle(E) O, satisfies ve(E) ==0.
EXAMPLES
34.5) Let fl be the measure in the square X  [0, 1; 0, IJ obtained
by extension of the linear Lebesgue measure of horizontal cells (cf.
sec. 10, Example 6). Show that X is the direct sum of the sets Xa
(OaI), where Xa{(X, y): Oxl, y==a}. Note, however, that Theo-
rem 4 in the present section does not hold, since fl does not possess
the finite subset property.

262
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 35
34.6) Let fl be the measure in the square X == [0, 1; 0, IJ obtained
by extension of the linear Lebesgue measure of horizontal and vertical
cells (cf. sec. 10, Example 7), and let fle be the corresponding con-
tracted measure. Let Xa=={(x, y): Oxl, y==ex} for Oexl, and Y p ==
{(x, y) : x=={J, Oy I} for O{J 1. Show that the collection consisting
of all X and all Y; is a maximal collection for the measure fle in the
sense of the present section. Show also that it is not necessarily true
that if one selects measurable sets P ()(, c X a, then P== U a P ()(, satisfies
P*==sup P. It is possible to select the sets Pa such that U a Pa is not
even measurable.
35. Localizable Measure
In the preceding section it was shown that fl has the finite subset
property if and only if for any fl-measurable set E we have E*==sup F*
for all subsets F c E of finite measure. It is not true in this case, how-
ever, that for every collection {F p} of fl-measurable sets of finite
measure the least upper bound sup F; necessarily exists. The present
section is devoted to an investigation of the consequences resulting
from the additional hypothesis that sup F; exists for any such col-
lection {F p}.
DEFINITION. The measure fl is called localizable whenever, for any col-
lection {F p} of fl-measurable sets F p of finite measure, the least upper
bound sup F; exists.
We shall assume in the remainder of this section that the measure fl
induced by ff==Jfdfl has the finite subset property, and we let {X}
be the same maximal collection of disjoint equivalence classes as in the
preceding section. The collection {X} will be kept fixed.
LEMMA ex. The measure fl is localizable if and only if, for any choice
of the measurable sets P()(,cX a , the least upper bound P*==sup P exists.
In addition, if P is then one of the sets in the equivalence class P*, the
sets P X a and P a are almost equal for all ex.
PROOF. Assume that, for any choice of PacX a , the least upper
bound P* ==sup P exists, and let the collection {F p} of measurable sets
of finite measure be given. For any ex, the least upper bound P==

Ch. 7, 935J
LOCALIZABLE MEASURE
263
supp(XaF p )* exists by Theorem 1 in the preceding section, and we
may obviously select the sets P aEP such that Pais entirely included
in X ex for any cx. Then, by hypothesis, P* ==sup P; exists; hence P* ==
sUPa,p(XexF p )*. On the other hand F===supex(XexFp)* for any fJ by
Lemma cx in the preceding section, so that in view of the existence of
supex,p(XaFp)* we may conclude that sup F exists. In addition, sup F
==suPex,p (X ex F p) * ==P*. This shows that # is localizable.
Conversely, it is evident that localizability of # implies the existence
of P* ==sup P; for any choice of the measurable sets Pac X ex.
Assume now that # is localizable, PacX a , and P*==sup P. Let P
be one of the sets in P*. From P*P, holding for all cx, it follows
easily that Pex-PX ex is a null set for any cx. If for some cxo the set D o ==
PXexo-P ao would fail to be a null set, the set Q=P-Do would satisfy
Q*P; for all cx, so Q*sup P;==P*, contradicting the fact that #(P-Q)
==#(Do) >0. Hence, #(Pex-PXex)==#(PXex-Pex)==O for all cx, i.e., the
sets P X ex and P ex are almost equal for all cx.
THEOREM 1. If there exist #-summable sets X ex such that X is the
direct sum of the sets X a, then # is localizable.
PROOF. Follows by combining the preceding lemma and Theorem 4
in the preceding section.
LEMMA fJ. Assume that the non-negative numbers aI, . . " an and, for
each cx, the #-measurable disjoint subsets Pal, . . . , P exn of X ex are given.
Let, for each cx, the step function tex(x) be defined by tex(x) == = 1 akXexk(x) ,
where Xexk denotes the characteristic function of P exk. Then, if # is local-
izable, the least upper bound t* ==sup t; exists and, if t(x) is one of the
functions in the equivalence class t*, the functions t(x) and ta(x) are almost
equal on X a for any cx.
PROOF. Let PZ=suPex P: k for k == 1, . . ., n, and select sets P k from
each PZ. Then, by Lemma cx, the sets PkX ex and P ak are almost equal
for all cx and all k; it follows that for k =l=-l the set PkPlX ex is a null set
for all cx, so PkPl is a local null set, and hence a null set. We set t(x) ==
=1 akXPk(x). Evidently, (akXpJ*==supa(akXexk)* for k===l, "', n, and
t*==SUPk (akXpJ* since the sets PI, . . " Pn are mutually almost dis-
joint. Hence
t* ==SUPex,k (akXak) * ==supa t.

264
THE RADON-NIKODYM THEOREM
[Ch. 7,  35
Finally, since for any cx the sets PkX rx and P ak are almost equal, the
functions t(x) and trx(x) are almost equal on X a.
THEOREM 2. The measure # is localizable if and only il every cross
section </> is determined by a function f, defined on the whole 01 X.
PROOF. (a) Assume that # is localizable, and let the cross section
<IE> be given. It is no restriction of the generality to assume for the
present purposes that <fE> is a non-negative cross section. We select
a function la(x)==fxrx(x) from each equivalence class l*xrx; it may be as-
sumed that fa(x)?:-O on X a , and Irx(x) ==0 outside Xa.
Let n 1 be an integer and let, for k==O, 1, 2, . . ., n. 2 n , the set P rxk
be defined by
P ak== {x: kj2 n <1 rx(x)(k+ 1) j2 n }.
Then the sets P ak (k==O, 1, 2, . . " n. 2 n ) are #-measurable disjoint
subsets of Xa. The step function tna(x) is now defined by
n.2 n
tna(x)==  (k/2 n )Xak(X) ,
k=O
where Xrxk denotes the characteristic function of P ak. By the preceding
lemma t ==suPrx trx exists, and for any ex the functions tn(x) and tnrx(x)
are almost equal on X rx. Furthermore, in view of t n + 1,rx(X) tnrx(x),
holding for any cx, we have also t+ l?:-t:, so 1* ==supn t: exists. This
shows that 1* ===suPn,rx t: a . On the other hand, we have tnrx(x) i I a(x) as
noo by the definition of tna(x), so f==suPn t: rx . It follows then, on
account of 1*==suPn,a t: a , that I*===suprx f. Since on any Xa the functions
t n and t na are almost equal, it is evident (n-+oo) that f and fa are
almost equal on X a, i.e., (IXxrx) * == l*xrx holds for any cx. It remains to
prove that (IXE)*==I; holds for any #-summable E. This, however, is
easily deduced from the fact that E is included, except for a null set,
in an at most countable union of sets X a.
(b) Assume that, given any cross section <I;>, there exists a function
I on X such that (IXE)*==I; for all fE in the cross section. In order to
prove that # is localizable, it is sufficient to show that for any choice
of the measurable sets Pac X rx, the least upper bound P* ==sup P
exists.

Ch. 7,  35J
LOCALIZABLE MEASURE
265
Given any fl-summable set E, this set is almost equal to i EX at
by Lemma cx in the preceding section. For this E, let PE(X) be defined
as the characteristic function of i EP ai' Then <p;> is evidently a
cross section, so there exists a measurable function P(x) such that
(PXE) * p; for any p; in the cross section. In particular, setting E ==X a,
we obtain the result that (PXxJ*==X(X for all cx. It follows easily that
the set on which p does not assume the values zero or one is a local
null set, and therefore a null set. Hence, we may assume that p is the
characteristic function of some measurable set P, and we have then
that (PXa)*==P; for all cx. Finally, by Theorem 3 in the preceding
section, P*SUp(PXa)*==SUp P. This completes the proof.
Combining the last theorem and sec. 33, Theorem 5, the following
theorem, essentially due to I. E. SEGAL (1951, [IJ), is obtained.
THEOREM 3 (SEGAL'S THEOREM). The measure fl, induced by ff==
J fdfl and possessing the finite subset property, is localizable if and only
if fl has the Radon-Nikodym property.
Exercises
A PROPERTY OF LOCALIZABLE MEASURE
35.1) Show that if fl is localizable and has the finite subset proper-
ty, and {fp(x)} is an arbitrary collection of non-negative measurable
functions, then f* ==sup fp exists.
LOCALIZABILITY WITHOUT THE FINITE SUBSET PROPERTY
35.2) Let fl be the measure induced by ff J fdfl, and let fle be the
corresponding contracted measure. Show that if fle(E) <00, there ex-
ists a fl-summable set FeE such that fl(F)fle(E), so that E-F is
then a local fl-null set. Show that the set F is fl-uniquely determined.
35.3) As before, let f* be the equivalence class of all functions fl-
almost equal to f, and f** the equivalence class of all functions locally
fl-almost equal to f, that is, all functions fle-almost equal to f. Show
that there exists a one-one correspondence between all cross sections

266
THE RADON-NIKODYM THEOREM
[Ch. 7, 9 35
<I;> (for fl(E) <00) and all cross sections <1;'*> (for fle(F) <00), in the
following way: Given <I;> and the set F satisfying /-le(F) <OC>, we set
F ==E + F' with fl(E) ==fle(F) and fle(F') ==0 (cf. the preceding exercise),
then select a function g from I; such that g vanishes outside E, and
set 1;'* equal to the equivalence class g**. Conversely, given <1;'*> and
the set E satisfying fl(E)<oc>, we select a function h from 1;* such that
h vanishes outside E, and set I; equal to h*. Show that if <I;> and
</*> are thus corresponding, and one of them is determined by the
measurable function 1 defined on the whole of X, then the other is
determined by the same function I.
35.4) Show that if fl is localizable, then fle is localizable. Show also
that the following three statements are equivalent.
(a) fle is localizable.
(b) Every cross section <I;> is determined by a function 1 defined
on the whole of X.
(c) fl has the Radon-Nikodym property.
35.5) Assume that fl does not have the finite subset property, and
X is the direct sum of the fl-summable sets X ex. Show that X is with
respect to the measure fle still the direct sum of the sets Xex. Derive
from this fact that fle is localizable, so that the assertions (a), (b) and
(c) in the preceding exercise hold in this case.
EXAMPLES
35.6) Let fl be the measure in the square X==[O, 1; 0, IJ obtained
by extension of the linear Lebesgue measure of horizontal cells (cf.
sec. 10, Example 6). Show that fl does not have the finite subset proper-
ty, and fl is not localizable. Show also that the contracted measure fle
is localizable.
35.7) Let fl be the measure in the square X==[O, 1; 0, IJ obtained
by extension of the linear Lebesgue measure of horizontal and vertical
cells (cf. sec. 10, Example 7). Show that fl does not have the finite
subset property, and fl is not localizable. Show by means of Exercise 2.6
that, if we assume the continuum hypothesis to hold, then the con-
tracted measure fle is localizable.

CHAPTER 8
DIFFERENTIA TION
This chapter is devoted to differentiation, the inverse process of integration.
There is an extensive literature on this subject; we shall restrict ourselves, how-
ever, to some of the main points. It is important to observe that the natural
extension of the classical differentiation procedure leads to a procedure which
is applied to set functions instead of to point functions, and in sec. 36 this
procedure will be introduced and discussed for set functions in an abstract point
set. In sec. 37 the point set is specialized to Euclidean space, and we shall ob-
tain Lebesgue's theorem that the indefinite integral of a Lebesgue summable
function f has, almost everywhere, f as its derivative.
36. Differentiation in Abstract Sets
It is well-known that if f(x) is a continuous function on the linear
interval [a, bJ, and F(x)==J: f(t)dt for axb, then the derivative
F ' (x) of F(x) equals f(x), i.e.
. F(X') - F(x) f )
11m I == (x.
x'x X -x
Introducing Lebesgue measure #(E) and the set function a(E) ==
fEfd#, this formula may be written in the form
. a(l)
11m == f(x);
x'x #(1)
:and it is not difficult to prove that the slightly more general formula
I == [ x, x'] or I == [ x', x],
lim (1(1) = f(x) ;
x'X,X"-'1'X #(1)
is also true. This suggests immediately how to extend the differenti-
ation formula to real k-dimensional number space Rk for k> 1, and
I == [x', x"], x' < x" ,

268
DIFFERENTIATIO
[Ch. 8, 9 36
even to abstract point sets in which a measure # is defined. The present
section is devoted to the realization of this program in abstract sets;
differentiation in real number space will be discussed in the next
section.
Let # be a a-finite measure in the point set X, such that #(X) >0.
By a net .AI in X we shall mean a sequence of disjoint measurable
subsets of X, each subset of finite measure, such that X is the union
of these subsets. A sequence  of nets in X is called monotone if each
set of +1 is a subset of some set of .Al n , i.e., if each net J+l is a
refinement of its predecessor .
N ext, given the monotone sequence of nets  (n== 1, 2, . . .) in X,
let a(E) be a real set function, defined at least for every set E of every
net , such that a(E) is finite whenever #(E) is finite. We introduce
the point functions dn(x) on X as follows. For each value of n the
given point x belongs to one (uniquely determined) set I nE; we
define
! a(I n) /#(1 n) if #(1 n) >0,
dn(x) = +00 f ft(l n) =0, (1(1 n)O,
-00 If #(In) ==0, a(In) <0.
The upper derivative D a(x) and the lower derivative D a(x) of a(E) at
the point x, with respect to the measure # and the sequence of nets.
, are now defined by
D a(x) ==lim sup dn(x), D a(x) ==lim inf dn(x).
Note that D a(x) and D a(x) are #-measurable point functions on X
since all dn(x) are #-measurable. If
-oo< D a(x) == D a(x) < +00
at the point x, we shall say that a is differentiable at x (with respect
to # and J); the common value of D a(x) and D a(x) is now called
the derivative of a at x, and denoted by Da(x).
LEMMA 1. If  is a monotone sequence of nets, if J1IO==Ur.Al n (that
is, the elements of J1IO are all the sets of all ), and if vii is a collection
of sets from J1IO, then there exists a sequence In of disfoint sets from vii
such that  I n == U I EA I.

Ch. 8,9 36J
DIFFERENTIATION IN ABSTRACT SETS
269
PROOF. Let viiI be the countable collection of all different I Evil
such that I EJVi; let vii 2 be the countable collection of all different
I E_A, not already included in vii 1, such that I EJV2, and so on. Then
U lEe/It I == U 1 {U IEAt I} may be written as  In, where all In are dis-
join t.
In order to obtain for the abstract case a parallel of the differenti-
ation formula in real number space, we shall impose upon the sequence
 of nets an additional condition which will be defined now.
DEFINITION. Let  be a sequence of nets in X, and JVO==Ur .
The sequence  is called regular with respect to the measure # whenever,
given e > 0 and any #-measurable set E, there exists a sequence I n of sets
from JVO such that
00
00
#(E - "'L In) ==0,
1
#( I n-E)e.
1
Observe that if, in addition, the sequence  is monotone, we may
assume (in view of Lemma 1) the sets I n in the regularity condition to
be disjoint. Also, if the regularity condition is satisfied for each E of
finite measure, then it is satisfied for each #-measurable E. More gener-
ally, if the regularity condition is satisfied for each of the #-measur-
able sets En (n== 1, 2, . . .), then it is satisfied for E ==  En. Indeed,
given e>O, there exist sets IniEJVO such that #(En-ilni)==O and
#(i I ni-En)e/2n, hence
#(E-"'L Ini)==#{ (En-  I ni )}
n
#{ (En-  I ni)} #(En-  I ni)==O,
n i n i
and
#("'L  I ni-E)#{ ("'L I ni-En)} #( I ni-En)e.
n i n i
We assume now that v is a second measure in X, defined for all #-
measurable sets (at least), and satisfying the condition that, given
e>O, there exists a number £5>0 such that #(E)£5 implies v(E)e.
It follows then immediately that #(E)==O implies v(E) ==0. It is not

270
DIFFERENTIATION
[Ch. 8, 9 36
true, conversely, that every measure v for which #(E) ===0 implies
v(E) ===0 satisfies the condition that, given e>O, there exists a number
>O such that #(E) implies v(E)e. The example where X consists
of the points Xl, X2, . . ., and #({x n })==n- 1 , v({x n })=== 1, is a simple
counterexample. On the other hand, if #({x n })===1 and v({Xn}) ===00 for
all n, then v satisfies the mentioned (e, )-condition, and hence also the
condition that #(E) ==0 implies v(E) ==0, so v is #-absolutely continu-
ous in the sense defined in the preceding chapter, but the corresponding
integrals are not extensions of elementary integrals with the same in-
itial domain of definition. Evidently, if v(E) is finite whenever #(E) is
finite, then the (e, )-condition implies that v is #-absolutely continuous.
LEMMA 2. Let the measure v, defined for all #-measurable sets, satis-
fy the condition that, given e>O, there exists a number >O such that
#(E) imPlies v(E)e, and let the sequence of nets  .be regular with
respect to #. Then any #-measurable set E satisfies the regularity con-
dition with respect to  and the measure v.
PROOF. Let e>O, and choose >O such that #(A) implies v(A)e.
Given the #-measurable (and hence v-measurable) set E, there exists a
sequence InE.AD==Ur.Alk such that #(E- In) ===0 and #( In-E)
. It follows that v(E -  In) ==0 and v( I n-E)e.
THEOREM 1. Let  be a monotone sequence of nets in X, regular with
respect to #, and let v be a measure in X, defined for all #-measurable sets,
such that v(E) is finite whenever #(E) is finite, and satisfying the con-
dition that, given e>O, there exists a number >O such that #(E) im-
plies v(E)e. Then, if E is #-measurable, and a is a finite real number
such that D v(x) a at all xEE, we have v(E) ?::-a#(E).
PROOF. For aO the theorem is clearly true; we may assume, there-
fore, that a>O. Let first #(E) <00, and hence v(E) <00, and let e>O.
Then, by Lemma 2, there is a sequence In of sets from .AD=== Ur .AIk
such that
00
00
#{ (I nnE)}==#(E) ,
1
v( I n)v(E) +e.
1
For fixed n, we now consider an arbitrary point xEI nnE. Since D v(x)
a, we have (by the definition of D v) that lim sup di(x)?::-a as ioo,

Ch. 8, 9 36J
DIFFERENTIATION IN ABSTRACT SETS
271
i.e., if O<b<a, then di(x) >b for infinitely many values of i. It follows
that if Ii,x is the set from JVi which contains x, then v(Ii,xrb#(Ii,x)
for infinitely many i. In addition Ii,x c In for i sufficiently large. The
set I nnE is therefore covered by the union of a collection of such
Ii,x c In, and since on account of Lemma 1 we may replace this col-
lection by a sequence I ni (i== 1, 2, . . .) from the collection, we obtain
InnEc InicI n ,
i
v(I nirb#(I ni)
for i == 1, 2, . . . .
Once more on account of Lemma 1, the double sequence I ni (i, n==
1, 2, . . .) may be replaced by a single sequence] k of disjoint sets, such
that each] k is one of the I ni, and ] k==   I ni (so ] k=>  (I nnE)).
Then
v(E) +EV( I n)v(  I ni) ==v( ] k) ==  v(] k)
b  #(]k)==b#( ]k)b#{ (InnE)}==b#(E),
hence v(E)b#(E) for all b such that O<b<a. This shows that v(E)
a# (E) .
If #(E)==cx>, then E==lim En, where En is an ascending sequence of
sets of finite #-measure, hence
v(E)=lim v(En)lim a#(En)==a#(E).
COROLLARY. (a) If #(E) >0, and D v(x) >0 on E, then v(E) >0.
(b) If P is #-measurable, and v(P) ==0, then Dv(x) ==0 holds #-almost
everywhere on P.
PROOF. (a) We have E==lim En, where En=={x: Dv (x)n-1} for n==
1,2, . . . . Then v(En)n-1#(En), and the numbers #(En) are positive
from some index no on (since lim #(En) ==#(E) >0). It follows that
v(Eno) >0, so v(E)v(Eno) >0.
(b) If #(P) ==0, the statement is true; let, therefore, #(P) >0, and
assume that D v(x) >0 on a subset E cp of positive #-measure. Then
v(E) >0 by (a), hence v(P) >0, a contradiction. It follows that D v(x) =0
almost everywhere on P, and this implies that Dv(x) ==0 almost every-
where on P (since O D v(x) D v(x) for all x).
THEOREM 2 (DENSITY THEOREM). Let  be a monotone sequence of
nets in X, regular with respect to #, and let F be a fixed #-measurable set.

272
DIFFERENTIA TION
[Ch. 8,  36
Then, if the measure v is defined for any #-measurable E by v(E) ==
JEXpd#==#(EnF), we have Dv(x)==Xp(x) almost everywhere on X.
PROOF. Since v(E)#(E), the measure v satisfies the conditions of
the preceding theorem. Furthermore, if Fe==X-F, we have v(Fe) ==0,
so Dv(x) ==0 almost everywhere on Fe by Corollary (b). This shows al-
ready that Dv(x)==Xp(x) holds almost everyvvhere on Fe. Introducing
the measure v1(E)==#(EnFe), we have similarly DV1(X)==0 almost
everywhere on F. Since v(E) +v1(E) ==#(E) for all #-measurable E, and
D#(x) == 1 almost everywhere on X (the points where D#(x) "* 1 are the
points x contained in a set of #-measure zero from  for infinitely
many values of n), it follows therefore that Dv(x) ==D#(x) -DV1(X) == 1
holds almost everywhere on F. The final result is that Dv(x)==Xp(x)
holds almost everywhere on X.
LEMMA 3. If f(x) is non-negative on X and #-summable over X, then
the measure v(E) == JEfd# satisfies the condition that, given 8>0, there
exists a number £5>0 such that #(E)£5 imPlies v(E)e.
PROOF. Assuming the lemma to be false, there exists a number 8>0
and a sequence En eX such that #(En)2-n and v(En) >8 for all n.
If E==lim sup En, we have EeEn+En+1 +. .. for all n, hence #(E)
2-(n-1) for all n, i.e., #(E) ==0. On the other hand v(E)==v(lim sup En)
lim sup v(En)e (we use here that v(X)<oo). This is a contradiction.
THEOREM 3 (DIFFERENTIATION OF AN INDEFINITE INTEGRAL). Let
 be a monotone sequence of nets in X, regular with respect to #, and
let f(x) be real and #-summable over X. Then, if a(E)==JEfd#, we have
Da(x) ==f(x) almost everywhere on X.
PROOF. If the theorem is true for f1 and f2, then it is true for f1 -f2;
we may assume therefore that Of<oo on X. Let rand p be rational
real numbers such that r<p, and let Erp=={x: r<f(x) <p}. Then a(E)==
a(EErp)+a(EEp), and by Lemma 3 and Corollary (b) of Theorem 1 the
derivative of a(EEp) is zero on E rp , except on a subset Hp of measure
zero. For a(EE rp ) == JEErp fd# we have
r# (EE rp ) a(EE rp) p# (EE rp ) ,
and by the density theorem the derivative of # (EE rp ) is equal to one on
E rp , except on a subset H;p of measure zero. It follows that the upper

Ch. 8, 936J
DIFFERENTIATION IN ABSTRACT SETS
273
and lower derivatives of a( E Erp) lie between rand p on Erp-H;p, and
this implies that r D a(x)Da(x)p on Erp-H rp , where H rp===Hp+H;p'
Consider now all rational pairs r, p such that r<p, and the countable
collections of corresponding sets E rp and H rp. Setting H == U H rp, we
have #(H) ===0. Given any XEX -H, we choose the rational pair r, p
satjsfying r<f(x)<p. It follows then that xEErp-HcErp-Hrp, so r
pa(x) D a(x)p. This holds for all pairs r, p such that r<f(x) <p, hence
Da(x)===f(x). This holds for all xEX-H, so Da(x)===f(x) holds #-almost
everywhere on X.
We conclude this section by making several additional remarks con-
cerning the regularity condition for a sequence of nets  in X. As
observed earlier in this section, it is sufficient to require that the regu-
larity condition is satisfied for any set of finite measure. In some cases
a further simplification may be made.
LEMMA 4. If the measure # is generated by extension of a measure
having the semi-ring r as initial domain of definition and if the regu-
larity condition is satisfied for each set A E r of finite measure, then the
sequence of nets  is regular with respect to #.
PROOF. We have already proved that if the condition is satisfied
for each of the sets En (n== 1,2, . . .), then it is satisfied for  En. It
follows, in view of the present hypothesis, that the condition is satis-
fied for each a-set of finite measure. Let now E be #-measurable, and
of finite measure. Then, given e>O, there is a a-set 5 such that E c 5
and #(S)#(E) +le, and corresponding to 5 there is a sequence I nEJVO
=UrJVk such that #(S-In)==O and #(In)#(S)+le. Hence
#(E -  In) ==0 and #( I n)#(E) +8. This shows that the regularity
condition is satisfied for each set of finite measure, i.e., the sequence
 is regular.
Applying the lemma in the particular case that # is Lebesgue measure
in k-dimensional real number space Rk, it follows that  is regular
with respect to Lebesgue measure whenever the regularity condition is
satisfied for each (bounded) cell in Rk. Since any (bounded) cell is the
union of a sequence of (bounded) closed intervals, it is also true that
 is regular with respect to Lebesgue measure whenever the regu-
larity condition is satisfied for each (bounded) closed interval in Rk.

274
DIFFERENTIATION
[Ch. 8, 9 36
We shall say that the sequence of nets  in Rk is indefinitely fine
whenever, given any xER k and 8>0, there is a set in JVO== Ur ,
containing x and of diameter smaller than 8. The following lemma is
now of practical use.
LEMMA 5. If the sequence of nets  is indefinitely fine, then  is
regular with respect to Lebesgue measure #.
PROOF. It will be sufficient to prove that the regularity condition is
satisfied for each (bounded) closed interval F. Let xEF, and let
B(x, m- 1 ) be the open sphere of centre x and radius m- 1 , where m is a
positive integer. Then, since the sequence  is indefinitely fine, there
is a sequence In of sets from JVO== Ur JVi such that
Fe  In c U xEF B(x, m- 1 )==Om.
The set Om is open and bounded, and hence of finite measure. Further-
more, the sequence Om (m== 1, 2, . . .) is descending, and lim Om==F, so
# (F) ==lim #(Om)' Hence, given 8>0, we have for m sufficiently large
that #(F -  In) ==0 and #( I n-F)E.
Differentiation of a-additive set functions in abstract point sets was
introduced by R. DE POSSEL (1935, [IJ).
Exercises
LEBESGUE DECOMPOSITION FOR MEASURES
36.1) Let the measures # and v, initially defined on the same ring A
of subsets of X, generate a-finite measures in X. Note that the ex-
tended measures # and v may be regarded as extensions of measures
initially defined on the subring Al of A, consisting of all A EA such
that #(A) and v(A) are finite. Show that v has a unique decomposition
V==V1 +V2, where the measure VI is #-absolutely continuous, and the
measure V2 satisfies v2l..#, i.e., there exists a decomposition of X into
disjoint sets F and H such that v2(F) ==0 and #(H) ==0. The equality
V==V1 +V2 is to be understood in the sense that the set E c X is v-
measurable if and only if E is vI-measurable as well as v2-measurable.

Ch. 8, 9 36J
DIFFERENTIATION IN ABSTRACT SETS
275
THE INTEGRAL OF THE DERIVATIVE
36.2) Let the measures # and v, initially defined on the same ring A
of subsets of X, be such that (after extension) # is a-finite and v(X) is
finite, and let  be a monotone sequence of nets in X, regular '\vith
respect to #. The decomposition V==V1 +V2, and the sets F and H ==
X -F such that v2(F) #(H) O, are the same as in the preceding exer-
cise. Show that the derivative DV1(X) with respect to # and  exists
as a finite number #-almost everywhere on X, and
v(E) ==v(EH) + I XE(X) DV1 (x) d#
for any v-measurable set E.
36.3) In general DV1(X) may not be replaced by Dv(x) in the last
formula of the preceding exercise. Show this by considering the follow-
ing example. X  ( -1, 1 J ; # is Lebesgue measure on (0, 1 J and zero on
(-1, OJ, whereas v is Lebesgue measure on (-1, OJ and zero on (0, IJ,
and the monotone sequence of nets  is indefinitely fine in (0, IJ and
such that each set I EJVO Ur JVk consists of two cells (a, b J c (0, 1 J and
(-b, -aJ C (-1, OJ respectively. Show that the sequence  is regular
with respect to # as well as with respect to v. Show also that VI is
identically zero, so DV1(X) o at all x, but Dv(x)  1 at all x. Show
finally that JV;1, is not regular with respect to the measure #+v.
36.4) Let # and v satisfy the same hypotheses as in Exercise 36.2,
and let  be a monotone sequence of nets in X, regular with respect
to the measure #+v. The measures VI and V2, and the sets F and H,
are the same as in Exercise 36.2. Show that the derivative Dv(x) with
respect to # and  exists as a finite number #-almost everywhere on
X, and
v(E) v(EH) + I XE(x)Dv(x) d#
for any v-measurable set E. Note that although v is not necessarily
defined for all #-measurable sets, the notion of Dv(x) makes sense since
v is defined for all sets of any .
36.5) Let # be Lebesgue measure in k-dimensional real number
space Rk, and let v be a measure in Rk, initially defined on the semi-
ring of all cells, such that after extension V(Rk) is finite. Let  be a
monotone sequence of nets in Rk, such that each set from JVO Ur 
is #-measurable as well as v-measurable, and let the sequence JV n be

276
DIFFERENTIATION
[Ch. 8,  37
indefinitely fine. Show that the sequence  is regular with respect to
#+v, so that, consequently, the result of the preceding exercise holds.
37. Differentiation in Finite-dimensional Number Space
In the preceding section we have obtained the result that if f(x) is
real and #-summable over X, and  is a monotone sequence of nets,
regular with respect to #, then a(E)==/Efd# has everywhere on X, ex-
cept on a set H of #-measure zero, the derivative f(x) with respect to #
and. The exceptional set H, however, depends on the sequence,
and it has been proved that in general it is not true that there exists
a set H of #-measure zero, depending only on f(x), such that Da(x)
== f(x) for all x EX - H and all monotone regular sequences  simul-
taneously. The question may be raised, however, what will happen if
we restrict the choice of the sequences  by a suitable additional
condition. The answer is of importance in the case that # is Lebesgue
measure in Rk, since in the one-dimensional case (k=== 1) the existence
of such a set H, of #-measure zero and independent of the particular
sequence , is equivalent to the property that the indefinite integral
F(x)==J: f(t)d# satisfies F'(x)==f(x) almost everywhere, where F'(x) is
the derivative in the classical sense.
Let # be Lebesgue measure in Rk (k?;:-I), and let a(E) be a real set
function, defined and finite at least for every E ==K, where K is a
closed cubic (non-degenerate) interval. For any xER k , the symmetric
upper derivative D sa(x) and the symmetric lower derivative D sa(x) of a
at x are defined by
- a(K)
Dsa(x) = lim sup ( ) '
p,(K)-+O # K
a(K)
D sa(x) == lim inf ,
p,(K)-+O #(K)
where K is a closed interval of centre X. Whenever -oo< D sa(x)===
D sa(x) <00 at the point x, the common value of D sa(x) and D sa(x) is
called the symmetric derivative of a at x, and denoted by Dsa(x). In
that case we have
I . a(K n)
1m === Dsa(x)
p,(Kn)-+O #(K n)

Ch. 8, 9 37J
FINITE-DIMENSIONAL NUMBER SPACE
277
for any sequence Kn of closed cubic intervals of centre x, satisfying
lim #(Kn) ==0.
We shall consider more general sets than only cubic intervals around
the point x. Assume 110W that a(E) is a real set function, defined at
least on a a-field Al of #-measurable sets containing all cells, and such
that a(E) is finite whenever #(E) is finite. Given again the point
xER k , let En be a sequence of sets in AI, possessing the following
properties.
(a) #(En) >0, and (En) -+0, where (En) is the diameter of En with
respect to the usual Euclidean distance in Rk (hence #(En)-+O).
(b) The sets En are, in a certain sense, not too flat, and they tend
to the point x; precisely, there exists a number ex such that lex<oo,
and there exists a sequence Kn of closed cubic intervals of centre x,
such that En cK n and #(Kn)/#(En)ex for all n.
Any sequence En, satisfying these conditions, is said to tend regu-
larly to the point x, and the number ex is sometimes called a parameter
oj regularity of the sequence En. We consider now all sequences En
tending regularly to x (different sequences may have different para-
meters ex). The numbers
- . a(En)
Dra(x) = sup hm sup (E) '
{En} noo # n
D ( ) . f I . . f a(E n)
_ ra X == In 1m In (
{En} noo # En)
are called the regular upper derivative and regular lower derivative of a
at x respectively. If they are equal and finite at x, their common
value is called the regular derivative of a at x, and denoted by Dra(x).
In that case we have
. a(E n)
11m == Dra(x)
noo #(En)
for each sequence En which tends regularly to x. In particular, the
symmetric derivative Dsa(x) exists then, and Dsa(x) ==Dra(x). In the
presence of an additional hypothesis, the converse is also true, as shown
by the following lemma.

278
DIFFERENTIATION
[Ch. 8,  37
LEMMA 1. If v(E) is a measure in Rk, defined on a a-field of #-measur-
able sets containing all cells, such that v(E) is finite whenever #(E) is
finite, and it Dsv(x) ===0 at the point x, then Drv(x) ==0.
PROOF. Let the sequence En tend regularly to x, and let Kn be a
corresponding sequence of closed cubic intervals of centre x, such that
En cK n and #(Kn)/#(En)Ci for all n. Then
v(En)
O
#(En)
v(Kn)

#(En)
#(Kn)
#(En)
v(Kn) v(Kn)
Ci -+0
#(Kn) #(Kn)
as n-+oo.
We consider now a very special sequence of nets in Rk. For k== 1, the
net JVi consists of cells of length one, and each net +1 is made by
dividing each cell of its predecessor  into three disjoint cells of
equal length. The sequence  is, therefore, n10notone and indefi-
nitely fine, so it is regular with respect to # by Lemma 5 in the pre-
ceding section. For k> 1, we proceed on each coordinate axis as we
did in R1, and the so obtained sequence  in Rk is again monotone,
and regular with respect to #. .
Returning to the case that k==l, we make a second sequence of nets
/, in the following way. For each value of n, the centres of the cells
of  are the endpoints of the cells of /. Then the sequence ' is
also monotone and regular with respect to #; the sequences  and
' are called conjugate sequences of nets. We shall prove now that if K
is a closed interval of length A<!, then K is included in a cell of length
not exceeding 6A from one at least of the sequences  and /. In-
deed, there is an index no such that the length l of the cells of o and
: satisfies 2A<l6A; the distance between the endpoints of the cells
of o and /: is then !l, and !l>A. It follows that if K contains an
endpoint of a cell of o' then K contains no endpoint of a cell of :,
and similarly with o and : interchanged. Hence, K is included in
a cell of length not exceeding 6A of either Ur  or Ur /.
In Rk (k> 1) we have on each coordinate axis the two conjugate
sequences; selecting now on each axis either the initial or the conjugate
sequence we obtain 2 k different sequences of nets; these 2 k sequences
form a system ot conjugate sequences of nets. If K is a closed cubic inter-
val [aI, b 1 ; . . . ; ak, bkJ such that A==b 1 -a1 == . . . ==bk-ak<!, it follows

Ch. 8,  37J
FINITE-DIMENSIONAL NUMBER SPACE
279
now immediately that K is included in a cell of measure not exceeding
(6A)k from one of the sequences in the system of conjugate sequences.
LEMMA 2. If v(E) is a measure in Rk, defined on a a-field of #-measur-
able sets containing all cells, such that v(E) is finite whenever #(E) is
finite, and if at the point x the derivative of v with respect to each of the
conjugate sequences is zero, then Dsv(x)==O (so Drv(x)==O by Lemma 1).
PROOF. Let K be a cubic interval of centre x and measure #(K)==Ak.
We may assume that A<!. Then K is included in a cell C of measure
not exceeding 6 k #(K) from one of the conjugate sequences. Hence
v(K) ::( v(C) = ft(C) _ (C) ::( 6 k v(C) -+ a
#(K) #(K) #(K) #(C) #(C)
as #(K) o. It should be observed for this final conclusion that if K
runs through a sequence Kn such that #(Kn) O, then the correspond-
ing sequence C n satisfies #(Cn)O, and this implies that, no matter
from which of the 2 k conjugate sequences AI?) (i== 1, . . " 2k) the cell
C n comes, the index i==i(n) tends to infinity.
The remainder of the discussion is very much as in the abstract case.
We first prove the parallel of Theorem 1, Corollary (b) in the preceding
section.
LEMMA 3. Let v be a measure in Rk, defined for all #-measurable sets,
such that v(E) is finite whenever #(E) is finite, and satisfying the con-
dition that, given 8>0, there exists a number £5>0 such that #(E)£5
imPlies V(E)8. Then, if P is #-measurable, and v(P) ==0, we have
Dsv(x) ==0 almost everywhere on P (so Drv(x) ==0 almost everywhere on P).
PROOF. It follows from Theorem 1, Corollary (b) in the preceding
section that, for each sequence in the system of conjugate sequences,
Dv(x) ==0 holds almost everywhere on P, where Dv(x) is the derivative
with respect to that particular sequence. Hence, since the number of
sequences in the system is finite, there exists a set H of #-measure zero
such that, at each xEP-H, the derivative of v with respect to each
of the conjugate sequences is zero. This implies, by the preceding
lemma, that Dsv(x) ==0 for all xEP-H.

280
DIFFERENTIATION
[Ch. 8,  37
THEOREM 1 (DENSITY THEOREM). II F is a lixed #-measurable set,
and
v(E)== J Xpd# ==# (EnF) ,
E
then Drv(x) == XF(X) almost everywhere on Rk.
PROOF. By the preceding lemma Drv(x)==O almost everywhere on
Fe==Rk-F. Similarly, if v1(E)==#(EnFe), then DrV1(X)==0 almost
everywhere on F. Hence, since V+V1 ==#, and Dr,u(x) == 1 on the whole
of Rk, we have Drv(x)== 1 almost everywhere on F. The final result is
therefore that Drv(x) ==Xp(x) almost everywhere on Rk.
THEOREM 2 (LEBESGUE'S THEOREM ON DIFFERENTIATION OF AN IN-
DEFINITE INTEGRAL). III(x) is real and #-summable over Rk, and a(E)==
/ E I d#, then Dra(x) == I(x) almost everywhere on Rk.
PROOF. Exactly as the proof in Theorem 3 of the preceding section,
replacing everywhere the derivative with respect to the sequence 
by the regular derivative.
The Theorems 1 and 2 are due to H. LEBESGUE (1904, [IJ, for k==l;
1910, [4J, for k>I); the device of using a system of conjugate nets in
the proof is due to CH. J. DE LA VALLEE POUSSIN (1915, [2J).
ExamPle in R 1 . If hn!O, the sequences I n== [x, x+hnJ and] n==
[x-h n , xJ tend regularly to x. Hence, if I(x) is #-summable over [a, bJ,
F(x)==/: I(t)d# for axb, and a(E)==/Eld# for Ee [a, bJ, then
a(I n) F(x+hn)-F(x)
#(In) h n
a(J n) F(x) - F(x - h n )
--
#(J n) h n
F(x-h n ) -F(x)
-h n
so that it follows from Theorem 2 that F' (x) == f(x) almost everywhere
on [a, bJ, where F'(x) is the derivative of F(x) in the classical sense.
Similarly, if xoE[a, bJ and F1(X)==/x/(t)d# (this means for X<XO that
F 1 (x)==-/:o I(t)d#==-a([x, xoJ) in view of the usual conventions),
then Fi(x)==/(x) almost everywhere on [a, bJ.

Ch. 8,  37J
FINITE-DIMENSIONAL NUMBER SPACE
281
Exercises
THE INTEGRAL OF THE REGULAR DERIVATIVE
37.1) Let v be a measure in Rk, initially defined on the semi-ring
of all cells, such that (after extension) V(Rk) is finite. Show that Drv(x)
exists as a finite number #-almost everywhere on Rk (where # is Le-
besgue measure), and show that there exists a set H of #-measure zero
such that
v(E) ==v(EH) + I XE(X) Drv (x) d#
for any v-measurable E. It is understood here that the sets in any
sequence, tending regularly to x, are v-measurable as well as #-measur-
able.
DERIVATIVE OF AN INCREASING FUNCTION
37.2) Let g(x) be a non-decreasing function on R1, i.e., g(X2)g(X1)
for X2>X1. Show that the derivative g'(x) exists as a finite number
almost everywhere on R 1 . Assume first that g is right continuous and
g(oo) -g( -00) is finite, and then remove these restrictions.
INCREASING SEQUENCE OF MEASURES
37.3) Let #, v and Vn (n==l, 2, ...) be measures in the point set X,
extensions of finite-valued measures on the ring A of subsets of X, and
let Vn iv on A. Assume that, after the extension, # is a-finite and v is
finite (so all Vn are finite, since Vn i v holds for all v-measurable sets by
Exercise 9.21). The Lebesgue decompositions of v and Vn with respect
to # are V==V01 +V02 and Vn==Vn1 +Vn2 respectively, where VOl and Vn1
are #-absolutely continuous, and V02 and Vn2 are orthogonal to # (cf.
Exercise 36.1). Show that Vn1 iV01 for all vOl-measurable sets, and
Vn2 i V02 for all v02-measurable sets. Let In and I be the #-summable
functions satisfying Vnl (E) == J XEI n d# and VOl (E) == J XEI d# respectively.
Show that In i I holds #-almost everywhere on X.
37.4) Let v and Vn (n== 1, 2, . . .) be measures in Rk, initially defined
on the semi-ring of all cells, such that Vn iv on this semi-ring. Assume
that, after the extension, V(Rk) is finite (so all Vn(Rk) are finite, since
vniv holds for all v-measurable sets by Exercise 9.21). Finally, let #

282
DIFFERENTIATION
[Ch. 8, 9 37
be Lebesgue measure in Rk. Show that the regular derivatives Drvn(x)
and Drv(x) with respect to #, existing and finite #-almost everywhere
by Exercise 37.1, satisfy Drvn(x) i Drv(x) #-almost everywhere.
DERIVATIVE OF A SERIES OF INCREASING FUNCTIONS
37.5) Let gn(x), for n== 1,2, . . ., be non-decreasing on Rl, and let
g(x) == = 1 gn(x) be finite on R 1 . Show that the derivatives satisfy
g'(x)==r g;"(x), except on a set of Lebesgue measure zero. This result
is due to G. FUBINI (1915, [2J). Give one proof by combining the re-
sults in the Exercises 37.2 and 37.4, and give an "elementary" proof
by using only the result in Exercise 37.2, as follows. We may restrict
ourselves to a bounded interval [a, bJ, and we may assume without
loss of generality that gn(a) ==0 for all n. Let sn(X)==i=l gi(X) on [a, bJ,
so Sn i g on [a, bJ. The derivatives s;" and g' exist almost everywhere on
[a, bJ as finite numbers, and S;"S;"+lg' for all n; hence, s;" converges
almost everywhere to a sumfunction which does not exceed g' (x). In
order to prove that g' -s;"-*O almost everywhere, it is sufficient to
prove this for a subsequence g' -Sk' Select the indices nk such that
g(b)-snk(b)<2- k , and show that g'-Sk-*O.
LEBESGUE POINTS
37.6) If j(x) is Lebesgue summable over R1, the differentiation
theorem may be formulated by stating that, for almost every xER 1 ,
the expression J {j(x+t)-j(x)}dtjh tends to zero as h-*O. Prove
the stronger result that, for almost every xER 1 , the expression
J Ij(x+t)-j(x)ldtjh tends to zero as h-*O. The points where this
holds are called the Lebesgue points of j. Start the proof by observing
that if ex is any real number, then it follows from the differentiation
theorem that
h
I . f Ij(x+t) -exl dt I j( ) I
1m h == X-ex
h-+O
o
for all xER 1 -EO(" where EO(, is a null set. Hence, if ex runs through the
rational numbers, and E ==  EO(" then E is a null set. Show now that
any xER 1 -E is a Lebesgue point of j.

Ch. 8,  37J
FINITE-DIMENSIONAL NUMBER SPACE
283
DIFFERENCE SET OF TWO SETS
37.7) Let E 1 and E 2 be Lebesgue measurable subsets of Rk having
positive measure. Show that there exists a point cER k such that
#{E 1 (E 2 -c)}>0, where E 2 -c is the set of all points x-c, xEE 2 . For
the proof, observe that if v1(E)==#(EE 1 ) and v2(E)==#(EE 2 ) for all E,
then there exist (in view of the density theorem) points aEE 1 , bEE 2
such that DrV1(a) == 1 and DrV2(b) == 1. Hence, if c==b-a, there exists an
open sphere A of centre a such that #(E 1 A»!#(A) and #{E 2 (A+c)}
>!#(A). It follows that also #{(E 2 -c)A}>!#(A), and so #{E 1 (E 2 -c)}
>0.
37.8) Let E1 and E 2 be Lebesgue measurable subsets of Rk having
finite positive measure, and let f(x)==#{E 1 (E 2 -x)} for all xER k . Show
that f(x) is continuous. Observe that f(x) == J XEI (Y)XE2(Y+X) dy, so
It(x+h) -f(x) 1/lxE2(y+x+h) -XE2(Y+X) I dy-+O
as h-+O by Exercise 30.17.
37.9) Let E1 and E 2 be Lebesgue measurable subsets of Rk having
positive measure. The difference set J(E 1 , E 2 ) is the set of all points
X==X2-Xl, X2EE2, Xl EEl. Show that J(E 1 , E 2 ) contains an open sphere,
thus generalizing the result in Exercise 10.9. For the proof we may
evidently assume that E 1 and E 2 are of finite positive measure. By
Exercise 37.7 there is a point cER k satisfying #{E 1 (E 2 -c)}>0, and by
the preceding exercise the function f(x)==#{E 1 (E 2 -x)} is continuous.
Hence, there is an open sphere K of centre c such that f(x) >0 for all
xEK. It follows that, for any given xEK, the set E 1 (E 2 -x) is not
empty, so there exist Xl EEl and X2EE2 such that X1==X2-X, i.e., x==
X2-X1. The present proof is due to S. KUREPA (1960, [IJ).

CHAPTER 9
NEW VARIABLES IN A LEBESGUE INTEGRAL
In this chapter we shall derive the transformation formula which describes
the effect of a change of variables in a Lebesgue integral. Such a change of
variables is represented by a certain transformation, and what the formula ex-
presses is, essentially, the theorem that a certain set function has, almost
everywhere, as its derivative the absolute value of the Jacobian of this transfor-
mation. We shall prefer, however, to make no use of differentiation theory,
since it is possible to present an independent proof which is neither longer nor
more difficult. The exercises deal with an extension to more general cases and
with some examples.
38. A Simple Example in HI
In this section we present a simple example in order to show what
happens if a "new variable" is introduced in a Lebesgue integral over
R 1 . The example covers the greater part of the cases met in practical
applications.
Let # be Lebesgue measure in the real line R1, and let cp(x), defined
on R 1 and increasing in the strict sense (i.e., cp(X1) <cp(X2) for Xl <X2)
have the continuous derivative cp'(x). Hence, cp'(x)O on R 1 . Obviously
the numbers cp(-oo)==limcp(x) as x!-oo and cp(+oo)==limcp(x) as
xjoo exist, where it is possible that cp( -00) == -00 and (or) cp( +00)
== +00. The Stieltjes-Lebesgue measure in R1, generated by cp(x), will
be called v, and we shall prove first that J f dv== J fcp' dx for all v-
summable f (as on previous occasions, we write dx instead of d# for
the Lebesgue measure #). In order to do so we observe that, given the
bounded cell (a, b], there exists a constant M>O such that cp'(x)M
on (a, b] (this is due to the continuity of cp'(x) on [a, b]); hence, if A ==
(a', b'] is a cell contained in (a, b], we have
v(A) ==cp(b') -cp(a')M(b' -a') == M# (A ),

Ch. 9, 938J
A SIMPLE EXAMPLE IN Rl
285
so the exterior measures v* and #* satisfy v*(E)M#*(E) for any set
E c (a, bJ. In particular, any #-null set in (a, bJ is a v-null set. Since #
is a-finite, it follows then that any #-null set in R 1 is a v-null set, and
this implies immediately that any #-measurable set is v-measurable,
and also that v is #-absolutely continuous. Hence, by the Radon-
Nikodym theorem, there is a #-measurable lo(x)O satisfying! Idv=
! Ilodx for all v-summable I (i.e., equivalently, for all I for which 110 is
fl-summable). In order to show now that lo(x)=q/(x), it will be suf-
ficient to verify that! I dv=! Ip' dx holds for each characteristic func-
tion I==XA, where A=(a, bJ is an arbitrary cell (cf. the remark (3) in
sec. 32). This, however, is evident in view of
I XAdv== I dv=p(b) -p(a) = I p'(x) dx== I XAP' dx.
.A .A.
Our next purpose is to get rid again of the measure v, and to express
jldv as a Lebesgue integral over (p(-oo), p(+oo)). By xx'=p(x) we
define a one-one mapping of R 1 on the open interval {x': p(-oo)<
x'<p(+oo)} of the real line Ri, and we show first that EcR 1 is v-
measurable if and only if p(E) c Ri is #-measurable, and also that in
this case we have v(E)==#{p(E)} (by p(E) is meant the image of E, i.e.,
the set of all x' =p(x) for xEE). The assertion is true if E is a cell
(since E is a cell if and only if p(E) is a cell), so that the assertion is
also true if E (or, equivalently, p(E)) is a a-set, or the limit of a de-
scending sequence of a-sets. This shows, in addition, that v(E) =0 if
and only if #{p(E)}==O, and the desired result follows. It is an im-
mediate consequence that, given the arbitrary v-step function I(x) =
l cnXEn(X), we have
p p
Ildv== cnv(En)== cn#{p(E n )}= 11{-l p (X')}dx',
1 1
where x=-l p (x') is the inverse function of x'=p(x). The extension of
the formula! Idv==! 1{-l p (X)}dx to all v-summable I (i.e., equivalently,
to all I such that 1{-l p (X)} is #-summable over the interval p(R 1 ) ==
{x: p( -00) <x<p( +oo)}) is then not difficult (approximation by step
functions). Combining this result and the former result that! Idv=
! Ip' dx, we obtain
I 1{-l p (x)}dx= J I(x)p'(x) dx
(lll) ill

286
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9, 9 38
for all j such that j{-lcp(X)} is #-summable over cp(R 1 ) , i.e., equiva-
lently, for all j for which jcp' is #-summable over RI'
If, instead of increasing, cp(x) is decreasing in the strict sense, we
get a similar result if the measure v(A) of the cell A === (a, bJ is now
defined by v(A)===cp(a)-cp(b); the image cp(R 1 ) of R 1 is now the open
interval {x': cp(+oo) <x'<cp(-oo)}. Hence, if cp(x) is monotone in the
strict sense, and cp'(x) is continuous, we have
I j{-lcp(x)}dx== I j(x) Icp'(x) I dx.
(ill) ill
Setting j{-lcp(x)}===F(x) , we have j(x) === F{cp (x)}, so that we may as well
write
I F(x) dx== I F{cp(x)}lcp' (x) I dx,
(ill) ill
and this holds for all F such that F is #-summable over cp(R 1 ), i.e.
equivalently, for all F such that F {cp(x) }Icp' (x) I is #-summable over R1
A similar result holds if R 1 is replaced by an open interval (a, b) ;
cp(x) is then merely defined on (a, b).
ExamPles. If A=¥:O is a real constant, and cp(x) ===AX, we obtain
I F(x) dx== IAII F(AX) dx.
ill ill
If A is a real constant, and cp(x) ==X+A, we obtain
I F(x) dx=== I F(X+A) dx.
ill ill
In these examples a more direct argument is of course available (ap-
proximation by step functions).
Exercises
AN EXTENSION
38.1) Show that the obtained result may be extended as follows
Let 1p(x) be non-negative on Rl, and #-summable over each (bounded)
cell, with the additional property that for each cell A the integral of 1p'
over A is positive. Set cp(x) === It 1p(t) dt for all xER 1 ; hence, cp(x) is

Ch. 9, 939J
LINEAR TRANSFORl\1ATIONS IN Rk
287
strictly increasing on R 1 . Then
I F(x) dx== I F {p(x) }1p(x) dx
cp(RI) Rl
holds for all F such that F is #-summable over p(R 1 ) or, equivalently,
for all F such that F{p(x)}1p(x) is #-summable over R 1 . Note that it is
not necessary to use in the proof that p' (x) ===1p(x) holds #-almost every-
where.
39. Linear Transformations in Rk
Let y==Tx be a linear transformation in k-dimensional space R k ,
i.e., if X==(X1, . . ., Xk), Y===(Y1, . . ., Yk), and (tij) is the matrix of T, then
! l =tllXl +t12 X 2+ · · · +tuxk,
Yk==tk1 X 1 +tk2 X 2+ . . . +tkkXk.
Of course, the numbers tij (i, i == 1, . . " k) are real. As is well-known,
T is non-singular (i.e., T has an inverse transformation T-1) if and
only if the determinant det (T) *0. Instead of y==Tx we shall write
occasionally (Y1, . . . , Yk) == T(X1, . . ., Xk), since this is an efficient way
to indicate a transformation of one of the following types T a, T (3 or T y :
T a(X!, . . . ,Xi, . . . , Xk) === (Xl, . . . ,AXi, . · · , Xk),
where A=FO,
T {3 (x 1, . . . , Xi, . . " X k) == (X 1, . . . , Xi + AX j, · · " X k) ,
where i =F i , and A is real,
Ty(X1, ...,Xi, ...,Xj, ...,Xk)==(X1, ...,Xj, ...,Xi, ...,Xk).
Evidently each of these transformations is non-singular, and det (T rx)
===A, det (T (3) == 1, det (T y) === -1. We recall the multiplicative property
of the transformation determinant; if T 1 and T 2 are linear transfor-
mations in Rk, then det (TIT 2) ==det (T 1) det (T 2)' The unit transfor-
mation E is defined by Ex===x for all xER k .

288
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9,9 39
THEOREM 1. Every non-singular linear transformation T in Rk is the
product of a finite number of transformations of the kinds T a, T {3 or T y.
PROOF. Since the inverse of a transformation of one of the kinds
T a, T {3 or T y is of the same kind, it will be sufficient to prove that the
unit transformation E may be written as E  TIT 2. . . T pT, where each
of the transformations T 1, . . " T p is of one of the kinds T a, T {3 or T y
(indeed, if E  TIT 2. . . T pT, then T == T;l . . . TilT!l). Observing now
that mat (TaT) is obtained by multiplying the i-th row of mat (T) by
A, mat (T (3T) is obtained by addition of A times the j-th row to the i-th
row in mat (T), and mat (T yT) is obtained by interchanging the i-th
and j-th rows in mat (T), we proceed as follows (take k==3 for sim-
plicity): In the first column of mat (T) there is at least one number
*0; applying a transformation T y, we can make t11 *0 if it is not al-
ready so, and next, applying transformations T {3, we can make t21 ==
t31 o (it will cause no confusion if we use the same notation tii for all
the matrices appearing thus). At this point in the procedure we have
t22 t23 =F 0
t32 t33 '
since det (T) continues to differ from zero; applying a T y we can make
t22*0 if it is not already so, and next we can make t12t32==0 (all
this without changing the numbers in the first column). Then t33*0,
and we make t13t230 (without changing the numbers in the first
and second columns). Finally, applying transformations T a, we make
t11 t22t33 1.
THEOREM 2. Let T be a non-singular linear transformation in Rk, and
let J  Idet (T) I. Then, if fl is Lebesgue measure in Rk and f(x) is Lebes-
gue summable over Rk, we have
I f(x) dfl== I f(Tx)J dfl.
Rk Rk
PROOF. On account of the preceding theorem and the multiplicative
property of the transformatjon determinant it will be sufficient to give
the proof for transformations of the types T a, T {3 and T y.

Ch. 9,  39J
LINEAR TRANSFORMATIONS IN Rk
289
(ex) J f(x) d#==
Rk
== J . . . J {J f(X1, · · · ,Xi, . . " Xk) dXi} d(X1, · . . , Xi-I, Xi+ 1, · · " Xk)
== IAIJ. . · J{Jf(X1' . . " AXi, · · " Xk) dXi} d(X1, · · " Xi-I, Xi+1, . . " Xk)
== J f(T (XX)] dfl
Rk
by Fubini's theorem and one of the examples in the preceding section.
(fJ) J f(x) d#==
Rk
== J . . · J {J f (x 1, . . . , Xi, · . " X k) dXi} d (X 1, . . . , Xi -1, Xi+ 1, . · · , X k)
== J. . · J {J f (X 1, · · " Xi + AX j, . · " X k) dXi} d (X 1, . · ., Xi -1, Xi+ 1, . . . , X k )
== J f(T pX)] d#,
Rk
once more by Fubini's theorem and one of the examples in the pre-
ceding section.
(y) Follows immediately from Fubini's theorem.
COROLLARY. If E c Rk is Lebesgue measurable, and we denote by T(E)
the set of all points Tx, xEE, then
#{T(E)}==] #(E).
PROOF. Follows by choosing f==XT(E)'
THEOREM 3. The real linear vector space of all points x== (Xl, . . " Xk),
where Xl, . . " Xk are real numbers, is a Banach space M k with respect to
the norm IIxll ==maxlik IXil.
PROOF. It is evident that IIxil ==0 if and only if x== (0, . . " 0), IIx+YII
llxlI+ IIyll, and IIaxII== lal' IIxll for any real constant a. The completeness
of the space is derived from the fact that limllxm-xnll==O for m, noo
implies convergence for each of the coordinates separately.
THEOREM 4. If A is a linear transformation with matrix (aiP) in the
Banach space M k of the preceding theorem, then the norm IIAII=
sup!!x!I=lI1 Ax ll atisfies IIAII==maxlik P laipl.

290
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9,9 40
PROOF. Denoting maxlik p laipl by Q, it follows from Yi===
p aipx p that IYillIxll p laiPIQllxll, hence IIYIIQllxll, i.e., IIA IIQ.
On the other hand, if maxlik p laipl is attained for i===j, then
p lajpl ===Q, and we define x== (Xl, . . " Xk) by x p == + 1 if ajp?::-O, and
x p ==-1 if ajp<O. Then IIxll==l, and Yj==pajpxp===plajpl==Q, so
IIylI?::-Q==Qllxll, and it follows that IIAII?::-Q. Hence, IIAII===Q. .
Any linear transformation on M k into M k is therefore bounded, so
that, by the results in sec. 26 (in particular Theorem 2), the collection
of all linear transformations A on M k into M k is a Banach space with
respect to the norm IIAII==maxlik p laipl. The unit transformation
E clearly satisfies liE II == 1, and the inequality
IIA1A2XIlIIA111' IIA2xllIIA111 'IIA211' IIxll
implies that IIA 1 A 211IIA III 'IIA 211.
40. Change of Variables in a Lebesgue Integral
We assume in this section that the real functions hI (x), . . ., hk(x)
are defined on the open set 0 1 c Rk, hence h(x) == (hI (x), . . . , hk(x)) on
0 1 , and that all partial derivatives jiP(X) ==ohi(x) jox p exist as continu-
ous functions on 0 1 . In addition, we shall assume that det (jiP(X)) *0
at all points XE01, and we shall denote by O 2 the image h(Ol) of 0 1
(the set O 2 is, therefore, the set of all points Y=== (Y1, . . " Yk) such that
Yi==hi(x), i== 1, . . " k, for some XE01). Then, as well-known, the
mapping y===h(x) of 0 1 onto O 2 is locally invertible, that is, if XOE01
is given and yo===h(xo), there exist open neighbourhoods of Xo and Yo,
the points of which are in one-one correspondence by means of the
mapping y==h(x). We shall assume now, finally, that the mapping is
even invertible in the large, in other words, we assume that y==h(x)
establishes one-one correspondence between the points of 0 1 and O 2 .
It is a well-known property of such a mapping that O 2 is now also an
open set, and the inverse mapping x==-lh(y), defined on O 2 , has also
continuous first partial derivatives. In addition, if yo==h(xo), the matrix
product
( Ohi ) ( o-lh )
mat · mat
oX p Xo oYp yo

Ch. 9,  40J CHANGE OF VARIABLES IN A LEBESGUE INTEGRAL 291
is the unit matrix. The image of any open subset is open (for the
mapping h as well as for the inverse mapping -lh), and the image of
any compact (i.e., bounded and closed) subset is compact. The function-
al determinant o(h) /o(x) ==det (fip(X)) is usually called the Jacobian of
the mapping h, and we shall denote its absolute value by J(x).
I t will be our purpose to prove the transformation formula
I f(y) dfl== I j{h(x)}](x) dfl,
02 01
where fl is Lebesgue measure in Rk.
LEMMA cx. If C is a closed interval contained in 0 1 , and h(C) is its
(compact) image in O 2 , then
fl{h(C)}1 J(x)dfl.
G
PROOF. We denote by f(x) and -If(y) the linear transformations in
the Banach space M k (introduced in the preceding section) with matrix
elements fiP(X)==ohi(x)/ox p and -lfiP(y)==o-lh i (y)/oyp respectively.
Since all functions fip(X) and -lfiP(Y) are continuous, the norms IIf(x)II
and 1I-1f(y) II of the transformations f(x) and -If(y) are continuous
functions on 0 1 and O 2 respectively; in particular, II-1f(y) II is bounded
on h(C). The continuity of the functions fiP(X) and -lfiP(Y) implies also
that, given 8>0, there exists a number £5>0 such that IIf(x)-f(x')II<8
for any pair of points XEC, x' EC, satisfying IIx-x'II<£5.
Assuming first that C is a closed cube (i.e., a closed cubic interval),
let Xo== (X01, . . " XOk) be its centre and 2s the length of its edges; hence,
C=={x: Ilx-xolls} and fl(C)==(2s)k. For any X1==(X11, ..., X1k)EC we
have
k
h i (X1) -hi(xo) ==  fiP[XO+(Ji(X1) (X1- X O)J (X1p- X O P )
p=l
for some (Ji(X1) between 0 and 1, so
Ilh(X1)-h(xo)lls maxllf(x)ll.
XEG
In other words, h(C) is contained in the cube
{z: Ilz-h(xo) "s max IIf(x) II},
XEG

292 NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9,9 40
and it follows that
#{h(C)}{max Ilf(x) II}k#(C).
XEC
(1)
Next, let A be a non-singular linear transformation (the matrix ele-
ments aij of A are therefore real constants), and consider the transfor-
mation l(x)==A- 1 h(x), defined on 0 1 . It has the same continuity and
invertibility properties as h(x) itself, and the relations 8l i (x) f8x p ==
=1 a;l)jnp(x) show that the role of j(x) is taken over by the transfor-
mation A -lj(X). Hence, applying (1),
#{A -lh(C)}{max IIA -lj(x)ll}k#(C).
XEC
(2)
Also, by the Corollary of Theorem 2 in the preceding section,
#{A -lh(C)}== Idet (A -1) I' #{h(C)}.
Combining (2) and (3), we obtain therefore
#{h(C)}ldet (A) I {max IIA -lj(x)II}k#(C).
XEC
(3)
(4)
Now, given some £5>0, we divide C into a finite number of non-
overlapping closed cubes C 1, .. " C m with centres x(l), . . " x(m), such
that all edges of all C i are smaller than £5 in length. We then apply (4)
to each C i in succession, choosing A ==f(x(i»), so A -l==-lf(y(i»), where
y(i) =h(x(i»), and we sum over i, obtaining in this way
m
#{h(C)}  Idet j(x(i») I {max II-1j(y(i»)f(x) II}k#(C i ).
i= 1 XECt
Next, if M is such that 11-1j(Y)IIM on h(C), and 8>0 is given, we
choose the above £5>0 such that XEC, x' EC, IIx-x'II<£5 implies
IIj(x)-j(x')II<efM. Then, for XEC i and E the unit transformation, we
have
I II-1j(y(i»)j(x)II- l l==1 II- 1 j(y(i»)j(x)II-IIEil I
11-1j(y(i»)j (x) -Ell == 11- 1 j(y(i») {j(x) _j(x(i») }II
11-1f(y(i»)II'llj(x)-j(x(i»)II<e,
hence
{max 11-1j(y(i»)j(x) lI}k( 1 +e)k
XECt

Ch. 9,  40J CHANGE OF VARIABLES IN A LEBESGUE INTEGRAL 293
for i== 1, . . " m, and this implies that
m
fl{h(C)}(1 +e)k  Idet i(x(i») Ifl(C i ),
i=l
Letting !O (which is permitted for fixed e>O), the sum on the right
tends to the Riemann integral /c](x) dfl, so
fl{h(C)}(1 +e)k J ](x) dfl.
C
This holds for all e>O, hence, finally,
fl{h(C)}J ](x) dfl.
C
I t is evident that the same method of proof yields the same result if
C is a closed interval (not necessarily a cube), contained in 0 1 .
We extend the result, proved in Lemma ex for a closed interval, to
an arbitrary measurable set.
LEMMA fl. If EeOl is fl-measurable, then its image h(E) is fl-measur-
able, and
fl{h(E)}J ](x) dfl.
E
PROOF. By Lemma ex, the assertion is true if E is a closed interval
C, contained in 0 1 . We shall prove first that it is also true if E is a
null set contained in the closed interval C. Let M ==max ](x) on C,
and let e>O. Then there exists a set S==r C n , where the sets C n are
closed intervals, such that EeSeC, and r fl(C n ) <elM. Indeed,
cover E first by disjoint cells of total measure smaller than elM, and
let C n be the intersection of the closure of the n-th cell with C. Hence
h(E) eh(S), so
00
00
00
fl*{h(E)}fl{h(S)} fl{h(Cn)} J ] (x) dflM  fl(C n ) <e.
1 1 Cn 1
This shows that fl{h(E)}==O, Le., fl{h(E)}==O== /E](X) dfl. Denoting by
A the cell having the same vertices as the closed interval C e 0 1 , it is
an immediate consequence that h(A) is fl-measurable (note that C-A
is a null set), and fl{h(A)}==fl{h(C)}==/c](x)dfl==/A](x)dfl. The as-

294
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9, 9 40
sertion to be proved holds therefore for the cell A, and then also for
any cell contained in A. It follows that the assertion holds for any a-
set E c A, and then also for any E c A which is the limit of a descending
sequence of a-sets. Finally, the assertion holds if E c A is a null set,
and this shows then that it holds for any measurable E cA. Since 0 1
is a countable union of disjoint cells A of the type that the closure of
A is still contained in 0 1 , the assertion is true for any measurable
E cO l .
Finally, we derive the desired transformation formula. The proof
consists, essentially, in showing that the inequality in the last lemma
is an eq uali ty .
THEOREM 1. The junction j(y) is Lebesgue summable over O 2 ij and
only ij j{h(x)}](x) is Lebesgue summable over 0 1 , and then
Ij(y)dfl== Ij{h(x)}](x)dfl'
02 01
PROOF. Since the mapping y==h(x) of 0 1 on O 2 is invertible, and
since the inverse mapping x==-lh(y) has the same properties as y==
h(x), it follows from the preceding lemma that the mapping generates
a one-one correspondence between the measurable subsets of 0 1 and
O 2 . Every measurable subset of O 2 is, therefore, the image h(E) of
some measurable E c 01, and, if j== Xh(E)' we have
I j(y) dfl==fl{h(E)},
02
I j{h(x)}](x) dfl== I XE (x)] (x) dfl=== I ](x) dfl,
01 01 E
hence
I j(y) dfl1 j{h(x)}](x) dfl
02 01
(5)
by Lemma fJ. It follows that (5) holds for any step function on O 2 , and
this implies that (5) holds for any measurable j(y)O on O 2 , since there
exists a sequence of step functions jn i j.
Similarly, writing ]*(y) === Idet -ljiP(y) I on O 2 , we have
I g(x) dfl1 g{-lh(y)}]*(y) dfl
01 02

Ch. 9,9 40J CHANGE OF VARIABLES IN A LEBESGUE INTEGRAL 295
for any measurable g(x)O on 0 1 , so
I f{h(x)}](x) dfl1 fCh{-lh(Y)}J]{-lh(y)}]*(y) dfl= I f(y) dfl, (6)
01 02 02
since ]{-lh(y)}]*(y)== 1 for all YE02. Combining (5) and (6), we obtain
If(y)dfl= If{h(x)}](x)dfl
02 01
(7)
for any measurable f(y)O on 02, and the same is then evidently true
for any f(y) which is Lebesgue summable over 02. It remains to prove
that (7) holds also for any f such that f{h(x)}](x) is Lebesgue summable
over 0 1 .
Similarly as in (7), we have, of course,
I g(x) dfl== I g{-lh (y)}]* (y) dfl
01 02
for any g(x) which is Lebesgue summable over 0 1 ; hence, assuming now
that f{h(x)}](x) is Lebesgue summable over 01, and setting g(x) ==
f{h(x)}](x), we find
I f{h(x)}](x) dfl== I f(y) dfl.
01 02
This completes the proof. The method of proof based upon Lemma ex,
as we have presented it here, is due to J. SCHWARTZ (1954, [IJ).
ExamPle (TRANSFORMATION FROM RECTANGULAR TO POLAR COORDI-
NATES). Let 0 1 be the open set in the (r,cp)-plane, defined by 0 1 ==
{(r, cp): O<r<oo, 0<cp<2n}, and let 02 be the entire (x, y)-plane with
the exception of the points (x, 0), xO. By
x==h 1 (r, cp)==r cos cp,
y==h 2 (r, cp) ==r sin cp,
we define a one-one mapping of 0 1 onto 02 with continuous partial
derivatives, such that ](r, cp)=r*O on 0 1 . Hence
I f(x, y) dfl== I f(r cos cp, r sin cp)rdfl,
02 01
or, in the more usual notation (integration over 02 is the same as inte-

296
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9,9 40
gration over the entire (x, y)-plane),
00 00 00 2n
J J f(x, y) dxdy== J J f(r cos p, r sin p)rdrdp.
-00 -00 0 0
Similarly
00 00 00 in
J J f(x, y) dxdy== J J f(r cos p, r sin p)rdrdp.
o 0 0 0
Exercises
EXTENSION OF THE TRANSFORMATION FORMULA TO NON-INVERTIBLE
MAPPINGS
40.1) Let 0 1 c Rk be open, and let h(x) == (hI (x), . . . , hk(x)) be defined
on 0 1 such that all partial derivatives fiP(X)==8h i (x)f8x p are continuous
on 0 1 and det (fip(X)) *0 on 01. The image h(Ol) of 0 1 is denoted by
O 2 . As observed before, the mapping yh(x) of 0 1 onto O 2 is locally
invertible; we shall not assume in this exercise that the mapping is
invertible in the large. Given XOE01 and writing yo==h(xo), there exist
open neighbourhoods %(xo) and %(yo) of Xo and Yo respectively, the
points of which are in one-one correspondence by means of the mapping
y==h(x). Since 0 1 is the union of all %(xo) for Xo running through 0 1 ,
it follows that O 2 is the union of the corresponding ,;V(Yo), so O 2 is
open. For any YE02, let N(y) be the number of different XE01 satis-
fying h(x)y. Hence, IN(y)oo. The function Idet (fiP(X)) I will be
denoted by ](x), and Lebesgue measure in Rk by #. Show that N(y)
is Lebesgue measurable on O 2 , and also that for any f(y), non-negative
and Lebesgue measurable on O 2 , the equality
J N(y)f(y) d# J f{h(x)}](x) d#
02 01
holds. Show, finally, that if f(y) is complex and Lebesgue measurable
on O 2 , and either f{h(x)}](x) is #-summable over 0 1 or N(y)f(y) is #-
summable over O 2 , then the same equality holds.
40.2) We make the same assumptions on 0 1 and the mapping h as
in the preceding exercise, except that we admit now that ] (x) ==0 for

Ch. 9,9 40J CHANGE OF VARIABLES IN A LEBESGUE INTEGRAL 297
some points XE01. Show that if J(a) ==0 at a== (al, . . " ak) E01 and
e>O is given, then there exists a closed cube U(a) of centre a such that
#{h(C)}e#(C) for every closed interval C satisfying C c U(a).
40.3) Show that if U(a) is the same closed cube as in the preceding
exercise, and E is an arbitrary measurable subset of U(a), then the
exterior measure #*{h(E)} of the image h(E) satisfies #*{h(E)}e#(E).
Note in particular that if A is a compact subset of U(a), then its image
h(A) is also compact, so #{h(A)}e#(A).
40.4) Let 0 1 c Rk be open, and let h(x) === (hI (x), . . " hk(x)) be de-
fined on 01 such that all partial derivatives iiP(X)==ohi(x)jox p are con-
tinuous on 0 1 . For any YEh(Ol), let N(y) be the number of different
XE01 satisfying h(x)==y, and let J(x) denote the function Idet(iip(x))I.
Show that if f(y) is non-negative and measurable on h(Ol), then
I N(y)f(y) d#== I f{h(x)}J(x) d#.
h( OI) 01
A SUFFICIENT CONDITION FOR INVERTIBILITY
40.5) Let 0 c Rk be open and convex (i.e., if a and b are two arbi-
trary points in 0, then the segment from a to b lies wholly in 0), and
let h(x) == (h 1 (x), . . " hk(x)) be defined on 0 such that all partial de-
rivatives iiP(X)===ohi(x)jox p are continuous on O. Assume, furthermore,
that for all XEO and all cx== (CX1, . . " CXk) ERk, cx=¥=(O, 0, . . " 0), the quad-
ratic form = 1 = 1 iiP(X)CXiCXp is positive. Show that J(x) == Idet (iiP(X)) I
=¥=O at every point XEO, and the mapping of 0 onto its image h(O) is
invertible, i.e., show that h(a) =¥=h(b) for a=¥=b. Consider in particular
the case that k==2, and h(x, y)==(u(x, y), v(x, y)) with oujox===ovjoy,
oujoy=== -ovjox, and oujox>O at all points (x, y) EO (oral communi-
cation by W. A. J. Luxemburg).
AREAS
40.6) Evaluate, in terms of Gamma functions, the twodimensional
Lebesgue measure of the open set 0 in the (x, y)-plane, defined by
O=={(x, y): lax+by+cI P + la1x+blY+C1IQ< 1,
LI ==ab 1 -a 1 b=¥=0, P>O, q>O}.

298
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9,  40
Check the cases that a==b 1 == 1, b==c==a1 ==C1 ==0, and p==q==2 or P==
q== 1.
40.7) Let O<a 1, and let 0 be the open set in the (x, y)-plane,
defined by
O=={(x, y) : O<x< tn, O<y< tn, sin y<a cos x}.
Note that for a== 1 the boundary of 0 consists of straight line segments.
By the relations u(x, y)==sin y/cos x, v(x, y)==cos y the set 0 is mapped
in a one-one way onto an open set 0 1 in the (u, v)-plane. Write the
Lebesgue measure of 0 as a double integral over the set 01, and develop
the double integral as a power series in a. Check the result for a== 1
(oral communication by H. D. Kloosterman).
AN INTEGRAL OF DIRICHLET
40.8) Let f(x) be bounded and Lebesgue measurable on [0, 1J, and
let PI, . . . , Pn be positive. Furthermore, let O 2 be the open set in Rn
defined by
O 2 == { (y 1, . . . , Y n) : Y i > 0 for i == 1, . . . , n) y 1 + . . . + y n < 1 }.
Show that the integral
D== /f(Y1+... +Yn)Yfl-ly2-1.. .yn-ldY1.. .dYn
02
sa tisfi es
1
D == T(P1)'" T(Pn) f f(X)XPl+P2+...+pn-l dx.
T(P1+" .+Pn)
o
Start the proof by mapping O 2 one-one onto
01 =={(X1, . . " Xn) : Xi >0
for i== 1, . . " n-l, Xl + . . . +X n -1 < 1, O<xn< I}
by means of Yi==Xi for i== 1, . . ., n-2, Yn-1 ==Xn-1Xn,
Yn==X n -1 -X n -1 X n.
40.9) Let f(x) be bounded and Lebesgue measurable on [0, IJ, and
let al, . . . , an; PI, . . . , Pn; q1, . . " qn be positive. Furthermore, let 0

Ch. 9, 940J CHANGE OF VARIABLES IN A LEBESGUE INTEGRAL 299
be the open set in Rn defined by
n
O=={(X1, "', x n ): Xi>O for i==l, ..., n,  (xi/ai)Pt<I}.
i=1
Show that
f IH x 1/ a 1)Pl+ · · · + (Xn/an)pn}xi' -1. . .x;n-1 d X 1. · · dXn=
o 1
ail. · · an r (r1) · · · r (r n) f + + _ 1
==' f(x)xTI ... Tn dx
P1...Pn r(r1+...+ r n) ,
o
where ri==qi/Pi for i== 1, . . . ,n. Show, in particular, that the n-di-
mensional Lebesgue measure of the subset of Rn, bounded by the
-ellipsoidal surface  (Xi/ai)2== 1, equals a1... ann!n/ r(!n+ 1 ). Note
that if V n is the n-dimensional volume of the unit sphere  x;< 1,
then V n is maximal for n==5, and V n tends to zero as noo.
POSITIVE DEFINITE MATRICES
40.10) In this exercise, and the next ones, the reader is assumed to
have some familiarity with linear algebra. Let (aij); i, i == 1, . . . , n, be
.a real matrix which is symmetric (i.e., aij==aji for all i, i) and positive
,definite (Le., j=1 aijCXiCXj>O for all cx== (CX1, . . " cxn) *(0, . . .,0)). The
,determinant of (aij) will be denoted by det (A). Show, immediately
from the definition of positive definiteness, that det (A) >0. Show,
next, that
f ntn
/ = e - 'E.a'jX'X; dft = { det (A)}l ,
Rn
where # denotes Lebesgue measure.
40.11) Let (aij) and (b ij ); i, i == 1, . . " n, be symmetric and positive
.definite matrices, and let OAI. We set Cij==Aaij+(I-A)bij, so that
.(Cij) is also symmetric and positive definite. Show that
det (C){det (A)}A{det (B)}l-A.
'This result is due to Kv FAN (1949, [IJ).
40.12) Given the determinant det (A) with n rows and columns, we
denote by det (A)1,k the determinant formed by the first k ro\vs and

300
NEW VARIABLES IN A LEBESGUE INTEGRAL [Ch. 9, 9 40
columns; similarly, the determinant formed by the remaining rows
and columns is denoted by det (A)k+l,n' Obviously, if (aij) is symmetric
and positive definite, then det (A)l,k and det (A)k+l,n are positive for
1 <kn. Show that
det (A)det (A)l,k det (A)k+l,n,
and derive from this result that det(A)rri=l aii.

CHAPTER 10
MEASURES AND FUNCTIONS ON THE REAL LINE
It will be proved in sec. 41 that there exists a one-one correspondence be-
tween the collection of all measures v, initially defined (and finite) on the semi-
ring of all cells in Rl, and the collection of all functions g(x) on Rl, increasing
on Rl and vanishing at the origin. In addition, it will be shown that v is Le-
besgue absolutely continuous if and only if the corresponding g(x) is the inte-
gral (between 0 and x) of its derivative g'. In the exercises one may find the
Lebesgue decomposition theorem for an increasing function in its original
version. The next section is devoted to functions of finite variation (also called
functions of bounded variation). Additional properties of continuous functions
of finite variation, and applications to arc length, are discussed in the exercises.
41. Measures and Functions on the Real Line
We assume in this section that # is Lebesgue measure in the real
line R1, and v is another measure in R1, obtained by applying the ex-
tension procedure for measures to a measure which was initially de-
fined only on the semi-ring r of all cells A==(a, bJ. For simplicity we
shall also assume that v(A)<oo for all (bounded) cells A==(a, bJ.
THEOREM 1. The measure v, introduced above, is a Stieltjes-Lebesgue
measure generated by the right continuous function g(x), defined by
g(x) == ! v, xJ
-v(x, OJ
for x==O,
for x>O,
for x<O.
The function g is continuous at the point xER 1 it and only if v ({x}) ==0
for the set {x} consisting of the single point x.
PROOF. The proof that v is generated by g(x) was presented already
in sec. 5, Example 8. In order to prove the second statement, let X n i x.

302 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10,9 41
Then {x}== rr;;=l (xn, xJ; hence
v{{x}) ==lim v(xn, xJ ==lim {g(x) -g(x n )},
and it follows that g is continuous at x if and only if v({x}) ==0.
Any real or complex function g(x), defined on R1, is called absolutely
continuous on R 1 whenever, given e>O, there exists a number £5>0
such that for all finite unions  (ai, biJ of disjoint cells satisfying
#{ (ai, b i J}<£5, we have  Ig{b i ) -g(ai) I <e. If P is a closed interval,
and g(x) is defined on P, a similar definition holds for absolute con-
tinuity on P. Evidently, absolute continuity of g implies continuity of
g at all points x.
THEOREM 2. The measure v is #-absolutely continuous if and only if
the corresponding function g(x), introduced in the preceding theorem, is
absolutely continuous on each bounded closed interval.
PROOF. We assume first that v is #-absolutely continuous, i.e., if
#(E) ==0, then E is v-measurable, and v(E) ==0. We prove first that,
given e>O and the bounded closed interval P, there exists a number
£5>0 such that EcP and #(E)<£5 imply v(E)<e. Indeed, assuming
this to be false, there exists a number e>O and a sequence En cp
such that #(En)1/2n and v(En)e, and then E==lim sup En satisfies
00
#(E)#(En+En+1+"')<  2-p==2-(n-1)
p=n
for all n,
so #(E) ==0, whereas on the other hand v(E)lim sup v(En)?::-e (we use
here that En cp for all n, and v(P) <00). This is a contradiction, and
it follows, therefore, that if f= 1 (ai, biJ is a finite union of disjoint
cells in P such that #{ (ai, b i J}<£5, then
n n n
 {g(b i ) -g(ai)}==  v(ai, biJ==V{ (ai, biJ}<e.
1 1 1
In other words, g(x) is absolutely continuous on P.
Assume now, conversely, that g(x) is absolutely continuous on each
bounded closed interval. Let P==[P, qJ, P 1 ==[P-l, q+ IJ, e>O, and let
£5 >0 be such that  (ai, biJ c PI (where all (ai, biJ are disjoint) and
#{ (ai, b i J}<£5 imply  {g(bi)-g{ai)}<e. We wish to show that v is

Ch. 10,  41J MEASURES AND FUNCTIONS ON THE REAL LINE 303
#-absolutely continuous on P, and for this purpose we assume given
the set E c P such that #(E) ===0. Then there is a a-set 5 == r (ai, b i ]
c PI, consisting of disjoint cells, such that E c 5 and #(5) === r #(ai, biJ
<£5, and this implies that  {g(b i ) -g(ai)}<e for all n, so
r {g(bi)-g(ai)}e. It follows that the exterior measure v*(E) of E
sa tisfi es
00
00
v*(E) v(ai, biJ ===  {g(b i ) -g(ai)}e,
1 1
so E is v-measurable, and v(E) ==0.
If, finally, E is an arbitrary subset of R 1 such that #(E)===O, then
E ===  En, where each En is included in a bounded closed interval, and
each #(En)==O; consequently, v(En) ==0 for all n, so v(E) ===0. The final
result is, therefore, that v is #-absolutely continuous.
The preceding theorems show that if g(x), defined and increasing
(i.e., non-decreasing) on R1, is absolutely continuous on every bounded
closed interval, then the corresponding Stieltjes-Lebesgue measure v is
absolutely continuous with respect to Lebesgue measure #. Hence, by
the Radon-Nikodym theorem, there exists then a #-measurable function
fo(x)O on R 1 such that v(E)==J XEfod# for all v-measurable sets E. In
particular, g(b) -g(a) ==1: fod# for any pair of real numbers a<b. In
addition, by the theorem on differentiation of a indefinite integral,
the derivative g'(x) exists and is finite at almost every xER 1 , and
g'(x)==fo(x) at almost every xER 1 . Summing up: If g(x) is absolutely
continuous on every bounded closed interval, then g(b) -g(a) ==
J: g'(x)d# for any pair of real numbers a<b.
Conversely, if fo(x)O is #-measurable on R 1 and #-summable over
every bounded closed interval, then v(E) === J XEfod# is a #-absolutely
continuous measure in R1, and the corresponding function g(x) is abso-
lutely continuous on every bounded closed interval, satisfying
g(b)-g(a)===J: fod# for a<b, and g'(x)===fo(x) for almost every xER 1 .
In connection with Theorem 2, the question may be raised whether
any function g(x), increasing and continuous on R1, is not already
absolutely continuous on every bounded closed interval. The answer is
negative, as shown by the following example.
We define g(x) on [0, IJ as follows:

304 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10,9 41
g(x) ==! on (1, i);
g(x) ==1 on (!' j), and g(x) ==! on (, ) ;
g(x)==l, i, i, ! (from the left to the right) on the open middle third
parts of the four still remaining closed intervals; and so on.
The union of the open intervals, on which g(x) is then already de-
fined, is an open set; its complement C (called Cantor's ternary set) is
therefore closed, and C contains no open interval since, by the con-
struction procedure, the open intervals of the complement of C invade
into each subinterval of [0, IJ. It follows that the domain [0, IJ-C of
the increasing function g(x) is dense in [0, IJ, and it may be noted
incidentally that C is, therefore, a boundary set, so that, being closed
in addition, C is even a nowhere dense set. The range of g(x) is also
dense in [0, IJ; hence limtx g(t) exists for all XE[O, 1J, so that, if we
set g(x)==limtx g(t) for all XEC, the function g(x) is now increasing
and continuous on [0, IJ. Defining g(x)==O for x<O, and g(x)== 1 for
x> 1, the function g(x) is increasing and continuous on R 1 .
Cantor's ternary set C satisfies #(C) ==0 since, after n steps of the
construction, the measure of the union of the remaining closed inter-
vals is (i)n, so, given e>O, there exists an open set 0 =:> C such that
#(0) <e. But 0== r 1 n, where the sets 1 n are disjoint open intervals;
hence, by the Heine-Borel covering theorem, we have already C c
o 1 n for some index no. Let, from the left to the right, 11== (aI, b 1 ),
. . ., 1 no== (a no ' b no ), so a1 <0 and b no > 1. Since no point of C is con-
tained in any of the intervals [b i - 1 , aiJ, we have g(b i - 1 ) ==g(ai) , so
no
 {g(b i ) -g(ai) }==g(b no ) -g(a1) == 1,
1
whereas
no 00
 #(ai, biJ #(1 n) ==#(0) <e.
1 1
This shows that g(x) is not absolutely continuous on [ -1, 2J.
We note finally that if v is the Stieltjes-Lebesgue measure generated
by g(x), and C is Cantor's ternary set, then ,u([0, IJ-C)==1 and
v([O, IJ-C)==O, whereas #(C)==O and v(C)==I. All the power of the
measure v is concentrated, therefore, in the set C of Lebesgue measure
zero. This shows, incidentally, that the number of points in C is un-

Ch. 10,9 41J MEASURES AND FUNCTIONS ON THE REAL LINE 305
countable (since the v-measure of each point, and hence of each count-
able et, is zero). The derivative g'(x) is zero on [0, IJ-C, so g'(x) ==0
holds #-almost everywhere on [0, IJ, and the function g(x) is, therefore,
an example showing that it may happen that
1
g( 1) -g(O) == 1 >0== I g' (x) d#.
o
Exercises
LEBESGUE'S ORIGINAL DECOMPOSITION THEOREM
41.1) Let g(x) be increasing on R1, and g(O) ===0. Show that g(x)==
gl(X) +g2(X), where gl(O) ==g2(0) ==0, gl(X) and g2(X) are increasing, and
gl(X) is absolutely continuous on every bounded closed interval (so
gl(X) ==10 gi(x) d#) with, in addition, g' (x) ==gi(x) holding almost every-
where on R1 (so g2(X) ===0 almost everywhere on R 1 ). This is Lebesgue's
decomposition theorem (presented in general formulation in Exercise
32.5, and for measures in Rk in Exercise 37.1) in its original version
(H. LEBESGUE, 1904, [IJ).
GENERAL CANTOR SETS
41.2) Cantor's ternary set is a particular example of a Cantor set in
the interval [0, IJ. In order to define such a more general Cantor set,
we shall say that the C-procedure is applied to a bounded closed inter-
val A of positive Lebesgue measure #(A) whenever an open interval B,
satisfying O<#(B)<#(A) and with the same centre as A, is removed
from A. A general Cantor set is constructed by first applying the C-
procedure to [0, IJ, then to the two remaining closed intervals (such
that the two removed open subintervals are of the same measure),
next to the four still remaining closed intervals, and so on. The point
set C which remains after a countable number of steps is a Cantor set.
By its construction, any Cantor set is a closed boundary set, and hence
nowhere dense. Show that, given any real cx such that Ocx< 1, there
exists a Cantor set C satisfying #(C) ==cx.
It will be evident how to construct Cantor subsets of an arbitrary

306 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10,9 41
bounded closed interval [a, bJ, and how to extend the construction to
higher dimensions.
A STRICTLY INCREASING CONTINUOUS FUNCTION WITH DERIVATIVE ZERO
ALMOST EVERYWHERE
41.3) Let us say for brevity that the function g(x) on R1, presented
as an example at the end of the present section, is the Cantor function
corresponding to [0, IJ. The Cantor function for an arbitrary bounded
closed interval [a, bJ is defined similarly. The derivative of any such
Cantor function is zero almost everywhere, and this is due of course
to the fact that the function is piecewise constant on intervals the sum
of whose lengths is equal to the measure of the entire closed interval.
One might believe that this is also a necessary condition for the de-
rivative to be zero almost everywhere. Show, however, by suitable
superposition of Cantor functions, that there exists a function g(x),
continuous and strictly increasing on [0, IJ, such that g'(x) is zero
almost everywhere.
MEASURABLE FUNCTION OF A MEASURABLE FUNCTION
41.4) Let I and g be real Lebesgue measurable functions on [0, IJ
such that the range of I is a subset of [0, IJ. Then g{f(x)} is not neces-
sarily Lebesgue measurable, not even if I is absolutely continuous.
Show this by means of the following example. Let C and C 1 be Cantor
sets in [0, IJ such that #(C»O and #(C 1 )==0, where # is Lebesgue
measure, and let I be the function which maps each interval of [0, 1 J-C
linearly on the corresponding interval of [0, IJ-C 1 (the graph of the
function mapping the interval (a, b) linearly on the interval (c, d) is the
line segment, minus endpoints, joining the points (a, c) and (b, d) in
R 2 ). The domain and range of I are dense in [0, IJ, and f is extended
onto the whole of [0, IJ by continuity. Then I is continuous and strictly
increasing, so the derivative I' exists and is finite almost everywhere
on [0, IJ, in particular at all points of D==[O, IJ-C. Show that I is
absolutely continuous. By Exercise 10.7, the set C has a non-measur-
able subset E; let g be the characteristic function of the image E 1 ==
I(E) of E. Show that g is measurable, but g{/(x)} is not measurable.
Finally, show that if g is an arbitrary measurable function on [0, IJ,
then g{f(x) }f' (x) is measurable (cf. Exercise 38.1).

Ch. 10,9 41J MEASURES AND FUNCTIONS ON THE REAL LINE 307
A MEASURABLE FUNCTION WITH A NON-MEASURABLE INVERSE
41.5) Let f be the same function on [0, 1 J as in the preceding exer-
cise, and let E be the same non-measurable set. The function h(x) on
[0, 2J is defined as follows: h(x) ==f(x) for XE [0, IJ -E and h(x) ==f(x) + 1
for xEE; h(x)==f(x-l)+1 for x-lE[O, IJ-E and h(x)==f(x-l) for
x-I EE. Then h maps [0,2J in a one-one manner onto [0, 2J, so that
the inverse function -lh exists. Show that h is non-measurable, but
-lh is measurable.
MEASURE AND CATEGORY
41.6) All sets in this exercise are subsets of X == [0, 1 J, and # is Le-
besgue measure in X. Show that #(E)== 1 implies that E is dense in X
(with respect to the distance d(x, Y)==lx-YI), but the converse does
not hold. Show that #(E)==O implies that E is a boundary set in X,
but the converse does not hold. Show that if E is closed in X and #(E)
==0, then E is nowhere dense, but #(E) ==0 alone does not imply that
E is nowhere dense. Show that, given ex such that Oex< 1, there exists
a nowhere dense #-measurable set C satisfying #(C) ==ex. Finally, show
that there exists a set E of the first category satisfying #(E) == 1, and a
set F of the second category satisfying #(F) ==0.
DERIVATIVE OF A CONVEX FUNCTION
41.7) In Exercise 25.2, convex functions were introduced. We recall
that f(x), defined on the open interval (a, b) cR 1 , is called convex
whenever
f{ exX 1 + (l-ex)X2}exf(X1) + (l-ex)f(x2)
for all Xl, X2 in (a, b)and all ex satisfying Oex 1. Furthermore, if we
denote the difference quotient {f(X2) -f(X1)}/(X2-Xl) by q(X1, X2), then
f is convex if and only if q(X1, X2)q(X2, X3) for all triples Xl <X2<X3;
also, f is convex if and only if q(X1, X2)q(X1, X3) for all triples Xl <
X2<X3. Let f be convex in (a, b), and show that the right derivative
D+f(x)==lim hto q(x, x+h) exists at every xE(a, b) as a finite number.
Similarly, show that the left derivative D-f(x) ==limhtO q(x-h, x) exists
at every xE(a, b) as a finite number, and D-f(x)D+f(x) on (a, b).
Finally, show that D-f and D+f are non-decreasing functions of x on
(a, b).

308 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10, 9 42
41.8) We use the same notations as in the preceding exercise. Let I
be convex on (a, b). Since D-f(x) is non-decreasing, the number of
discontinuity points of D-f is at most countable. Show that D-f(x) ==
D+/(x) at every point of continuity of D-I. Hence, the derivative I'(x)
exists except on an at most countable subset of (a, b), and I' is non-
decreasing.
41.9) Let I be convex on (a, b). Show that f is absolutely continuous
on every closed subinterval of (a, b).
41.10) Let I(x) be convex on (a, b), and let xOE(a, b). Show the ex-
istence of a non-decreasing function g(x) defined almost everywhere
(with respect to Lebesgue measure) on (a, b), namely g(x) ==1' (x) , such
that I(x) == I(xo) + Ix g(t) dt. We recall that for x <XO the integral Ix is
to be interpreted as - 1:°.
41.11) Conversely, let g(x) be finite-valued and non-decreasing on
(a, b), and let xoE(a, b). Show that l(x)==lx g(t)dt is convex on (a, b),
and I' ==g except on an at most countable set. For the convexity proof,
observe that if X1<X2 and X3==exX1+(I-ex)X2 for some ex (OexI),
then the inequality to be proved can be written as
X3 X2
I g(t) dt(I-ex)1 g(t) dt,
Xl
Xl
that is,
X3 X2
ex I g(t) dt (I-ex) I g(t) dt.
Xl
X3
42. Functions of Finite Variation
In order to show of which part of classical analysis the theory of
signed measures in the next chapter is a generalization, we shall
present in this section a short survey of the main properties of functions
of finite variation; the discussion will be restricted to real functions.
Let I be a real finite-valued function, defined on the bounded closed
interval [a, b] c R 1 . Given the partition &P=={a==xO<x1 < . . . <xn==b}
of [a, b], we set S&JI==f=11/(Xi)-/(Xi-1)1. The number VI==V':)==
sup S&JI is called the total variation of lover [a, b]; this number may be
either finite or +00. Note that the least upper bound is taken over all
possible partitions &P of [a, b], where the number n of subintervals in

Ch. 10,9 42J
FUNCTIONS OF FINITE VARIATION
309
g; is of course variable. If Vf<oo, the function f is said to be of linite
variation (or of bounded variation) on [a, bJ; we shall denote the col-
lection of all (real) I of finite variation on [a, b J by BV[ a, b J, or shortly
by BV.
THEOREM 1. 1 ° We have V/== V+ VI for any c satisfying a<
c<b.
2° If fEBV, then f is bounded; more precisely, If(x)Ilf(a)l+ Vf for
all x E [a, b J .
3° We have V(a/)==laIVI and V(/1+f2)Vf1+ V12. Hence, BV is a
linear collection.
4° If f1 and 12 are in BV, then 11f2 EBV .
5° If I is monotone on [a, bJ, then fEBV and Vf==lf(b)-f(a)l.
6 ° I I f is absolutely continuous on [a, b], then f E B V. I t is not true,
however, that every continuous function in BV is absolutely continuous.
PROOF. 1 ° The union of any partition of [a, c] and any partition of
[c, bJ is a partition of [a, bJ, so V+ VfV':J follows by taking least
upper bounds. Conversely, note that for any partition fJJ of [a, b] the
corresponding SflJf is not decreased if c is added as one of the partition
points. Hence, denoting the new partition by fJJ 1 , we have SflJfSflJlf,
and SflJ1f may be decomposed into parts corresponding to [a, cJ and
[c, bJ respectively. Taking least upper bounds, we obtain V':J
V+ V/.
2° If fEBV and xE[a, bJ, then
I/(x) -f(a) I + If(b) -/(x) IVf
by the definition of VI, so I/(x)-/(a)IVf, and this implies that If(x)l
I/(a)l+ Vf.
3° It is evident from the definition of Vf that V(cx/)== IcxlVf for any
I, and
V(f1 +f2)Vf1 + Vf2
is an immediate consequence of
SflJ(/1 + f2)SflJf1 +SflJf2,
holding for all partitions fJJ.
4° Let 11 and f2 be in BV. Then 11 and f2 are bounded, Le., there
exists a constant M>O such that 111 (x) IM and If2(X) IM for all

310 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10, 9 42
XE[a, b]. It follows easily that
S(f1/2)M{S/1 +Sf2}
for all partitions flJ, so
V(f1/2)M{Vf1 + Vf2} < 00.
5° If I is monotone on [a, b], then Sf:::= If(b) -f(a) I for all partitions
flJ, so Vf== If(b) -f(a) I <00.
6° Let f be absolutely continuous on [a, bJ. Then, by the definition
of absolute continuity, there exists a number £5>0 such that for all
finite unions  (ai, b i ] of disjoint cells in [a, b], satisfying  (bi-ai)
<£5, we have  If(bi)-f(ai) 1< 1. Select a fixed partition flJ1==
{a==yo<Y1 < . . . <YN:::=b} such that Yi-Yi-1 <£5 for i== 1, "', N, and
let flJ be an arbitrary partition with corresponding sum Sf. In order to
prove the existence of a fixed number M such that S/M for all
partitions flJ, we may just as well assume that all points Yi (i==O, 1,
. . ., N) are among the partition points of flJ, since addition of new
points does not make Sf smaller. It follows then immediately that
Sf<N. Hence, VfN.
The function g(X) , defined in the final paragraphs of the preceding
section, is continuous and of finite variation, but not absolutely con-
tinuous.
Denote in 5 f== i= 1 I/(Xi) - f(Xi-1) I by Sf the sum of the terms for
which f(Xi)-f(Xi-1)0, and by -SPJf the sum of the terms for which
f(Xi) -f(Xi-1) <0. Then Sf==Sf-SPJf, and
n
Sf+SPJf:::=  {f(Xi) -f(Xi-1)}==f(b) -f(a).
i=l
Just as the total variation Vf ot f was defined by Vf:::=sup Sf, we
define now the positive variation V+f and the negative variation V-f of
f by V+f:::=sup Sf and V-f:::=inf SPJf.
THEOREM 2. If fEBV, then
Vf== V+f- V-f
and
f(b) -f(a):::= V+f+ V-f.

Ch. 10,9 42J
FUNCTIONS OF FINITE VARIATION
311
PROOF. Given e>O, we. select a partition flJ such that SflJf> Vf-e,
I.e.,
Sf+(-Sf» Vf-e.
It follows that V+f- V-(;?:; Vf.
Conversely, if flJ 1 and flJ 2 satisfy Slf> V+f-!e and -S2f>
- V-f-!e, then the partition flJ, having as partition points all partition
points of flJ 1 and flJ 2 taken together, satisfies the same inequalities
(since under refinement of a partition flJ the sums Sf and -Sf do
not decrease). Hence
Sf-Sf> V+f- V-f-e,
I.e.,
SflJf> V+f- V-f-e.
It follows that Vf V+f- V-f.
The equality f(b)-f(a)== V+f+ V-f follows immediately, by taking
least upper bounds, from
Sf=={f(b) -f(a)}-Sf.
Note that, with little change in the proof, the equalities Vf==
V+f- V-f and V+f=={f(b)-f(a)}- V-f hold as well in the case that
Vf==+oo. Evidently, Vf==+oo if and only if V+f==- V-f==+oo.
Still assuming that the real and finite-valued function f is defined on
[a, bJ, we can apply all the above definitions to any subinterval [a, xJ
of [a, b]. Evidently, since any partition of [a, xJ can be extended to a
partition of [a, X/J for x<x/b, the total variation v(x)== Vf of f over
[a, xJ is a non-decreasing function of X. Similarly, v+(x) and -v-(x)
are non-decreasing, where v+(x) and v-(x) denote the positive and nega-
tive variations respectively of f over [a, xJ. Defining v(a)==v+(a)==
v-(a) ==0, we have therefore, for any fEBV[a, bJ and any xE[a, b], that
v (x) ==v+(x) -v-(x),
f(x) -f(a) ==v+(x) +v-(x).
This shows that any f of finite variation on [a, bJ is the difference
/1 - f2 of two non-decreasing functions (namely, fl (x) == f(a) +v+(x) and
12(X) == -v-(x)), and since for each of these functions fi (i== 1, 2) the
right limit fi(X+ )==lim fi(X+h) as h!O exists as a finite number for

312 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10,  42
all xE[a, b), the same holds for f(x+). Similarly, the left limit f(x-)
exists as a finite number for all xE(a, b]. Furthermore, since the
number of discontinuity points of f1 and f2 is at most countable (since
f1 and f2 are monotone), the number of discontinuity points of f is at
most countable.
Before proceeding with the continuity properties of functions of
finite variation, we prove that the decomposition f(x) - f(a) ==
v+(x) -{ -v-(x)} is, so to say, the most efficient way of writing f(x) - f(a)
as the difference of two 110n-decreasing functions which vanish at the
point a.
THEOREM 3. If fEBV on [a, b], and f(x)-f(a)==w1(x)-W2(X) for all
xE[a, b], where WI and W2 are non-decreasing and W1(a)==W2(a) ==0, then
v+(X)W1(X) and -v-(X)W2(X) on [a, b].
PROOF. Let Xo satisfy a<xob. Then, if  (ai, b i ] is any finite
union of disjoint cells contained in [a, xo], it follows from
f(b i ) - f(ai) =={W1 (b i ) -WI (ai) }-{w2(b i ) -W2(ai)}
that
n n
 {f(b i ) - f(ai) } {WI (b i ) -W1(ai) }W1 (xo).
1 1
Assuming now that flJ is an arbitrary partition of [a, xo], the corre-
sponding sum S;f is exactly of the form  {f(b i ) -f(ai)} for some
finite union  (ai, b i ] of disjoint cells contained in [a, xo]; hence S;f
W1 (xo). Taking the least upper bound for all such flJ, we obtain v+(xo)
W1(XO)' It follows then from W1-W2==V+-( -v-) that -v-(xo)
W2(XO) .
THEOREM 4. If f is of finite variation on [a, bJ and axo<b, then
f(x) is right continuous at Xo if and only if v(x) is so, and v(x) is right
continuous at Xo if and only if v+(x) and v-(x) are so. A similar statement
holds jor left continuity. Hence, f(x) is continuous at any xoE[a, b] if and
only if v(x) is so, that is, if and only if v+(x) and v-(x) are continuous at
Xo. Furthermore, at each point xoE[a, b) one at most of the functions v+(x)
and v-(x) is right discontinuous (and similarly for left discontinuity).
PROOF. We first prove the last statement. Assume that xoE[a, b)
and that v+(xo+) -v+(xo) and -v-(xo+) -{ -v-(xo)} are both positive.

Ch. 10,9 42J
FUNCTIONS OF FINITE VARIATION
313
Let M be the smallest of these two numbers, and set W1(X) equal to
v+(x) on [a, xoJ and equal to v+(x) -M on (xo, bJ. Similarly, let W2(X)
be equal to -v-(x) on [a, xo] and equal to -v-(x)-M on (xo, b]. Then
WI and W2 are non-decreasing, w1(a) ==w2(a) ==0 and f(x) -f(a) ==
Wl(X)-W2(X) for all xE[a, bJ. Since W1(X) <v+(x) for all XE(XO, b], this
contradicts the result obtained in the preceding theorem. A similar
proof holds for left continuity of v+ and V-.
The other statements follow now easily. Since v (x) ==v+(x) +{ -v-(x)}
with v+ and -v- non-decreasing, it is evident that v is right continu-
ous at Xo if and only if v+ and v- are so. Furthermore, since f(x) - f(a) ==
v+(x) +v-(x), and one at least of v+ and v- is right continuous at Xo by
what has already been proved, it is also evident that f is right continu-
ous at Xo if and only if v+ and v- are so. A similar proof takes care of
left continuity.
The case that the function f of finite variation is right continuous at
all points of [a, bJ is of particular interest, since in this case the corre-
sponding variations v+ and v- are also right continuous. It follows that
v+ and -v- are increasing (i.e., non-decreasing) and right continuous,
and therefore generate Stieltjes-Lebesgue measures #1 and #2 in [a, bJ,
defined by #l(A) ==V+(X') -v+(x) and #2(A) == -V-(X') +v-(x) for any
cell A == (x, x'] in [a, b]. This implies that #1 (A) - #2(A) == f(x ' ) - f(x) for
A == (x, x'], i.e., the set function V==#1-#2 is generated by the point
function f of finite variation. This set function v, defined on the a-
ring of all subsets of [a, b] which are #l-measurable as well as #2-
measurable, is evidently countably additive and has, therefore, one of
the main properties of a measure although it may assume negative
values. Any set function of this type is called a signed measure, and
the theory of such signed measures will be developed in the next
chapter.
THEOREM 5. If f is of finite variation on [a, b], then f is absolutely
continuous if and only if the total variation v is so, and this is the case
if and only if both v+ and v- are absolutely continuous.
PROOF. Let f be absolutely continuous on [a, b]. This means that,
given 8>0, there exists a number £5>0 such that for all finite unions
 (ai, b i ] of disjoint cells in [a, b], satisfying  (bi-ai) <£5, we have

314 MEASURES AND FUNCTIOS ON THE REAL LINE [Ch. 10,9 42
 I/(bi)-/(ai)l<e. Let  (ai, b i ] be such a finite union. Now, sub-
dividing each (ai, b i ] into a finite number of disjoint cells, we get
 (ai, bi]==i (Cj, dj], so i I/(dj)-/(cj)l<e holds as well for the union
i (Cj, d j ]. Keeping the cells (ai, b i ] fixed, but varying the subdivisions
and taking least upper bounds for all such subdivisions, we obtain
 V:/e, i.e.,  {V(bi)-v(ai)}e, and this shows that v is absolutely
continuous. Conversely, since
I/(b i ) -/(ai) Iv(bi) -v{ai)
tor aai<bib,
the absolute continuity of v implies the absolute continuity of I. Simi-
larly, since
[v+(b i ) -v+(ai)] + [{ -v-(b i ) }-{ -v-{ai)}] ==v(b i ) -v(ai),
and the square brackets are non-negative, the absolute continuity of v
implies the absolute continuity of v+ and V-. The converse is obvious
since it follows immediately from the definition of absolute continuity
that the sum or difference of two absolutely continuous functions is
absolutely continuous.
The extension of the definitions and theorems in this section to the
case that I is defined on the whole real line R 1 and of finite variation
on every bounded interval will be evident.
Exercises
BANACH SPACE OF FUNCTIONS OF FINITE VARIATION
42.1) Show that the collection BV 0 of all real I of finite variation on
the bounded interval [a, b], which satisfy the additional condition I(a)
==0, is a Banach space with respect to the norm 11/11== V:/. More gener-
ally, show that the collection BV of all real I of finite variation on
(a, bJ is a Banach space with respect to II/II==I/(a)l+ V.
CONTINUOUS FUNCTIONS OF FINITE VARIATION
42.2) Let BV 0 be the same as in the preceding exercise, and let
CBV o and AC o be the subcollections of all continuous functions and

Ch. 10,9 42J
FUNCTIONS OF FINITE VARIATION
315
all absolutely continuous functions respectively in BVo. Evidently,
ACocCBVocBVo. Show that CBV o and AC o are closed linear sub-
spaces of BVo.
42.3) Let C be the Banach space of all real continuous functions f
on the bounded interval [a, b] with the uniform norm Ilfll=max If(x) I
for axb, and let CBV and AC be the subcollections of all functions
of finite variation and all absolutely continuous functions respectively
in C. Show that AC, and hence CBV, is dense in C, and show also that
CBV, and hence AC, is of the first category in C. Derive from this
result that the set Aa of all fECBV satisfying V(X (where ex is given,
O<ex<oo) is nowhere dense in C.
42.4) We use the same notations as in the preceding exercise. In
addition, if f is real and contjnuous on [a, bJ and if [c, d] is a subinterval
of [a, bJ, we shall denote by Qf the oscillation of f in [c, d], that is, Qf
is the difference between the maximum and minimum of f in [c, dJ.
Show that if fECBV and e>O are given, there exists £5>0 such that
for any partition &P=={a=xo<x1 < . . . <xn==b}, satisfying
maxi (Xi-Xi-1) <£5,
we have
n
V-e<S&Jf  Q:_lfV.
i=l
In other words,
n
lim S&Jf==lim  Q:-lf== V
i=l
as maxi (Xi-Xi-1) o. Note that for fEBV, but f not continuous, this
result is not generally true.
THE BANACH INDICATRIX OF A FUNCTION IN CBV
42.5) With the same notations as in the preceding exercises, let
fECBV on [a, bJ. For any real number y, let Nf(y), finite or infinite,
denote the number of times that f assumes the value y in [a, bJ. The
function Nf(Y) is called the Banach indicatrix of f (introduced by S.
BANACH, 1925, [3J). Show that Nf(y) is Lebesgue summable, and
Joo Nf(y) dy== V, where dy denotes Lebesgue integration. Derive
from this result that every value y, except for a set of y-values of Le-
besgue measure zero, is assumed by f only a finite number of times.

316 MEASURES AND FUNCTIONS ON THE REAL LINE [Ch. 10,9 42
A NECESSARY AND SUFFICIENT CONDITION FOR A CONTINUOUS FUNCTION
TO BE OF FINITE VARIATION
42.6) Show, by a simple modification of the argument, that the
final result in Exercise 42.4 holds as well for any continuous function
I on [a, b] satisfying V':) +00. Similarly, show that the main result
in Exercise 42.5, namely 1 Nf(y) dy V':), holds for any such I. Derive
from this result that the continuous function I on [a, b J is of finite
variation if and only if its Banach indicatrix Nf(y) is Lebesgue
summable.
DIFFERENTIATION OF FUNCTIONS OF FINITE VARIATION
42.7) Let I(x) be of finite variation on the bounded interval [a, bJ.
Show by means of the Lebesgue decompositions of v+(x) and v-(x) in
Exercise 41.1 that l(x)-/(a)==/1(x)+/2(X), where 11(a)==/2(a)0, 11 is
absolutely continuous (so 11 (x) ==1: d(t)dt for d(t)/l(t)) and 12 is of
finite variation with 12(X)0 almost everywhere. It follows that I'(x)
==d(x) holds almost everywhere; show that v' (x)  Id(x) I holds almost
everywhere, where v is the total variation of I.
COMPLEX FUNCTIONS OF FINITE VARIATION
42.8) Let I(t) be a complex function on the bounded interval [a, bJ;
I(t)g(t)+ih(t), g and h real. For each partition fJJ of [a, b] the sum
S&JI is defined exactly as for real I, and the total variation V':) is again
the least upper bound of all S&JI. Show that I is of finite variation if
and only if g and h are so, and I is absolutely continuous if and only if
g and h are so.
42.9) Assume that I is complex and absolutely continuous on [a, b],
i.e., I(t) -/(a) == I I'(u) du, and let v(t) == Vt,j. Show that v(t) is abso-
lutely continuous and v' (t)  II' (t) I for almost every t.
42.10) Let I be complex and of finite variation on [a, b], and let
v(t)  V/. Show that v' (t)  II' (t) I for almost every t. Consider first the
case that I' (t)  0 almost everywhere.
ARC LENGTH
42.11) Let g(t) and h(t) be real continuous functions on the bounded
interval [a, b]. The set of points (g(t), h(t)), atb, in the plane R 2 of

Ch. 10,9 42J
FUNCTIONS OF FINITE VARIATION
317
points (x, y) is called a continuous curve defined by the equations x==
g(t), y==h(t), atb, and the total variation Vj of the complex function
j(t)===g(t)+ih(t) on [a, bJ is called the length of the curve. Similarly, if
v(t) == V, then v(t) is called the length of the arc of the curve between
the points (g(a), h(a)) and (g(t), h(t)). If the curve has finite length it
is called a rectijiable curve. Show that the curve is rectifiable if and
only if g and h are of finite variation on [a, bJ. Show that in this case
1J(t));J {(g')2+(h')2}ldu for all t, and equality holds for all t if and only
if g and h are absolutely continuous.
42.12) Let C be Cantor's ternary set in [0, 1J, and let g(x) be the
-corresponding Cantor function, presented as an example at the end of
sec. 41. Show that the graph of g(x), Oxl, has length equal to 2.
More generally, let C 1 be a general Cantor set in [0, 1 J of given
measure ex, Oex<1 (cf. Exercise 41.2), and let gl(X) be the correspond-
ing Cantor function (that is, gl(O) ==0, gl(I)== 1, gl(X)==! on the middle
open interval of [0,.IJ-C 1 , and so on). Show that the graph of gl(X),
'Oxl, has length equal to l-ex+(I+ex 2 )!. Note for the proof that
for a complex continuous function j(x) of finite variation on [a, bJ the
sum S&Jj tends to V':J as the maximal length of the subintervals of fJJ
tends to zero.
42.13) Assume that g(x) is continuous and non-decreasing on [0, IJ,
g(O) ==0, g(I)==ex. By L g we denote the length of the graph of g(x).
Show that Lg== 1 +ex if and only if g' (x) ==0 almost everywhere. Note
that, by Exercise 41.3, such a function g may still be strictly increasing.
42.14) Let g(x) be continuous and strictly increasing on [0, 1J,
g(O) ==0, g( 1) ==ex, and let G(y) be the inverse function. Hence, G(y) is
continuous and strictly increasing on [0, exJ. Show that g'(x) ==0 almost
,everywhere on [0, 1 J if and only if G' (y) ==0 almost everywhere on [0, ex J.

CHAPTER 11
SIGNED MEASURES AND COMPLEX MEASURES
The present chapter is devoted to some concepts generalizing the concept of
a measure. More specifically, we shall consider signed measures (differences of
two finite measures) and complex measures v' +iv" (where v' and v" are signed
measures). Among the most important properties are the fact that the col-
lection of all signed (or complex) measures is a real (or complex) Banach space,
and Nikodym's theorem that any (finite) limit of a converging sequence of
signed (or complex) measures is a signed (or complex) measure. Nikodym's
theorem is shown to be an immediate consequence of the Hahn-Saks theorem,
asserting that if IEfndf-t converges (as n-+oo) to a finite number v(E) for every
measurable E, then the set functions IE IfnI df-t are uniformly f-t-absolutely con-
tinuous, and v(E) = IEfdf-t for some f.
43. Signed Measures
We assume in this section that A is a fixed a-ring of subsets of the
non-empty point set X, and we shall call the sets of A measurable sets
(although this does not mean that there is necessarily given in advance
a measure # in X such that the sets of A are #-measurable). If EEA
and if, corresponding to every measurable set DeE, there exists a.
finite real number v(D), we say that v(D) is a finite real set function,
defined on the a-ring A E of all measurable subsets of E (note that
DEA E if and only if D==D 1 nE with D1EA). The finite set function
v(D) is said to be a-additive (or completely additive) on A E whenever
v( Dn)= v(Dn) for every sequence of disjoint DnEAE' Note that
v(0)=0 holds in that case.
If # is a measure in X, if A is the a-field of all #-measurable sets,.
and the real function f is #-summable over EEA, then Jnfd# is a finite
a-additive set function on A E . In particular, if #(E) < 00, then #(D) is
finite and a-additive on A E . More generally, if #1 and #2 are measures
on A, and a, b are real constants, then v(D)=a#1(D)+b#2(D) is finite

Ch. 11,  43J
SIGNED MEASURES
319
and a-additive on A E for any E EA for which #l(E) and #2(E) are
finite. Conversely, if v(D) is a finite non-negative a-additive set function
on A E (i.e., Ov(D)<oo for all DEA E ), then v is a finite measure on
A E (by a finite measure we mean here that v(E) <00). One of the main
purposes in this section will be the proof that every fInite real a-
additive set function on A E is the difference of two such finite measures.
In view of this property finite real a-additive set functions are also
called signed measures.
THEOREM 1. If v(D) is a signed measure on AE, and Dn is a monotone
sequence of sets of AE, then v(lim Dn) ==lim v(Dn).
PROOF. Aln10st exactly as the proof of the corresponding theorem
for measures in sec. 8, Theorem 3.
If v(D) is a signed measure on AE, the number v+(E) ==sup v(D) for
all DE A E is called the positive variation of v over E, and v-(E) ==inf v(D)
for all DEA E is called the negative variation of v over E. Since v(0) ==0,
we have v-(E)Ov+(E). The number
Ivl (E) ==v+(E) + Iv-(E) I ==v+(E) -v-(E)
is now called the total variation of v over E.
THEOREM 2. If v(D) is a signed measure on AE, the variations v+(E)
and v-(E) are finite numbers. In addition, v(D) is bounded on A E ,' more
precisely, Iv(D)Ilvl(E) for any DE.L1 E .
PROOF. Assume that Ivl(E) ==00. Then there exists a descending
sequence DnEAE such that
Ivl (Dn) ==00,
Iv(Dn) In-l
for n == 1, 2, . .. .
(1)
Indeed, choose D 1 ==E, and assume that the existence of D1, . . ., D k ,
satisfying (1), has already been proved. Since Ivl(Dk) ==00, there is a set
A cD k such that Iv(A)Ilv(Dk)l+k. If now Ivl(A)==oo, we may choose
Dk+1==A, and then this Dk+1 satisfies (1) for n==k+l. If, however,
Ivl(A)<oo, then Ivl(Dk-A)==OO and, in view of Iv(Dk-A)I==
Iv(Dk)-v(A)Ilv(A)I-lv(Dk)lk, we may choose Dk+1==Dk-A in this
case. For the so obtained descending sequence Dn, satisfying (1), we have
now by Theorem 1 that Iv(lim Dn) I ==1im Iv(Dn) I ==00, which is impossi-

320 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 43
ble since v(lim Dn) is finite by definition. It follows that Ivl (E) is finite,
and any DEA E satisfies
Iv(D) Imax {v+(E), Iv-(E) I} Ivl (E).
THEOREM 3 (JORDAN DECOMPOSITION OF A SIGNED MEASURE). If
1J(D) is a signed measure on AE, then v+, -v- and Ivl are finite measures
on AE, and v==v++v- on A E . Any signed measure on A E is, therefore,
the difference of two finite measures on A E . In addition, if #1 and #2
are finite measures on A E such that V=#1-#2 on AE, then V+#l and
-V-#2 (in other words, v==v+-(-v-) is the most efficient manner of
writing v as the difference of two finite measures).
PROOF. Since Ov+(D)<oo for any DEA E by the preceding theo-
rem, it will be sufficient to show (in order to prove that v+ is a finite
measure) that v+( Dn) ==  v+(Dn) for every sequence of disjoint
DnEAE' For any D c  Dn we have v(D) ==  v(DDn) v+(Dn), hence
1J+( Dn) v+(Dn). Conversely, if An cD n (and AnEAE), then
1J+( Dn)v( An)= v(An), so v+( Dn) v+(Dn). The desired re-
sult follows by combining the two inequalities. The proof for -v- is
similar, and in view of Ivl=v++( -v-), the set function Ivl is then also
a finite measure on A E .
Furthermore, if A cD is measurable, then
v(A) ==v(D) -v(D - A )v(D) -v-(D),
so v+(D)v(D)-v-(D). Also
v(A) ==v(D) -v(D - A) v(D) -v+(D),
so v-(D)v(D)-v+(D). Hence v(D)==v+(D)+v-(D).
Finally, if #1 and #2 are finite measures on A E such that V==#1-#2,
and if A cDcE (where A and D are measurable), we have
v(A) =#l(A) - #2(A )#l(A )#l(D),
so v+(D)#l(D). This shows that V+#l on A E . It follows then from
}l1-#2=V+-( -v-) that -V-#2 on A E .
THEOREM 4 (HAHN DECOMPOSITION OF A MEASURABLE SET). If v(D)
is a signed measure on AE, there exists a decomposition E==E++E- of E

Ch. 11, 9 43J
SIGNED MEASURES
321
into disjoint measurable sets E+ and E- such that v-(E+) ==v+(E-) ==0. In
other words, v(D)O lor every measurable D cE+, and v(D)O lor every
measurable D c E-.
PROOF. The extension procedure for measures may be applied to
the measure Ivl on AE, and Ivl is thus extended to the a-ring A * of all
lvi-measurable subsets of E. It may happen then that A* contains more
sets than AE, but since Iv I (E) is finite, any set DELi * is the difference
of a set of A E and a lvi-null set. Similar remarks hold for extension of
v+ and -v-. The measures v+ and -v- are lvi-absolutely continuous, so
that by the Radon-Nikodym theorem there exist lvi-measurable
functions 100 and I'OO such that v+(D) == !nlodlvl and -v-(D) ==
!nl'Odlvl for all DEA*, in particular for all DEA E . If g==min(/o, 1'0) and
a(D)==!ngdlvl on A*, the measures #l==v+-a and #2==-v--a satisfy
#1-#2==V on A E ; it follows therefore from the preceding theorem that
a is identically zero, i.e., the set Do=={x: g(x) >O} is of lvi-measure zero.
Setting 10==/'0==0 on Do does not alter the values of v+ and V-, so that
we may assume, therefore, that at each point of E one at least of 10
and t'O is zero. Setting now E+=={x: lo(x) >O} and E-==E -E+, we ob-
tain the desired result provided E+EA E . If this is not so, addition of a
suitable lvi-null set yields a set E+EA E with the desired properties.
Evidently, the Hahn decomposition of E is not uniquely determined,
since any set DEAE, satisfying Ivl(D)==O, may be added to or sub-
tracted from the above E+.
For an ((elementary" proof of the Hahn decomposition theorem,
which does not make use of the Radon-Nikodym theorem, cf. Exer-
cise 43.1.
We note at this point that it is possible to characterize the total
variation Ivl (E) of a signed measure v over E in a different way which
is independent of the positive and negative variations v+ and V-. In
fact, Ivl(E)==sup  Iv(En) I for all possible decompositions E== En of
E into a countable number of disjoint sets EnEAE' For the proof, note
that  Iv(En)I Ivl(En)==lvl(E) for any such decomposition, and that
on the other hand, if E==E++E- is a Hahn decomposition, then
Ivl (E) == Ivl (E+) + Ivl (E-) ==v+(E+) -v-(E-) == Iv(E+) I + Iv(E-) I.

322 SIGNED lVIEASURES AND COMPLEX MEASURES [Ch. 11, 9 43
The same argument shows that Ivl (E) ==sup  Iv(E n) I for all possible
decompositions E== En into a finite number of disjoint sets EnEAE,
and this is evidently the parallel of the definition of the total vari-
ation of a function as introduced in the preceding section.
In fact, we prove now that if the real function 1 is of finite vari-
ation and right continuous on E==[a, b], then the total variation VI
as defined in the preceding section is exactly the total variation Ivl (E)
of the signed measure v generated by 1 as defined by v{(c, d]}==/(d) -/(c)
. for all cells (c, d] in E. Indeed, let v+(x) and v-(x) be the positive and
negative variations of lover [a, x], and let #1 and #2 be the Stieltjes-
Lebesgue measures generated by v+ and -v-. Then#l(A) -#2(A) ==v(A)
for any cell A cE, and so (by extension) #1-#2==V on A E . It follows by
Theorem 3 that V+#l on A E . For an arbitrary cell A==(c, d] in E, let
f!}J be a partition of A and consider S/. Since SI is a finite sum of
terms I(Xi) - I(Xi-1), there exists a finite union D of subcells of A such
that S/==v(D), andsoS/v+(A). But then v+(d)-v+(c)==sup&, Sf
v+(A), i.e., #l(A)v+(A). It follows (by extension) that #lV+ on AE,
and combining this with the earlier result that V+#l on A E we obtain
#l==V+. But then #2==-V-, and so Ivl==v+-v-==#1+#2' Observing
finally that #1 + #2 is generated by v+(x) -v-(x) ==v(x), we find V/==
v(b) -v(a) == (#1 +#2) (E) == Ivl (E).
In the remaining part of this section we consider, instead of one
single signed measure, the collection of all signed measures on AE, and
we shall prove first that this collection is a Banach space with respect
to the total variation as norm.
THEOREM 5. The collection B 01 all signed measures on A E is a real
Banach space with respect to the norm IIvll==lvl(E).
PROOF. In order to prove that B is a normed linear space, it will be
sufficient to show that IIvil satisfies the triangle inequality, the other
norm properties being evident. Let, therefore, VI and V2 be in B. Then
(VI +V2) (D) ==v1(D) +v2(D) vt (E) +vt (E)
so (VI +v2)+(E)vt (E) +vt (E).
Also (VI +V2) (D) ==v1(D) +v2(D)vl (E) +v 2 (E),
for all DEAE,

Ch. 11, 9 43J
SIGNED MEASURES
323
so (v1+v2)-(E)vl(E)+v2(E). It follows that IV1+V21(E)lv11(E)+
IV21(E). In order to prove that the normed linear space B is complete,
let us assume that IVn-vml(E)-+O as m, n-+oo. Since l(vn-vm)(D)I
Ivn-vml(E) for every DEAE, the limit v(D)==lim vn(D) exists as a
finite number, and v is finitely additive on A E . We have to show that
v is a-additive on AE, i.e., that Dn i D implies v(Dn) -+v(D). The proof
of this is similar to the proof that the limit function of a uniformly
converging sequence of continuous functions is continuous. Given E>O,
choose k such that IVm -vnl (E) < is for all m, nk (so, in particular,
IVk(A)-v(A)Iis for all AEA E ), and then choose no such that
IVk(D) -vk(Dn) I < is for all nno. It follows that, for n?::-no,
Iv(D) -v(Dn) I  Iv(D) -vk(D) 1+ IVk(D) -vk(Dn) 1+ IVk(Dn) -v(Dn) I <E.
Finally, since l(vn-v)(D)Iis for all nk and all DEAE, we have
(vn-v)+(E)!E and I(vn-v)-(E)I!E, hence IVn-vl(E)<;E for all nk,
showing therefore that limllvn-vll==O.
Let v be a signed measure on AE, and let I(x) be a complex function,
defined on E and measurable on E (that is, if I==g+ih with g and h
real, then the sets {x: g(x) >a} and {x: h(x) >a} are in A E for every real
a). Then I is v+ -measurable as well as (-v-)-measurable. If, in ad-
dition, I is v+-summable as well as (-v-)-summable, we write by defi-
nition
I Idv== I Idv+-I Id( -v-)
D D D
for every DEA E . Note that measurability and boundedness of I(x) is
sufficient for the existence of Inldv. The final theorem in this section
shows the connection between the so defined integral with respect to a
signed measure v and the total variation Ivl.
THEOREM 6. II v is a signed measure on AE, then
IIvll== Ivl (E) == max Illdvl.
If(x)ll E
PROOF. If I is measurable on E and I/(x) I  1 on E, then
II Idvl/1/1 dv++ I1II d( -v-)v+(E) -v-(E) == Ivl (E).
E E E

324 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 43
If E ==E+ + E- is a Hahn decomposition of E, and f== 1 on E+, f= -1
on E-, then
I fdv==v{E+) -v(E-) ==v+(E+) -v-(E-)
E
==v+(E) -v-(E) == Ivl (E).
Here it has been used several times that v-(E+) ==v+(E-) ==0.
Exercises
HAHN DECOMPOSITION
43.1) Show the existence of a Hahn decomposition of the set E with
respect to the signed measure v on A E by arguing as follows. Since
v+(E)==supv(D) for all DcE, there exists a sequence DncE such
that v+(E) -2-nv(Dn)v+(E) for n== 1, 2, . . . . Since v(Dn)v+(Dn)
v+(E), it follows easily that v-(Dn)-2-n and v+(E -Dn)2-n. Show
that E+==lim inf Dn satisfies v-(E+) ==0, and E-==E -E+ satisfies
v+(E-) ==0.
SUPREMUM AND INFIMUM OF SIGNED MEASURES
43.2) The notation flv for the signed measures fl and v denotes
that fl(D)v(D) for all DEA E . Show that the collection B of all signed
measures on A E is partially ordered with respect to .
For all DEA E and fl and v in B, let 7(D)==sUP{fl(F)+v(D-F)} for
all measurable F cD. Show that 7 is a signed measure, and 7 is the
least upper bound of fl and v in the introduced partial ordering. In-
stead of 7 we usually write flVv for this least upper bound. Show that
if 71==(-fl)V(-V), then -71 is the greatest lower bound of fl and v,
usually denoted by flAv. Note that flVO==fl+ and flAO=fl-. Show,
finally , that
flVV==!{(fl+ V ) + Ifl- v l}
and
flAv==!{(fl+ V ) -Ifl- v l}.
43.3) Redevelop the results in the preceding exercise in a less ele-
mentary but more intuitive manner as follows. Given the signed
measures fl and v, let A be a finite measure on A E such that Ifll and

Ch. 11, 9 43J
SIGNED MEASURES
325
Iv I are A-absolutely continuous (e.g., A== Ifll+ Ivl). Then, by applying
the Radon-Nikodym theorem to fl+, -fl-, v+ and -V-, we obtain real
A-measurable functions (dfljdA) (x) and (dvjdA) (x) such that fl(D) ==
In(dfljdA)dA and v(D)==ln(dvjdA)dA for all A-measurable sets D, so in
particular for all DEA E . Show that
f ( dfl dv )
(flVV)(D) == max dA ' dA dA,
D
f ( dfl dv )
(flAv) (D) == min -, - dA
dA dA
D
for all DEA E . For the interpretation of the final result in the pre-
ceding exercise, note that Ifl-vl (D) == In IdfljdA-dvjdAI dA.
ORTHOGONALITY OF SIGNED MEASURES
43.4) The signed measures fl and v on A E are called orthogonal (no-
tation fl 1..v) whenever Ifll A Ivl ==0, where 0 denotes here the measure
which is identically zero. Show, for flO and vO, that this definition
is equivalent to the earlier definition in Exercise 32.5. Show that fl 1..v
if and only if Ifl+vl==lfl-vl. Finally, show that fl1..v if and only if
Ifl+vl == IfllVlvl.
ARC LENGTH
43.5) We reconsider the problem on arc length in Exercise 42.13,
i.e., we assume that g(x) is continuous and non-decreasing on [0, 1J,
g(O) ==0, g( 1) ==cx, and by Lg we denote the length of the graph of g(x).
Writing f(x) ==x+ig(x), it is trivial that Lg== VAf== VA{x+ig(x)} 1 +
VAg== 1 +cx== VA{x+g(x)}; the last equality holds since x and g(x) are
non-decreasing. Show first that Lg VA{x-g(x)}. Hence V{x-g(x)}==
IIfl-vllLg;;jlfl+vll== V{x+g(x)}, where fl is Lebesgue measure (gener-
ated by the function x) and v is the Stieltjes-Lebesgue measure gener-
ated by g(x). Now assume that g'(x)==O holds fl-almost everywhere,
and show that in this case fl 1..v, that is, IIfl-vll== Ilfl+vll, which implies
then immediately that Lg== 1 +cx.

326 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 44
LEAST UPPER BOUND OF A COLLECTION OF SIGNED MEASURES
43.6) Let {Va} be an arbitrary (not necessarily countable) collection
of signed measures on ./1 E , bounded above by the signed measure #,
i.e., Va# for all CX. Show that the collection {va} has a least upper
bound (denoted by sup Va) in the sense of the partial ordering.
44. Complex Measures
The theory in the preceding section may be extended to complex a-
additive set functions. If, corresponding to every DEAE, there exists a
(finite) complex number v(D) such that v( Dn) ==  v(Dn) for every
sequence of disjoint DnEAE, then v(D) is called a complex a-additive
set function on AE, or also a comPlex measure on A E . In that case we
have clearly v(D)v'(D)+iv"(D), where v' and v" are real signed
measures on A E . Furthermore, v(D) is bounded on A E in view of
Iv(D)Ilv'(D)I+lv"(D)Ilv'I(E)+lv"I(E) for all DEA E . It is, of course,
not possible to introduce positive and negative variations of the com-
plex measure v over the set E, but we may define the total variation
Ivl(E) of v over E by Ivl(E)==sup  Iv(En) I for all possible decompo-
sitions E== En of E into disjoint sets EnEAE' For real v the new
definition gives the same value as the previous definition in the pre-
ceding section. The total variation Ivl (D) is bounded on E, since for
D DncE with disjoint Dn and vv'+iv" we have
 Iv(Dn)I Iv'(Dn)I+ Iv"(Dn)I Iv'I(Dn)+ Iv"I(Dn)
== Iv'l (D) + Iv" I (D)  Iv'l (E) + Iv" I (E).
Also, Ivl is a-additive on A E . Indeed, if D  Dn is a decomposition
of DEA E and if 8>0, there exist decompositions Dn==i Ani such that
Ivl(Dn)<i Iv(Ani)l+ef2n, hence
 Ivl(Dn)  Iv(Ani)I+8lvl(D)+e,
and so  Ivl(Dn)lvl(D). On the other hand, if D== Ak is another
decomposition of D, and Akn==AknD n , then
 Iv(Ak)I I V(Akn)I  Iv(Akn)I Ivl(Dn),
k n n
so that Ivl(D) Ivl(Dn). It follows that Ivl is a finite measure on A E .

Ch. 11, 9 44J
COMPLEX MEASURES
327
The collection B of all complex measures v on A E is now a Banach
space with respect to the norm IIvll == Iv I (E). The triangle inequality for
Ilvll follows by observing that  l(v1+v2)(En)I IV1(En)I+ IV2(En) I
for any decomposition E ==  En, and the completeness of the space is
immediately derived from the real case in view of the inequalities
max{lv'l (E), Iv"l (E)}lvl (E) [v' I (E) + Iv"l (E)
f ,. "
or v==v +v .
Finally, let v==v' +iv", and let for any measurable complex function
I for which JEldv' and JEldv" exist as finite numbers, the integral of I
over E with respect to the comPlex measure v be defined by JEldv==
JEldv' +i JEldv". Then
IIvll==lvl(E)== sup Ifldvl.
If(x)ll E
In order to prove this, we note first that I==XD (where DEA E ) satisfies
If Idvl == Iv(D) I Ivl (D)f III dlvl,
E E
so that the same is true if I is a complex measurable step function.
Hence, since any bounded measurable I is the uniform limit of a
sequence of such step functions, the inequality IJEldvlJE III dlvl holds
for all I which are bounded and measurable. In particular, for I/(x) I  1,
we obtain
If Idvlf III dlvllvl (E).
E E
On the other hand, if E>O and E ==  En is a decomposition of E such
that Ivl(E) Iv(En)I+E, let I(x)==e- icpn on En for v(En)== Iv(En) I e icpn .
Then JEldv==  JEnldv==  Iv(En) 1lvl (E) -E.
Exercises
TOTAL VARIATION
44.1) Show that the total variation Ivl (E) over E of the complex
measure v on A E satisfies Ivl (E) ===sup  Iv(En) I for all decompositions
E ==  En of E into a finite number of sets EnEAE'

328
SIGNED IVJ:EASURES AND COMPLEX MEASURES [Ch. 11, 9 45
TOTAL VARIATION OF A COMPLEX FUNCTION OF FINITE VARIATION
44.2) Let I==g+ih (g and h real) be a complex function on the
bounded interval E==[a, b], such that I is right continuous on E and
of finite variation over E. Then g and h are right continuous and of
finite variation; let Vg and Vh be the signed measures on A E generated
by g and h respectively (where A E is the a-ring of all subsets of E which
are vg-measurable as well as vh-measurable). It can be stated now that
the complex measure v==Vg+ivh on A E is generated by I, in the sense
that v{(c, d]}==/(d)-/(c) for all cells (c, d] in E. Show that the total
variation VI of I, as defined in Exercise 42.8, is exactly the total
variation Ivi (E) of v over E.
45. Absolutely Continuous Signed and Complex Measures
In this section we assume that A is a a-field of subsets of X, and #
is a fixed measure on A. The signed or complex measure v on A is
called #-absolutely continuous whenever, for any E EA satisfying #(E)
==0, we have also v(E) ==0.
THEOREM 1. The signed or comPlex measure v on A is #-absolutely
continuous il and only ii, given 8>0, there exists £5>0 such that EEA,
#(E) <£5 imPlies Ivl(E)<e (so that, then, Iv(D)I<e lor all DeE in A).
PROOF. If, given any 8>0, there exists such a number £5>0, then
EEA, # (E) ==0 implies Ivl(E)==O, so v(E) ==0.
Let now, conversely, v be #-absolutely continuous, and assume that
for some 8>0 such a corresponding £5>0 does not exist. Then there is
a sequence EnEA such that#(En) 1 /2 n and Ivl(En)e, so E ==lim sup En
is in A, and satisfies
00
#(E)#(En+En+1 +. . .)  2-p==2-(n-l)
p=n
for all n,
i.e., #(E) ==0. On the other hand, since Ivl is a finite measure (i.e.,
Ivl(X)<oo), we have
Ivl(E)==lvl(lim sup En)lim suplvl(En)e,
contradicting the hypothesis that v is #-absolutely continuous. Indeed,
since #(E) ==0 and hence #(D) ==0 for all DeE in A, it follows that
v(D) ==0 for all DeE in A, and so Ivl (E) ==0.

Ch. 11, 9 45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 329
COROLLARY. If the signed measure v on A is #-absolutely continuous,
then the measures v+, -v- and Ivl are #-absolutely continuous in the
previously defined sense (cf. sec. 32). Similarly, if the comPlex measure
v==v' +iv" (v' and v" signed measures) is #-absolutely continuous, then the
measures Iv' I, Iv"l and Ivl are #-absolutely continuous in the previous sense.
PROOF. We have to show that, starting from the measures # and Ivl
on A, and extending them by means of the extension procedure for
measures, the extended measures # and Ivl have the property that
every #-null set is a lvi-null set. Let, therefore, #(E) ==0 and, given
e>O, let £5 >0 be such that A EA, #(A) <£5 implies Ivl (A) <c. Since there
exists a a-set A (with respect to A) such that A =>E and #(A)<£5, and
since A EA (note that A is a a-field), this implies Ivl(A)<e, so Ivl(E)==O.
The #-absolute continuity of the other measures follows immediate-
ly, since in the real case v+lvl and -v-Ivl, and in the complex case
Iv'llvl and Iv"Ilvl.
For simplicity, we shall assume from here on that A is the a-field
of all #-measurable sets. As far as we concern ourselves with sets of
finite or a-finite #-measure, this amounts to the possible addition of a
collection of #-null sets to the "old" A, and the last corollary shows
that these sets are lvi-null sets as well, so that the assumption in
question is no serious restriction of the generality.
We recall (cf. sec. 11) that the collection Al of all sets of finite #-
measure (under identification of #-almost equal sets) is a metric space
with respect to the distance function p1(E, F)==#(E -F) +#(F -E). In
sec. 20, Theorem 1 it was proved that the metric space Al is complete.
THEOREM 2. Any #-absolutely continuous signed or comPlex measure
v on A is a continuous function on the metric space Al of the sets of finite
fl-measure (more precisely, given E>O, there exists £5 >0 such that p1(E, F)
<£5 imPlies Iv(E)-v(F)I<E). Hence, given aO, the sets EEA 1 satisfying
Iv(E) I <a form an open subset of AI, and the sets E EA 1 satisfying Iv(E) Ia
form a closed subset of AI.
PROOF. Given e>O, let £5>0 be such that #(E) <£5 implies Ivl(E)<!E.
For any pair of sets E, FEA1, satisfying
p1(E, F) ==#(E -F) +#(F -E) <£5,

330 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 45
we have then
Iv(E) -v(F) 1== Iv(E - F) -v(F - E) I Ivl (E - F) + Ivl (F - E) <E.
We will show next that, for a a-finite measure fl, the collection of
all fl-absolutely continuous signed or complex measures is the same as
the collection of all fl-summable real or complex functions respectively.
More precisely, the collection of all fl-absolutely continuous signed (or
complex) measures, considered as a subcollection of the Banach space
of all signed (or complex) measures on A, may be identified with the
Banach space of all fl-summable real (or complex) functions on X.
THEOREM 3. If the measure fl in X is a-finite, and the signed (or
comPlex) measure v on the a-field A 01 all fl-measurable sets is fl-abso-
lutely continuous, there exists a real (or comPlex) fl-summable function
fo(x) such that V(E)==/Elodfl and Ivl(E)==/El/oldfl lor all EEA. Con-
versely, if fo(x) is real (or comPlex) and fl-summable over X, then v(E)==
/Efodfl is a signed (or comPlex) measure on A, and Ivl(E)==/Elfoldfl for
all E EA.
PROOF. Assume first that v is a signed measure. Then v==v+-(-v-),
where v+ and -v- are finite fl-absolutely continuous measures on A.
Note that the (extended) integrals with respect to fl and v+ (or -v-)
may be regarded as extensions of elementary integrals defined initially
only on the fl-step functions. It follows that the Radon-Nikodym theo-
rem is applicable, and hence there exist fl-summable functions It (x)O
and -/o(x)O such that v+(E)==/Eftdfl and -V-(E)==/E (-/0) dfl for
all EEA. Hence, writing fo==ft +/0, we have V(E)==/Efodfl for all
E EA. If v is complex, v==v' +iv", we apply this result to v' and v" sepa-
rately, obtaining thus real fl-summable functions fo and I such that
v'(E)==/Efodfl and v"(E)==/E/dfl for all EEA, and then V(E)==/Efodfl
for fo== lo+i/.
In order to prove that Ivl (E) == /E 1/01 dfl, in the real as well as in the
complex case, we observe that in view of the results in the preceding
sections we have Ivl(E)==supl/Ehdvl for all fl-measurable h(x) satis-
fying Ih(x)I1 on X. If h(x) satisfies these conditions, then
If hdvl == If hlodfllJ Ihfol dflf Ifol dfl,
E E E E

Ch. 11,  45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 331
hence Ivl(E)/El/oldfl for all EEA. In order to prove the inverse ine-
quality, let g(x) ===fo(x)/l/o(x) I at all XEX where fo(x) =1=-0, and g(x)==1
at all XEX where fo(x) ==0. Then g(x) is fl-measurable, and Ig(x) 1==1 at
all XEX. It follows that the complex conjugate function g(x) is also fl-
measurable, and satisfies Ig(x) I == 11 jg(x) I === 1 at all XEX. Note that
g(x)fo(x) == I/o(x) I at all XE X. Hence,
Ivl (E)II gdvl === II gfo dfl I == 11/01 dfl.
E E E
The final result is, therefore, that Ivl (E) === IE 1/01 dfl for all E EA.
The proof of the last statement in the theorem is now trivial.
In the case that X===R1 or X==[a, bJ cR 1 , fl is Lebesgue measure in
X, and v is a fl-absolutely continuous signed measure, the measures v+
and -v- are also fl-absolutely continuous and they are generated by
right continuous and non-decreasing functions g+(x) and -g-(x), abso-
lutely continuous on X (cf. sec. 41, Theorem 2), such that v+(c, dJ==
,g+(d) -g+(c) and -v-(c, dJ == -g-(d) +g-(c) for every cell (c, dJ eX.
Setting g(x) ==g+(x) +g-(x) , we have therefore v(c, dJ===g(d)-g(c) for
every cell (c, dJ, i.e., v is generated by g(x). The so obtained function
g(x), as a sum of absolutely continuous functions, is absolutely con-
tinuous.
Hence, if v is now a fl-absolutely continuous complex measure in X,
V===V1 +iV2, then VI and V2 are fl-absolutely continuous signed measures,
so they are generated by absolutely continuous real functions gl(X) and
g2(X) respectively. Then v is generated by the absolutely continuous
-complex function g==gl +ig 2 .
Conversely, let the real and right continuous function g(x) on X ==
[a, bJ be absolutely continuous. Then g is of finite variation (cf. sec. 42,
Theorem 1), and its positive variation v+(x) and negative variation
v-(x) are right continuous (cf. sec. 42, Theorem 4) and absolutely con-
tinuous (cf. sec. 42, Theorem 5). It follows that v+(x) and -v-(x) gener-
ate ,u-absolutely continuous measures v+ and -v- in X, and the signed
measure v===v++v-, generated by g(x) , is then also fl-absolutely con-
tinuous.
Finally, if the complex and right continuous function g==gl +ig 2 (gl
and g2 real) is absolutely continuous on X==[a, bJ, then gl and g2 are
also right continuous and absolutely continuous (the absolute conti-

332 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 45
nuity of gl follows from Ig1(y) -gl(X) 1lg(y) -g(x) I, holding for all
x, YEX), so gl and g2 generate #-absolutely continuous signed measures
VI and V2 respectively. Then g generates the #-absolutely continuous
complex measure V===V1 +iv2.
The theorems on the integration of increasing sequences and on
dominated convergence supply sufficient conditions for a sequence of
functions fn(x) in order that fn-+f imply / fn d #-+ / fd#. Our next theo-
rem goes, to a certain extent, in the converse direction. For Lebesgue
measure on a bounded interval in R 1 the theorem is due to H. HAHN
(1922, [2J), who based his proof upon a theorem of H. Lebesgue (1909).
The general formulation and the foundation of the proof upon Baire's
category theorem is due to S. SAKS (1933, [2J).
THEORE1\{ 4. Let the comPlex functions f n(x) (n== 1, 2, . . .) be #-
summable over X (we do not assume that # is a-finite) and let, for
every EEA, the sequence of numbers 'J,'n(E)==/Efnd# converge to a finite
number v(E). Then:
(a) The measures IVnl (E) are uniformly #-absolutely continuous, i.e.,
given any 8>0, there exists £5 >0 such that #(E) <£5 imPlies IVnl (E)8
lor all n simultaneously.
(b) II the descending sequence 01 sets EpEA (P== 1, 2, . . .) converges
to the set Eo 01 #-measure zero, and 8>0 is given, there exists an index
Po such that IVnl (E p)8 lor all PPo and all n.
(c) There exists a comPlex #-summable lunction I such that v(E) ==
/Eld# lor all EE.!i. II all In are real, then I is real.
PROOF. Although one might believe for a moment that the statement
in (b) follows immediately from the statement in (a), this is not true.
It is true that if Ep!Eo with #(Eo) ==0, and #(Ep) <00 for P==P1 (so
#(Ep) <00 for all PP1), then (b) is a consequence of (a) for this par-
ticular sequence E p, since #(E p) ! 0 in this case. It may very well
happen, however, that for some sequence Ep we have Ep!Eo with
#(Eo) ==0 and #(E p) ==00 for all p.
It is of some interest to observe that, on the contrary, (a) follows
quickly from (b). Indeed, assume that (b) holds and (a) does not hold.
Then there exists a number 8>0 and a sequence of sets Ep such that
#(Ep)1/2P and IVnpl(E p ) >8 for some index np corresponding to the

Ch. 11, 9 45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 333
index p. The sequence of sets Sp==Ep+Ep+1 + . .. descends to 5 0 ==
rr;=l Sp, and since #(Sp)1/2P-1 for all p, we ha've #(5 0 )==0. Hence,
by (b), IVnl(5p)e for all ppo and all n, contradicting the fact that
IVnpl(5P)lvnpl(Ep) >E for all p.
These remarks show that (b) implies (a), and (a) implies (b) if
#(X) < 00. Since the proof of (b) in the general case will essentially be
reduced to this special case, we shall prove (a) first.
To start the proof, note that F n=={x: /n(x) =¥=O} is of a-finite #-
measure since / n is #-summable. Hence, X' == r F n is of a-finite #-
measure. Furthermore, for any E c X -X' we have vn(E) ==0 for all n,
so v(E) ==0. We may, therefore, restrict ourselves in the entire proof to
sets included in X'; in other words, we may and shall assume that X
is of a-finite #-measure.
(a) If Al is the complete metric space of all sets of finite #-measure,
if E>O and m, n are arbitrary indices, then the sets EE.l11, satisfying
IVn(E)-vm(E)I l2 E, form a closed subset of Al by Theorem 2. It
follows that
00 00
Sk== II II {E: IVn(E) -vm(E) l l2 e}
n=k m=k
is a closed subset of AI. Furthermore, since vn(E) converges for all
EEA1, every EEA1 belongs to some Sk, so r Sk==A1. By sec. 3,
Theorem 7 (a theorem which is an immediate consequence of Baire's
category theorem) Al is of the second category, so one at least of the
closed sets 5 k contains an open sphere. Let 5 ko contain the sphere
{E: p1(E, Eo) <ro}, and let, then, £5>0 have the properties that £5<ro
and that #(E) <£5 implies IVn(E) 1< l2 E for n== 1, . . " ko.
Now, let EEA be such that #(E) <£5. For n==l, '.', ko we have al-
ready IVn(E) I < l2 E< ie. Furthermore,
p1(Eo-E, Eo) ==#(EonE) <£5<ro,
P1(E +Eo, Eo) ==#(E -Eo) <£5<ro,
so Eo-E and E+Eo are points in the open sphere {E: p1(E, Eo) <ro},
and hence they are points of Sko' Observing that E == (Eo+E) -(Eo-E)
with Eo-E included in Eo+E, we have therefore for n>k o that
Ivn(E) 1lvko(E) I + IVn(E) -vko(E) I
IVko(E) I + IVn(Eo+E) -vko(Eo+E) I + IVn(Eo-E) -vko(Eo-E) I <i e

334 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 45
by the definition of Sko' Hence, if #(E)<, then IVn(E)I<!e for all n.
It follows that if Vn==Vn1 +iVn2 (Vn1 and Vn2 signed measures), then
I V n1(E)I<!e and IVn2(E)I<!e for #(E)<. But then v1(E)!e and
IV;1 (E) I!e, and similarly v;2(E)!e and IV;2(E) I!E, so IVn11 (E)!E
and IVn21(E)!e. On account of IVnllvn11+lvn21 this implies finally
that if #(E)<, then IVnl(E)e for all n.
(b) Since X is of a-finite #-measure, we have X==r X n , where all
X n are disjoint and of finite #-measure. For any EEA we define now
00 #(EX n )
T(E) = n"El 2 n {1 +,u(Xn)} ,
and then T is a measure on the a-field A such that T(X) <00. Further-
more, the a-field AT of all T-measurable sets satisfies A T ==L1. Indeed,
it is clear that A CAT and that any set in AT is the difference of a set
in A and a T-null set. It will be sufficient, therefore, to show that
every T-null set is in A. Given such a T-null set E, there exists a
sequence AkEA satisfying Ak=>E and lim T(Ak) ==0, so that for every
fixed n we have AkXn=>EXn, AkXnEA and limk #(AkXn) ==0. This
shows that the exterior measure #*(EX n ) is zero, so EXnEA for all n.
But then E EA, and this shows that AT==A. Since T(E) ==0 if and only
if #(E) ==0, the measures IVnl (E) are T-absolutely continuous on AT==A,
and even uniformly so by (a).
Let now Ep!Eo with #(Eo)==O, so that in view of T(Eo)==O and
T(X) < 00 we have T(E p) ! o. Furthermore, by the uniform T-absolute
continuity of the measures IVnl, if e>O, there is a number >O such
that T(E)< implies IVnl(E)e for all n. It follows that if the index Po
is determined such that T(Ep)< for all PPo, then IVnl(Ep)E for
PPo and all n.
(c) In view of the uniform T-absolute continuity of the sequence
IVnl and the convergence of vn(E) to v(E) there exists, corresponding to
any e>O, a number >O such that T(E)< implies Iv(E)Ie. Hence, if
#(E)==O, then T(E)==O, so v(E)==O. Furthermore, it is a trivial conse-
quence of the additivity of all Vn that v is finitely additive, i.e., if E ==
l E i and E1, "', Ep are disjoint, then v(E)==l V(Ei)' We shall
prove that v is a-additive. Let, for this purpose, E == r En where all
En are #-measurable and disjoint. Then T(Rn)!O for Rn==+1 Ek;
hence, given e>O, there exists an index no such that T(Rn)< for all

Ch. 11,  45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 335
n?::-no, that is, Iv(Rn) Ie for all n?;:;no. It follows that
n
Iv(E) -  V(Ek) 1 == Iv(Rn) IE
1
for n?::-no,
so v(E)==r v(En). To summarize, v(E) is a #-absolutely continuous
complex measure on A. But then, by the preceding theorem, there
exists a complex #-summable function f(x) such that v(E)==JEfd# for
all E EA. Evidently, if all Vn are signed measures, then v is a signed
measure, so that in this case the function f is real.
By means of the last theorem we can prove now an important result
concerning converging sequences of complex measures. The result is
due to O. M. NIKODYM (1933, [2J), and the original proof was inde-
pendent of the Hahn-Saks theorem.
THEOREM 5. Let A be a a-ring of subsets of x}. the sets of A will be
called measurable sets, although there is not assumed to be given in ad-
vance a measure on A. If Vn is a sequence of comPlex measures on the a-
ring A E of all measurable subsets of the measurable set E, converging on
A E to the finite set function v, then v is a comPlex measure on .liE.
PROOF. We introduce the finite measure # on A E by
D _ 00 IVnl(D)
p,( ) - nl 2n{1+lvnl(E)}
for all DEAE, and then (after extension) all Vn become #-absolutely
continuous on the a-ring All of all #-measurable subsets of E. Hence
by the Radon-Nikodym theorem, every Vn has a representation vn(D) ==
fnfnd# for all DEAIl' with fn(x) #-summable. Note also that con-
vergence of Vn to v holds not merely on A E but also on the possibly
larger a-ring All' since any set in All differs from some set in A E only
by a #-null set. The desired result follows, therefore, from the pre-
ceding theorem.
It is of some interest to add some remarks and examples. In con-
nection with Theorem 4 the problem arises whether the convergence
of vn(E)==JEfn d # to v(E)==JEfd# for all EEA implies the pointwise
convergence of fn(x) to f(x) for #-almost every XEX, or at least the

336
SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 45
pointwise convergence of some subsequence of In(x) to f(x), like in the
situation where lim Ix Iin-/l dfl==O. The following example will show
that this is not necessarily so.
Let X==[O, 2nJ and fl Lebesgue measure in X. The functions pn(x)
==e inx are fl-summable over X for n== 1, 2, . . " and IPn(x) 1==1 for all n
and x. Hence, if f is an arbitrary fl-summable function, then Ixfpndfl
exists as a finite number for each n. We will show first that lim Ix IPn dfl
==0 as n-+oo (RIEMANN-LEBESGUE THEOREM). Since the Lebesgue inte-
gral over X may be regarded as the extension of the elementary Lebes-
gue integral on the collection of the step functions = 1 CnXAn(x) where
all An are cells, there exists for any given e>O a step. function s(x) of
this kind satisfying Ix I/-si dfl<e. Hence
Iflpndfl-f sPndfll<e
for all n.
It will be sufficient, therefore, to show that limlsPndfl==O as n-+oo
for any step function s(x) of the indicated type, and for this purpose
it is sufficient to show that lim I pndfl==O for 0a<b2n. This
follows immediately from I/ Pndfll2In.
Applying the thus obtained result to the special case that I==XE,
where E is an arbitrary fl-measurable subset of X, we obtain limlEPndfl
==0, so vn(E) == /Epndfl converges to v(E) ==0== IEO. dfl for all fl-measur-
able E eX. But neither pn(x), nor any of its subsequences, converges
to zero for any XEX, since IPn(x)l== 1 for all n and all XEX.
We may ask also whether the convergence of vn(E)==IElndfl to
1J(E)==/Eldfl implies convergence of Ivnl(E)==/Ellnldfl to Ivl(E)==
JE III dfl. Once more, the answer is no as the example presented above
shows. Indeed,
IVnl (E) == fleinxi dfl== f dfl
E E
for all n,
and Ivl(E)==/EO'dfl. The answer is negative even for real In. It follows
from the presented example that
lim f sin nxdfl==lim f cos nxdfl==O
E E
for any fl-measurable E c [0, 2nJ, but it is not true that
lim flsin nxl dfl==lim flcos nxl dfl == 0,
E .hJ

Ch. 11,  45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 337
for this would imply that
lim 1 sin 2 nxdfl==lim 1 cos 2 nxdfl==O,
E E
contradicting the fact that
Isin 2 nx dfl + Icos 2 nxdfl=fl(E)
E E
for all n.
More precisely, we have
lim 1 f(x) Isin nxl dfl== (2/n)1 fdfl
x x
for any fl-summable f. The proof is almost the same as the proof of the
Riemann-Lebesgue theorem. As above, it is sufficient to consider the
special case that f(x) is a step function = 1 CnXAn(X) where all A n are
cells, and so it is sufficient to give a proof for the special case that f is
the characteristic function of an interval. In other words, we have to
prove that J Isin nxldfl tends to (2/n)(b-a). This follows easily by
observing that the integral of Isin nxl over any interval of length n/n
equals 2/n.
Combining all results, we have now the following statement: If g(x)
is an arbitrary Lebesgue summable function on [0,2nJ and if fn(x)==
g(x) sin nx for n==l, 2, ..., then vn(E)==JEfndfl tends to V(E)=JEO'dfl
==0 for all fl-measurable sets Ec[O, 2nJ, but [vnl(E)=JElfnldfl tends to
v1(E) == JE (2/n) Igi dfl.
Assuming again (for a general measure fl) that fn and f are real, and
JEfndflJEfdfl for all fl-measurable sets E, one may ask now (in view
of the negative results in the preceding paragraphs) whether there
exists any pointwise connection at all between the sequence fn(x) and
the function f(x). There is some connection, as may be expected,
namely that
lim inf fn(x)f(x)lim sup fn(x)
holds fl-almost everywhere on X. It will follow then that if fn(x), or a
subsequence of fn(x), converges pointwise fl-almost everywhere on X
to some function g(x), then g(x) is necessarily equal (fl-almost every-
where) to f(x). In order to prove generally that
f(x)lim sup f n(x) =P(x),

338
SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11,  45
we introduce the functions gn==sup (In, In+1, . . .) and we observe that
gn!P. Since # is a-finite, we have X==r X n with #(Xn)<oc>, and it
will be sufficient to prove that IP holds on each X n separately. In
other words, we may and shall assume that #(X)<oc>. We first look at
the subset P=={x: P(x)==-oc>}. Since gn!P, we have p==r P n , where
Pn=={x: XEP, gn(X) <O}.
On each P n the theorem on integration of monotone sequences may
be applied to the sequence gn, gn+1, ..., hence lim/pgkd#==/FPd# for
all sets F c P n. Since, therefore,
I Id#==lim I Ikd#Jim I gk d #== I pd#
I? F F F
for all Fe P n, we obtain the result that fP holds #-almost every-
where on P n , and hence on p==r Pn. Since I==P==-oc> on P, this
shows incidentally that #(P) ==0. It is obvious that IP on the set
where P(x) == +OC>, and it remains to consider the set where -OC><
P (x) < OC>. This set is the union (for N == 1, 2, . . .) of the sets
QN=={X: -NP(x)N},
and it is sufficient to prove that IP on each QN. Assume, therefore,
that we have a fixed QN. The set QN is the union of the sets
Sn=={x: x E QN, gn(x) -P(x) < I},
and it is sufficient to prove that IP on each Sn. Any Sn is of finite
measure, and the functions P(x), gn(x), gn+1(X), . .. are uniformly
bounded on Sn, so lim/pgkd#==/ppd# for all FcS n . But then, again,
I Id#==lim I Ikd#lim I gk d #== I pd#
F F F F
for all Fe Sn, and it follows that IP holds almost everywhere on Sn.
This finishes the proof that Ilim sup In holds almost everywhere on
X. The proof of the inequality lim inf In1 is then immediately derived
by observing that lim inf I n== -lim sup ( - In).
We finally observe that the situation is much simpler if all In are
real and vn(E)==/Elnd#iv(E)==/Eld# on the collection A of all #-
measurable sets E. We may assume then that all Vn are (non-negative)

Ch. 11,9 45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 339
measures, since otherwise we replace Vn and v by Vn-V1 and V-VI re-
spectively. Now we have In+1/n almost everywhere (for the proof,
consider the set E=={x: In+1<ln}), so g==liml n exists, and
vn(E)== JlnXEdfli JgXEdfl== Jgdfl
x x E
on A.
It follows that IEldfl==IEgdfl for all EEA, so g==1 almost everywhere
on X. In other words, we have now that In i I almost everywhere on X.
Exercises
EXTENSION OF THE RIEMANN-LEBESGUE THEOREM
45.1) Let cp(x) be real, Lebesgue measurable, bounded and of period 1
on the real line, and let Il cp(x) dX==Ci. Show that 10 1 I(x)cp(nx) dx-+
Cill I(x)dx as n-+oo for any I which is Lebesgue summable over [0, 1J.
Observe that we may assume for the proof that Ci==O (if necessary, re-
place cp(x) by cp(x) -Ci). Under this assumption, the integral of cp(nx)
over any interval of length n- 1 is zero, and it follows easily that for any
interval [a, bJ C [0, IJ we have II: cp(nx) dxlfJn-1-+0, where fJ==
Illcp(x) I dx. Then Il s(x)cp(nx) dx-+O for any finite linear combination
s(x) of characteristic functions of intervals. Given the Lebesgue
summable function I(x) and e>O, there exists such a function s(x)
satisfying IlI/-sl dx<e, and the desired result follows. Note that for
cp(x)==sin 2nx, sin 2 2nx or Isin 2nxl we have Ci==O, ! or 2/n respectively.
45.2) The result in the preceding exercise can be extended to
unbounded cpo Let cp(x) be real, Lebesgue measurable, of period 1,
and Lebesgue summable over [0, IJ with Il cp(x) dX==Ci. Show that
Il I(x)cp(nx) dX-+Ci Il I(x) dx for any I which is Lebesgue measurable
and (essentially) bounded on [0, IJ. As in the preceding exercise, we
may assume that Ci==O. Denoting Lebesgue measure by fl, there exists
for any given 8>0 a number £5>0 such that E c [0, 1J, fl(E)£5 implies
fElcp(x)ldxe. It follows that if the positive integer n is given and F is
included in one of the intervals [kin, (k+ 1)/nJ for k==O, 1, . . ., n-l,
then fl(F)£5/n implies IF Icp(nx) I dxe/n. Hence, if E c [0, IJ andfl(E)£5,
then IE Icp(nx) I dxe for all n simultaneously. Denoting cp(nx) by In(x),
it is true therefore that the integrals IE Ilnl dx are uniformly absolutely

340 SIGNED MEASURES AND COMPLEX MEASURES [Ch. 11, 9 45
continuous, the integral of In over any interval of length n- 1 is zero,
and Illlni dxfJ for all n, where fJ is a positive constant.
Let I be measurable and (essentially) bounded; I/(x)IM on [0, IJ.
Select e>O such that EfJ, and then £5>0 such that #(E)£5 implies
IE Ifni dXE for all n. We may assume that £5e/fJ. Similarly as before,
there exists a step function of intervals s(x) such that IlI/-sl dx£52;
it may be assumed that Is(x)IM. Let E1 and E 2 be the subsets of
[0, IJ where I/-si >£5 and I/-sl£5 respectively. Then #(E1)£5, and so
II (f-s)lndxl2M Illnldx2ME,
El El
1
II (/-s)lndxl£5/llnl dx£5fJe.
E2 0
Hence
1 1
1IIIndx-ISlndxl(2M + I)E
o 0
It remains to prove that Il slndxO as noo, and it is sufficient for
this purpose to prove that I: IndxO for any interval [a, bJ. This
follows from the fact that the integral of In over any interval of length
n- 1 is zero and from the uniform absolute continuity of IEllnldx.
Note that if cp(x)==x--1 on [0, IJ, then cx==l; if cp(x)==log sin x on
[0, inJ and cp has period in, then cx== -in log 2.
45.3) Let I n(x), for Ox 1 and n== 1, 2, . . " be a non-negative
function whose graph (or more precisely, the part of the graph where
In(x) >0) consists of 2 n (narrow and high) congruent isosceles triangles
standing upon the x-axis around the centres of the intervals [k/2 n ,
(k+l)/2nJ, k==O, 1, "', 2 n -l. These triangles are such thatll fndx==l
and the sum of the lengths of their bases equals 1/2n+1. Since the inte-
gral of In over any interval of length 1/2n equals now 1/2 n , it is easy
to derive that I: Indxb-a== I: dx as noo for any interval [a, bJ C
[0, IJ. Show that it is not true that IElndx/Edx for every Lebesgue
measurable set E. This is due to the fact that the integrals IElndx==
IE Ilnl dx are not uniformly absolutely continuous.
for all n.
POINTWISE EQUALITIES AND INEQUALITIES
45.4) Let cp(x) be real, Lebesgue measurable and of period 1 on the
real line, and let A be the essential least upper bound of cp(x); it may be

Ch. 11,9 45J ABSOLUTELY CONTINUOUS SIGNED MEASURES 341
that A=== -00 or A=== +00. Finally, let In (x) ===p(nx) for n== 1, 2, . . . .
Show that lim sup In(X)===A at almost every point x. More generally,
show that lim sup fnk(x) ===A almost everywhere for any subsequence Ink'
A similar statement holds for lim inf In(x) and the essential greatest
lower bound of p(x). Note that, in view of this result, we have
lim sup sin nx== 1 and lim inf sin nx== -1 almost everywhere; lim sup
Isin nxl === 1 and lim inf Isin nxl ===0 almost everywhere; if p(x) is of
period 1 and equal to x--1 on [0, IJ, then limsupp(nx)===+oo and
liminfp(nx)==O almost everywhere; if p(x) is of period in and equal
to log sin x on [0, inJ, then lim sup p(nx) ===0 and lim inf p(nx) === -00.
45.5) Let p(x) be a real function of period 1 on the real line, Lebes-
gue summable over [0, IJ, such that the essential least upper bound
and greatest lower bound of pare +00 and 0 respectively (e.g., p(x)==
x-!-1 on [0, IJ). For k==l, 2, ... and 1==1, ...,kthefunctiongkl(x)
is defined on [0, 1 J as follows: gkl(X) ==p(kx) on [(1-1) /k, l/k J and
gkl(X) ==0 elsewhere. Hence, gk1, . . " gkk are formed by splitting up
p(kx) into k disjoint parts. Arranging all the functions gkl(X) into a
single sequence In (n===l, 2, .. .), show that fElndxfEfdx for every
Lebesgue measurable E, where
O===lim inf I n(x) === I(x) <lim sup I n(x) === + 00
for almost every point XE[O, IJ.
45.6) Let I(x) and g(x) be real and Lebesgue summable over [0, 2nJ,
where If(x)lg(x) on [0,2nJ. Show the existence of a sequence of Le-
besgue summable functions In (x) such that fElndx fEldx and fE ifni dx
fEgdx for every Lebesgue measurable set Ec[O, 2nJ.

CHAPTER 12
CONJUGATE SPACES
AND WEAK SEQUENTIAL CONVERGENCE
The main results in this chapter are the representation theorems for bounded
linear functionals on the normed linear spaces CK and Lp (l<,p < 00), where CK
is the space of continuous functions f(x) on Euclidean space Rk having a bounded
carrier .dj, and with norm the uniform norm. In other words, the conjugate
spaces of C K and Lp' (1<,p < 00) are determined, and it turns out that these
conjugate spaces can be identified with the Banach space of all complex
measures on the Borel sets in Rk and with the space Lq (P-l+q-l= 1) respective-
ly. In sec. 51 the Banach-Steinhaus theorem is proved, sec. 52 is devoted to a
discussion of weak sequential convergence in Lp spaces, and in the final sec. 53
the notion of a reflexive Banach space is introduced.
46. Borel Sets in Euclidean Space
Let :T be the collection of all a-fields A of subsets of k-dimensional
Euclidean space Rk such that every A E:T contains all bounded closed
intervals. The collection :T is not empty, for the a-field of all subsets
of Rk is an element of :T (and the same is true of the a-field of all
Lebesgue measurable subsets of Rk). Evidently, the intersection A b ==
nAE.r A is also a a-field; more precisely, Ab is (in view of its definition)
the smallest a-field containing all bounded closed intervals. The sets
in Ab are called the Borel sets in Rk.
THEOREM 1. The a-field Ab of all Borel sets is the smallest a-field
containing all bounded open intervals as well as the smallest a-field con-
taining all (bounded) cells. It is also the smallest a-field containing all
bounded open sets as well as the smallest a-field containing all compact
(i.e., bounded and closed) sets.

Ch. 12,  46J
BOREL SETS IN EUCLIDEAN SPACE
343
PROOF. Since every bounded open interval I is the union of a
sequence of bounded closed intervals, we have I EAb, and this implies
that Ab is one of the a-fields containing all bounded open intervals.
Hence A' cAb, if A' is the smallest a-field containing all bounded open
intervals. Conversely, every bounded closed interval leis the inter-
section of a sequence of bounded open intervals, so IeEA', which im-
plies that Ab c A'. It follows that A' ==Ab. The proof for cells is similar,
and the proofs for the remaining cases are based on the observation
that any open set is a union of a countable number of bounded open
intervals, so that any open set, and hence any closed set, is a set in Ab.
The real function f(x), defined on Rk, is called a Borel function if
{x: f(x) >a} is a Borel set for every real a. Note that, in this case, the.
sets {x: f(x)a}, {x: f(x)a} and {x: f(x) <a} are Borel sets just as well.
The complex function f==g+ih (g and h real) is called a Borel function
whenever g and h are Borel functions. Every continuous function on
Rk is a Borel function, since for real f the set {x: f(x) >a} is open for
any real a.
Let C<t2 be the collection of all real continuous functions f on Rk
having a compact carrier L1f. We recall that the carrier L1f of f is the
closure of the set {x: j(x) =¥=O}, so that L1f is automatically closed. The
assumption that L1f is compact is equivalent, therefore, to the as-
sumption that L1f is bounded. Note that L1f depends on f. Let A K be
the smallest a-field containing all sets {x: f(x) >a}, where f runs through
C<[2 and a runs through the real numbers. Then AK==Ab. It is evident
from the definition of A K that AKcA b , and in order to prove the
converse it is sufficient to show that every bounded open interval I is
in A K . If Ie is the complement of I, then f(x)d(x, Ie), where d de-
notes Euclidean distance, belongs to C) and the set {x: j(x) >O} is
exactly I; hence I EA K . Compare this result with the definition of
abstract Borel sets in Exercise 32.13. Note that A K is also the smallest
a-field containing all sets {x: f(x) >a}, where f runs through the non-
negative funtions in CCf) and a runs through the positive numbers.

344 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 47
47. Function Spaces
This section is devoted partly to a recapitulation of previously dis-
cussed subjects. We shall denote by C K the collection of all complex
continuous functions f(x) on Rk having a compact carrier L1f. The col-
lection C K is evidently a linear collection, and it is a normed linear
space with respect to the uniform norm Ilfll==max XERk If(x) I. The sub-
collection C) of all real fEC K is a real normed linear space with re-
spect to the same norm. We recall the Riesz representation theorem for
non-negative linear functionals on C<J2 (cf. sec. 21, Theorem 3):
Any non-negative linear functional J(f) on C) is an elementary inte-
gral on Ccp, so J(f)===IRkfdv for all fEC<[2, where v is the measure induced
in Rk by the extended integral J(f). The a-field Av of all v-measurable
subsets of Rk contains all open sets, so AvAb (where Ab is the a-field
of all Borel sets). Furthermore, every bounded set in Av is of finite v-
measure.
We make three additional remarks. In the first place, if J(f) is
bounded on Ccp, i.e., if IJ(f) IMllf" for some MO and all fEC),
then the corresponding induced measure v is a finite measure. Indeed,
given the bounded closed interval Ie, there exists a function fEC)
such that ot.1 on Rk and f== 1 on Ie. Hence v(I e)J(f)M. Since
this holds for every bounded and closed Ie, it follows that V(Rk) M.
Conversely, if V(Rk) <00, then J(f) ==1 fdv is a bounded linear functional
on C<;2. Indeed,
IJ(f) 1/lfl dv{v(Rk)}"fll.
Secondly, as observed in sec. 20, the collection CK==C)+iC) is
dense in the Banach space of all complex v-summable functions, that
is, given the complex v-summable function g(x) and the number e>O,
there exists a function fECK satisfying 1 Ig-fldv<e.
Thirdly, as is already stated and proved in sec. 21, Theorem 3, we
have the converse property that if v is a measure on Ab such that v(E)
is finite for all bounded EEAb, then /(f)===lfdv is a non-negative
linear functional on C<f). Hence, by the representation theorem, /(f)
is an elementary integral on C), so that /(f) induces by extension a
measure VI in Rk satisfying /(f) ===1 fdv1 for all fEC<;2. The problem
arises, of course, whether this new measure VI is, at least on Ab, again
the initial measure v. Evidently, this is the problem of uniqueness of v

Ch. 12, 9 47J
FUNCTION SPACES
345
in the representation f (I) === J Idv. In the terminology of measure theory
the problem may be restated as follows. Given the measure # and a
semi-ring r contained in the a-field AJl of all #-measurable sets, and
denoting by #1 the restriction of # on r, what is the relation between
the extended measure #1 and the initial measure #? In the present
situation # is then the measure in the Cartesian product Rk X Rt corre-
sponding to the integral J Idv generated by the v-step functions (i.e.,
if F is the ordinate set of the v-measurable function IO, then #(F) ==
J Idv), and r is the semi-ring of all differences of ordinate sets of non-
negative functions in C). An answer, sufficient for our purposes, is
easily available: Any #l-measurable set E of finite #l-measure is #-
measurable, and #l(E)===#(E). The proof follows by observing first
that this is true by definition if E E r, so that the same holds if E is
a a-set (with respect to r) or the limit of a decreasing sequence of
such a-sets of finite measure. Furthermore, the result is true for #1-
null sets (covering by a-sets of arbitrarily small measure). The con-
clusion may be drawn, therefore, that #1==# for sets of finite #1-
measure. Applied to the present situation this means that J Idv1 ==
! Idv for any v1-summable I, in particular for I==XE, where E is an
arbitrary bounded Borel set. Hence VI ==v for the bounded Borel sets,
and this implies that VI ==v on the whole of Ab. The measure v in the
representation f(/)==Jldv is, therefore, unique on Ab.
Finally, we recall the definition of a Banach function space in sec. 30.
If # is a measure in the non-empty point set X, and M+ is the col-
lection of all non-negative #-measurable functions, we assume that
corresponding to every IEM+ there exists a number p(/) such that
Op(/)oo, and
(a) p{/) ===0 if and only if 1==0 (almost everywhere); p(/1 +/2)p(/1)+
p(/2) ; p(al) ===ap(/) for all aO;
(b) if InEM+ (n== 1,2, . . .) and In i I (almost everywhere), then
p (I n) i p (I) ·
As shown in the section referred to, the collection Lp of all complex
,u-measurable functions I, satisfying II/II===p(I/I)<oo, is a Banach space
with respect to the norm 11/11. Examples of such Banach spaces are the
spaces Lp (IPoo), where II/lIp==(Jx I/I P d#)l/ p for lp<oo and 11/1100
is the essential least upper bound of I/(x)/. If Lp is a Banach function

346 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 48
space, we shall denote by Lr) the real Banach space consisting of all
real fEL p .
Exercises
THE CLOSURE OF C K
47.1) Let Coo be the collection of all complex functions, defined and
continuous on Rk, and tending to zero as x tends to infinity (that is,
given fEC oo and e>O, there exists a compact set L1f,e such that If(x)l<e
outside L1f,e). Show that Coo is a Banach space with respect to the norm
IIfll ==max xERk If(x) I, and show also that the normed linear space C K is
dense in Coo'
BOREL SETS
47.2) Let the non-negative linear functional f(f) on C) induce the
measure v in Rk. Show that every v-measurable set E c Rk is included
in and v-almost equal to a Borel set.
47.3) Let the non-negative linear functional f(f) on C) induce the
measure v in Rk, and let Av be the a-field of all v-measurable sets. The
restriction of v on the a-field Ab of the Borel sets is denoted by Vb, and
the extension procedure for measures is applied to (Rk, Ab, Vb), where-
by the extended measure VI on the a-field Al is obtained. Show that
Al ==Av and VI ==v on Al ==Av.
48. Decomposition of a Real Bounded Linear Functional
By V we shall denote in this section either the real normed linear
space C) or the real Banach function space Lr). As defined previ-
ously, any real linear function F(f) on V into the set of all real numbers
is called a real linear functional on V, and F(f) is said to be bounded
if IF(f) IMllfll for some MO and all fE V. The norm IIFII of F(f) is
defined by IIFII==sup IF(f) I for Ilfll 1.
We shall prove now a decomposition theorem for real bounded
linear functionals on V which is similar to the Jordan decomposition
theorem for signed measures. The proof is also similar.

Ch. 12,9 48J
REAL BOUNDED LINEAR FUNCTIONAL
347
THEOREM 1. Every real bounded linear functional F(f) on V is the
difference of two non-negative bounded linear functionals, that is, F(f) ==
P+(f)+F-(f), where F+ and F- are bounded and linear, and where
F+(f)O and F-(f)O for OfE V.
PROOF. For any non-negative fEV, let F+(f)==sup F(g) for all gEV
satisfying Ogf. Then F+(f)O in view of F(O) ==0, and IF+(f) I
IIFII'llfll in view of IF(g) 1IIFII'lIgllIIFII'lIfll for Ogf. Furthermore,
it is evident that F+(af)==aF+(f) for any constant aO. Now, let O
f1EV and 0f2EV, and choose gl, g2EV such that 0glf1 and O
g2f2' Then
F+(f1 +f2)SUp F(gl +g2) ==sup F(gl) +sup F(g2) ==F+(f1) +F+(f2).
Conversely, if 0gf1 + f2, we have Ogl ==min (f1, g)f1 and 0g2==
g-min (f1, g)f2, so
F(g) ==F(gl) + F(g2)F+(fl) + F+(f2) ,
and hence
F+(f1 +f2)==SUP F(g)F+(f1) +F+(f2)'
I t follows that F+(f1 + f2) ==F+(f1) + F+(f2)' For an arbitrary fE V we
write f==f++f-, where f+ and f- are the positive and negative parts of
f, and we define F+(f) by F+(f)==F+(f+)-F+(-f-). In order to show
now that F+ is linear on V, assume that the same f is written also as
f==f1-f2, where f1 and f2 are non-negative functions in V. Then f1-f+
==f2-(-f-)0, so
F+(f1) -F+(f+) ==F+(f2) -F+( - f-),
and this shows that
F+(f) ==F+(f+) - F+( - f-) ==F+(f1) - F+(f2)'
It follows easily that F+ is linear on V, and F+ is bounded on account of
IF+(f) 1== IF+(f+) -F+( -f-) IF+(f+) +F+( -f-) ==F+(lfl)IIFII'llfll.
Setting now F-(f)==F(f)-F+(f), the linear functional F-(f) is clearly
bounded on V, and -F- is non-negative since F+(f)F(f) for all fO
in V. Hence F==F++F-, where F+ and -F- are bounded and non-
negative on V.

348 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 48
In order to present a more complete picture of the situation, we
finally prove that, for O/EV, we have F-(/)inf F(g) for all gEV
satisfying Og/. For this purpose, let O/E V and denote inf F(g) for
Og1 by fJ. Select a sequence gn (Ogn/) such that F(gn) fJ. Then
it follows from F(/)F(gn)+F(I-gn) that F(/)fJ+F+(/). Selecting
another sequence g (Og;"/) such that F(g)F+(/), it follows simi-
larly that F(/)?;::-F+(/) +fJ. Hence
fJF(/) -F+(f) ==F-(f).
It follows then also, similarly as for F+, that IF-(/) IIIFII 'lIfil for all
fEV.
COROLLARY. The decomposition F==F+-(-F-) is the most efficient
decomposition 01 F in the sense that if FF1-F2 is any other decom-
position 01 F into non-negative bounded linear functionals F 1 and F 2,
then F1-F+F2-(-F-) is non-negative.
PROOF. If OfE V and if gE V satisfies Og/, then
F(g) ==F 1 (g) -F 2(g)F l(g)F l(f),
so F+(/)sup F(g)F1(/).
Exercises
THE ABSOLUTE VALUE OF A BOUNDED LINEAR FUNCTIONAL
48.1) Let V be as above and let F be a bounded linear functional
on V. We consider the non-negative bounded linear functionaIIFI(f)=
F+(/) - F-(f). Show that
(a) IF(/)IIFI(III) and "FI(f)IIFI(lfl) for all fEV;
(b) IIIFIII==sup IFI (f) for all f?;::-O satisfying IIfll 1 ;
(c) for any f);O we have IFI(f)sup F(f1-f2) for all decompositions
f==f1+/2, 0/1EV, 0/2EV;
(d) for any 1);0 we have 1}1"I(/)supIF(g)1 for all gEV satisfying
Igl/;
(e) IIIFIIIIIFII, and IFIF whenever F is non-negative.
48.2) Let V be either the complex normed linear space C K or the
complex Banach function space L p , and let F be an arbitrary bounded

Ch. 12,  48J
REAL BOUNDED LINEAR FUNCTIONAL
349
linear functional on V. For any non-negative fEV we define IFI(f)==
sup IF(g) I for all gE V satisfying Iglf. For any real fE V the number
IFI(f) is defined by IFI(f)==IFI(f+)-IFI(-f-), and for every complex
j==g+ihEV by IFI(f)==IFI(g)+iIFI(h). Show that IFI(f) is a non-nega-
tive bounded linear functional on V such that" IFII[==IIFII. Note that
if F(f) is real for all real fE V, then IFI (f) ==sup IF(g) I for all real gE V
satisfying Iglf. In particular, if F(f)O for all fO in V, then IFI==F
by the preceding exercise.
PARTIAL ORDERING IN THE SPACE OF BOUNDED LINEAR FUNCTIONALS
48.3) Let V be the same real space as in Exercise 48.1. Given the
bounded linear functionals F 1 and F 2 on V, write F 1 F 2 whenever
F 2 -F 1 is non-negative. Show that this defines a partial ordering in
the conjugate space V*. Given FE V*, show that F+ and F- are the
least upper bound and the greatest lower bound respectively of F and
the null functional (with respect to the partial ordering). Given F 1 and
F 2 in V*, show that the least upper bound of F1 and F2, denoted by
F 1 V F 2, exists and satisfies
F 1 V F 2==F 1 + (F 2- F l)+==F 2+ (F 1- F 2)+.
Show similarly that the greatest lower bound F 1AF 2 exists and satisfies
F 1AF2==F 1 + (F 2 -F 1)-==F 2 + (F 1-F2)-'
Show now that F1AF2==F1-(F1-F2)+, and derive the result that
(F 1 VF 2 ) + (F 1 A F 2) ==F 1 +F2.
48.4) Let V be as in the preceding exercise, and assume that F and
F1, "', Fn are non-negative elements of V* satisfying F'L/i=l Fj.
Show the existence of non-negative elements G1, . . . , G n of V* such
that GjFj for i == 1, . . . , nand F == '):/j= 1 Gj.
48.5) Let the real space V and the real bounded linear functionals
F 1 and F 2 be as in Exercise 48.3. Show that
2(F1 VF 2 )==F 1 +F2+ I F 1- F 21
and
2(F1 AF 2)==F1 +F 2 -I F 1- F 21.

350 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 49
Furthermore, show that IF1 +F21<IF11+ IF21, i.e.,
IF 1 +F21 (f)<IF 11 (f) + IF21(f)
for every fO In V.
Show that this triangle inequality continues to hol when F 1 and F 2
are complex bounded linear functionals on the complex space V con-
sidered in Exercise 48.2. Finally, show that laFI == lal'IFI for any
complex number a and IIF11-IF211<IF1-F21.
49. The Conjugate Space of CK
As shown previously (cf. sec. 26), the collection of all bounded linear
functionals F(f) on an arbitrary normed linear space V is a Banach
space V* with respect to the norm I[FII==sup IF(f) I for IIfll< 1. The space
V* is called the conjugate space of V. The present section will be de-
voted to the proof that the conjugate space C of the normed linear
space C K may be identified with the collection of all (finite) complex
measures v on the a-field Ab of the Borel sets; in view of the results
in sec. 44 this collection is a Banach space with respect to the norm
Ilvll== Ivl(Rk)' Thus we obtain the Riesz representation theorem (a par-
ticular case of which was considered in sec. 21, Theorem 3 and in sec. 47)
in more general formulation.
THEOREM 1 (RIESZ REPRESENTATION THEOREM FOR BOUNDED LINE-
AR FUNCTIONALS ON C K). There exists a linear one-one correspondence
between the Banach space C of all bounded linear functionals F on C K
and the Banach space of all comPlex measures v on the a-field Ab of all
Borel sets in Rk. This correspondence is given by
F(f)== ffdv,
Rk
holding for all fEC K . The correspondence is norm preserving, i.e., if F
and v correspond, then IIFII==llvll==lvl(Rk)'
PROOF. (a) We consider first the case that F(f) is a real bounded
linear functional on C). Then F==F++F-, where F+ and -F- are
bounded and non-negative, so that (by the remarks in sec. 47) there
exist finite measures #1 and #2 in Rk, defined and unique on the a-

Ch. 12,  49J
THE CONJUGATE SPACE OF C K
351
field Ab of all Borel sets, such that
F+(f)== f fd#l,
Rk
-F-(f)== f fd#2
Rk
for all fEC). It follows that V#1-#2 is a signed measure on Ab,
and if v+ and v- are the positive and negative variations of v on Ab,
then the difference #1-V+#2-(-V-)# is a finite measure on Ab
(cf. sec. 43, Theorem 3). Hence,
F(f)F+(f)+F-(f) ffd#1-ffd#2
{ffdv++ ffd#}-{ffd(-v-) + ff d #}
== ffdv+-ffd(-v-)== ffdv
on C) (cf. the definition of J fdv in sec. 43). In order to prove the
uniqueness of v, let us assume that there exists a signed measure T on
Ab such that
F(f) == f fdT== f fdT+-f fd( -T-) F 1 (f) -F 2 (f)
on C). Then F 1 - F+==F 2 - (- F-) is a non-negative bounded linear
functional on C) in view of the corollary in the preceding section, and
it follows that there exists a finite measure a on Ab which represents
F 1 -F+==F2-(-F-). The functionals F1 and F 2 are represented,
therefore, by #1 +a and #2+a respectively. Since, on the other hand,
the same functionals are represented by T+ and -T- respectively, and
since the representation for non-negative functionals is unique, we
have T+==#l+a and -T-#2+a, so T==T+-(-T-)==#1-#2==V. This
shows that the signed measure v in F(f) == J fdv is unique on Ab. Further-
more, since T+ - (-T-) is the most efficient manner to express T==V as
a difference of two finite measures, and since T==V==#1-#2(T+-a)-
(-T--a) is another similar decomposition, we obtain the additional
result that a is identically zero on Ab. Hence, #1 v+ and #2== -V-, so
F+(f)==Jfdv+ and -F-(f)Jfd(-v-) on C).
(b) Now, let F(f) be a bounded linear functional on C K . Then F(f)==
Fr(f) +iFim(f) , where Fr(f) and Fim(f) are real bounded functionals on
C K which are real linear, that is, Fr(af1 +bf2) ==aF r (f1) +bF r (f2) for real
a and b, and similarly for F im . In particular, Fr(f) and Fim(f) are real
bounded linear functionals on C), so that in view of (a) there exist

352 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  49
unique signed measures v' ==fl1 -fl2 and v" ==fl3-fl4 on Ab such that
fli (i==l, 2, 3, 4) is a finite measure on Ab, and Fr(f)==ffdv' and Fim(f)
== f fdv" on CCf). Hence, if the complex measure v on Ab is defined by
v==v' +iv", and f==g+ih (g and h real) is an arbitrary function in C K ,
we have
F(f) ==F(g) +iF(h) ==Fr(g) +iFim(g) + i{Fr (h) +iFim(h)}
== I gdv' +i I gdv" +i{1 hdv' +i I hdv"}== I fdv' +i I fdv" == I fdv.
Evidently, v is unique since v' and v" are unique.
It remains to prove now that IIFII==lvl(Rk). In view of the inequality
If fdvlf If I dlvl, holding for all bounded Borel functions f (cf. sec. 44),
we obtain immediately the result that
IF(f) I == II fdvlJ If I dlvl{lvl (Rk)} 'lIfll
for all fEC K , hence "FIIlvl(Rk)' In order to prove the converse ine-
quality, we recall (cf. sec. 44) that Ivl(Rk)==sup  Iv(En) I for all decom-
positions Rk== En into a sequence of disjoint Borel sets En. Let
therefore, if e>O is given,  En be a decomposition such that  Iv(En) I
> Ivl (Rk) -8, and let v(En) == Iv(En) I e iCPn . Then
g(x) ==  e- icpn XEn (x)
is a Borel function satisfying f gdv==  Iv(En) I. Furthermore, since g(x)
is bounded, g(x) is summable over Rk with respect to each of the finite
measures fli (i==l, 2, 3, 4), and hence also with respect to the finite
measure P:==fl1 +fl2+fl3+fl4. By the remarks in sec. 47, the collection
C K is dense in the Banach space of all complex p:-summable functions,
i.e., there exists a function f* EC K such that f If* -gl dp:<e. It may
happen that If*(x) I> 1 on a subset of Rk, and we wish to avoid this.
For that purpose we observe that If*(x) I is continuous, so that, conse-
quently, the function max(lf*l, 1) is also continuous. Then
f == f* /max (I f* I, 1) E C K,
and If(x) I 1 for all X. Furthermore, since Ig(x) 1==1 on Rk, we have
If(x) -g(x) I If*(x) -g(x) I
for every x,

Ch. 12, 9 49J
THE CONJUGATE SPACE OF C K
353
so f Ij-gl d{l<E. This implies that
If (j-g) dvllf (f-g) dv'l + If (j-g) dv"l
4
  If (f-g) dflil f If-gl dfli== f Ij-gl dfl<e,
i=l
so
Ilf fdvl-If gdvlllf (j-g) dvl <e,
and hence
Iffdvllf gdvl-e==  Iv(En)l-e>lvl(Rk)-2e.
It follows that Ivl (Rk)SUp If jdvl for all fEC K satisfying Iljll 1, that is,
Ivl(Rk)IIFIi. The final conclusion is, therefore, that IIFII== Ivl(Rk).
It is obvious that if, conversely, v is a complex measure on Ab (ob-
serve that this implies, by definition, that v(E) is a finite complex
number for every EEA b ), then F(j) == f fdvis a bounded linear functional
on C K, so that, by what we have already proved, there exists a unique
complex measure VI on Ab such that IIFII== I V 11 (Rk) and F(f) == f jdv1 on
C K. The uniqueness implies that VI ==v, so IIFII == Ivl (Rk). This completes
the proof.
Note that the proof of IIFII==lvl(Rk) is also valid for: a real bounded
linear fu'nctional F(j) == f fdv on C), since the relation Ivl (Rk) ==
sup  Iv(En) I for all decompositions R k =  En with disjoint Borel sets
En holds also for a signed measure V on Ab (cf. sec. 43).
Exercises
THE CONJUGATE SPACE OF THE CLOSURE OF C K
49.1) If Coo is the Banach space of all complex continuous functions
on Rk which tend to zero as x tends to infinity (cf. Exercise 47.1), show
that the conjugate spaces C: and C are identical. In other words, C:
may be identified with the Banach space of all complex measures on
Ab, in the following sense: If FEC: and the complex measure v corre-
spond, then F(f)==ffdv for all fEC oo and IIFII==lvl(Rk)'

354 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  49
NORM OF A BOUNDED LINEAR FUNCTIONAL
49.2) In the proof of the Riesz representation theorem it was shown
that if FEC and the complex measure v on Ab correspond, then
IIFII==lIvll==lvl(Rk)==sup  Iv(En) I
for all possible decompositions of Rk into disjoint Borel sets En. Show
that IIvll may as well be defined by IIvil ==sup  Iv(E n) I for all decom-
positions of Rk into disjoint v-measurable sets En, where by a ')J-
measurable set we mean a set which is fli-measurable (after extension
of the measures) for i==l, 2, 3, 4 if V==(fll-fl2)+i(fl3-fl4) in the no-
tations of the proof referred to.
PARTIAL ORDERING
49.3) Show that if F1 and F 2 are real bounded linear functionals on
. C), corresponding to the signed measures VI and V2 respectively, then
F1 VF 2 and F1 A F 2 correspond to V1VV2 and V1Av2 respectively (for the
notations, cf. Exercises 48.3, 43.2 and 43.3). Show also that if F corre-
sponds to v, then IFI (as defined in Exercise 48.1) corresponds to Ivl.
The functionals F 1 and F 2 are called orthogonal whenever IF 11A IF 21
==0, where 0 is the null functional. Show that F1 and F 2 are orthogonal
if and only if the corresponding signed measures VI and V2 are ortho-
gonal, i.e., if and only if IV1IAlv21==0, where 0 is the null measure (cf.
Exercise 43.4). Show that F1 and F 2 are orthogonal if and only if
IF 1 +F21 == IF 1-F 21 or, alternatively, if and only if IF1 +F21 == IF IIVIF 2 1.
ABSOLUTE VALUE OF A COMPLEX BOUNDED LINEAR FUNCTIONAL
49.4) Let F be a bounded linear functional on C K corresponding to
the complex measure v. Show that IFI (as defined in Exercise 48.2)
corresponds to Ivl. Similarly as in the real case, F 1 and F 2 are called
orthogonal whenever IF1IAIF21==0. Show that if F1 and F 2 are ortho-
gonal, then IF1+F21==IF1-F21 and IF1+F21==IF1IVIF21. Finally, show
that neither of these conditions is sufficient for orthogonality of F 1 and
F 2 .

Ch. 12,9 50J
THE CONJUGATE SPACE OF Lp
355
50. The Conjugate Space of Lp
Given the measure fl in the non-empty point set X, we shall now
consider the Banach function space Lp(X, fl) for values of p satisfying
lp(X). For lp<(X) the space Lp consists, therefore, of all complex
fl-measurable f(x) such that J If(x) IP dfl is finite, and Lp is a Banach
space with respect to the norm
pp(f) ==pp(lfl) == IIfllp== (I If (x) I P dfl)l/p.
For p==(X) the space Lp==Loo consists of all complex fl-measurable f(x)
such that the essential least upper bound of If(x) I is finite, and Loo is
a Banach space with respect to the norm Poo(f) ==Poo(lfl) == IIflloo==
esssuplf(x)l. For lp(X) the subcollection L) of all real fEL p is a
real Banach space with respect to the same norm IIflip. It will be shown
in this section that, in general, the conjugate space L; (lp<(X») may
be identified with Lq, where q is determined by p-1+ q -1= 1, whereas.
L 1 is, in general, a proper linear subspace of L. Several lemmas will
be proved first.
LEMMA cx. (a) Let 1 <p(X), p-1+q-1 1, and fELq. Then
IIfllqmax II fgdfll ==max Ilfgl dfl
for all gEL p satisfying IIgllp 1. If fELr), then the same holds for g re-
stricted to all gEL) satisfying Ilgllp 1.
(b) Let fl have the finite subset property (i.e., any set of positive
measure has a subset of finite positive measure), and let fELoo. Then
IIflloo==sup II fg dfll sup Ilfgl dfl
for all gEL 1 satisfying IIg"ll. If fEL<:O), then the same holds for g re-
stricted to all gELi r ) satisfying IIg111 1. If fl does not have the finite subset
property, there exists a function fELoo such that IIflloo== 1 and J Ifgl dfl=O
for all gEL 1 .
PROOF. (a) Given that 1 <p(X), p-1+ q -1= 1 and fEL q , Holder's
inequality shows that
II fgdfll/lfgl dflllfllq
for all gEL p satisfying Ilgllpl. It will be sufficient, therefore, to show
the existence of a particular function gEL p satisfying IIgllp== 1 and

356 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  50
If fgdfll ===llfllq. Obviously, we may assume for that purpose that Ilfllq>O.
For any complex number ex, let the number sgn ex be defined by sgn ex===
ex/lexl for 0< lexl <00, and sgn ex== 1 for ex==O or lexl ===00. Then
h(x) == l/sgn f(x) ELoo,
and
Ilhlloo === 1.
Hence, setting g==h for the case that p===oo, we have Ilgllp==lIhllp==
IIhlloo == 1 and
II fgdfll == II fhdfll == Ilfl dfl=== IIfl11 === IIfllq.
This completes the proof for P===oo. For 1 <p<oo, we have
k(x) == If(x) Iq-1/ sgn f(x) ELp
on account of IkI P === Iklq/(q-l)== Ifl q , and I[kll p === I[fll/p. Hence, defining
g(x) by g(x) ==k(x) /IIfllg/ P , we have gELp and Ilgllp== 1. Furthermore,
II fgdfll == II fkdfll/llfllg/ P
= J Ifl q dfl/llfll/P== Ilfll-q/P === IIfllq.
Note that if fELr), then the maximizing function g belongs to L).
(b) Let fl have the finite subset property, and let fELoo. Given e>O,
the set {x: If(x)I>lffiloo-e} contains a subset E of finite positive
measure. Hence, if we define g(x) == {fl (E) sgn f(x) }-1 on E and g(x) ==0
on X-E, then g(x) satisfies IIgl[l===1 and
Ilfgdfll== Ilfldfl/fl(E)lfflloo-e.
E
It follows that IIflloosup If fg dfl I for IIglll 1. The inverse inequality
follows, similarly as in part (a), from Holder's inequality. Note that if
fEL<:O), then the introduced "almost maximizing" function g(x) belongs
to Li r ).
Finally, if fl does not have the finite subset property, there exists a
set E such that fl(E) ==00 and fl(F) ==0 for all measurable sets FeE of
finite measure (and hence for all FeE of a-finite measure). It follows
that any gEL 1 vanishes almost everywhere on E, and so f===XE satisfies
IlxEI100 == 1, but f IXEgl dfl==O for all gEL 1 .
The next lemma is an irnmediate consequence.

Ch. 12,9 50J
THE CONJUGATE SPACE OF Lp
357
LEMMA fJ (EMBEDDING THEOREM). (a) Let 1 <poo, p-1+ q -1== 1,
and fELq. Then the function f determines a bounded linear functional F
on Lp by means of the defining relation F(g) == f fg dfl, holding for all
gELp. The functional F satisfies IIFI[==IIfl[q, which imPlies immediately
that different fELq determine different FEL;. Hence, the relation F(g)==
f fg dfl defines a one-one linear and norm preserving correspondence be-
tween the functions fELq and the elements F of a certain sub collection of
L; (this subcollection is then, in view of the correspondence, a closed linear
subspace). In other words, Lq may thus be identified with a closed linear
subspace of L;. Similarly, Lr) may be identified with a closed linear
subspace of (L»)*.
(b) Let fl have the finite subset property, and let fELoo. Then f de-
termines a bounded linear functional F on L I by means of the defining
relation F(g) == f fgdfl, holding for all gEL 1 . The functional F satisfies
IIFII==IIflloo' which imPlies that different fELoo determine different FELi.
Hence, similarly as above, Loo may be identified with a closed linear sub-
space of Li. Similarly, L) may be identified with a closed linear subspace
of (L'») *.
PROOF. (a) Given that 1 <poo, p-1+ q -1== 1 and fEL q , we have
II fgdfll/lfgl dflllfllqllgilp
for all gEL p , and hence F(g) == f fgdfl is a bounded linear functional on
Lp. Furthermore, the preceding lemma shows that
IIfl!q== max Ilfgdfll== sup IF(g)I==IIFII.
Ilgllpl Ilgllpl
(b) Given fEL oo , we have If fgdfllf Ifgl dflIIfll001lgll1 for all gEL 1 ,
and hence F(g)==ffgdfl is a bounded linear functional on L I . The pre-
ceding lemma shows that
IIflloo== sup Iffgdfll== sup IF(g)I==IIFII.
'1gI111 Ilg!lll
In the next lemma we extend, for the case that X is of a-finite fl-
measure, the results proved in Lemma a, and we also prove a property
of the spaces Lp and Lq (P-1+q-1== 1) which, to some extent, may be
considered as a converse of Holder's inequality. Given the non-nega-
tive measurable function f(x) on X, we write pp(f) == (f f p dfl)l/V when

358 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 50
lp<(X), and poo(/)==ess sup I(x). It is very well possible, therefore,
that pp(/) == (X) for some or all values of p (1 p(X)) ; if pp(/) < (X) , then
IELp and pp(/)==ll/llp.
LEMMA y. (a) Let X be of a-finite measure, lp(X), p-1+ q -1== 1,
and let f(x) be a comPlex measurable function on X. Then pq(lfl) ==
sup J Ifgl d# for all gEL p satisfying IIgllp 1.
(b) Let X be of a-finite measure, lp(X), p-1+q-1==1, and let the
comPlex measurable function f(x) have the property that fg is summable
over X for every gEL p . Then fELq.
PROOF. (a) If pq(lfl) <(X), the desired result follows from Lemma ex;
we may assume, therefore, that pq(lfl) ==(X). Since X is of a-finite
measure, there exists an ascending sequence of measurable subsets
X n eX such that X n i X and #(X n ) <(X) for all n. Let fn==min (III, nx Xn )
for n== 1, 2, . . . . Then fnELq and fn i If I , so IIlnllq i pq(lfl) as n-+(X). By
Lemma ex there exist functions gnELp (n== 1, 2, . . .) such that IIgnllp== 1
and J Ifngnl d#== IIlnllq. Hence J Ifgnl d#lIfnllq for n== 1, 2, . . " which
shows that
sup Ilfgl d#lim IIfnllq==pq(lfl) ==(X).
IIg!!p:::::;;;l
(b) Given the measurable function f, having the property that
J Ifgl d# is finite for every gEL p , we have to show that pq(l/l) <(X).
Assume that pq(lfl)==(X). Then, on account of the result in (a), there
exist functions gnELp such that IIgnllp== 1 and f I/gnl d#>n 3 for n==
1,2, ... . Let g==r n- 2 Ignl. The partial sums Sn of this series satisfy
"snllpr n- 2 , and since IIsnilp i pp(g) as n-+(X), it follows that pp(g)
r n- 2 , so gEL p . But then, by hypothesis, the integral J Ifgl d# is
finite. On the other hand, J I/gl d#n-2 J Ifgnl d# for n== 1, 2, . . " so
J Ifgl d#>n for n== 1,2, · · " i.e., J Ifgl d#==(X). This is a contradiction.
It follows that pq(lfl) <(X).
For an extension of this result to the non-a-finite case, cf. Exercises
50.1 and 50.2.
Given the bounded linear functional F(/) on the Banach function
space L p , we write F(/)==Fr(f)+iFim(f) for all fELr), where Fr(/) and
F im (I) are the real and imaginary parts , respectively, of F (I). Then

Ch. 12,  50J
THE CONJUGATE SPACE OF Lp
359
Fr(f) and Fim(f) are real bounded linear functionals on Lr), and on
account of sec. 48, Theoren1 1 we have Fr==F: +F; and Fim===F+
Fim, where F:, -F;, F and -Fim are non-negative bounded linear
functionals on Lr). We recall that, for OfELr), the number F:(f) is
defined by F:(f)===sup Fr(g) for all gELr) satisfying Ogf. Similarly
for F(f). The functional F;, defined on Lr), is extended on L p by
defining F;(f)===F:(g)+iF;(h) for f==g+ih (g and h real) in L p . Simi-
larly for F;, F and Fim. The final result is that F===(F;+F;)+
i(F+Fim) on L p . We shall call this the standard decomposition of F
into non-negative bounded linear functionals.
LEMMA £5. Let X be of a-finite measure, and let F be a bounded linear
functional on the Banach function space L p , having the property that O
fnELp (n==l, 2, .. .), fn!O, imPlies F(fn)-+O. Furthermore, let
F ===Fr+iFim==(F; +F;)+i(F+Fim)
be the standard decomposition of F. Then OfnELp (n=== 1,2, . . .), fn!O,
imPlies F:(fn)!O, as well as -F;(fn)!O, F(fn)!O and -Fim(fn)!O.
In other words, these functionals are elementary integrals on Lr).
PROOF. It will be evident that, in view of the hypotheses made,
OfnELp (n== 1, 2, . . .), fn! 0, implies Fr(fn) -+0 as well as Fim(fn) -+0,
and so it will be sufficient to prove, e.g., that F; (fn)! O. For this
purpose, let gEL p satisfy 0gf1. Then g-min (fn, g)f1-fn for n==
1, 2, . . . . Indeed, if gfn at the point XEX, then
g-min (fn, g) ===g-g==0f1-fn,
and if g>fn at the point XEX, then
g-min(fn, g)==g-fnf1-fn.
Hence, by the definition of F:, it follows from this inequality that
Fr{g-min (fn, g)}F; (f1-fn),
so
F; (fn) -Fr{min (fn, g)}F; (f1) -Fr(g).
Since min(fn, g)!O, we have Fr{min(fn, g)}-+O by hypothesis. Further-
more, since F; is non-negative, F; (f n) is non-increasing, so lim F; (f n)
exists. It follows that lim F; (f n) F; (f1) - F r(g). This holds for all
g E L p satisfying 0gf1; hence, taking in the last inequality the least

360 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  50
upper bound of Fr(g) for all such g, we obtain lim F: (fn)F: (f1)-
F:(f1)==0. This shows that F:(fn)!O.
In Lemma fJ it was proved that, under certain conditions which are
automatically satisfied whenever X is of a-finite measure, Lq is a
closed linear subspace of L; (P-1+q-l== 1). In the next theorem we
present, for the case that X is of a-finite measure, a necessary and
sufficient conditio for an element F EL; in order that it be contained
in this closed linear subspace.
THEOREM 1. Let X be of a-finite measure, IP(X), p-1+q-1== 1,
and let FEL;. Then F(g) is of the form F(g)==J fgdfl for some fELq and
all gEL p if and only if it follows from gnELp (n== 1,2, .. .), gn! 0, that
F(gn)-*O. In this case, F and f determine each other tniquely (in fact,
IIFII==IIfllq by Lemma ex). If FE(L»)*, then f is real.
PROOF. If F(g)==Jfgdfl for some fELq and all gEL p , and if gn!O
for some sequence gnELp (n==l, 2, .. .), then Ifg11 is summable, Ifgnl
lfg11 for all n, and Ifgnl!O. Hence J Ifgnldfl!O by dominated con-
vergence, and so F(gn)-*O on account of
IF(gn) I == II fgndfll/lfgnl dfl.
Evidently, F is uniquely determined by f.
Assume now, conversely, that FEL; satisfies F(gn)-*O whenever
OgnELp (n== 1,2, . . .), gn! 0, and let
F==(F: +F;)+i(F+F)
be the standard decomposition of F. In view of the preceding lemma
it follows that F: is an elementary integral on L). Let Lb be the col-
lection of all real bounded and measurable functions g(x) vanishing
outside some set L1g of finite measure (in general, different L1g for differ-
ent g). Evidently, fg==Jgdfl is finite for every gEL b , and fg is an
elementary integral on Lb. Applying the extension procedure for inte-
grals to this elementary integral, we obtain the same extended inte-
gral as when the extension procedure is applied to the elementary
integral J gdfl, defined initially only on the collection Ls of the fl-
step functions. This is due to the fact that Ls cL b cLi r ), where Lr)
is the collection of all realfl-summable functions (cf. sec. 17, Lemma a).
Furthermore, Lb c Lr), and so F: is also an elementary integral on

Ch.12,9 50 ]
THE CONJUGATE SPACE OF Lp
361
Lb (since F: is an elementary integral on L»). We shall denote the
corresponding extended integral by 1. Hence, f and 1 are ex-
tensions of elementary integrals, initially defined on Lb. Let X n (n==
1, 2, . . .) be an arbitrary sequence of measurable subsets of X such
that X n i X and #(X n ) <00 for all n. Since all XXn are in Lb (and hence
in Lp), we have Ol(XXn)==F:(XXn)<oo for all n, and this shows
that the measure induced in X by 1 is a-finite. Furthermore, if
f(l/l) ==0, then 1==0 holds ,u-almost everywhere, and so IEL b ; it follows
that (/)==F:(/)==F:(O)==O, which shows that  is f-absolutely
continuous. Hence, the integral version of the Radon-Nikodym theo-
rem can be applied; i.e., there exists an f-measurable function 11(x)0
such that (g)==Jllgd,u for all l-summable g.
I t will be shown now that 1 (g) is finite and equal to F;- (g) for all
gEL p ; it will evidently be sufficient to prove this for non-negative
gEL). Writing, in this case, gn==min (g, nx Xn ) where X n is the same
set as above, we have gnELb and Ogn i g. Hence g-gnEL) for n==
1,2, ... and g-gn!O, which implies F:(g-gn)!O, since F: is an ele-
mentary integral on L). In other words, F:(gn)i F:(g). We have also
l(gn) i l(g) by the theorem on integration of monotone sequences.
But l(gn)==F:(gn), since  and F;- coincide on Lb. Hence (g)==
F: (g) for any gEL p , and since F: (g) is finite for all gEL p , it follows
in particular that every gEL p is l-summable. The final result is now
that F:(g)==1(g)==Jl1gd# for all gEL p . It remains to prove that
11 ELq. Since we know now that 11g is summable for all gEL p , it follows
from part (b) in Lemma y that I1ELq.
The same argument shows the existence of functions 12, 13, 14ELq
such that F;(g)==JI2gd#, Ftm(g)==JI3gd# and F"hn{g)==JI4gd# for all
gEL p , so
1== (II + 12) +i(/3+ 14) ELq,
and F(g) == J Igd# for all gEL p . This is the desired result. Note that I is
real whenever F E (L») *.
In order to show, finally, that I is uniquely determined by F, as-
sume that I', l"EL q are such that F(g)==JI'gd#==JI"gd# for all gEL p .
Then J (I' -1")gd#==O for all gEL p , and in view of Lemma ex this im-
plies that III' -1"I[q==O, so I' ==1".
It is now easy to complete the picture for the a-finite case.

362 CONJUGATE SPACES AND WEAK COKVERGENCE [Ch. 12,9 50
THEOREM 2. Let X be oj a-jinite measure, 1 poc> and p-1+ q -1== 1.
For 1 p < oc>, there exists a one-one linear and norm preserving corre-
spondence between the junctions jEL q and the elements FEL;; the corre-
spondence is determined by F(g)==J jgdfl, holding jor all gEL p . In other
d L * - L 5 . . l 1 (L (r» ) * - L (r)
wor s, p- q. m ar y, p - q .
For p==oc>, there exists a one-one linear and norm preserving corre-
spondence between the junctions tELl and the elements FEL: satisjying
F(gn)O whenever gnELoo (n==l, 2, ...) and gn!O; the correspondence
is determined by F(g)==Jjgdfl, holding jor all gELoo. In general, L 1 is
thus a proper linear subspace oj L:O J . more precisely, ij X is the union oj
a countably injinite number oj disjoint measurable sets oj positive
measure, then L 1 is a proper subspace oj L:O. In particular, the sequence
space II is a proper subspace oj 1':0. Ij, however, X==k=l X k , where each
oj the sets Xl, . . " X p is oj jinite measure and contains only the empty
set and itselj as measurable subsets, then L 1 ==L:.
PROOF. Combining Lemma fJ and Theorem 1, we have immediately
the result that, for 1 poc>, Lq may be identified with the closed linear
subspace of L; consisting of all FEL; for which F(gn)O whenever
gnELp (n== 1,2, . . .) and gn!O. Hence, for lp<oc>, we have to prove
that every FEL; has this property. This is easy. If gnELp (lp<oc>;
n==I,2, ...) and gn!O, then IIgnllp!O, and so IF(gn)IIIFII'''gn''p!O.
N ext, we consider the space L: under the assumption that X ==
r X k , where the sets X k are disjoint and fl(X k ) >0 for all k. The col-
lection C of all gEL oo which are constant on each Xk, and for which
lim g(Xk) exists as koc>, is a closed linear subspace of Loo, and F(g)==
lim g(X k ) is a bounded linear functional on C. By the Hahn-Banach
theorem F may be extended to all gEL oo without change of norm, and
the thus extended F is then an element of L:. Let gn (n== 1, 2, . . .) be
defined by gn(x)==O on  X k and gn(x)==1 on +1 Xk. ThengnEC
cLoo for all n, and gn!O, but F(gn)O is not satisfied since F(gn)==1
for all n. This shows that F is not of the form F(g) == J jg dfl for some
fELl.
Finally, let X == k= 1 X k, where each of the sets Xl, . . " X p is of
finite measure, and contains only the empty set and itself as measur-
able subsets. Given the arbitrary element FEL:, let cxk==F(XxJ for
k== 1, . . ., p. Since any gEL oo is of the form g== '1. CkXXk' we have
F(g) == l CkCXk== J jgdfl for j== l CikXXk/fl(X k ). Evidently, tELl.

'Ch. 12,  50J
THE CONJUGATE SPACE OF Lp
363
For the case that X is a bounded interval on the real line and fl is
Lebesgue measure, the theorem is due to M. FRECHET (1907, [IJ) for
p=2, to F. RIESZ (1910, [2J) for 1 <P<oc>, and to H. STEINHAUS (1918,
[lJ) for p== 1.
The next step is to free ourselves from the restriction that X is of
.a-finite measure. For 1 <P<oc> this can be done without additional
hypotheses, but for p== 1 a new condition will be imposed on fl in ex-
.change for the a-finiteness.
THEOREM 3. Let 1 <P<oc> and p-1+ q -1== 1. Then L;==L q , in the
sense exPlained in the preceding theorem. Similarly, (L»)*==Lr).
PROOF. On account of the result in Lemma fJ it is sufficient to
prove that every FEL; is of the form F{g)==flgdfl for some function
.fELq and all gEL p . If X does not possess any subsets of finite positive
measure, then Lp and Lq consist exclusively of the null function, and
L; contains only the null functional. It follows that the theorem holds
in this case. We may assume, therefore, that X has at least one subset
X 1 of a-finite positive measure, and we consider the subspace Lp(X 1, fl),
-consisting of all gEL p vanishing outside Xl. Let FEL;, and let F1 be
the restriction of F on Lp(X 1, fl). Then F 1 is a bounded linear functional
-on Lp(X!, fl) satisfying IIF 111IIFII, so that by the preceding theorem
there exists a unique function I1ELq(X1, fl) such that F 1 (g)==fl1g d fl
for all gEL p (X1, fl), and II/111q==IIF111IIFli. If X 2 is another subset of
X of a-finite measure, we obtain in the same manner a function
f2ELq(X2' fl) representing F on Lp(X2, fl), and on account of the
uniqueness we have 11 == 12 almost everywhere on the intersection of
Xl and X 2 . Let now {X T } be the collection of all subsets X T of X of
a-finite measure, and let b==sup IIITllq, where IT represents F on X T .
Then bIIFjl, since IIITllqIIFIl for all T. It follows that there exists an
ascending sequence of subsets X n C X, all X n of a-finite measure, such
that the corresponding functions In satisfy Il/nllq i b. Furthermore, for
m>n,
I'fm-fnll== !Itm-fnlqdfl== ! Ilmlqdfl==
x Xm-X n
== !Ifmlqdfl- !lfnlqdfl==lltmll-llfnll-+O
X m Xn

364 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  50
as m, n-+oo, showing that f n is a fundamental sequence in Lq; its
limit function fELq vanishes, therefore, outside XO===l X n , and
Ilfllq===b. On account of the last equality it is impossible that any
function f-r (representing F on some set X-r of a-finite measure) can
differ from zero on a subset E of X -X o of positive measure, since
.."'I
otherwise the function t representing F on Xo+X-r would satisfy
/I
Iltllq>b. For each X-r we have therefore that f-r==fXx.'
Given now the arbitrary function gEL p , the set {x: g(x) #O} is of a-
finite measure, so this set is one of the sets X-r' and hence g==gXx. for
this particular T. This implies that F(g)===Jf-rgdfl===JfXx.gdfl===Jfgdfl,
so fELq has the property that F(g)===J fgdfl for all gELp, which is the
desired result.
Finally, it is evident from the proof that fELr) whenever FE(L»)*,
since all f-rELr) in that case.
It is of some importance to observe that, for p==2 (and hence q==2),
the theorem is an immediate consequence of the fact that any bounded
linear functional F on a Hilbert space H is of the form F(x) == (x, y)
for an appropriate YEH depending on F, and IjFII===IIYII. Applied to the
present situation, this means that for any given F EL'2 there exists a
function f1EL2 such that F(g)==Jg/1dfl for all gEL 2 , and IIFII==llflI12.
Writing 11===f, the result for p==2 is obtained.
For p== 1, as the example presented in the proof of Lemma ex shows,
the space Loo can be considered as a closed linear subspace of Li only
under the condition that fl has the finite subset property. Hence, this
is surely a necessary condition in order that Li ==Loo shall hold. One
might hope now that, if fl has the finite subset property, the above
proof can be repeated. The proof fails, however. It remains true that
for any FELi there exists an ascending sequence XncX, all X n of a-
finite measure, and a corresponding sequence fnELoo such that fn
represents F on X nand Ilfnlloo i b===sup IIf-rlloo' but fn is not necessarily
a fundamental sequence in Loo. If the additional condition is now im-
posed that fl is localizable (cf. sec. 35), the situation becomes different,
and the desired result holds.
THEOREM 4. Necessary and sufficient in order that Li ==Loo shall hold
is that fl is localizable and has the finite subset property.

Ch. 12,  50J
THE CONJUGATE SPACE OF Lp
365
PROOF. Assume first that # is localizable and has the finite subset
property. In view of # having the finite subset property, part (b) in
Lemma fJ may be applied, and it follows that Loo is a closed linear
subspace of Li. It will be sufficient to show, therefore, that every
FELi is of the form F(g) == J Igd# for some IELoo and all gEL 1 . Let
{E T } be the collection of all #-summable subsets ETcX. Corresponding
to every E T there exists, by Theorem 2, a unique function ITELoo,
vanishing outside E T and representing F on ET' i.e., F(g) == J ITgd# for
all gEL 1 (E T , #) and IIITllooIIFII. In view of the uniqueness, we have
IT1==IT2 almost everywhere on any intersection ETIET2. Hence, denoting
by I: the equivalence class of all functions almost equal to IT' the col-
lection of all I: is a cross section <I:> according to the definition in
sec. 33. Since # is localizable and has the finite subset property, we
may apply sec. 35, Theorem 2, and it follows that </> is determined
by a measurable function I, defined on the whole of X, i.e., IT==lxET
for all T. Hence F(g)==JIgd# for every gEL 1 vanishing outside some
set of finite measure. This result may immediately be extended to an
arbitrary gEL 1 , since for any gEL 1 the set {x: g(x) *O} is of a-finite
measure. It remains to prove that IELoo. Assume, for this purpose,
that ess sup I/(x) I> IIFII. Then III> IIFII on some set of positive measure,
and hence on some set E T of finite positive measure (by the finite
subset property). On the other hand, 1/1IIITllooIIFIi on this particular
set ET' and so we obtain a contradiction. Hence IELoo.
Let now, conversely, Li ==Loo. As observed already, this implies that
# has the finite subset property (since otherwise there exists a function
IELoo, representing the null functional and satisfying 11/1100== 1). In order
to show now that # is localizable, it will be sufficient (by sec. 35, Theo-
rem 2) to prove that every non-negative cross section <I;> is de-
termined by a function I, defined on the whole of X. Assuming first
that O/E(X) 1 almost everywhere for every IE, it is evident that
<I;> determines a bounded linear functional F on L1, i.e., F(g)==
J IEgd# for every gEL 1 (E, #), and since by hypothesis F is repre-
sented by a function IELoo on X, it follows that <I;> is determined by
this I. Evidently, since IXE==IE for every #-summable E and IE(X)O,
the set {x: I(x) <O} is a local null set, and hence a null set. Given now
an arbitrary non-negative cross section <I;>, let In,E(X) ==min{IE(x) , n}
for n== 1, 2, . . . . Then </:,E> is determined by a measurable function

366 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch..12, 9 50
fn(x)O in view of what has already been proved, and it follows easily
that 11f2' . ., except on a null set. But then <I;;> is determined by
f==lim In. This completes the proof.
Exercises
CONVERSE OF HOLDER'S INEQUALITY
50.1) As above, we write pp(/)==(ff P dfl)l/p for lp<(X) and any
non-negative measurable f, and Poo(/) ==ess sup f(x) for any non-nega-
tive measurable I. It is trivial that Pl(I/I)==supflfgldfl for all gEL oo
satisfying IIglloo 1 (take g(x) == 1 for all x). In order to extend this
property to pq(l/l) for values q* 1, it is necessary to impose an ad-
ditional condition on fl. Show that if 1 <q(X), p-1+ q -1== 1, then in
order that
pq(lfl) == sup /lfgl dfl
Ilullp:::::;;;l
shall hold for all measurable f, it is necessary and sufficient that fl has
the finite subset property.
50.2) It is trivial that if f is measurable and fg is summable for
every gELoo, then fELl (take g(x)== 1 for all x). Show that, for 1 <
q(X), p-1+ q -1== 1, the measurability of f and the summability of fg
for every gELp implies fELq if and only if fl has the finite subset
property.
THE BANACH SPACE a-Loo
50.3) Let fl be non-a-finite, and denote by a-Loo the subcollection
of all fELoo such that {x: f(x) *O} is of a-finite measure. Show that
a-Loo is a proper closed linear subspace of Loo. Show also that the
subset of (a-L oo )* consisting of all FE (a-Loo) * satisfying F(gn) O when-
ever gnEa-Loo (n== 1,2, . . .), gn!O, may be identified with L 1 .
L 1 AS A PROPER SUBSPACE OF L:
50.4) In order to show in Theorem 2 that L 1 is, in general, a proper
subspace of L:, we have made use of the Hahn-Banach extension

Ch. 12, 9,50J
THE CONJUGATE SPACE OF Lp
367
theorem. The question may be raised if, for the special case that we
have Lebesgue measure in the bounded interval [-1, IJ of the real
line, this can be avoided by defining explicitly a suitable bounded
linear functional F on L<:O) such that F is not of the form F(g)  J fg d#
for some fELl and all gEL<:O). In this connection one might think of
the functional F(g)==limeto{ess sUPlxl<e g(x)}. Show, however, that this
functional is not linear. Show also that this defect cannot be remedied
by defining for example
F1(g)lim{ess sup g(x)+ess inf g(x)}.
etO Ixl <e Ixl <e
NON-NEGATIVE LINEAR FUNCTIONAL ON THE COLLECTION OF ALL
BOUNDED CONTINUOUS FUNCTIONS
50.5) The present exercise is related to the Riesz representation
theorem, but methodically it belongs in the present section since in
the proof we have another application of the Hahn-Banach extension
theorem, similar to the proof that L 1 is, in general, a proper subspace
of L:O. We recall that if C) is the collection of all real continuous
functions with a compact carrier in k-dimensional Euclidean space Rk
and f(j) is a non-negative linear functional on C), then f(j) is an
elementary integral on C). Furthermore, it was proved in Exercise
21.23, that if C is the collection of all real continuous functions on Rk,
then any non-negative linear functional on C is an elementary integral
on C. In the next exercise (Exercise 21.24) it was shown then that an
elementary integral f(f) on C is, essentially, not more general than an
elementary integral on C), since the measure induced in Rk by f(f) is
concentrated in a compact subset of Rk. Consider now a class of
functions intermediate between C) and C, namely the class C B of all
real bounded and continuous functions on Rk. For simplicity, let k== 1,.
ap.d show the existence of a non-negative linear functional F on CB
such that F is not an elementary integral on C B . Use the result in
Exercise 28.2 on the existence of a Banach-Mazur limit.
PARTIAL ORDERING
50.6) Let 1 poc>, P-1+ q -1== 1, and for the case that p== 1, assume
that # has the finite subset property. Hence, it follows from Lemma fJ

368 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,  50
that for all these values of p the space Lr) is a closed linear subspace
of (L»)*. Consider elements F, F1, F2, . .. in this subspace and show
that if F1, F 2 correspond to 11, 12ELr), then F1VF2 and F 1 AF 2 corre-
spond to max (11, 12) and min (11, 12) respectively (for the notations, cf.
Exercise 48.3). Show also that if F corresponds to I, then IFI (as de-
fined in Exercise 48.1) corresponds to III. It follows now that F 1 and
F 2 are orthogonal (i.e., IF 11A IF 21 ==0) if and only if min (1/11, 1121) ===0.
Show, finally, that F 1 and F 2 are orthogonal if and only if 111 + 121 ==
1/1-/21 or, alternatively, if and only if 111 +/21 ===max (1/11, /121).
50.7) Let p, q and # satisfy the same hypotheses as in the preceding
exercise. Hence, by Lemma {J, Lq is a closed linear subspace of L;.
Consider elements F, F 1, F 2, . . . in this subspace, and show that if F
corresponds to IELq, then IFI (as defined in Exercise 48.2) corresponds
to III. Similarly as in Exercise 49.4, the orthogonality of F 1 and F 2
(i.e., IF1IAIF21==0) implies that IF1+F21===IF1-F21 and IF1+F21==
JF 11VIF 21, but neither of these conditions is sufficient for orthogonality
of F1 and F2.
THE SUBSPACE (co) OF loo
50.8) Show that the collection of all 1=== (11, 12, . . .) El oo , satisfying
fnO as noo, is a closed linear subspace of loo. This subspace is usu-
ally denoted by (co). Show that (co)* may be identified with II in the
following sense: GE(Co)* if and only if G(/)==r Ingn for some g==
(gl, g2, . . .) Ell and all IE (Co), and in this case IIG"=== "gill.
THE SUBSPACE (c) OF loo
50.9) Show that the collection of all 1=== (11, 12, . . .) El oo , for which
fo==lim In as noo exists as a finite number, is a closed linear sub-
space of loo. This subspace is usually denoted by (c). Show that (c)* may
be identified with II in the following sense: GE (c) * if and only if G(/) ==
o Ingn for some g=== (go, gl, g2, · · .) Ell and all IE (c) with lo==lim In,
and in this case IIGII== IIg1l1.

Ch. 12, 9 51 J
THE BANACH-STEINHAUS THEOREM
369
51. The Banach-Steinhaus Theorem
In the present section the Banach-Steinhaus theorem will be proved,
wherein a (necessary and) sufficient condition is stated in order that
a collection of bounded linear transformations on a Banach space V be
uniformly bounded. It is essential for the theorem that V is a Banach
space, and not merely a normed linear space (cf. Exercise 51.1). We
recall that, by the definitions in sec. 3, the subset A of the metric
space X is said to be dense if its closure A satisfies A==X, and A eX
is called a boundary set if its complement A e==X -A is dense. Further-
more, A e X is called nowhere dense if A is a boundary set, and A eX
is said to be of the first category if A is a countable union of nowhere
dense sets. Any set A e X which is not of the first category is said to
be of the second category. Baire's category theorem (sec. 3, Theorem 6)
shows that in a complete metric space any set of the first category is
a boundary set. In particular, every complete metric space is of the
second category, showing thus that a complete metric space has at
least one subset of the second category. Since any Banach space V is
a complete metric space with respect to the norm as distance function,
the space V has, therefore, at least one subset of the second category.
THEOREM 1 (BANACH-STEINHAUS THEOREM; 1927, [lJ). Let G be a
set of the second category in the Banach space V, and let {TT} be a col-
lection of bounded linear transformations, defined on V (the range of TT
may be different for different T). Furthermore, let the collection of numbers
{IITTXII} be bounded for each xEG separately. Then the transformations TT
are uniformly bounded, i.e., there exists a finite number M;::.O such that
IITTXIIMllxll for all TTE{TT} and all XEV.
PROOF. Assume that the transformations TT are not uniformly
bounded. Then there exists a sequence T nE{TT}' n== 1,2, . . " such
that lim liT nil ==00, and hence there exists a corresponding sequence
XnEV such that IIxnll1 for all nand limIITnxnll==oo. Let G be the
set of all XE V for which the sequence liT nxil is bounded. Then G  G,
so G is of the second category. Finally, let Gk, for k== 1, 2, . . " be the
set of all XE V satisfying liT nxllk for all n. Then all sets G k are closed,
and G==r G k . It follows from Baire's category theorem, since G is of
the second category, that one at least of the sets G k , say G ko ' is not

370 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,951
nowhere dense. Hence, since G ko is closed, Gko contains a closed sphere
5 (centre xo, radius p>O). If, therefore, YE V satisfies IlylIp, then x===
XO+YES, so IITnYIIIITnxll+IITnxoll2ko for all n. Hence, lIylll im-
plies that liT nYII2kolp for all n; in particular liT nXnll2kolp for all n,
contradicting lim liT nXnll ==00. This shows then that the transformations
T T are uniformly bounded.
COROLLARY. Let Tn (n=== 1, 2, . . .) be a sequence of bounded linear
transformations on, the Banach space V, and let there exist a sequence
XnE V stch that IIxnll 1 for all nand lim sup liT nXnll===oo. Then the set G
of all XE V for which the sequence liT nxil is bounded is of the first category
in V, and so Ge===X -G is of the second category by Baire's category
theorem.
PROOF. The assumption that G is of the second category yields the
result that the transformations T n are uniformly bounded, and this is
not so since lim sup liT nXnll===oo. Hence, G is of the first category.
Exercises
A COUNTER EXAMPLE
51.1) In the sequence space ll, let en (n=== 1, 2, . . .) be the point with
n-th coordinate equal to one and all other coordinates zero, and let V
be the linear subspace of all x== l Ciiei (Cii complex, p finite but vari-
able). For n=== 1, 2, . . " the bounded linear functional F n on V is de-
fined by F n(e m ) ==nmn (where mn is the Kronecker delta: rnn=== 1 for
m==n and mn==O for m=j=.n), and Fn(l Ciiei)==l CiiFn(ei). Show that
the sequence IF n(x) I is bounded for each XE V separately, but IIF nll===n
for all n. This shows that the conclusion of the Banach-Steinhaus theo-
rem is not necessarily valid if V is merely a normed linear space.
CONDENSATION
51.2) Let lq<oo, and p-1+ q -1==1. Show that if the sequence
fnELq satisfies lim supllfnllq==oo, then there exists a function gEL p
such that lim sup If fngdfll===oo.

Ch. 12,  52J WEAK SEQUENTIAL CONVERGENCE IN Lp
371
51.3) Assume that fl has the finite subset property, and show that
if the sequence fnELoo satisfies lim suPllfnlloo==oo, then there exists a
function gEL 1 such that lim sup If fngdfll ==00.
52. Weak Sequential Convergence in Lp
If X n (n== 1, 2, . . .) and Xo are elen1ents of the normed linear space
V, the sequence X n is said to converge weakly to Xo if, for every bounded
linear functional F on V, the sequence F(xn) converges to F(xo). The
sequence X n E V is called a weak fundamental sequence (or a weak Catchy
sequence) if, for every bounded linear functional F on V, the sequence
F(x n ) is a fundamental sequence of numbers. If X n converges to Xo in
norm, i.e., if IIxn-xoll-+O as n-+oo, then X n converges weakly to Xo on
account of IF(xn)-F(xo)I==IF(xn-xo)IIIFII'lIxn-xoll, but the con-
verse is not necessarily true. If V is the Hilbert space L 2 ([0, 2nJ, fl),
where fl is Lebesgue measure, and pn(x) ==e inx for n== 1, 2, . . " then
! fpndfl-+O as n-+oo for every fEL 2 in view of the Riemann-Lebesgue
theorem proved in sec. 45 (it was proved that f fpndfl-+O for every
fELl; observing that fEL 2 implies fELl since the interval [0, 2nJ is of
finite measure, we obtain f fpndfl-+O for every fEL 2 ). In other words,
(Pn, f) -+0 for every fEL 2 , i.e., F(pn) -+0 for every bounded linear
functional on L 2 . This shows that the sequence pn (n== 1, 2, . . .) con-
verges weakly to the null element of L 2 ; on the other hand, since
Ilpnll2== (2n)! for all n, the sequence pn does not converge in norm to
the null element.
Obviously, if X n converges weakly to xo, the sequence Xn is a weak
fundamental sequence. If, conversely, every weak fundamental se-
quence in V has a weak limit, then V is said to be weakly sequentially
comPlete. It is to be observed here that the weak limit of the sequence
X n , if it exists, is uniquely determined. Indeed, assuming that Xn con-
verges weakly to Xo as well as to xo, where yo==xo-XO=FO, we have
F(yo) ==0 for every bounded linear functional F, and this is impossible
since the linear functional G(exyo) ==exIIYoll, defined on the linear sub-
space of all elements exyo (ex complex), may be extended to a bounded
linear functional F on V satisfying F(yo) == IIYoll =FO.

372 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12, 9 52
For 1 <P<oo, weak convergence of the sequence fnELp to fOELp
means that F(fn)-+F(fo) for every FEL;. But L;Lq (P-1+q-1 1) by
sec. 50, Theorem 3, so weak convergence of fnELp to fOELp simply
means that ffngd#-+ffogd# for every gELq. For p1 it is not true
that Li Loo without additional hypotheses on the measure #, but we
shall prove that, nevertheless, the sequence fnEL1 converges weakly to
fOEL 1 if and only if f fngd#-+ f fogd# for every gELoo, without any ad-
ditional hypothesis on #. In order to obtain a more complete result,
connecting weak convergence in L 1 with the Hahn-Saks theorem in
sec. 45, Theorem 4, a lemma will be proved first.
The #-measurable set E e X of positive measure, having the property
that any #-measurable subset FeE satisfies either #(F)O or #(E -F)
==0, is called an atom for the measure # (ExamPle: If # is discrete
measure, i.e., if every point XEX is of measure one, then every point
XEX is an atom). Obviously, identifying almost equal sets as usual,
different atoms may be considered as disjoint, and it follows that any
set of finite measure (and hence any set of a-finite measure) contains
an at most countable number of atoms. Hence, any set E of a-finite
measure is the disjoint union of two sets A and B, where A is a count-
able union of atoms, and B does not contain any atoms.
LEMMA cx. If E is of finite positive measure, and E does not contain
any atoms, then there exists for any given number fJ, satisfying O<fJ<#(E),
a subset F of E such that #(F) fJ.
PROOF. E does not contain any atoms, so E js the disjoint union of
two sets of positive measure. Hence, one of these sets is of measure at
most equal to t#(E). Continuing with this method of decoIIlposition,
it follows that E has subsets of arbitrarily small positive measure, so
that the collection C of all sets A eE, satisfying O<#(A)fJ, is not
empty. The collection C is partially ordered by inclusion; let {AT} be
a chain with respect to the partial ordering, and let fJ1suPT #(A T ), so
0<fJ1 fJ. Picking a suitable ascending sequence An (n 1, 2, . . .) from
the chain {AT} such that #(A n )ifJ1, it follows easily that Aor An
is an upper bound of the chain. Hence, since every chain in C has an
upper bound, Zorn's lemnla shows that C has a maxinlal element F. We
assert that #(F) fJ. Indeed, if #(F) <fJ, there exists a subset G of

Ch. 12,9 52J WEAK SEQUENTIAL CONVERGENCE IN Lp
373
E -F of sufficiently small positive measure such that F +G still satis-
fies #(F+G)<fJ, and this contradicts the maximality of F.
THEOREM 1. The lollowing statements concerning the lunctions InEL1
(n=== 1, 2, . . .) and IOEL1 are equivalent.
(a) In converges weakly to 10,
(b) I Ingd# I logd# lor every gEL oo ,
(c) IElnd# IElod# lor every #-measurable set E eX.
PROOF. It will be proved first that (a) and (b) are equivalent. Let
(a) hold, and assume that gELoo. Then G(/) ===1 Igdfl is a bounded linear
functional on L1 in view of IG(/) 1lIglloo '1[/111, holding for all IEL1 (note
that it may very well happen, however, that IIGII<lIglloo; cf. the ex-
ample in sec. 50, Lemma ex). Hence G(ln)G(/o) by hypothesis, i.e.,
I Ingd# I logd#.
Conversely, let (b) hold, and assume that G is a givell bounded linear
functional on L1. Since InEL1 (n== 1, 2, . . .) and IOEL1, there exists a
subset Xo eX of a-finite measure such that 10 and all In vanish outside
Xo. Denoting the restriction of G to L1(XO, #) by Go, the bounded
linear functional Go on L1(XO,'#) is represented by a function gOELoo
(cf. sec. 50, Theorem 2), so it follows from I Ingod# J. logo d# that
Go(ln)Go(/o). But G===G o on L 1 (.Y O , #), so G(ln)G(/o). Since G is
arbitrary, it follows that In converges weakly to 10.
It is trivial that (b) implies (c), and so it remains only to prove that
(c) implies (b). It follows from IElnd#/Elod#, holdjng for any #-
measurable set E, that the Hahn-Saks theorem (sec. 45, Theorem 4)
may be applied. Denoting IElnd# by vn(E), and observing that IVnl(E)
===IEllnl d # by sec. 45, Theorem 3, the Hahn-Saks theorem states that,
for any sequence E p (p=== 1, 2, . . .) descending to the empty set and
for any 8>0, we have IE p Ilnl d#E for all PPO(E) and all n. In par-
ticular it follows that, corresponding to any E>O, there exists £5>0
such that #(E) <£5 implies IE Ilnl d#<E for all n simultaneously. Intro-
ducing, as above, a set Xo of a-finite measure such that all In vanish
outside Xo, and denoting by X n (n===l, 2, ...) an arbitrary sequence
of sets of finite measure such that X n i X o , we have Xo-Xn! 0, so
there exists N such thatlx_xNllnld#1 for all n simultaneously. Con-
sidering now the set X N for this fixed value of N, we have XN==A +B,
where A is an at most countable union of atoms and B does not contain

374 CONJUGATE SPACES AND vVEAK CONVERGENCE [Ch. 12,9 52
any atoms. We recall the existence of a number >O such that fl(E)
implies IE II nl dfl 1 for all n simultaneously. On account of Lemma ex
the set B may be decomposed into {fl(B) + 1}/ disjoint subsets of
measure at most , and hence
Illnl dfl{fl(B) + 1 }/===M 1
B
tor all n simultaneously. Finally, A === r Ai, where the sets Ai are the
disjoint atoms of A, so r fl(Ai)==fl(A) <00, and hence  fl(Ai)<
for some P. It follows that, for all n, the integral of Ifni over  Ai
does not exceed one. On each of the atoms AI, . . ., A p - 1 the functions
In are constant, and IAilndfl converges as noo. It follows that In
converges on each Ai, so Ilnl converges on each Ai, and this shows the
existence of a constant M20 such that, for all n, the integral of Ilnl
over f-l Ai does not exceed M 2 . Adding up, we have
JllnldflI+Ml+1+M2=M
for all n.
Starting again from IElndfl/Elodfl, holding for all fl-measurable
sets E, we obtain immediately that I Ingdfl I logdfl for every function
g(x) assuming only a finite number of different finite values, each of
these values on a fl-measurable set. Let now Og(x) ELoo' Then there
exists a sequence gn of functions, each gn assuming only a finite number
of finite non-negative values on fl-measurable sets, such that gn i g
uniformly, so IIg-gnllooO. Also,
II (/o-ln)gdfll11 IO(g-gN) dfll + II (/o-ln)gNdfll + Illn(g-gN) dfll
 II (fo-ln)gNdfll +{II/oll1 +M}. IIg-gNlloo,
where M is the number satisfying IIln111M for all n, the existence of
which was established above. Now, given E>O, we may first choose N
so large that the last term does not exceed iE, and then the index no
so large that the preceding term does not exceed iE for nno. It
follows that ling dfl  I log dfl for Og(x) E Loo, and the same is then
true for every gELoo. Thus, (b) holds.
THEOREM 2. For lp<oo, the space Lp is weakly comPlete. In other
words (in view 01 the results established above), given the sequence InELp

Ch. 12,9 52J WEAK SEQUENTIAL CONVERGENCE IN Lp
375
such that J fngdfl converges to a finite number for every gEL q (P-1+q-1
===1), there exists a unique /oELp such that Jfngdfl--*Jfogdfl for every
gELq.
PROOF. Let first 1 <P<(X), so 1 <q<(X). Writing Fn(g)==Jfngdfl for
gEL q , the functional F n is linear and bounded on Lq, and !IF nll==llfnllp.
For each gEL q the sequence F n(g) is convergent by hypothesis; the
sequence of numbers IF n(g) I is, therefore, bounded. It follows, by the
Banach-Steinhaus theorem, that the sequence IIF nil is bounded; IIF nil
M for all n. Writing F(g) ==1im F n(g), it is evident that F is a linear
functional on Lq, and IF n(g) IMllgll, holding for all n, implies IF(g) I
Mllgll, showing that F is bounded. But then, by sec. 50, Theorem 3, F
is represented by a function fOEL p , i.e., F(g)==J fogdfl on Lq. It follows
that
lim If ng dfl==lim F n(g) ==F(g) == I fog dfl.
For p== 1 it follows from the hypotheses that J fngdfl converges for
every gEL oo , so in particular JEfndfl converges (to a finite number) for
every fl-measurable set E. Then, by the Hahn-Saks theorem, there
exists a function fOEL 1 such that JEfndfl--*JEfodfl for every measur-
able E, and the result in the preceding theorem shows now that Jfngdfl
--* J fogdfl for every gELoo.
For p==(X) there exists a variant of the last theorem.
THEOREM 3. If fl is localizable and has the finite subset property (so,
in particular, if fl is a a-finite measure), and if the sequence fnELoo has
the property that J fngdfl converges to a finite number for every gEL 1 ,
fhen there exists a unique function fo E Loo such that J f ng dfl --* J fog dfl for
every gEL 1 .
PROOF. By sec. 50, Theorem 4, the functionals Fn(g)==Jfngdfl are
bounded and linear on L1, and IIF nil == Ilf nlloo. By hypothesis the sequence
F n(g) converges for every gEL 1 , so it follows from the Banach-Stein-
haus theorem that IlfnllooM for some constant MO and all n. Then
F(g) ==lim F n(g) is also a bounded linear functional on L1, so that, once
more by sec. 50, Theorem 4, we have F(g)=Jfogdfl for some unique
/oELoo' Hence, J /ngdfl--* J fogdfl for every gEL 1 .

376 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 52
There is another remarkable theorem about weak convergence in L1,
stating essentially that in the sequence space II weak convergence and
norm convergence are identical notions.
THEOREM 4. Let every set of finite positive measure be a union of
atoms (note that this condition is satisfied if # is discrete measure). Then
any sequence fnEL1 converges weakly if and only if it converges in norm.
For 1 <p<oo, however, there are sequences in the sequence space lp which
converge weakly without converging in norm. Also, if X is the interval
[0,2nJ and # is Lebesgue measure, weak convergence in L 1 (X, #) does not
always imPly convergence in norm.
PROOF. Assume that every set of finite positive measure is a (neces-
sarily countable, or even finite) union of atoms. If there are no sets of
finite positive measure (e.g., if X is a union of atoms of infinite measure),
then L1 consists only of the null function, so weak convergence and
norm convergence are trivially the same. Assume, therefore, the ex-
istence of at least one set of finite positive measure, and let the sequence
fn EL 1 (n== 1,2, . . .) converge weakly. On account of Theorem 2 there
is a weak limit, and so it is no loss of generality to assume that fn con-
verges weakly to the null function. There exists a set X 0 c X of a-
finite measure such that all fn vanish outside Xo, and by hypothesis
Xo is an at most countable union of atoms (of necessarily finite measure),
say XO==l Ai with #(Ai)==cxi. We shall write fn==(fn1, fn2, .. .),
where fni is the value of fn at the points xEA i . Since JEfnd#O for
every measurable set E, the Hahn-Saks theorem shows that if E p ! 0,
and E>O is given, there exists an index Po such that JE p Ifni d#<!E for
ppo and all n. Choosing for E p the set ;+1 Ai, we obtain the result
that po+l Ifnilcxi<!e for all n. Furthermore, since fniO as noo
for each fixed i (weak convergence on each atom separately), we may
choose no so large that fl Ifnilcxi<!e for all nno. Hence J Ifni d#<E
for all nno, and this shows that f n converges in norm to the null ele-
ment.
If 1 <P<oo, aIld fn== (15 n 1, 15 n2 , . . .) with 15 nk == 1 for k==n and 15 nk ==O
for k=l=-n, then fnElp for n== 1, 2, . . " and fn converges weakly to the
null element of lp since k 15 nk g k ==gn O for any g== (gl, g2, . . .) Elq
(P-1+q-1== 1). On the other hand, Ilfnllp== 1 for all n, so fn does not
converge in norm to the null element.

Ch. 12,9 52J WEAK SEQUENTIAL CONVERGENCE IN Lp
377
Finally, if X = [0, 2n J and # is Lebesgue measure, the sequence I n(x)
==e inx (n= 1, 2, . . .) converges weakly in L1(X, 11) to the null function
in view of the Riemann-Lebesgue theorem, but Illnl11==2n for all n.
Exercises
WEAK INCOMPLETENESS OF C K AND Coo
52.1) Show that the spaces CK and Coo, introduced in sec. 47, are
not weakly complete.
WEAK INCOMPLETENESS OF Loo
52.2) Show that if VIis a closed linear subspace of the Banach space
V, and the sequence X n EV1 converges weakly to XOEV, then XOEV1.
Derive from this result that if VIis not weakly complete, then V is
not weakly complete.
52.3) . Show that if X is a countably infinite union of disjoint sets
of positive #-measure, then Loo(X, #) is not weakly complete.
WEAK CONVERGENCE AND NORM CONVERGENCE IN 1 1
52.4) It was proved in Theorem 4 that in the sequence space 1 1 weak
convergence and norn1 convergence are the same. The proof was based
on the Hahn-Saks theorem. There is also an "elementary" proof. As-
sume that I n= (I n1, I n2, . . .) converges weakly to the null element of
1 1 , and denote k Iinkl, where the summation is from k==k 1 + 1 to k=k 2 ,
by IIln(k 1 , k 2 )II. If III nil does not tend to zero, we have lim suPlllnll>E for
some 8>0, and we may assume without loss of generality that Il/nll>E
for all n. Choose n1 arbitrarily, and then k 1 such that Il/n 1 (k 1 , 00)11<-18
(hence IIln1(0, k 1 )II>!E). Next, choose n2>n1 such that Il/n 2 (0, k1)1I<-1e
(this is possible, since InkO as noo for fixed k), and then k 2 >k1
such that IIln 2 (k 2 , oo)lI<iE (hence Illn 2 (k 1 , k 2 )1I>!E). This procedure is
repeated. Let
! l/sgn In1k
b k =
1 /sgn I nrk
for
for
1 kk1,
k r - 1 <kkr.

378 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 53
Then b==(b 1 , b 2 , .. .)Eloo, so kfnrkbk-+O as r-+oo by the weak con-
vergence. Show, however, that Ik fnrkbkl >/08 for all Y.
POINTWISE INEQUALITIES
52.5) For p== 1, it was proved in sec. 45 that if the real functions
/nEL1 (n== 1,2, . . .) converge weakly to fOELl, then lim inf fn(x)
/o(x)lim sup fn(x) holds almost everywhere. Show that the same re-
sult holds for weak convergence in Lp (1 <P<oo). Show also that the
result continues to hold for weak convergence in Loo if the measure fl
has the finite subset property. More precisely, show that if fl has the
finite subset property, and the real functions fnELoo (n== 1,2, . . .) and
/oELoo satisfy J fngdfl-+ J fogdfl for every gEL 1 , then lirn inf fnfo
lim sup fn almost everywhere.
53. The Second Conjugate Space and Reflexivity
The discussion of weak sequential convergence in the preceding
section will keep some unsatisfactory aspects if we do not introduce
some further notions, important in themselves for the general theory
of normed linear spaces.
Let V be a normed linear space (with elements x, y, . . .), V* the
conjugate space of V (with elements F, G, .. .), and V** the conjugate
space of V* (with elements P, G, .. .). The space V** is, therefore, the
Banach space of all bounded linear functionals P(G) on V*, and the
norm of P is I[PII==supIP(G)1 for all GEV* satisfying IIGIII. Given a
fixed element XOEV, we define the functional Po on V* by Po(G)==
G(xo), holding for all GEV*. Obviously, Po is linear, i.e.,
P o(aG1 + bG 2 ) ==aP O(G1) +bP o(G 2 ),
and Po is bounded since
IP o(G) I = IG(xo) 1lIxoll' IIGII
for all G.
Hence POEV**, and IIPolllIxoll. We shall prove that IIPoll==lIxoli. For
that purpose we observe that, by the Hahn-Banach extension theorem,
there exists GoEV* such that IIGoll== 1 and Go(xo)==IIxoli. Hence Po(Go) =
Go(xo)==lIxoll for this particular Go satisfying I!G o lI==l, so IIPolIllxoli. It

Ch. 12,9 53J SECOND CONJUGATE SPACE AND REFLEXIVITY 379
follows that IIFoll==lIxoll, and this shows that different elements XoEV
yield different elements FOEV**. Considering now for every XOEV the
thus corresponding FOE V**, we have a one-one linear and norm pre-
serving correspondence, and so (identifying F 0 with xo) the space V
may be considered as a linear subspace of V**. Evidently, the linear
subspace is closed if and only if V is a Banach space. If, under this
identification, V is the whole of V**, then V is called reflexive. In view
of the preceding remark, a reflexive normed linear space is a Banach
space. The converse is not true; there exist non-reflexive Banach spaces,
as will be shown in the next paragraph.
For 1 <P<(X) and p-1+ q -1== 1, the space L; may be identified with
Lq, that is, if GEL; and gEL q are corresponding, then IIGII==llgllq and
G(f)==Jfgd# for all fEL p . Hence, L;* and L may be identified, i.e.,
if F is an arbitrary element of L;*, then F may also be considered as
an element Fq of L; (and conversely), and F(G)==Fq(g) whenever
GEL; and gEL q correspond. Since 1 <q<(X), L may be identified
again with Lp, that is, corresponding to F qEL there exists fEL p such
that I!fllp== IIF qll== IIFII and F q(g) == J fgd# for all gELq. The final result
is that F(G)==Fq(g)==Jfgd#=G(f) for all GEL;, and this shows that
every element of L;* is already contained in Lp, when Lp is considered
as a closed linear subspace of L;* in the manner explained above. In
other words, Lp is reflexive for 1 <P<(X).
Considering the simple case that X is of a-finite measure and X is a
countably infinite union of sets of positive measure (such as, e.g., the
case that X is a bounded or unbounded interval, and the measure is
Lebesgue measure), we have Li==Loo by sec. 50, Theorem 4, and L 1 is
a proper subspace of L: by sec. 50, Theorem 2. Hence, L 1 is a non-
reflexive Banach space in this case; in particular, the sequence space II
is not reflexive. Other examples will follow below.
If V is a normed linear space, and FEV*, then IIFII=suPIIXIIlIF(x)1
by definition. If V is a Banach space, F nE V* for n== 1, 2, . . " and the
sequence IF n(x) I is bounded for each XE V separately, then the sequence
IIF nil is bounded by the Banach-Steinhaus theorem. The introduction
of the second conjugate space makes it possible to prove the dual
statements.

380 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 53
THEOREM 1. (a) If V is a normed linear space, and XEV, then IlxlI==
sup IF(x) I for all FE V* satisfying IIFII 1.
(b) If V is a normed linear space, X n E V for n== 1, 2, . . " and the
sequence IF(xn) I is bounded for each FE V* separately, then the sequence
Ilxnll is bounded.
PROOF. (a) Let F be the element of V** corresponding to the given
element XEV, so F(F)==F(x) for all FEV*. Then IIFII==llxll by the corre-
spondence, and IIFII==supIF(F)1 for all FEV* satisfying IIFI!I. Hence
IIxlI==sup IF(F) I ==sup IF(x) I for all F satisfying IIFII 1.
(b) Let F n be the element of V** corresponding to X n E V. Then, by
hypothesis, the sequence IFn(F)I==IF(xn)1 is bounded for each FEV*
separately, so that, since V* is a Banach space, the Banach-Steinhaus
theorem may be applied. It follows that the sequence IIF nil is bounded,
i.e., the sequence IIxnll is bounded.
THEOREM 2. (a) Every weak fundamental sequence in the normed line-
ar space V is bounded.
(b) If the sequence X n in the normed linear space V converges weakly
to XOE V, then IIxolllim inf IIxnll.
(c) Every reflexive Banach space is weakly sequentially complete.
PROOF. (a) Follows immediately from part (b) of the preceding
theorem.
(b) Given e>O, there exists by part (a) of the preceding theorem an
element FOEV* such that IIFolI==1 and IFo(xo) I >lIxoll-e. On the other
hand, F o(xo) ==lim F o (x n ) and IF o(x n ) IIIF oll'lIxnll== IIxnll, hence IF o (xo) I
lim inf IIxnll. It follows that IIxolllim inf IIxnll.
(c) Let Xn (n==l, 2, ...) be a weak fundamental sequence in the
reflexive space V, and let F n be the element of V** corresponding to
x n , hence F n(F) ==F(xn) for all FE V*. Since F n(F) ==F(xn) converges
to a finite limit as n-+oo for every FEV*, and since the sequence
IIFnll==llxnll is bounded by part (a), it follows easily that Fo(F)==
lim F n(F) is a bounded linear functional on V*. Then, on account of
the reflexivity, there exists an element XOEV such that Fo(F)==F(xo)
for all FE V*, and hence F(xo) ==lim F n(F) ==lim F(xn) for all F, so Xn
converges weakly to Xo.

Ch. 12,  53J SECOND CONJUGATE SPACE AND REFLEXIVITY 381
The last part of the theorem shows again that Lp is weakly complete
for 1 <P<oo. It is very well possible, however, that a Banach space is
weakly complete without being reflexive. Indeed, as is shown in the
preceding section, L 1 is weakly complete, but in general L1 is not re-
flexive.
THEOREM 3. (a) If VI is a closed linear subspace of the reflexive
Banach space V, then VIis reflexive.
(b) The Banach space V is reflexive if and only if its conjugate space
V* is reflexive.
PROOF. (a) To each FE V* we assign its restriction f on V!, i.e., f is
.only defined on VI, and f(x) ==F(x) for all XE VI. Then f is obviously a
bounded linear functional on VI, and IlfllIIFII. The function p, defined
by f===p(F) , is therefore a normdecreasing linear transformation on V*
into Vi. The range of p is the whole of Vi, since by the Hahn-Banach
extension theorem every fEVi has an extension FEV*. Algebraically,
A
<p is a homomorphism of V* onto Vi. Given now any !EV;*, we define
A
the functional F on V* by F(F)===!{p(F)}. Obviously, F is linear, and
F is bounded since
IF(F) 1lIfll'llp(F)IIlIfll'IIFIi.
Hence FEV**, and by the reflexivity of V there exists XOEV such that
P(F)==F(xo) for all FEV*. We shall prove that XOEV 1 . Indeed, if Xo
is not in VI, there exists FOE V* such that F 0==0 on VIand F o (xo) == 1.
,14 A
But then p(Fo) ===0, so 1 ==Fo(xo) ==F(Fo) ===!{p(Fo)}==!(O) ==0, a contra-
diction. Hence XOEV 1 , so it follows now from F(F) ===F(xo) , holding for
all FEV*, that f(f)==f(xo) tor all fEVi. This shows that VI is reflexive.
(b) It is evident from the definition of reflexivity that if the Banach
space V is reflexive, then V* is reflexive. Conversely, if V* is reflexive,
then V** is reflexive, so the closed linear subspace V of V** is reflexive.
The present theorem has several consequences. It was proved al-
ready that if X is a countably infinite union of sets of finite positive
p-measure, then L 1 (X, #) is not reflexive. It follows now immediately
that the same holds if X contains a subset Xo which is a countably
infinite union of sets of finite positive measure. Indeed, the non-re-
flexive space L 1 (X O , /1) is now a closed linear subspace of Li(X, #).

382 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 53
Hence, L 1 (X, fl) is reflexive if and only if there exists only a finite
number of disjoint sets of finite positive measure, i.e., if and only if
L1(X, fl) is finite-dimensional.
As observed earlier, the sequence space II is not reflexive, and hence
its conjugate space loo is not reflexive. This shows immediately that if
X contains an infinite number of different sets of (finite or infinite)
positive measure, then Loo(X, fl) is not reflexive, since in this case Loo
has a closed linear subspace which may be identified with loo' It follows
easily that Loo is reflexive if and only if Loo is finite-dimensional (this
does not mean, then, that necessarily L:L1, since X may consist of
a finite number of disjoint atoms of infinite measure).
We can state now, for 1 <P<oo, a necessary and sufficient condition
in order that a sequence fnELp converges weakly to fOELp.
THEOREM 4. For 1 <p<oo, the sequence fnELp converges weakly to
fOELp if and only if
(a) the sequence Ilf nil p is bounded,
(b) fEfndfl-+fEfodfl for any set E of finite measure.
If Lp does not consist exclusively of the null function, then (a) alone is
not sufficient for weak convergence. If there exists a subset Xo of a-finite
infinite measure, then (b) alone is not sufficient for weak convergence.
For p== 1, the conditions (a) and (b) are necessary, but if there exists a
subset Xo of a-finite infinite measure, (a) and (b) together are not suf-
ficient for weak convergence.
For p==oo, the conditions (a) and (b) are necessary, and it follows,
conversely, from (a) and (b) that f fngdfl-+ f fogdfl for every gEL 1 , but in
general (a) and (b) together are not sufficient for weak convergence.
PROOF. For lpoo, it is trivial in view of Theorem 2 that weak
convergence of fn to fo implies (a) and (b).
Conversely, let 1 <p<oo, and let (a) and (b) hold. TheIl f fngdfl-?-
f fogdfl for every fl-step function g (we recall that a step function, in
our terminology, is a finite linear combination of characteristic functions
of sets of finite measure). Given gELq, where p-1+ q -1= 1 (so 1 <q<CXJ),
there exists a sequence of step functions gn such that IIg-gnllq-+O, and
it follows exactly as in the last part of the proof of Theorem 1 in the pre-
ceding section that f fngdfl-+ f fogdfl. Hence, fn converges weakly to fo.

Ch. 12,9 53J SECOND CONJUGATE SPACE AND REFLEXIVITY 383
If Lp is of dimension at least one, there exists a function IELp which
is not the null function, and the sequence I, 0, I, 0, . . . satisfies condition
(a), but is not weakly convergent.
Assume now that Xo is a subset of a-finite infinite measure, so X o ==
r X n , where all X n are disjoint and of finite measure, but r fl(Xn)
F - 1 2 I t 2/(q-l) d d t . P "",nl X
==00. or n- , ,"', e Yn==n ,an e ermIne 1==1 k
such that fJ1==fl(P 1 r;::'Y1, then P2==+1 X k such that fJ2==fl(P 2 )?;:;Y2,
and so on. Let I n(x) == 1 on P n and zero elsewhere. Given the set E of
finite measure and the number 8>0, there exists N such that the inter-
section EP n is of measure smaller than 8 for all n?;:;N, and so fElndfl-+O
(i.e., the sequence In satisfies condition (b) for 10==0). Nevertheless, it
is not true that In converges weakly to zero. Indeed, let cxn==fJ;l, and
set g(x) equal to CXn on Pn for n== 1,2, . . . . Then gEL q , sincef Iglqdfl==
 cxfJn==  fJ-q n- 2 <00, but f Ingdfl==cxnfJn== 1 for all n, so f Ingdft
---+0 is not satisfied.
Now, let p== f, and assume again that X o == r X n with disjoint sets
X n of finite measure, but r fl(Xn) ==00. Determine P 1 == l Xk, P 2 ==
:+ 1 X k , and so on, such that fJn==fl(P n) 1 for all n, and let In (x) ==fJ;l
on Pn and zero elsewhere. Then Illnll1== 1 for all n, and fElndfl-+O for
every set E of finite measure (i.e., the conditions (a) and (b) are satis-
fied). Nevertheless, In does not converge weakly to zero, since for
g(x)==l we have gEL oo , but flngdfl==1 for all n.
Finally, let p==oo, and assume that (a) and (b) are satisfied. The
proof that this implies f Ingdfl---+ f logdfl for every gEL 1 is exactly as
above for the case that 1 <P<oo. In order to show that, in general
(a) and (b) together do not imply weak convergence, we consider the
simple example in Zoo that l(n)EZ oo has its first n coordinates zero and
all other coordinates equal to one. Then 11/(n)IIoo== 1 for all n, and
fEI(n)dfl-+O for every set E of finite measure. Nevertheless, I(n) does
not converge weakly to zero. Indeed, if (c) is the closed linear subspace
of Zoo consisting of all 1== (11, 12, . . .) for which lim In exists, then F 0(/)
===lim In is a bounded linear functional on (c), and Fo may therefore
be extended to a bounded linear functional on Zoo (which we still de-
note by F 0). Evidently F o(/(n») == 1 for all n, so I(n) does not converge
weakly to zero.

384 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 53
Exercises
THE SPACE Coo
53.1) Show that the Banach space Coo of all complex continuous
functions on Rk, which tend to zero as x tends to infinity, is not re-
flexi ve.
UNIFORMLY CONVEX BANACH SPACE
53.2) The Banach space V is called uniformly convex whenever
Ilxnll= 1, IIYnll== 1 for n== 1, 2, . . " and IIxn+Ynll-+2 as n-+oo, implies
that Ilxn-Ynll-+O. Show that if in the uniformly convex space V the
sequence XnEV converges weakly to XOEV, and in addition IIxnll-+
Ilxoll, then IIxn-xoll-+O. Observe that, by the parallelogram law, every
Hilbert space is uniformly convex.
UNIFORM CONVEXITY OF Lp
53.3) Let 1 <P<oo, and f(x)==(I+x)Pf(l+xp) for Oxl. Show
that, given any a such that O<a< 1, there exists 15(a) >0 such that
f(x){I-15(a)}2P-1 for Oxa.
53.4) Let 1 <p<oo. Show that, given any a such that O<a< 1,
there exists 15(a) >0 such that
( U+bU ) P { } uP+(bu)p
2 < 1 -15(a) 2
for all u?;::-O and all b such that Ob<a.
53.5) Let 1 <P<oo. Show that, given any e>O, there exists y(e) >0
such that
Xy P  {I _ y(e)} IXIPIYIP
for all complex numbers x, Y satisfying Ix-yl?;::-e max(lxl, IYI). Show
first that
l+z P { } 1+lzlP
 1 - y(e)
2 2
for all complex z such that Izl 1 and 11-zl?;::-e.

Ch. 12,9 53J SECOND CONJUGATE SPACE AND REFLEXIVITY 385
53.6) Let 1 <P<(X), and O<e< 1. Show the existence of q(e) >0
such that for all pairs I, gEL p satisfying 1I/IIp==llgllp== 1 and I I/-gIPdfl
e we have I 1!(/+g) IPdfl l-q(e). For this purpose, let cx==!e and
E=={x: If-glcx max (III, Igl)}, and show first that IE If-gIPdfl!e and
IE If-glpdfl2p-1 IE{lfI P + IgIP}dfl. Next, prove that
I-jl!(f+g) IPdfl!y(cx)j {I/IP+ Iglp}dfl,
E
where y(cx) has the same meaning as in the preceding exercise.
53.7) Show that Lp, for 1 <P<(X), is uniformly convex. Observe
that it follows now that if InELp converges weakly to 10ELp, and
Ilfnllpll/ol!p, then In converges in norm to 10.
53.8) Let fl be Lebesgue measure in the interval X==[O, 2nJ. Show
the existence of a sequence In ELI (X, fl) converging weakly to a function
IOEL1, such that in addition IIlnIl1==1I/01!1 for all n, but In does not
converge in norm to 10. Hence, L 1 (X, fl) is not uniformly convex. Com-
pare the results in this exercise and the preceding exercise to the result
in Exercise 30.21.
NORM CONVERGENCE, WEAK CONVERGENCE AND CONVERGENCE IN
MEASURE
53.9) We recall Exercise 18.14, stating that if In (n==l, 2, ...) and
f are finite and measurable on the set X of a-finite measure, then every
subsequence of In has a subsequence converging pointwise to I if and
only if In converges to I in n1easure on every set of finite measure. By
means of this result it was proved in Exercise 30.21, that for lp<(X)
we have 11/-lnllpO if and only if (a) In1 in measure on every set of
finite measure, and (b) IIlnllpI!/llp' The preceding exercises in the
present section show that, for 1 <P<(X) but not for p== 1, we have
1!/-fnllpO if and only if (b) IIlnllpll/lIp and (c) In1 weakly. Show
now that, for p== 1 but not for 1 <P<(X), we have "1-lnllpO if and
only if (a) and (c) hold.
THE DUAL THEOREM OF THE BANACH-STEINHAUS THEORElVI
53.10) Let {x T } be a collection of elements of the normed linear
space V such that, for each FE V* separately, the collection of numbers
{F(x T )} is bounded. Show that the elements X T are of uniformly.bounded

386 CONJUGATE SPACES AND WEAK CONVERGENCE [Ch. 12,9 53
norm, i.e., show the existence of a finite number MO such that "xT"M
for all T.
REPEATED APPLICATION OF THE BANACH-STEINHAUS THEOREM
53.11) Let {T T} be a collection of bounded linear transformations,
defined on the Banach space V and with the range of every T T in the
Banach space W. Show that if the collection of numbers {F(TTx)} is
bounded for each pair XE V, FE W* separately, then the transformations
TT are uniformly bounded.

CHAPTER 13
FOURIER TRANSFORMATION
In this chapter we shall first derive some properties of unitary transfor-
mations and projections in Hilbert space (a unitary transformation is the
generalization of an orthogonal linear transformation, Le., a rotation around
the origin, in real finite-dimensional Euclidean space). We shall then apply the
so obtained results to the Hilbert function space L2(X, It), and in particular we
shall discuss the Fourier transformation for the case that X is the real line Rl
and It is Lebesgue measure. Some of the results are extended to higher di-
mensional Fourier transformations.
54. Unitary Transformations and Projections in Hilbert Space
If H is a complex or real Hilbert space, any bounded linear transfor-
mation U on H into H (that is, Ux is defined for all XEH and UXEH ,
for all xEH) such that IIUxll==llxll for all xEH is called an isometric
linear transformation. An isometric linear transformation, therefore,
leaves the null element and all distances invariant.
THEOREM 1. A n isometric linear transformation U leaves the inner
product in H invariant, i.e., (Ux, Uy)==(x, y) for all x, YEH.
PROOF. It follows from II,U(x+y)11 2 ==llx+yI12 that
(Ux+Uy, [lx+Uy) " (x+y, x+y),
so
(Ux, Ux)+(UX, Uy)+(Uy, Ux)+(Uy, Uy)==
== (x, x) + (x, y) + (y, x) + (y, y).
Since (Ux, Ux) == (x, x) and (Uy, Uy)==(y, y) by hypothesis, it follows
that
(Ux, Uy)+(Uy, Ux) == (x, y)+(y, x).

388
FOURIER TRANSFORMATION
[Ch. 13,9 54
N ow, (x, y) and (y, x) are conjugate complex, and the same applies to
(Ux, Uy) and (Uy, Ux). It follows, therefore, that the real parts of
(Ux, Uy) and (x, y) are equal. Replacing x by ix, we find that i(Ux, Uy)
and i(x, y) have also the same real part, that is, (Ux, Uy) and (x, y)
have the same imaginary part. Hence, (Ux, Uy)==(x, y).
Given the isometric linear transformation U, the equation Ux=O
has x==o as the only solution; in other words, we have UX1==UX2 if
and only if Xl ==X2. This shows that, on the range of U (the range of
U is the set of all y==Ux), the inverse transformation U-1 of U is
defined, linear and isometric. If the range of the isometric linear
transformation U is the whole space H, then U is said to be a unitary
transfornlation (if H is a real Hilbert space, a unitary transformation is
also sometimes called an orthogonal transformation). Evidently, if U is
unitary, the inverse transformation U-1, defined on the whole space H,
is also unitary.
THEOREM 2. (a) If T is an arbitrary bounded linear transformation
on H into H, there exists a unique bounded linear tranformation T* (also
on H into H) such that (Tx, y)==(x, T*y) for all x, YEH. Furthermore,
T* satisfies II T* II == II T II. The trans formation T* is called the adjoint
transformation of T, and it is obvious that T is the adjoint transformation
of T*.
(b) The bounded linear transformation U is unitary if and only if
UU*== U*U =E, where E is the unit transformation satisfying Ex==x for
all xEH.
PROOF. (a) For a fixed YEH and variable XEH the functional F(x)
== (Tx, y) is linear and bounded on H (the boundedness follows from
IF(x) I  liT xll.l[yl! IITII.I[yil · IIxll),
hence there exists a unique Y*EH such that F(x)==(x, y*) for all xEH
(cf. sec. 29, Theorem 5). Note that IIY*II== IIFIIIITII' IIyll. Denoting this
element y* by T*y, we have therefore (Tx, y) == (x, T*y) for all xEH.
Doing this for every YEH, the transformation y*==T*y is clearly linear,
and the inequality
II T*YII = IIY*IIIITI! 'llyll

Ch. 13,  54J
UNITARY TRANSFORMATIONS
389
implies that T* is bounded; more precisely, IIT*IIIITII. In order to
show that T* is unique, let us assume that (x, T*y) == (x, T' y) for all
x, YEH. Then, if y is fixed, we have (x, T*y-T'y) ==0 for all xEH, in
particular for x==T*y-T'y, so T*y-T'y==O. Since this holds for all
YEH, it follows that T'T*. Finally, since (Ty, x)(y, T*x) for all
x, YEH, we have also (T*x, y)(x, Ty) for all x, YEH, showing that T
is the (unique) adjoint transfornlation of T*, so that IITIIIIT*II in view
of what has already been proved above. This, in combination with
IIT*IIIITII, yields IIT*IIIITII.
(b) Let U be unitary, and let x, YEH. Then, for z U-1 y , we have
y Uz, so
(Ux, y)(Ux, Uz)(x, z)(x, U-1 y ),
which shows that U*== U-1, that is, U*x U-1 x for all xEH. Then
UU*x UU-1 x ==x
and
U*Ux== U-1UXX
for all xEH, so UU*U*UE.
Conversely, if U is a bounded linear transformation satisfying UU* ==
U*U ==E, we have
(Ux, Uy)==(x, U*Uy) == (x, y)
for all x, y E H,
showing already that U is isometric. In addition, the range of U is the
whole of H, since any YEH may be written in the form y== U(U*y).
It follows that U is unitary.
Given the closed linear subspace M c H, we recall that the element
YEH is said to be orthogonal to M (notation y M) whenever (y, x) ==0
for all xEM. The set M 1- of all y M is clearly a closed linear subspace
of H, called the orthogonal comPlement of M (M 1- is closed since y n --+y,
YnEM1- implies (y, x)(Y-Yn, x) for all xEM, so I(y, x)IIIY-Ynll'llxll
O). The orthogonal complement M1-J.. of M1- is again M. Indeed, it
is evident that McM1-1-, and conversely, if xEM1-1-, we have x==
y+z, YEM, zEM1-, by sec. 29, Theorem 4, so (x, z)O. But also (y, z)
O, so (z, z) == (x-y, z) O, that is, z==o. Hence xy EM, i.e., M 1-1- eM.
The closed linear subspaces M and M 1- are, therefore, each other's or-
thogonal complements. Evidently, if M is the whole space H, then M 1-
contains only the null element.

390
FOURIER TRASFORMATION
[Ch.13,54
Any bounded linear transformation P (on H into H), satisfying
p2==P and IIPIIl (i.e., P(Px)==Px and IIPxllllxll for all xEH), is
called an orthogonal projection. The set N(P) of all elements y satisfying
Py==O is then evidently a closed linear subspace, called the null space
of the projection P. Furthermore, any element z in the range R(P) of
P has the property that Pz==z. Indeed, z==Pz 1 for some ZlEH, and
so PZ==P2 Z1 ==PZ 1 ==Z. It follows that IIPII==l, unless P is the null
transformation (then Px==O for all xEH, so IIPII==O).
THEOREM 3. (a) If P is an orthogonal projection, then the range R(P)
and the null space N(P) are orthogonal comPlements. Furthermore, the
adjoint transformation P* of P satisfies P*==P.
(b) If P is a bounded linear transformation (on H into H) satisfying
p2==P and P*==P, then P is an orthogonal projection.
PROOF. (a) We have noted already that every xER(P) satisfies
Px==x. Conversely, of course, if Px==x, then xER(P). In order to prove
that R(P) is the orthogonal complement of N(P) it is sufficient, there-
fore, to show that x==Px if and only if x N(P). Let first x==Px. By
sec. 29, Theorem 4 there is a unique decomposition x==y+z, YEN(P),
z N(P), so x==Px==Py+Pz==Pz. This shows that Pz==y+z with
(y, z) ==0. Hence
IlzI12>IIP z I12== Ily+zI12== Ify112+ Ilz112,
so y==O. It follows that x=z N(P). If, conversely, x N(P), we con-
sider the element y==Px-x. We have YEN(P) since Py==P2 x -Px==0,
and so (x, y)==O on account of x N(P). This shows that Px==x+y
with (x, y) ==0. Hence
IlxI1 2 >IIPxI1 2 == Ilx11 2 + Ily112,
so y==O. It follows that Px==x.
Once more on account of sec. 29, Theorem 4, any XEH has, there-
fore, a unique decomposition x==y+z, YEN(P), zER(P). Then Px=
Pz==z, so that the decomposition of x may be rewritten in the form x=
y+Px with YEN(P). It follows that, for arbitrary Xl, X2EH, we have
( PX 1, X2) == ( PX 1, Y2+ PX2) == ( PX 1, PX2),
and similarly (Xl, PX2) == (PX1, PX2). Hence (PX1, X2) == (Xl, PX2) for all
Xl, X2EH, so P*==P.

Ch. 13,  54J
UNITARY TRANSFORMATIONS
391
(b) We shall prove first that the range R(P) is contained in the
orthogonal complen1ent of the closed linear subspace N(P), i.e., we
shall prove that if z==Px and Py==O, then (z, y)==O. This follows im-
mediately from
(z, y) == (Px, y) == (x, Py) == (x, 0) ==0.
Next, we observe that for an arbitrary XEH we have y==x-PxEN(P),
hence x==Px+y with PXER(P) and YEN(P), so (Px, y)==O. It follows
that IlxI1 2 ==IIPxI12+llyI12, so IIPxI12llxI12, showing that IIPIIl. Hence,
since p2==P and IIPII 1, P is an orthogonal projection.
THEOREM 4. If PI and P 2 are orthogonal projections, then the ranges
R(P 1 ) and R(P 2 ) are orthogonal (i.e., R(P1) is contained in R(P 2 ) J.. and,
hence, R(P 2 ) in R(P 1 )J..) if and only if P 1 P 2 is the null transformation.
In that case P 2 P 1 is also equal to the null transformation, and PI +P 2 is
an orthogonal projection having for its range the set of all Xl +X2 with
X1ER(P1) and X2ER(P2).
PROOF. If R(P 1 ) and R(P 2 ) are orthogonal, then (PIX, P 2 y) ==0 for
all x, YEH. Hence (x, P 1 P 2 y) ==0 for all x, so P 1 P 2 y==0 for all YEH.
Sin1ilarly P2P1X==0 for all xEH.
Conversely, if P 1 P 2 y==0 for all YEH, then P 2 YEN(P 1 )==R(P 1 )J.. for
all YEH, so R(P 2 ) c R(P 1 ) J...
Furthermore, since
(PI +P2)2==Pr+P1P2+P2P1 +P==Pr+P==P1 +P 2
and
(PI +P 2 )*==Pi +P 2 ==P1 +P 2 ,
the transformation PI +P 2 is an orthogonal projection by part (b) of
the preceding theorem. Obviously, every element in the range of PI +P 2
is of the form Xl +X2 for some X1ER(P1) and X2ER(P2), and conversely,
if X is of that form, then
(PI +P2)X==P1X+P2X==P1X1 +P1 X 2+ P 2 X 1 +P 2 X 2
==P 1 X 1 +P2X2==X1 +X2==X
(in view of P1X2==P1P2X2==0 and, similarly, P2X1 ==0), so that
xER(P 1 +P 2 ).

392
FOURIER TRANSFORMATION
[Ch.13,54
Exercises
THE ADJOINT TRANSFORMATION IN HILBERT SPACE
54.1) It is immediately evident that, for any bounded linear transfor-
mation T (on the Hilbert space H into H), the null space N(T) is a
closed linear subspace, but it is not true that the range of T is always
equal to the orthogonal complement N (T) J... It is not even true in
general that the range R(T) is a closed linear subspace. Obviously,
however, R(T) is a linear collection, so its closure R(T) is a closed
linear subspace. Show that N(T) and R(T*), where "T* is the adjoint
transformation of T, are orthogonal complements.
54.2) Show that the operation of taking the adjoint transformation
has the follo,ving properties.
(a) T**==T, (T1+T2)*==Ti+T2' (OI.T)*==aT* for any complex num-
ber 01., and (TIT 2) * == T 2 Ti.
{b) IIT*TII==IITI12.
(c) E+T*T has a bounded inverse on H of norm not exceeding one
(where E is the unit transformation).
ANNIHILATOR
54.3) Let V be a Banach space and V* the conjugate space; the
elements of V and V* "viii be denoted by x and x* respectively, and
the value of x* E V* at the point XE V will be denoted by <x, x*). Given
the closed linear subspace L of V, the element y* E V* is said to be
orthogonal to L (notation y* L) whenever <x, y*)==O for all xEL. The
set LJ.. of all y* L is clearly a closed linear subspace of V*, and LJ.. is
called the annihilator of L. Given the closed linear subspace M of V*,
the element XE V is said to be inversely orthogonal to M (notation xT M)
whenever <x, y*)==O for all Y*EM. The set MT of all xT M is clearly a
closed linear subspace of V, and MT is called the inverse annihilator of
M. Show that (LJ..) T ==L for every closed linear subspace L of V. Evi-
dently, Me (MT) J.. for every closed linear subspace M of V*. Show
that the inclusion may be proper by considering the case that V is the
sequence space II and M is the closed linear subspace (co) of loo==li,
consisting of all sequences converging to zero (cf. Exercise 50.8). Note,
however, that if V is reflexive, then M == (MT) J.. for every M.

Ch. 13,  54J
UNITARY TRANSFORMATIONS
393
THE ADJOINT TRANSFO,RMATION IN BANACH SPACE
54.4) The notion of adjoint transformation may be extended to
bounded linear transformations operating from one Banach space into
another. Let VIand V 2 be Banach spaces, and let Vi and V 2 be their
conjugate spaces. The elements of VI, V 2, Vi and V 2 will be denoted
by x, y, x* and y* respectively, and the value of x* at x by <x, x*).
Show that if y==Tx is a bounded linear transformation, and y* is fixed,
then <x, x*)== <Tx, y*) is a bounded linear functional on VI. Sho\v next
that if the transformation T* on V 2 into Vi is then defined by T*y*==x*,
1. e.,
<x, T*y*)== <Tx, y*)
for all XE VIand all y* E V 2 , then T* is a bounded linear transformation
satisfying IIT*II==IITII. Finally, show that the property in Exercise 54.1
may be extended to the statement that
N (T*) =={R(T)}-L
and
N (T) =={R(T*)} T.
In view of the preceding exercise this implies that
R(T)=={N(T*)}T
and
R(T*) c{N(T)}-L.
ORTHOGONAL PROJECTIONS IN HILBERT SPACE
54.5) Show that if PI, . . " P k are orthogonal projections (on the
Hilbert space H into H), then the ranges R(P 1 ), . . ., R(Pk) are mutu-
ally orthogonal if and only if PiPj is the null transformation for i*i.
Show also that in this case PI +. . . +P k is an orthogonal projection
having for its range the set of all Xl +. . . +Xk for XiER(Pi). Finally,
show that if PI, . . " P k and P==P 1 + . . . +P k are orthogonal pro-
jections, then PiP j is the null transformation for i*i.
PROJECTIONS IN BANACH SPACE
54.6) Let P be a bounded linear transformation on the Banach
space V into V such that P2==P. Show that the range R(P) of P is the
null space N(E-P) of E-P, where E is the unit transformation. Show
next that every XE V has a unique decomposition x==y+z with YER(P)
and zEN(P), namely y==Px and z==x-Px. The transformation P is

394
FOURIER TRANSFORMATIO
[Ch. 13,  55
sometimes called the projection on R(P) along N(P). Note that E-P
is the projection 011 N(P) along R(P).
54.7) We use the same notations as in the preceding exercise. Show
that the adjoint P* of the projection P is also a projection with N(P*)
and R(P*) equal to {R(P)}l. and {N(P)}-L respectively.
54.8) Let PI and P 2 be projections on V into V. Show that PI +P 2
is a projection if and only if P 1 P 2 and P 2 P 1 are both equal to the null
transformation.
54.9) Let P be a projection on V into V. Show that the ranges R(P)
and R(P*) have the same dimension. To this end, show first that
dim R(P*»dim R(P) by selecting linearly independent bounded linear
functionals 11, . . " Ik on R(P), where k is any natural number less than
or equal to dim R(P). Set
<x, F i )== <Px, Ii),
i==l, ...,k,
for all XEV.
Then
I<X, Fi)Ill/ill'IIPII'llxll,
so F 1, . . " F k are linearly independent elements of V*. Show now that
F1, ..., Fk belong to {N(P)}l.==R(P*). It remains to show that if
dim R(P) ==k<oo, then dim R(P*) ==3k. For this purpose, let e1, . . " ek
be a basis in R(P), and choose the above 11, "', Ik such that <ei,lj)
-=f5ij, where f5ij is the Kronecker delta (f5 ij == 1 for i==j and f5ij==O for
i*j). It follows then easily that Px== <x, Fi)ei for all XEV, and
hence P*F== <ei, F)F i for all FEV*. The desired result follows.
55. Unitary Transformations in a Hilbert Function Space
We assume in this section that fl is a measure in the non-empty
point set X, and U a unitary transformation ill the Hilbert space
L2(X, fl). The collection of all sets EeX of finite fl-measure will be
denoted by L1 1 . Then XEEL 2 for every EEA1, so that the functions
H(E, X)==UXE(X),
K(E, x)== U-1 XE (X) ,
(1)
considered as functions of x, are in L 2 for every E EA 1 . If IEL2 is arbi-
trary, and g==UI, we have
(g, XE) == (UI, XE) == (I, U*XE) == (I, U-1 XE ) == f K(E, x)/(x) dfl,

Ch. 13,  55J TRANSFORMATIOS IN A HILBERT FUNCTION SPACE 395
so
f g(X) dfl= f K(E, x)f(x) dfl.
E x
(2)
Similarly,
so
(f, XE)=(U-1 g , XE)=(g , UXE) = f H(E, x)g(x)dfl,
ff(x)dfl== fH(E, x)g(x)dfl.
E x
(3)
Given any E'E.ii 1 , we substitute g(X)==XE'(X), and so f(x)=U-1 XE '=='
K(E', x), in (2) and (3), and we obtain
fl(EnE') = f K(E, x)K(E', x)dfl, (4)
x
f K(E', x)dfl== f H(E, x)dfl.
E E'
(5)
Similarly, substituting f(x) ==XE'(X) , so g(x) =H(E', x), we obtain (5) for
the second time, and also
fl(EnE') = f H(E, x)H(E', x)dfl.
x
(6)
The following theoren1 has thus been proved.
THEOREM 1. If U is a unitary transformation in L 2 (X, fl), the func-
tions H(E, x) and K(E, x), defined by (1) and depending upon the set
E E Li 1 and the point x EX, S atis fy the equalities ( 4) , (5) , ( 6 ) for all
E, E' EA 1 .
The main purpose in the present section is to prove a theorem in the
converse direction.
THEOREM 2. Let A 2 be a subcollection of Al such that the set of all
finite linear combinations s(x)== =1 cnXEn(x) for EnEA2 is dense in L2,
and let the functions H(E, x) and K(E, x), defined for all EEA 2 and all
XEX, be in L 2 (X, fl) for every EEL1 2 and satisfy (4), (5), (6) for all
E, E' E.i1 2 . Then there exists a unitary transformation U in L 2 (X, fl) such
that (1) is satisfied for all E EA 2 . It follows, therefore, that (2) and (3)
are satisfied for g= U f and all E, E' EA 2 .
PROOF. We define the transformations U and V on the set of all

396
FOURIER TRANSFORMATION
[Ch. 13,  55
characteristic functions XE (E EA 2 ) by
UXE==H(E, x), VXE==K(E, x). (7)
It follows then from (4), (5), (6) that
(U XE, U XE') == (XE, XE'), (V XE, V XE') == (XE, XE')'
(V XE, XE') == (XE, U XE') (8)
for all E, E ' EA2. Assume now that E1, "', E p EA 2 ad f=l CiXEi(X)
==0 for all XEX; we shall prove that this implies l CiVXEi==O for all
XEX. Indeed, we have
( Ci V XEi' XE') ==  Ci(V XEi' XE')
==  Ci(XEi' U XE') == ( CiXEi' U XE') ==0,
implying that (Ci V XEi' s) ==0 for any finite linear combination s(x)
of characteristic functions of sets in A2, and hence ( Ci V XEi' I) ==0 for
any IEL2 (since the set of all such finite linear combinations is dense
in L2)' It follows that l Ci V XEi==O. Similarly, we obtain l Ci U XEi==O.
Consequently, the transformation U may be extended unambiguously
to the functions s(x) == l CiXEi(X) by defining U s== f= 1 Ci U XEi' and the
same holds for V. Then U and V are linear on their new domain of
definition, and it follows from (8) that
(U Sl, U S2) == (Sl, S2), (V Sl, V S2) == (Sl, S2), (V Sl, S2) == (Sl, U S2)
for any two functions Sl, S2 in this domain. In particular, since U and
V are now isometric on their present domain of definition, it follows
that U and V are bounded on this domain. But then, the domain of U
and V being dense in L2, these bounded linear transformations may be
extended by continuity to the whole of L 2 such that linearity and
boundedness are preserved, and
( U I, U g) == (I, g), (V I, V g) == (I, g), (V I, g) == (I, U g)
hold for all I, gEL 2 . It follows from the first of these equalities that
U*U ==E (the unit transformation), from the third equality that V == U*,
and from the equality in the middle that V*V==E, so UU*==E. The
transformation U is therefore unitary, and V == U* == U-1; hence, (7)
shows that U satisfies (1) for all E EA 2 .

Ch. 13,  56J
THE FOURIER TRANSFORMATIO
397
The theorems in this section are essentially due to S. BOCHNER (1934,
[lJ).
56. The Fourier Transformation
We shall now specialize the discussion in the preceding section to the
case that X is the real line R 1 and fl is Lebesgue measure (and, as on
several previous occasions, we shall write f fdx instead of f fdfl). The
bounded cell having as endpoints the origin and the point s will be de-
noted by E 8 ; hence, E8 is {x: O<xs} or {x: s<xO} according as s>O
or s<O. Since any continuous function h{x) possessing a bounded carrier
may be approximated uniformly by finite linear combinations of charac-
teristic functions XEs (x) , and since the collection of such continuous
functions h(x) is dense in the Hilbert space L2 (cf. sec. 30, Theorem 6
and sec. 21, Theorem 2), the set of all finite linear combinations of
functions XEs (x) is also dense in L 2 ; in other words the collection A 2 of
all sets E8 satisfies the conditions imposed upon A 2 in Theorem 2 of
the preceding section. It follows that the theorem may be restated for
the present case as follows.
THEOREIVI 1. Let K 1 (s, x) and H 1 (s, x), defined for all sER 1 , xER 1 ,
belong to L 2 for every sER 1 , and let
f K 1 (s, x)K 1 (t, x)dx== f H 1 (s, X)H1(t, x)dx
Rl Rl
= ! mill (I, It I)
f K 1 (t, x)dx== f H1(S, x)dx
Es Et
for
for
sf;::; 0 ,
st<O,
for all s, t. Then there exists a unitary transformation g== U f in L 2 such
that
f g(x) dx== f K 1 (s, x)f(x) dx,
Es Rl
f f(x)dx== f H 1 (s, x)g(x)dx
Es Rl
for all s.
By a small modification, we obtain the following theorem.

398
FOURIER TRANSFORMATION
[Ch. 13,  56
THEOREM 2. Let K(s, x) and H(s, x), defined for all sER1, xER1, be-
long to L 2 for every sER1, and let
00 00
/ K(s, x)K(t, x) dx== J H(s, x)H(t, x) dx
-00
-00
= ! min (Isl, It I)
l 0
for
for
st?;;-O,
st<O,
s t
J K(t, x) dx== J H(s, x) dx
o 0
for all s, t. Then the relations
s 00
J g(x) dx== J K(s, x)f(x) dx,
o -00
s 00
J f(x) dx== J H(s, x)g(x) dx
o -00
define a unitary transformation g== Uf and its inverse f== U-1 g in L2.
PROOF. Apply the preceding theorem to K 1 (s, x) == (sgn s)K(s, x) and
H 1 ( s, x) == (sgn s) H ( s, x).
Note that, by Lebesgue's theorem on differentiation of an indefinite
integral, we have g(s) == (djds)/os g(x) dx for almost every s, so that the
unitary transformation g== Uf and the inverse f== U-1g are determined
by
00
00
d '
g(s) == - J K(s,x)f(x)dx,
ds
f(s) =  f H(S, x)g(x)dx.
ds
-00
-00
We shall now consider a further specialization.
THEOREM 3 (WATSON'S "GENERAL TRANSFORMS", 1933, [IJ). If cp(O)
==0, cp(x)jxEL 2 and
00
f cp(sx)cp(tx) dx == { min (Isl, Itl)
x 2 0
for
for
st?;;-O,
st<O,
-00
then the relations
00
00
d f cp(sx)
g(s) == - f(x) dx,
ds x
-00
d f cp(sx)
f(s) == ds x g(x) dx
00

Ch. 13,  56J
THE FOURIER TRANSFORMATION
399
define a unitary transformation g= Uf and its inverse f= U-1 g in L 2 .
PROOF. The functions
q;(sx)
K(s, x) = ,
x
H(s, x) = cp(sx)
x
satisfy the conditions of the preceding theorem.
For the purpose of the next theorem we shall write f(x)==Lim /n(x)
as an abbreviation for limfoo If-fnI2dx=0. In other words, the no-
tation Lim will denote convergence in the Hilbert space L 2 .
THEOREM 4 (PLANCHEREL'S THEOREM, 1910, [IJ). The formulas
00
g(s) =   f e- isx - ] f(x) dx,
.J2n ds -ix
-00
(1)
00
] d f e isx 1
f(s) = .J 2n ds ix g(x) dx
-00
define a unitary transformation g== Uf and its inverse f= U-1g in L2.
The defining formulas may also be written as
n
g(s) =  Lim f e-isx f(x) dx,
.J 2n noo -
-n
n
f(s) =  Lim f eisxg(X)dX.
.J 2n noo
-n
PROOF. If
1 e- ix - 1
cp(x) = .J-
2n -

400
FOURIER TRANSFORMATION
[Ch. 13,  S6
then cp(O) ==0, cp(x) fXEL 2 , and
00
00
f cp(sx)cp(tx) dx ==  f (eiSX - 1) (e- itX - 1) dx
x 2 2n x 2
-00
-00
00
==  f cos (s-t)x-cos sx-cos tx+ 1 dx
2n x 2
-00
00
1 f sin 2 y
== - {Isl + Itl-Is-tl} - dy == !{Isl + Itl-Is-tl}
2n y2
-00
= { min (I, It I)
for
for
stO,
st<O.
We have used here that
00 00 a
J (sin 2 yfy2) dy==2 J ==2lim J (sin 2yfy) dy==n
-00 0 aoo 0
(partial integration). It follows, therefore, from the preceding theorem
that the relations (1) define a unitary transformation g==Uf and its
inverse f== U-1 g . In order to prove the remaining part of the theorem,
let f(x) EL2, and denote by fn(x) the function, equal to f(x) on -n
xn and vanishing elsewhere. Writing gn== Ufn, we have
n
1 d f e-isx - 1
gn(S) = - - d . f(x) dx
2n s -x
-n
n
1 f e-i(s+ h)x - e- isx
_ lim . f(x) dx
2n hO -hx
-n
n
1 f sin !hx
_ lim e-!ihXe-iSXf(x)dx.
 2n hO !hx
-n
The function under the last sign of integration does not exceed If(x) I in

Ch. 13,  56J
THE FOURIER TRANSFORMATION
401
absolute value; hence
n
gn(s) =  f e-isx f(x) dx
2n
-n
by dominated convergence. Furthermore, since Ilf-fnllO as noo and
U is unitary, we have IIUf-Ufnll--+O, so Ilg-gnll--+ O for g==Uf, showing
that
n
g(s) =  Lim f e-iSxf(X)dX.
 2n n--+oo
-n
The proof for the inverse relation is similar.
The unitary transformation U in the last theorem is called the Fou-
rier transformation. Evidently, it has the special property that
Uf(X)==U-1f(-x)
for every fEL 2 ,
so U2f(x)==UU-1f(-x)=f(-x), implying that U4f(x)=f(x). It follows
that U4=E (the unit transformation) ; generally
Un+4== Un
(n==O, ='= 1, =,=2, . . .).
Let us consider now the bounded linear transformations
P O =!(E+U+U2+U3),
P 1 =!(E-iU-U2+iU3),
P 2 ==!(E- U + U2- U3),
P 3 ==!(E +iU - U2-iU3).
Simple computations, based on U*= U-1 and Un+4== Un, show that
P'k==P k and P%=P k for k==O, 1,2, 3; the transformations P k are there-
fore orthogonal projections (cf. sec. 54, Theorem 3). Furthermore, PkPZ
is the null transformation for k=l-l, showing that the projections P k
(k==O, 1, 2, 3) have mutually orthogonal ranges (cf. sec. 54, Theorem 4).
In addition, PO+P1 +P 2 +P 3 is exactly the unit transformation E.
Finally, UPk==ikP k for k=O, 1,2,3. It follows that any fEL 2 may be

402
FOURIER TRANSFORMATION
[Ch.13,56
written as
f==fof1f2f3
with
fk==Pkf
(k==O, 1, 2, 3),
(fk, fl) == 0
for k -::Il,
and
Ufo==fo, Uf1==if1, Uf2==-f2, Uf3==-if3,
so
Uf==foif1-f2-if3'
The preceding discussion may be extended to Rk for k> 1. For sim-
plicity, we shall write out the details only for k==2. The points XER2
will be denoted by x== (Xl, X2), and the bounded cell having the origin
and the point s== (Sl, S2), with Sl -::10 and S2 -::10, as opposite vertices will
be denoted by Es. It follows, simjlarly as in the one-dimensional case,
that the set of all finite linear combinations of functions XEs (x) is dense
in L2==L2(R2, # X #). Furthermore, for s== (Sl, S2), we introduce the no-
tationfg fdx==fgl fg2 fdx. By definition this is equal to (sgn SlS2)fEs fdx
for Sls2-::10, and to zero for SlS2==0. The parallel of Theorem 2 is then
as follows.
THEOREM 5. Let K(s, x) and H(s, x), defined for all sER 2 , xER 2 , be-
long to L 2 for every sER2, and let
f K(s, x)K(t, x) dx== f H(s, x)H(t, x) dx==
R2 R2
= { min (lsll, Itll)  min (ls21, It21)
for Sl t 1>0
otherwise,
and
S2 t 2>0,
s t
f K(t, x) dx== f H(s, x) dx
o 0
for all s, t. Then the relations
s
J g(x) dx== J K(s, x)f(x) dx,
o R2
s
J f(x) dx== f H(s, x)g(x) dx
o R2
define a unitary transformation g== Uf and its inverse f== U-1g in L 2 .
It is important for the applications to observe now that if K 1 (Sl, Xl)
and H l(Sl, Xl) is a pair of functions satisfying the conditions of Theo-

Ch. 13,  56J
THE FOURIER TRANSFORMATION
403
rem 2, and K 2 (S2, X2), H 2 (S2, X2) is a second pair satisfying the same
conditions, then K(s, x) ==K 1 (Sl, xl)K 2 (S2, X2) and H(s, x) ==H l(Sl, Xl) .
H 2(S2, X2) satisfy the conditions of Theorem 5. Hence, if 91l(X1) and
912(X2) both satisfy the conditions of Theorem 3, then
8
f g(x) dx == f tpl(SlXl)tp2(S2X2) f(x) dx,
x1 X 2
o R2
8
. J ' f(x) dx = f tpl(SlXl)tp2(S2X2) g(x) dx
X1 X 2
o R2
(2)
define a unitary transformation g== Uf and its inverse f== U-1 g in L 2 .
If gEL 2 , and E runs through all subsets of R 2 of finite measure, then
the set function fEgdx has, almost everywhere on R2, the function g
as its regular derivative. In particular, if h!O, we have
81+h 82+h
lim h- 2 I I gdx==g(s)
h+O ,1 82
(3)
at almost every sER 2 . Now, it is easy to check that, for h>O,
81 + h 82 + h 81 + h 82 + h 81 + h 82 81 82 + h .1 82
I I ==1 I -I I-I I +1 I,
81 82 0 0 0 0 0 0 0 0
regardless of the quadrant in which the point s is lying, and so we ob-
tain by combination of (2) and (3) that
g(s) == lim f [tpl{(Sl +h)Xl}-tpl(SlXl)] · [tp2{(S2+ h )X2}-tp2(S2 X 2)] f(x) dx
h+O hX1 hX2
R2
at almost every sER 2 . Applying this to the particular case that 91l(X)==
912 (x) ==91 (x) is the same function as in the one-dimensional Plancherel's
theorem, the following two-dimensional version is obtained.

404
FOURIER TRANSFORMATION
[Ch. 13,  56
THEOREM 6. The relations
n n
g(s) =  Lim f r e-i(SlXl+S2X2) f(x) dx,
.J 2n noo aI
-n -n
n n
Its) =  Lim f f ei(SlXl+s2X2)g(X)dX
.J 2n noo
-n -n
define a unitary transformation g== Uf and its inverse f== U-1 g in L2.
The unitary transformation U in the last theorem is the two-di-
mensional Fourier transformation. As before, we have Uf(x) == U-1f( -x),
so U4==E, and it follows that the same orthogonal projections P k (k==
0, 1, 2, 3) as before yield a decomposition f==  Pkf for any fEL2 such
that fk==Pkf satisfies U fk==ikfk for k==O, 1, 2, 3.
Exercises
WATSON'S GENERAL TRANSFORM
56.1) Let # be Lebesgue measure in R1, and let cp(x) be a Lebesgue
measurable function on R 1 such that, for any fEL 2 (R1, #), there ex-
ists a function g, summable over every bounded interval, with the
property that
S 00
f g(x) dx = f cpx) I(x) dx,
o -00
S 00
f I(x) dx = f cpX) g(x) dx
o -00
for all s, where it is part of the hypothesis that the integrals on the
right exist. Show that cp(x) satisfies the conditions of Theorem 3.
56.2) Show that if cp(x) == 1 for x 1 and cp(x) ==0 for x< 1, then cp
satisfies the conditions of Theorem 3. Determine explicitly the corre-
sponding unitary transformation g== Uf and its inverse f== U-1 g . Show
that U2==E, and that the orthogonal projections Po==t(E + U) and
P 1 ==t(E-U) have orthogonal ranges. Finally, show that if fo==Po/

Ch.13,56J
THE FOURIER TRANSFORMATION
405
and 11=P11 for an arbitrary IEL2, then 1=10+/1 and Ulo==-Io, Ul1
==-/1, so UI=lo-/1.
56.3) Show that if cp(x) =x for Ox< 1 and cp(x) =0 for all other
real x, then cp satisfies the conditions of Theorem 3. Determine ex-
plicitly the corresponding unitary transformation g= Uland its in-
verse 1= U-1g. Show that U2==E.
56.4) Show that if cp(x) ==log x-I for x> 1 and cp(x) ==0 for x< 1,
then cp satisfies the conditions of Theorem 3. Determine explicitly the
corresponding unitary transformation g== UI and its inverse 1== U-1 g .
Show that U2=E.
DETERMINATION OF A FUNCTION BY THE VALUES OF ITS INTEGRAL ALONG
LINES
56.5) Let I(x, y) be Lebesgue summable over R2, and let
F(x, y) == !C(a;x,y/(u, v) d(u, v),
where C (a; x, y) is the circle of centre (x, y) and radius a >0. Show that
F is summable over R2, bounded and continuous.
56.6) Let I and F be the same functions as in the preceding exercise,
and let foo I(x, y) dy==O for almost every x. Show that foo F(x, y) dy
==0 for every x.
56.7) Let I{x, y) be Lebesgue summable over R2, and let the integral
of I along almost any line in any direction be zero. This means that if
we choose cx arbitrarily, and we introduce x' and y' by
, .
x ==x cos cx-y SIn cx,
, .
y =x sin cx+y cos cx,
thenfoo Idy'=O for Lebesgue almost every x'. Show that 1=0 almost
everywhere. For the proof, observe first that the function F, intro-
duced jn the preceding exercises, has the property that the integral of
F along any line in R 2 is zero. Observe also that FEL2(R2, fl Xfl),
where fl is Lebesgue measure in R 1 . It is not difficult to show then that
the Fourier transform of F is identically zero.
State and prove the generalization to Rk for k>3.
56.8) Let g(x, y) == (1-x 2 -y2)-t in the interior C of the circle {(x, y) :
x 2 +y2== I}. Show that the integral of g along any chord of the circle
is constant. Note that fcg(x, y) d(x, y) is finite.

406
FOURIER TRANSFORMATION
[Ch.13,56
56.9) Let /(x, y) be Lebesgue summable over the interior C of the
circle {(x, y) : x 2 +y2== I}, and let the integral of / along almost any
chord in any direction have the constant value k. Show that /(x, y) ==
kf{n( 1-x 2 - y 2)t} at almost every point (x, y) EC.
56.1 0) We recall that the point set A in the linear space V is said
to be convex it xEA, YEA implies exx+(l-ex)YEA for every Oexl.
Let A be a non-empty, open, bounded and convex point set in the
plane R2. Any line in R2 having at least one point in common with A
will be called a line of intersection. Show that if L is a line of inter-
section, then the intersection LnA is an open interval on L. This
interval on L is called a chord of A. Next, consider all lines of inter-
section parallel to the y-axis, and show that their intersections with the
x-axis form an open interval {x: a<x<b} on the x-axis. The lines x==a
and x==b are called the lines 0/ support of A in the direction of the y-
axis, and the number b-a is called the width of A in the direction of
the x-axis. Any line of intersection parallel to the y-axis intersects A
in an open interval {y: /l(X) <y</2(X)}; show that /l(X) and -/2(X) are
convex in the open interval a<x<b, and hence continuous by Exer-
cise 25.4. Observe that A is a Lebesgue measurable set.
Show that A has two lines of support in every direction. If the width
in every direction is the same, A is said to be of constant width. Any
circle is of constant width, but a set of constant width is not neces-
sarily a circle.
56.11) Let A be a non-empty, open, bounded and convex set in R2,
and let /(x, y) be Lebesgue summable over A, such that the integral of
/ along almost any chord in any direction has the constant value k *0.
Show that A is the interior of a circle, and hence
/(x, y) ==kf{n(R2-r 2 )i},
where R is the radius of the circle and r the distance from (x, y) to the
centre. This result is due to J. W. GREEN (1958, [lJ).
56.12) Let A be a non-empty, open, bounded and convex subset of
R3. Show that the projection of A on any plane is non-empty, open,
bounded and convex. Let Az be the projection of A on the xy-plane,
and let any line of intersection parallel to the z-axis intersect A in
{z: /l(X, y)<Z</2(X, y)}. Show that /1 and /2 are continuous functions
on A z.

Ch. 13,  56J
THE FOURIER TRANSFORMATION
407
56.13) Let A be a non-empty, open, bounded and convex subset of
R3, and let f(x, y, z) be Lebesgue summable over A, such that the inte-
gral of f along almost any chord in any direction has the constant
value k =FO. Show that A is the interior of a sphere, and
f(x, y, z) ==kf{n(R2-r 2 )l},
where R is the radius of the sphere and r the distance from (x, y, z) to
the centre.
56.14) Let
B=={(x, y, z) : x2+y2+z2< I}.
Show that there exists no function f(x, y, z), Lebesgue summable over
B, such that the integral of f over every plane section of B equals a
constant k=FO.
n-TH ROOT OF THE UNIT TRANSFORMATION
56.15) Let T be a bounded linear transformation (on the Banach
space V into V) such that Tn==E for the natural number n?;::-2, where
E is the unit transformation, and let ck==e2nikln for k==O, 1, . . " n-1.
Show that
(i) Pk==n- 1 (E+c;lT +c;2T2+. . . +c;(n-l)Tn-1)
satisfies TPk==PkT==ckPk,
(ii) P k is a projection,
(iii) PkP l is the null transformation for k=Fl,
(iv) PO+P1 +. . . +Pn-1==E.
Show now that the decomposition x==Pox+. . . +P n - 1 x of XEV as
a sum of elements in the ranges of Po, . . " P n-1 is unique. Finally,
show that
Tx==coPox+. · · +cn-1 P n-I X .
Note that x+Tx+. . . + Tn-1 x ==nP o x, and so
m-l
m- 1  Tkx--+Pox
k=O
as m--+oo.

CHAPTER 14
ERGODIC THEORY
In the present chapter it will be assumed that !-t is a measure in the point set
X and T is a measure preserving transformation of X into jtself, Le., !-t{T-l(E)}
=!-t(E) for every measurable E, where T-l(E) is the set of all points XEX
satisfying TXEE. The main subject will be an investigation of the convergence
properties of the sequence Gn(x) =n- l : A f(Tkx) for a measurable function
f(x) , and we shall prove von Neumann's mean ergodic theorem and Birkhoff's
individual ergodic theorem, asserting that Gn(x) converges with respect to the
L2 norm for any fEL2 and Gn(x) converges pointwise almost everywhere on X
for any fELl. It will be shown also how the individual ergodic theorem can be
considerably extended.
57. Measurable Transformations
Let X be an arbitrary non-empty point set, and T a transformation
(or maPPing) of X into X itself, that is, to every point XEX is assigned
a point y==TXEX. The subset of X, consisting of all points YEX s'uch
that y=Tx for some XEX, is called the range of T (in accordance with
the definition presented jn sec. 1), and if the range of T is the whole
of X, the transformation T is said to be a transformation of X onto
itself. For any subset E of X, the image T(E) of E is the set of all
y==Tx for xEE, and the inverse image T-1(E) of E is the set of all
XEX satisfying TXEE. Hence, xET-1(E) if and only if TXEE. If T and
5 are transformations of X into itself, the transformation ST is defined
by STx=S(Tx). In particular, therefore, the powers Tn (n== 1, 2, .. .)
of T are inductively defined by Tn+1x=T(Tnx) for all XEX.
If T is a transformation of X into itself, and f(x) is a (real or complex)
function defined on X, we may consider the function g(x)=f(Tx). It is
convenient for future use to note already that if A is some subset of
the set of all (real or complex) numbers then
{x: g(x) EA}== T-1{y: f(y) E.£4}.

Ch. 14,  58J MEASURE PRESERVING TRANSFORMATIONS
409
Indeed, xET-1{y: f(Y)EA} holds if and only if TXE{Y: f(Y)EA} holds,
that is, if and only if g(x) ==f(Tx) EA is satisfied.
We shall assume now that # is a measure in X. The transformation
T of X into itself is called a measurable transformation (with respect to
#) if the inverse image of every #-measurable set is #-measurable. If
T is measurable and f(x) is a (real or complex) measurable function on
X, then g(x)==f(Tx) is also measurable. The proof is obtained by as-
suming first that f is real and noting then that in this case we have
{x: g(x) >a}== T-1{y: f(y) >a}
for every real number a. If Sand T are measurable transformations,
then ST is measurable. The proof follows by observing that (ST)-l(E)
==T-1{S-1(E)} for every subset E ot X.
If the measurable transformations Sand T have the property that
ST and TS are both equal to the unit transformation, we may note
first that the range of T is the whole of X (since x==TSx for every x).
It follows that any point YEX satisfies y==Tx for some XEX, and the
point x is then uniquely determined; in fact, Sy==STx==x, so x==Sy.
This shows at the same time that S is uniquely determined by T; the
transformation S is the inverse transformation of T, and T is called
invertible.
58. Measure Preserving Transformations
The #-measurable transformation T of X into X itself is said to be
measure preserving if #{T-1(E)}==#(E) for every #-measurable subset E
of X. It is convenient to perform some identifications in the collection
of measure preserving transformations. Explicitly, let T and T 1 be
transformations of X into itself such that T is m"easure preserving and
T 1 x==Tx almost everywhere on X. Then T 1 1 (E) and T-1(E) are almost
equal sets for every E eX; in particular, T I 1 (E) is measurable and
#{T 1 1 (E)}==#(E) whenever E is measurable. This shows that T1 is
measure preserving. Also, if f(x) is real or complex on X, the functions
gl(x)==f(T 1 x) and g(x)==f{Tx) are equal almost everywhere. Further-
more, if S is another measure preserving transformation, the transfor-
mations ST 1 and ST are equal almost everywhere, and the same holds
for TIS and TS. The first statement is obvious, and in order to prove

410
ERGODIC THEORY
[Ch.14,58
the second statement it is sufficient to observe that T 15x-=FT5x if and
only if 5x is a point of the null set Xo=={y: T1Y-=FTy}, Le., if and only
if xE5- 1 (X O ). Since #{5- 1 (Xo)}==0, the proof is complete. Note that
essential use is n1ade of the hypothesis that 5 is not merely measur-
able but measure preserving (or at least of the intermediate hypothesis
that #(E)==O implies #{5- 1 (E)}==0). It is now an immediate conse-
quence that if T and T 1 agree almost everywhere, and the same is
true of 5 and 51, then SIT 1x==5Tx almost everywhere. Hence, if we
identify almost equal subsets of X, we may also identify measure pre-
serving transformations which are equal almost everywhere on X.
It is evident that if a measure preserving transformation is in-
vertible then the inverse transformation is also measure preserving.
Some phenomena in statistical mechanics may be subjected to a
mathematical description in which measure preserving transformations
playa role. The different states which a certain system of particles
may possess are then represented by the points of a point set X in
which a measure #, satisfying #(X) == 1, is defined. The points of X are
moving (corresponding to the fact that the state of the system is
changing), and if the position of a certain point x at the time t (tO)
is denoted by x(t), the equation x(t)==Ttx(O) defines, for each value of
the non-negative parameter t, a transformation Tt of X into itself. We
assume that Ttl+t2==TtlTt2 for all t1, t 2 ?;:;0, which is obviously equivalent
to the assumption that the movement of the points in X, seen as a
whole, is time independent (this corresponds to the assumption that
the laws of nature which govern the change of state of the system of
particles are time independent). Such a movement is called a flow. If
A is an arbitrary #-measurable subset of X, the "probability" that a
point of X, say at the time t==O, is contained in A is #(A). Since the
points which at the time t are in A are exactly the points which at the
time t==o are in Tt 1 (A), the probability that a point of X at the time
t is contained in A is then #{Tt- 1 (A)}, so in order that this probability
be also time independent we have to assume that the transformations
Tt are measure preserving. This may usually be achieved by an ap-
propriate choice of the measure #. The mathematical object to be in-
vestigated is, therefore, a flow consisting of measure preserving transfor-
mations Tt in X, and among the questions of importance are questions

Ch. 14,  58J MEASURE PRESERVING TRANSFORMATIONS
411
about the asymptotic properties of Tt as t-+oo. For each to>O, we have
T nt o == (Tto)n for n== 1, 2, . . " so it is reasonable to expect that the
asymptotic properties of Tt are the same as those of (Tto)n as noo.
For this reason, we shall restrict our attention mainly to the sequence
of non-negative powers of a single measure preserving transformation T.
Examples. (1) X is real k-dimensional number space, and # is Le-
besgue measure in X. Any linear transformation T (on X into X) such
that det (T) ==::f:: 1 is then invertible and measure preserving. The proof
follows immediately from the corollary of sec. 39, Theorem 2.
(2) X and # are the same as above, c == (C1, . . . , C k) is a fixed point
of X, and Tx==x+c. The transformation T is invertible and measure
preservIng.
(3) X is the half open interval Ox< 1 with Lebesgue measure, and
T is defined by Tx==2x (mod 1), i.e., Tx==2x for Ox<i, and Tx==
2x-1 for ix<l. If A=={x: ax<b} is contained in X, then T-1(A)
is the union of {x: iax<tb} and {x: t(a+1)x<t(b+1)}, hence
#{T-1(A)}==#(A). It follows then that the inverse image of every
union 5 of such half open intervals has the same measure as 5 itself,
and this implies imn1ediately that T is measure preserving. Note, how-
ever, that T is not invertible. The same example is obtained if X is
the set of all complex numbers z satisfying Izl== 1 (i.e., X is the cir-
cumferellce of the unit circle), with # equal to 1f2n times Lebesgue
measure along the unit circle and T defined by TZ==Z2.
(4) X is again the half open interval Ox< 1 with Lebesgue measure,
c is a fixed point in X, and Tx==x+c (mod 1). Equivalently, X is the
circumference of the unit cjrcle in the complex plane, c is a point in X,
and Tz==cz. The transformation T is invertible and measure pre-
serVl ng.
(5) If # is discrete measure, any linear transformation T in Ex-
ample (1) with det (T) *0 is invertible and measure preserving. The
transformations T in (2) and (4) remain invertible and measure pre-
serving if # is discrete measure, but the transformation T in (3) is no
longer measure preserving.

412
ERGODIC THEORY
[Ch.14,59
59. Poincare's Recurrence Theorem
If T is a measure preserving transformation of X into itself, and E
is a measurable subset of X, then a point xEE is called recurrent (with
respect to E and T) if TnXEE for at least one positive integer n.
THEOREM 1 (POINCARE'S RECURRENCE THEOREM). If T is measure
preserving and #(X) is finite, and if E is a measurable subset of X, then
almost every point of E is recurrent. It is even true that, for almost every
xEE, there are infinitely many values of n such that TnXEE.
PROOF. We consider the set F of all points of E that never return
to E. The set F is measurable since
F==EnT-1(X -E)nT-2(X -E)n. . . .
Furthermore, if n is a positive integer, the sets F and T-n(F) are dis-
joint (indeed, if xEFnT-n(F), we should have xEF and TnXEF, contra-
dicting the definition of F). It follows that all sets F, T-1(F) , T-2(F) ,
are mutually disjoint, since
T-n(F) nT-<n+p)(F) == T-n{F nT-p(F)}.
Assuming the theorem to be false (in the restricted version), all the
sets T-n(F) , n== 1, 2, . . ., would be now of the same positive measure
and mutually disjoint, and this is impossible since #(X) is finite.
The strong version follows now by applying the weak version to
each power of T. Explicitly, given the positive integer n, let F n be the
set of all points xEE such that TknXEX -E for k== 1, 2, . . . . Then
fl(F n) ==0 for all n, so r F n is a null set. Take a point xEE - r F n.
Then xEE-F 1 , so TkxEE for some positive integer k. Similarly, since
xEE-F 2k for this particular k, we have T2kPXEE for some positive
integer p, and so on.
The recurrence theorem is due essentially to H. POINCARE (1890,
[lJ, p. 67-72); the first rigorous proof {along the lines already indicated
by Poincare) was given by C. CARATHEODORY (1919, [3J).

Ch. 14,  60J
MEAN ERGODIC THEOREM
413
Exercises
THE RECURRENCE THEOREM
59.1) Show that if T is measure preserving, #(X) finite, and E a
measurable subset of X, then ;;=o XE(Tnx) diverges for almost every
XEX, where Tn, for n==O, denotes the unit transformation. More gener-
ally, show that for any non-negative measurable function f(x) on X the
series ;;=o f(Tnx) diverges for almost every x in the set E =={x: f(x) >O}.
59.2) Show that for #(X) ==00 the conclusion of the recurrence theo-
rem does not necessarily hold.
THE RECURRENCE THEOREM IN EUCLIDEAN SPACE
59.3) Let # be Lebesgue measure in k-dimensional Euclidean space
Rk, and let X be a #-measurable subset of Rk of finite measure. Show
that if T is a measure preserving transformation of X into itself, then
almost every point XEX is a point of accumulation of the sequence of
its own images Tnx.
()o. Mean Ergodic Theorem
If T is a measure preserving transformation of X into itself, and if
p(X) is finite, the recurrence theorem states that almost every point
of any measurable subset E returns to E infinitely many times. The
question arises if such points have a mean time of sojourn in E, that
is, we ask if
n-l
lim n- 1  XE(Tk x )
noo k=O
exists, where Tk denotes the unit transformation for k==O. More gener-
ally, we may ask for which class of measurable functions f(x) the limit
(as n-+oo) of n- 1 : f(Tk x ) exists in some sense.
In order to deal with some aspects of the problem stated, it is useful
to consider more closely, for any measurable f(x) , the function g(x) ==
j(Tx). Irrespective of whether #(X) is finite or infinite, the function
g(x) is measurable, so that, writing g== Uf, the transformation U assigns
to each measurable f a measurable g. Obviously U is linear; further-

414
ERGODIC THEORY
[Ch. 14,  60
more, U is positive, i.e., g(x)==UI(x) is non-negative on X whenever
I(x) is so. Also, Ig(x) I == U{I/(x) I} whenever g(x) == U I(x). Finally, U is
norm preserving, as shown by the following theorem.
THEOREM 1. II lp<oo, and Ilfllp==(jxlflpdfl)l/P is the Lp norm of
any measurable f, then Ilgllp==ll/llp lor g== UI.
PROOF. If E is a measurable subset of X, and I==XE, then
g== UI==/(Tx) ==Xr-l(E)'
so
Ilgll==fl{T-1(E) }==fl(E) == II/II.
It follows that Ilgllp== 1I/IIp for every measurable I assuming only a finite
number of non-negative values. If I is measurable and non-negative,
there exists a sequence of measurable functions In, each assuming only
a finite number of non-negative values, such that In i I. Since gn== Uln
is a similar sequence, and since gn i g== UI, the theorem on integration
of monotone sequences shows that
Ilgllp==lim Ilgnllp==lim III nllp== Ilfllp.
The same result for an arbitrary complex measurable function I follows
now by observing that the Lp norms of I and III are the same, and that
g== Uf implies Igi == U{lfl}.
It has been proved in particular that if I(x) is a function of L 2 (X, fl),
then g(x)==Uf(x)==/(Tx) is also a function of L 2 (X, fl), and IlgI12==11/112.
In other words, U is an isometric transformation of the Hilbert space
L 2 into L 2 (cf. sec. 54 for the definition and main properties of an
isometric transformation). If T is invertible, Le., if there exists a
measure preserving 5 such that ST==TS is the unit transformation in
X, and if V is the isometric transformation in L 2 corresponding to 5,
then VU == UV is the unit transformation in L2. It follows that in this
case the range of U is the whole of L2, that is, U is a unitary transfor-
mation in L2, and V is the inverse transformation. The observation
that an invertible measure preserving transformation in X induces a
unitary transformation in the corresponding Hilbert space L 2 (X, fl) is
due to B. O. KOOPMAN (1931, [1 J).
The question raised at the beginning of this section is thus reduced,

Ch. 14,  60J
MEAN ERGODIC THEOREI
415
at least so far as it concerns functions fEL2, to the problem of existence
of the limit (as n-+oo) of n- 1 : Ukf(x), where U is an isometric
transformation in the Hilbert space L 2 (X, #). We will even consider a
slightly more geIleral class of linear transformations in L2, usually
called contractions. If the bounded linear transformation U (on the
complex Hilbert space H into itself) satisfies IIUII 1 (i.e., IIUfllllfll for
all fEH), then U is said to be a contraction. Obviously, any isometric
transformation is a contraction. Now, in a Hilbert space such as
L 2 (X, #) it is not only the pointwise convergence which is of importance
but also the convergence with respect to the norn1 (sometimes called
mean convergence), and for that reason the following theorem, due to
J. VON NEUMANN (1932, [3J) for the special case of unitary transfor-
mations, contains the answer to one of the main questions.
THEOREM 2 (l\iEAN ERGODIC THEOREM). If U is a contraction in an
arbitrary (complex) Hilbert space H, and if P is the orthogonal projection
on the closed linear subspace of all fEH satisfying Uf== f, then n- 1 : Ukf
converges (as n-+oo) to Pf for all fEH.
PROOF. It is perhaps appropriate to begin with the precise statement
of what is meant by the orthogonal projection on a closed linear sub-
space M of H (although it is exactly what it is intuitively felt to be).
If the closed linear subspaces M and M.1. are orthogonal complements,
each fEH has a unique decomposition f==f1 +f2 with f1EM, f2 EM .1.,
so that, if we set f1 ==Pf, the linear transformation P satisfies P2==P
and IIPflllltll. This shows that P is an orthogonal projection (cf. sec. 54),
and P is called the orthogonal projection on the subspace M. In other
words, the orthogonal projection on M is the orthogonal projection
having M as range.
In order to prove the theorem, we note first that Uf==f if and only
if U*f==f, where U* is the adjoint transformation of U (cf. sec. 54).
Observe that IIU*II==IIUIIl, and assume that Uf==f. Then
0IIU*f-fI12==IIU*fI12_(U*f, f)-(f, U*f)+llfI1 2
==IIU*fI1 2 -(f, Uf)-(Uf, f)+llfI1 2
== II U*fI1 2 - (f, f) - (f, f) + Ilf11 2 == II U*fl!2_lltI120,
so U*f==f. Similarly, U*f==f implies Uf==f.

416
ERGODIC THEORY
[Ch. 14,  61
Let now P be the orthogonal projection on the closed linear sub-
space M of all elements 1 satisfying UI==I. For any IEM the sums
n- 1 :A Ukl are all equal to I, and so converge to I==PI. For any 1 of
the special form I==-g- U g for some gEH, we have :A Ukl==g- ung,
so Iln- 1 :A Uk/ll21Igll/n0. The collection of these I==g- Ug is
linear, but not necessarily closed. Let 1 be an element in the closure K
of the collection (i.e., there is a sequence Ip==gp-Ug p tending to 1 as
Poo), and let An==n- 1 :A Uk. Since IIAnlll for all n, we have
IIA n/llIIA n(/- 1 p) II + IIA nl pll 11/- 1 pll + IIA nl pll,
so that, given e>O, we may first select p such that II/-Ipll<!e, and
then n such that IIAnlpll<!e. It follows that n- 1 :A Ukl converges
to the null element for every IEK.
The proof will be complete if we can show that the orthogonal
complement K 1- is exactly the subspace M of alii satisfying UI==I,
since it will follow then that every IEH is of the form 1==11 +12 with
/lEM, 12EK, so that n- 1 :A Ukl converges to 11+ 0 ==/1==PI. The
proof is easy: we have hEK 1.. if and only if (h, g- Ug)==O for all g
<?(h, g) - (U*h, g) ==O<?(h- U*h, g) ==0 for all g<?h== U*h
<=>h==Uh by what was noted above <?hEM.
The present proof is due to F. RIESZ and B. SZ.-NAGY (1943, [2J).
61. Individual Ergodic Theorem
The mean ergodic theorem proved in the preceding section implies,
in particular, that if T is a measure preserving transformation of X
into itself, and U is the norm preserving linear transformation of
L 2 (X, #) into itself defined by (U I) (x) == I(Tx) for all IEL2, then an(/; x)
==n- 1 :A (Ukl)(x) converges in the L 2 norm to a function I*(x) EL2.
It follows then that, for n1any subsequences ank(/; x), the subsequence
Gnk(/; x) converges pointwise to I*(x) almost everywhere on X (as k-+oo).
The original version of the individual ergodic theorem, due to G. D.
BIRKHOFF (1931, [lJ) asserts now that for any l(x)EL1(X, #) the
sequence an(/; x) itself, and not merely some subsequence, converges
pointwise to a limit function I*(x) almost everywhere on X. The first
proof for invertible measure preserving transformations has been simpli-

Ch. 14,  61J
INDIVIDUAL ERGODIC THEOREM
417
fied and the theorem itself has been considerably extended. In order to
indicate already into which direction the extensions go, note that if T
is a measure preserving transformation of X into itself, and U is the
corresponding linear transformation of L 1 (X, #) into itself defined by
(Uf)(x)==f(Tx) for all fELl, then UrO for f>O (i.e., f(x»O for almost
all x implies that (Uf)(x»O for almost all x) and IIUf111==llf111 for all
fELl' Furthermore, we have IIUflloo===llflloo for every fEL1nLoo. Instead
of considering now only transformations U corresponding thus to
measure preserving transformatjons T, we shall more generally con-
sider a bounded linear transformation A of L 1 (X, #) into itself such
that
Af>O
IIAf111llf111
IIAflloollflloo
for OfELI,
for all fELl,
for all fEL1nLoo.
Any A of this kind will be called a positive strong contraction in L I .
If A satisfies only the first two conditions, then A is called a positive
contraction in L 1 .
In what follows, the following fixed notations will be used:
sn(f; x) == f(x) + (Af) (x) + . . . + (A n-1f) (x),
an(f; x) ==n-1sn(f; x).
First we make some general remarks. Given fELl, the functions
A kf, k== 1, 2, . . " are all in L1, and hence all an(f; x) are in L 1 . It
follows that all sets X n=={x: an(f; x) *O} are of a-finite measure, and
so X'==;;=lXn is of a-finite measure. Since all an(f;x) vanish on
X-X', the convergence of the sequence an(f; x) on X-X' is obvious.
Hence, in order to prove that an(f; x) converges, we may assume with-
out loss of generality that X is of a-finite measure. From here on, we
will make this assumption.
For the second remark, denote /xfd# by J(f). Since, for f>O, we
have Af>O as well as J(Af).f(f), the linear functional cf{f)==J(Af)
is an J-absolutely continuous integral on L 1 such that Ocf(f)J(f)
for fO. Hence, by the Radon-Nikodym theorem, there exists a #-
measurable function fo on X such that Ofo(x)l and cf{f)==J{Af)==
/ ffod# for all fELl. Note that fo is uniquely determined by A, but A

418
ERGODIC THEORY
[Ch. 14,  61
is not uniquely determined by fo, unless in some trivial cases. If A
corresponds to a measure preserving T, i.e., if (Af) (x) ===f(Tx) , we have
/(f)===J{Af)== ff(Tx)dfl=== ffdfl
for every fELl, and hence fo(x)=== 1 on X. Returning to the more general
case, we will prove the following simple lemma concerning the function
fo.
LEMMA CX. Let A be a positive contraction in L 1 .
(1) Any to in L1, satisfying IIAfI11==llfI11, vanishs outside the set
{x: fo(x) == I}.
(2) If there exists a function gEL! invariant under A, i.e., Ag==g,
then g vanishes outside the set {x: fo(x) == I}.
(3) If there exists a function pEL1 such that P(x) >0 on X and IIAPII!
===IIPI11, then fo(x)== 1 on X, and so IIAf111===lIf111 holds now for every fO
in L 1 .
PROOF. (1) Let 0fEL1 and II A fI11===I!fI11, Le., f(Af)dfl===ffdfl. It
follows that /(f)===.f(f) for this f, i.e., f ffodfl===f fdfl. If f>O on a set
of positive measure disjoint to the set {x: fo(x)== I}, then we would
have f ffodfl<f fdfl. Hence f==O outside {x: fo(x)=== I}.
(2) Let Ag=g for some gEL!. Then the real and imaginary parts of
g are also invariant under A, and so we may assume that g is real. It
follows then from
g++g-==g==Ag==Ag++Ag-
that g+Ag+ holds pointwise. On the other hand we have f (Ag+) dfl
f g+dfl. Hence, Ag+==g+ holds pointwise. We may assume, therefore,
that gO on X. But then IIAgll!===llgI11 since g is invariant, and so the
result in part (1) above shows that g==O outside {x: fo(x) == I}.
(3) Follows immediately from part (1).
The next theorem belongs to the class of so-called maximal ergodic
theorems. The simple proof presented here, much simpler than earlier
proofs, is due to A. M. GARSIA (1965, [lJ), and is based upon the proof
of a related theorem by G. BAXTER (1964, [lJ).
THEOREM 1 (MAXIMAL ERGODIC THEOREM). If A is a positive con-
traction in L1, if fELl is real and E is the set of all points XEX where one

Ch. 14,  61J
INDIVIDUAL ERGODIC THEOREM
419
at least of the sums s n (f; x) is positive, the1
f fdfl?;:-O.
E
PROOF. We may assume that f(x) and all (Akf)(x) are finite every-
where on X. If P is a positive integer, and Ep is the set of all points x
where one at least of the sums sn(f; x) with np is positive, then
E p i E as p-+oo, so lim JEpfdfl= JEfdfl, and hence it will be sufficient
to prove that JEpfdfl?;:-O.
For this purpose, we set
t(x)= max s (f; x),
lnp
so Ep={x: t(x) >O}. Since f?;:-g implies Af?;:Ag, we have
f+At?;:-f+As n ===Sn+1 for n== 1, . . ., p,
and obviously also for n=O, so it follows now from f+Atsn (n=
1, . . " p ) that
(f+At)+?::-s; n== 1, . . " p.
This shows that (f+At) (x) >0 for xEE p , so
(1)
(f+At) (x) = (f+At)+(x)
on Ep,
and hence it follows from (1) that
(f+ At) (x)?::-s (x) for xEE p and n== 1, . . . , p.
But then (f+At) (x)?;:-t(x) on Ep, i.e.,
f(x)?;:-t(x) - (At) (x)
on E p.
Observing now that, for any x not in Ep, we have
t(x) - (At) (x) = - (At) (x)O,
the desired result follows from
f fdfl?;:- f (t-At) dfl?;:- f (t-At) dfl== Il t I11-II At I11?;:-0.
E p E p X
THEOREM 2. If A is now a positive strong contraction in L1, if b>O,
if fELl is real and Y is the set of all points XEX where an(f; x) >b for

420
ERGODIC THEORY
[Ch. 14,  61
at least one 01 the an(/; x), then
#(Y)b-1 fill d#.
x
PROOF. Let ZcY, #(Z)<oo, and set g(x)==/(x)-bxz(x), so gEL 1 .
For any fixed n and almost every XE Y, we have then that
n-l
sn(g; x) ==  {(A kl) (x) -b(A k XZ ) (x)}
k=O
n-l
  {(Akl)(x)-bXY(x)}.
k=O
Note that it has been used here that A is a strong contraction and not
merely a contraction. Now, for every point xEY, there exists an index
n==n(x) such that
n-l
 {(Akl)(x)-bXY(x)}>O,
k=O
and so it follows that Y, and hence Z, is contained except for a null
set in the set Y' of all points x where sn(g; x) >0 for at least one n.
Since ly,gd#O by the maximal ergodic theorem, we obtain
I Y' (/-bxz) d#O, so
b#(Z) f Id#f III d#.
Y' x
The set Y being of a-finite measure, there exists a sequence of subsets
Z key of finite measure such that Z k t Y, so it follows now from #(Z k) t
#(Y) that #(Y)b-1 Ix III d#.
THEOREM 3. Under the same hypotheses as in the preceding theorem,
the lunction 5(1; x) ==supn lan(/; x) I satislies
#{x: 5(1; x) >b}2b-1 fill d#,
x
and hence 5 (I; x) is linite almost everywhere on X.
PROOF. The set {x: 5(1; x) >b} is included in the union of the sets
{x: an(/; x) >b for at least one n} and {x: an( -I; x) >b for at least one
n}.

Ch. 14,  61J
INDIVIDUAL ERGODIC THEOREM
421
Before proving the next theorem, we introduce some additional fixed
notations. For any real fELl, we set
gl(f; x) ==lim sup an(f; x),
g2(f; x) ===lim inf an(f; x).
From the last theorem it follows immediately that gl and g2 are finite
almost everywhere on X. It is evident that pointwise convergence of
an(f; x) almost everywhere on X (as noo) is equivalent to gl(X)===
g2(X) holding almost everywhere on X.
It is not difficult now to prove pointwise convergence almost every-
where of an(f; x) in the case that A corresponds to a measure pre-
serving transformation T in X.
THEOREM 4 (INDIVIDUAL ERGODIC THEOREM FOR A MEASURE PRE-
SERVING TRANSFORMATION; G. D. BIRKHOFF, 1931 [lJ). Let T be a
measure preserving transformation of X into itself, and let (Af) (x)==f(Tx)
for all fELl' Then an(f; x) converges almost everywhere on X to a function
f*EL 1 , and f* is invariant under A, i.e., f*(Tx)===f*(x) holds almost
everywhere on X.
PROOF. As observed earlier, we may assume that X is of a-finite
measure. Furthermore, it is sufficient to present the convergence proof
for f finite and non-negative on X.
Under these assumptions, let a and b be positive numbers satisfying
O<a<b, and let Y === Y(a, b) be the set of all points x where
g2(X) <a<b<gl(X).
I t is easily verified that
lim sup an(Af; x)==lim sup an(f; x),
1.e.,
lim sup an(f; Tx) ==lim sup an(f; x),
and similarly for liminf. Hence, XEY if and only if TXEY. In other
words, Y is invariant under T. This has the important consequence
that in any application of the maximal ergodic theorem or corollaries.
thereof we may substitute Y for X (i.e., if the real function g is

422
ERGODIC THEORY
[Ch. 14,  61
summable over Y, and if E is the subset of Y on which one at least
of the an(g; x) is positive, thenfEgdfl>O).
Applying Theorem 2 (with Y substituted for X), we immediately
obtain that fl(Y)b-1 fy fdfl, i.e., fy{f(x) -b}dfl>O. Similarly, since
an(axy-f; x)==axy(x)-an(f; x)
is positive at each point XE Y for at least one value of n, another appli-
cation of the maximal ergodic theorem shows that fy{a-f(x)}dfl>O.
By addition we obtainfy(a-b)dfl>O, so that fl(Y) ==0 in view of a<b.
Since, therefore, fl{Y(a, b)}==O for all pairs of rational numbers a, b
satisfying O<a<b, we have gl==g2 almost everywhere on X, and so
an(f; x) converges to a limit function f*(x) almost everywhere on X.
The summability of f* for any, not necessarily positive, fELl follows
by observing that
n-l n-l
J lanl dfl==n- 1 J I  A kfl dfln-1  J IA kfl dfl== J If I dfl,
o 0
so that by Fatou's lemma
J If*1 dfllim inf J lanl dflJ If I dfl.
Finally, it follows from the already established equality
lim sup an(f; Tx)=lim sup an(f; x)
for non-negative fELl that f*(Tx)==f*(x) for any non-negative fELl,
and hence f*(Tx)==f*(x) for any fELl.
We return to the more general case that A is a positive strong
contraction in L1, not necessarily derived from a measure preserving
transformation in X. We recall that Li r ) is the set of all real fELl'
THEOREM 5. If D is a norm dense subset of Li r ) such that an(f; x)
converges pointwise almost everywhere on X for every fED, then a n(f; x)
converges almost everywhere on X for every fELl.
PROOF. We may restrict ourselves to real fELl. Writing
p(f; x) ==lim sup lam(f; x) -an(f; x) I
for fELl,
m, n---+oo

Ch. 14,  61J
INDIVIDUAL ERGODIC THEOREM
423
we have p(f; x) ==0 almost everywhere on X for all fED. Assume now
that, for some real fELl, the function p(f; x) is not identically zero
almost everywhere on X. Then there exist positive numbers e, y such
that
#{x: p(f; x) >e}>y >0.
Now, let flED satisfy f If-f1Id#<i e y. For XE{X: p(f; x»e} we have
e<p(f; x)==p(f-f1; x)25(f-f1; x),
where, as before, 5(f; x)==suPnlan(f; x)l. Hence
{x: p(f; x»e}c{x: 5(f-f1; x»!e}.
Comparing the measures of these sets, we obtain by Theorem 3 that
yft{X:S(f-h;x»!e} ; fll-h,dft
4
< -'ley == ly.
e
Contradiction.
The following simple lemmas will be useful.
LEMMA p. If 0f1f2'" t fgEL1, then 0Af1Af2... t Af,
where the arrows denote pointwise convergence almost everywhere on X.
Similarly for decreasing sequences.
PROOF. Since f (f-fn)d#!O by the theorem on dominated con-
vergence and since IIAIIl, we havef(Af-Afn)d#!O, and so there is
pointwise convergence to Af for a subsequence of Afn. But then, since
0(Af1)(X)(Af2)(X). .. at almost each x, the sequence Afn itself
converges pointwise to Af.
LEMMA y. Assume that Ag==g for the function 0gEL1, i.e., assume
that g is invariant under A. Then any 0fEL1, with the properties that
f(x) ==0 where g(x) ==0 and (Af)(x)f(x) holds pointwise, is also invariant
under A.
PROOF. Assume first that Of(x)kg(x) for some constant k>O.
Then
Okg-fkg-Af==A (kg-f),

424
ERGODIC THEORY
[Ch. 14,  61
so
f (kg-I) dflf {A (kg-/)}dfl.
The inverse inequality holds as well, so both integrals are equal, and
it follows that f (Af) dfl== f fdfl. Since A/I pointwise, this implies that
AI==f pointwise. If there is no such constant k, we introduce the
functions fn==min (f, ng) for n== 1, 2, . .. . Then Ofn i f pointwise, so
Afn i Af pointwise by the preceding lemma. Since AfnAff as well
as AfnnAg==ng hold pointwise, we have Afnmin(f, ng)==fn, and so
Afn==fn in view of what has already been proved. Hence we have now
that fn i I as well as fn i Af, and so Af==f.
LEMMA €5. Let fELl be real and bounded, and denote, for the fixed
positive integer p, the lunction ap(f; x) temporarily by g. Then
lim sup an(g; x) ==lim sup an(f; x),
n
n
lim inf an(g; x) ==lim inf an(f; x)
n n
holds almost everywhere on X.
PROOF. For n large compared to p, we have
1 f 2 Af p-l Ap- 2 f
an(g)==-.-+_o_+...+ · +
p n p n p n
Ap- 1 f+...+An- 1 f p-1 Anf 1 An+p- 2 f
+ +. +...+-.
n p n p n
f+Af+. . . +A n-1f
n
p-l
P
f
p-2 Af
P
'--
n
n
1 Ap- 2 f p-1 A nf 1 A n+ p -2f
--' + . +...+-. ,
P n P n P n
from which the desired result follows immediately by observing that
If(x)IM for all x implies that I(Akf)(x)IM holds for all k and almost
all x in view of the condition that IIAflloollflloo'
We will now present a reasonably simple proof of the individual
ergodic theorem for the case that fl(X) is finite.

Ch. 14,  61J
INDIVIDUAL ERGODIC THEOREM
425
THEOREM 6 (INDIVIDUAL ERGODIC THEOREM FOR A FINITE MEASURE).
Let #(X) be finite, and let A be a positive strong contraction in L 1 (X, #).
Then, given fELl, the sequence
n-l
an(f; x) ==n- 1  (A kf) (x)
k=O
converges pointwise almost everywhere on X to a function f*(x) ELI satis-
fying Af*==f*. Furthermore, an(f; x) converges also in norm to f*, i.e.,
f It*-anl d#O as noo.
PROOF. In order to show that gl(f; x) ==g2(f; x) holds for every real
fELl, it is sufficient by Theorem 5 to present the proof for f real and
bounded, and hence it is also sufficient to consider only the case that
f is non-negative and bounded. Hence, let fELl satisfy O<J(x)M on X.
(ex) We note first that O(Akf)(x)M for all k and almost all x, so
Oan(f; x)M for all n and almost all x, and hence
0g2(X)gl(X)M
almost everywhere on X. It follows that gl ELI and g2EL1; we use here
that #(X) <00.
Next, we observe that Pn(X)==SUP1nn am(f; x) satisfies Pn!gl,
and so APn!Ag1 by Lemma fJ. But APn>Aa 1n for all m>n, so APn?::-
sUPmn Aa m . It follows that
Ag 1 == lim APn > lim sup Aa m
noo mn
= lim sup AO'n = limsu p ( n:l O'n+1 -  )=gl'
i.e., Ag1gl holds pointwise almost everywhere on X. Similarly, Ag2g2
holds pointwise almost everywhere. It is immediately derived from
Ag 1 ?::-gl and! (Ag 1 ) d#! gl d# that Ag 1 ==gl, and hence it follows from
Ag2g2 by means of Lemma y that Ag 2 ==g2. In other words, gl and g2
are invariant under A.
(fJ) It was observed already that
Pn(x) ==sup am(f; x) !gl(X)
mn
pointwise, and so !FPnd#! !Fg1d# for every subset Fe X. In particular,

426
ERGODIC THEORY
[Ch. 14,  61
jf E=={x: gl(X»O} and Ee==X-E, then JEcpndfl!O. Hence, given e
such that O<e< 1, there exists an index no such that JEcpnodfl<e, and
so JEcano(f; x) dfl<e. On account of Lemma €5 we may assume therefore,
in the proof of gl==g2, that JEcfdfl<e.
Under this assumption, we consider the fUIlction f+eg1-g1, and we
note that
a n (f+eg1-g1) ==an(f) - (l-e)gl'
For any xEE, there is an index n==n(x) such that an(f; x) >(l-e)gl(x),
so an(f+egl-g1; x) >0. It follows that E is included in the set E* of
all points x where an(f+eg1-g1) >0 for at least one value of n. The
maximal ergodic theorem shows that
1 (f+eg1 -gl) dflO,
E*
.
I.e.,
1 fdfl>(l-e)1 gl d fl.
E* E
By the assumption that JEcfdfl<e we obtain
1 fdfl+e>{l-e)1 gl d fl,
E E
so
(l-e)1 gldfl-e<1 fdfl.
E E
(2)
Similarly,
a n (g2+ e g1-f)==g2+ e g1- a n(f),
and for any XEE this is positive for at least one n, whereas for XEEC
it is always non-positive. By the maximal ergodic theorem we have
1 (g2+ e g1 - f) dfl>O,
E
1. e.,
1 fdfl1 g2 d fl+ e 1 gldfl.
E E E
(3)
Comparing (2) and (3), we obtain
(1-2e)1 gldfl-e1 g2 d fl.
E E
(4)
Note that (4) is independent of the additional assumption that JEcfdfl
<e. Hence (4) holds for every e>O so JEg1dflJEg2dfl. Since gl>g2

Ch. 14,  61 ]
INDIVIDUAL ERGODIC THEOREM
427
holds pointwise and gl==g2==0 on Ee, the desired result that gl==g2
holds follows now. Denoting gl ==g2 by f*, we have proved, therefore,
that an(f; x) converges pointwise almost everywhere on X to f*(x). The
invariance of gl (and g2) under A implies that Af*==f*, and! If*-anld#
-+0 follows from the dominated convergence theorem. Note, however,
that these last results have been proved thus only for bounded f.
(y) If fELl is arbitrary (i.e., real or complex, bounded or un-
bounded), we note first that
n-l
flanld#n-1  fIAkfld#flfld#,
k=O
so that by Fatou's lemma f If*1 d#lim inf f lanl d#f If I d#. This shows
that f* ELI. Now, let f be unbounded. Given e > 0, there exists a bounded
gEL1 such that Ilf-gl11 <e, and so
n-l
Ilan-f*111lln-1  (Akf-Ak g )111 +
o
n-l
+lln- 1  Akg_g*111 +llg*-f*111
o
n-l
llf-gI11 +lln- 1  Ak g _g*111 +llg-fI11
o
n-l
lln-1  Akg_g*111+2e
o
for all n, so that, choosing no such that Iln- 1 -l Ak g _ g *111 <e for
n'?-no, we obtain Ila n -f*111 <3e for n'?-no, i.e., f lan-f*1 d#-+O as n-+oo.
Finally, if f is unbounded and gn is a sequence of bounded functions
in L 1 such that Ilf-gnI11-+0, we have Ilf*-g,:111 O, and so IIAf*-Ag:111
O. But Ag==g: in view of what has already been proved, so we have
1IAf*-g:111-+0 as well as Ilf*-g:lll-+O, which implies that Af*==f*. This
completes the proof.
We extend the theorem to the case that #(X) ==00. Given the subset
SeX such that #(5)<00, the projection operator Ps is defined on
L 1 (X, #) by (Psf)(x)==xs(x)f(x) for all fEL 1 (X, #). It is evident that
if A is a positive strong contraction in L 1 (X, #), then As==PsAPs is

428
ERGODIC THEORY
[Ch. 14,  61
a positive strong contraction in L1(5, fl). Furthermore, given 0fEL1,
we have OAsfAf pointwise almost everywhere. For any fELl we
will wrjte
an,s(f; x) ==n- 1 {(P s f) (x) + (A sf) (x) + . . . + (A-lf) (x)} (5)
for n== 1, 2, . . ., and by Theorem 6 the sequence an,s(f; x) converges
pointwise almost everywhere to a functio:J;1l(x) ELI, vanishing outside
5 and such that Asf;=== f;. The function f is invariant not only under
As, but also under A. For the proof we may assume that f'?::-O, and
hence f;'?::-O. It follows then that Of;===Asf;Af; holds pointwise. On
the other hand we have f (AI;) dflf I;dfl, and so Af;=== I;.
For what follows, note also that Oan,s(f; x)an(f; x) for any
0fEL1.
THEOREM 7 (INDIVIDUAL ERGODIC THEOREM). Let A be a positive
strong contraction in L 1 (X, fl). Then, given fELl, the sequence
n-l
an(f; x) ===n- 1  (A kf) (x)
k=O
converges pointwise almost everywhere on X to a function f*(x) ELI satis-
fying A f* === f* .
PROOF. We may assume that the given function fELl is non-nega-
tive. Then, given e>O, there exists (in view of Theorem 3) a set 5===5 8
of finite measure such that O(]n(f; x)e for all n and all x outside 5.
The functions an,s(f; x) are now defined by (5), and we will prove that
Oan(f; x) -an,s(/; x)e (6)
holds for all n and almost every x. This holds obviously for any x
outside 5, and for n=== 1 and XES we have
a1(f; x) -a1,s(f; x) == f(x) - (Psi) (x) === f(x) - f(x) ===0.
Assuming that (6) holds for a certain value of n, we obtain by means
of the condition IIAflloollflloo that
OA{an(f; x)-an,s(f; x)}e,
so
0[Af+A2f+. .. +AnjJ-[APsf+A(PsA)Psf+...
+A (PsA)n-1PsfJne

Ch. 14,  61 ]
INDIVIDUAL ERGODIC THEOREM
429
holds pointwise almost everywhere on X. By addition of f in both the
square brackets, and assuming that XES (for X outside S it was already
observed above that (6) holds for all n), we obtain the desired result
for n+ 1 upon dividing by n+ 1. Now, letting n-+oo in (6), and ob-
serving that an,s(f; x) tends to f;(x), we derive that
lim sup an(f; x) -lim inf an(f; x)2e
for almost every x. This holds for every e>O, so j*(x)==lim an(f; x)
exists pointwise almost everywhere on X. The proof that f* ELI holds
is the same as in Theorem 6.
It remains to prove that Aj*==j*. With the same notations as before,
we derive from (6) that Of*(x)-f;(x)e on X, and hence
O(Af*) (x) - (Af;) (x)e
on X. But Af;==f;, and so
If*(x) - (Af*) (x) Ie
on X. The last inequality holds for every e>O; it follows that Af*==f*.
We finally prove one more theorem for the case that #(X) is finite
and A corresponds to a measure preserving transformation.
THEOREM 8. If #(X) is finite and A corresponds to a measure pre-
serving transformation T, and if f*(x)==lim an(f; x) for some fELl, then
f f* d#== f fd#.
PROOF. Let p be an arbitrary integer and n a positive integer. The
set
X(P, n)=={x: PI2nf*(x)«p+1)/2n}
is invariant, and for XEX(P, n) and e>O one at least of the sums
: {(Akf)(x)-(PI2n-e)} is positive, so
f fd#>(PI2n-e)#{X(p, n)}
X(p,n)
by the maximal ergodic theorem, and hence
f fd#>(PI2 n )#{X(P, n)}.
X(p,n)

430
ERGODIC THEORY
[Ch. 14,  61
There is a similar inequality in the converse direction, so
Lp,{X(p,n)} f fdP, p+l p,{X(p,n)}.
2 n 2 n
X(p,n)
Obviously, the same inequalities hold for f*, so
1 f 1
- - #{X(P, n)}  (f-f*) d# - #{X(P, n)},
2 n 2 n
X(p,n)
and summation over p shows that Ilx (f-f*) d#I2-n#(X). This holds
for every n, and it follows that I f* d#== I fd#.
Exercises
A COUNTEREXAMPLE
61.1) Show that if #(X)=oo and A corresponds with a measure
preserving transformatiol). T, then I f* d#== I fd# and I If* -ani d#O
need not hold.
INDIVIDUAL ERGODIC THEOREM IN Lp
61.2) If #(X) is finite, te individual ergodic theorem for measure
preserving transformations can immediately be extended to functions
fELp(X, #) for 1 <P<oo. More precisely, show that if #(X) <00, T is
measure preserving and fEL p for some p satisfying 1 <p<oo, then
an(f; x) converges to a limit function f*EL p almost everywhere on X,
and lim Ilf* -anllp=O.
THE CONTINUOUS CASE
61.3) Let # be a a-finite measure in X, and for every non-negative
value of t let Tt be a measure preserving transformation of X into
itself such that T t1 + t2 == Tt1Tt2 for all t1, t2?;;0 (hence, To is the unit
transformation on X). Denote Lebesgue measure in the half line Ri+)==
{t: t>O} by tn, and write the Lebesgue integral It f(u) dm as It f(u) duo
Finally, denote by v the product measure of # and m in X X Ri + ). Let

Ch.14,62J
ERGODIC TRANSFORMATIONS
431
j(X)EL1(X,#) and assume that f(Ttx) is v-measurable on XxRi+).
Show that at(x) ==t- 1 fJ f(T u X ) du is #-summable for every t >0, and
converges (as t-+oo) to a #-summable limit function f*(x) almost every-
where on X. Show also that if #(X) is finite, then f f* d#== f fd# and
f If*-atld#-+O as t-+oo.
62. Ergodic Transformations
We begin by recalling some definitions. If T is a measure preserving
transformation of X into itself, the measurable subset E of X is called
invariant if T-1(E) and E are almost equal, and the measurable
function f(x) is called invariant if f(Tx)=f(x) holds almost everywhere.
Since XT-l(E)(X) ==XE(Tx) , it is evident that the set E is invariant if and
only if its characteristic function is so. Any measure preserving transfor-
mation with the property that every measurable invariant set E satis-
fies either #(E) ==0 or #(X -E) ==0 is called ergodic or metrically transi-
tive. In other words, T is said to be ergodic if X is not the union of
two disjoint invariant sets of positive measure.
THEOREM 1. The measure preserving transformation T is ergodic if
and only if every measurable invariant function is (almost everywhere) a
constant.
PROOF. If the functions almost equal to a constant are the only in-
variant functions, then any measurable invariant characteristic function
is either almost equal to the characteristic function of the whole set X
or almost equal to the characteristic function of the empty set. Hence,
in that case, every measurable invariant set is either almost equal to
the whole set X or to the empty set, i.e., T is ergodic. Conversely, if
-T is ergodic and f(x) is measurable and invariant, we may assume that
f is real, and we then consider the sets
X(p, n)=={x: pj2nf(x)«P+ 1)j2 n },
where p is an arbitrary integer and n a positive integer. The invariance
of f implies that XEX(P, n) if and only if TXEX(P, n) for almost
every x, i.e., X(P, n) is invariant. It follows that #{X(P, n)}==O or
4 U {X-X(P, n)}==O. Similarly, the sets X(-oo, n)=={x: f(x)==-oo} and

432
ERGODIC THEORY
[Ch.14,62
X(oo, n)=={x: f(x)==oo} are invariant. Since
00
X==XC-oo, n)+X(oo, n)+  X(P, n),
p= -00
we may conclude that for each fixed n all sets on the right are of
measure zero, except one, say X(Pn, n), whose complement is of
measure zero. Then the complement of X o == rr=l X(pn, n) is also of
measure zero, and f is constant on Xo. This shows that f is constant
almost everywhere.
COROLLARY. If #(X) is finite, and every summable invariant function
is (almost everywhere) a constant, then T is ergodic.
PROOF. Every measurable characteristic function is now summable
(since #(X) <00), so any measurable invariant characteristic function
is (almost everywhere) a constant. It follows as above that T is ergodic.
THEOREM 2. 1fT is an ergodic measure preserving transformation, if
f(x) is summable, and if .
n-l
f*(x)==lim n- 1  f{Tkx),
k=O
then f*(x) is (almost everywhere) equal to a constant. More precisely: if
#(X)==oo, then f*(x) ==0 almost everywhere on X, and if # (X) <00, then
f*(x)=={#(X)}-l f fd# almost everywhere on X. In other words, if #(X) is
finite, the space mean {#(X)}-l f fd# is almost everywhere equal to the
time mean lim n- 1 : f(Tk x ).
Conversely, if #(X) <00, and if the measure preserving transformation
T has the property that for each summable f(x) the function
n-l
f*(x) ==lim n- 1  f(Tk x )
k=O
is a constant, then T is ergodic.
PROOF. Since Tis n1easure preserving, it follows from the individual
ergodic theorem that
n-l
f*(x)==lim n- 1  f(Tk x )
k=O
is summable and invariant. If in addition T is ergodic, f*(x) is, there-

Ch. 14,  62J
ERGODIC TRANSFORMATIONS
433
fore, equal almost everywhere to a constant. For #(X) ==00 this constant
must be zero, since zero is the only summable constant. For #(X) <00
it follows from fl*d#==fld# that 1*(x)=={#(X)}-l fld# holds almost
everywhere.
Assume now, conversely, that # (X) is finite, T is measure preserving,
and that for each summable I the function
n-l
I*(x) ==lim n- 1  I(Tk x )
k=O
is a constant. In order to show that T is ergodic, it will be sufficient to
prove that every sumn1able invariant function is a constant. Let, there-
fore, I(x) be summable and invariant. Then I(Tkx) == I(x) almost every-
where for k==O, 1, 2, . . " so I*(x) == I (x) holds almost everywhere. Since
1* is constant by hypothesis, it follows that I(x) is constant almost
everywhere.
If U is a bounded linear transformation on the Hilbert space H (into
H itself), and if the element 1-=1-0 of H has the property that UI==AI
for some complex number A, then I is called an eigenelement (also proper
element or characteristic element) of the transformation U, and A is said
to be the corresponding eigenvalue (also proper value or characteristic
value). If the Hilbert space H consists of functions I(x) , one usually
speaks of eigenlunctions instead of eigenelements. Obviously, if Ul1==
All, U 12==A/2, and ex, fJ are complex, then
U( ex /1 +fJ/2) ==A(ex/1 +fJ/2) ;
also, if U I n==AI n for n== 1, 2, . .. and lim I n== I, then U I==AI. This im-
plies that the set of all eigenelements belonging to the san1e eigenvalue
A forms, together with the null element, a closed linear subspace of H,
called the eigenspace of the eigenvalue A. If the eigenspace of A is one-
dimensional, A is called a simPle eigenvalue.
If U is an isometric linear transformation (i. e., if II U III == 1II1I for all
IEH), then every eigenvalue A of U satisfies IAI== 1. Indeed, if UI==AI
for some 1-=1-0, then
IAI'II/II== IIA/II== II UIII== 11111,
so I A I == 1 .

434
ERGODIC THEORY
[Ch. 14,  62
We recall that if T is a measure preserving transformation of X into
itself, then T induces an isometric linear transformation U in the Hil-
bert space L2(X, #) by means of (Uf) (x)==f(Tx). If fEL2 is invariant and
not identically zero, then f(Tx)==f(x) holds almost everywhere, so Uf==
f-=l-O in L 2 (X, #), and this shows that A== 1 is an eigenvalue of U. In the
case that #(X) is finite, any constant is an invariant function in L2, so
that in this case the eigenspace of A== 1 contains all constants.
THEOREl'II 3. If #(X) is finite, the following statements hold.
(a) The measure preserving transformation T is ergodic if and only if
A== 1 is a simPle eigenvalue of the induced isometric transformation U.
(b) If T is ergodic, every eigenvalue of U is simPle, the absolute value
of every corresponding eigenfunction is constant, and the set of all eigen-
values ot U is a group with ordinary multiplication as group operation.
PROOF. (a) If #(X) is finite and T is measure preserving, then T is
ergodic if and only if every invariant function in L 2 is (almost every-
where) equal to a constant. The proof of this is similar to the proof of
the corollary to Theorem 1. On the other hand, the set of all invariant
functions in L2 is exactly the eigenspace of the eigenvalue A== 1 of U.
Indeed, t(Tx) == f(x) is equivalent to U f== f. Hence, T is ergodic if and
only if the constants are the only eigenfunctions of U belonging to the
eigenvalue A== 1. In other words, since the eigenspace of A== 1 contains
at least all constants, T is ergodic if and only if A== 1 is a simple eigen-
value of U.
(b) Let T be ergodic, and let A be an eigenvalue of U (hence IAI == 1).
If f(x) is a corresponding eigenfunction, then f(Tx) ==Af(x) , so
If(Tx) 1 == IAf(x) I == If(x) I,
i.e., If(x) I is invarjant. The ergodicity of T implies then that If(x) 1 is
constant, and since f is not the null element of L2, this constant is not
zero. If g(x) is another eigenfunction corresponding to the same eigen-
value A, then Ig(x) I is a non-zero constant, so f(x)/g(x) makes sense, and
(fIg) (Tx) == f(Tx) Ig(Tx) ==Af(x) lAg (x) == (fIg) (x).
Hence fIg is an invariant function, and so fIg is constant. This shows
that the eigenvalue A. is sinlple. Finally, if Al and A2 are eigenvalues ot'
U with corresponding eigenfunctions f and g, then fIg is an eigen-

Ch. 14,  62J
ERGODIC TRANSFORMATIONS
435
function with eigenvalue A1/A2; it follows that the eigenvalues of U
form a group with ordinary multiplication as the group operation.
We remark that the proof of part (b) is also valid if #(X)==oo. In
that case, however, the results are devoid of interest, since it is easy
to see that now U has no eigenvalues at all. Indeed, if the group of
all eigenvalues is not empty, it contains at least A== 1, so there exists
an invariant function fEL 2 which is not identically zero. Then f is
constant by the ergodicity of T, and this is impossible since L 2 con-
tains no non-zero constant (on account of #(X) ==00)..
ExamPles. (1) If X is the set of all integers with discrete measure,
and Tx==x+a for some fixed integer a, then T is ergodic for a==:t1,
and T is not ergodic for a-=l-:t 1. Indeed, for a==O every subset of X is
invariant, and for a-=l-O and -=I-:t 1 the subset of all multiples of a is
in variant.
(2) If X is real k-dimensional space, # is Lebesgue measure, and T
is a linear transformation with det (T) ==:t 1, then T is invertible and
measure preserving (cf. the examples in sec. 58), but T is not ergodic.
In order to prove this, we shall construct a non-constant invariant
function f(x) on X. Evidently, the set X (consisting of all points x==
(x(l), "', x(k») with X(l), .. . ,x(k) real) is a subset of complex k-di-
mensional space X e (consisting of all points z == (z(l), . . ., Z(k») wi thO
z(l), "', z(k) complex), and this complex space X e may be regarded as
a Hilbert space with inner product (Zl, Z2) == = 1 z)z). The transfor-
mation T may now be extended to Xe in an obvious way (the matrix
of the extended T is sin1ply the same as of the initial T). Then the
adjoint transformation T*, defined by (Tz, w)==(z, T*w) for allz, WEX e ,
is also a linear transformation in X e , and it is easily seen that if mat (T)
== (tij) and mat (T*) == (t i ;), then tij==lji, so det (T*) ==det (T) ==:t 1. In the
present case all numbers tij are real, so tij==tji' The complex number A
is an eigenvalue of T* if and only if T*Z-AZ==O for some z-=l-O, that is,
if and only if det(T*-AE)==O, where E is the unit transformation in
Xe. This equation det(T*-AE)==O, called the characteristic equation
of T*, is an equation of degree k in A; it has therefore k roots, and the
product of these roots is det (T*) == :t 1. We shall denote the different
roots (that is, the different eigenvalues of T*) by AI, . . " Ap, and their

436
ERGODIC THEORY
[Ch.14,62
multiplicities as roots of the characteristic equation by n1, . . " np
(hence, r=1 nr==k). Corresponding to each eigenvalue Aj of T* we
choose an eigenelement Zj, so T*zj==AjZj, and we consider the function
f(z)=== IT 1= 1 I(z, zj)nil. This function satisfies
f(Tz) == IT I (Tz, zj)nil === IT I (z, T*zj)nil
=== IT IAjln i IT I(z, zj)nil===f(z),
since IT IAjln i == I IT Ajil === Idet (T*) 1===1. It follows that I(Tx) === f(x) for all
XEX, that is, f(x) is invariant on X. The function I(x) on X vanishes on
the union of the sets N j === {x: (x, Zj) ===O}, and if zj===xj+iYj (Xj, Yj real
points), we have (x, Zj) ==0 if and only if (x, Xj) === (x, Yj) ==0, so Nr==
{x: (x, Xj) === (x, Yj) ==O} is a proper linear subspace of X (the subspace
is proper since Zj::l=O). Hence I(x) ==0 on a non-empty subset of measure
zero, and f(x)::I=O elsewhere on X. Assuming now that I(x) is almost
everywhere equal to a constant c, we must have, therefore, c::l=O by the
just obtained result. But then, since f(x) is continuous on X, we have
f(x)===c::l=O on the whole of X, contradicting the fact that 1 (x) ===0 on a
non-empty subset. It follows that I(x) is a non-constant invariant
function, and this shows that T is not ergodic.
In order to discuss the next examples, it will be useful to make first
some remarks about orthonormal sequences in Hilbert space. The
sequence f{Jn (n== 1, 2, . . .) of elements of the Hilbert space H is called
orthonormal if (f{Jrn, f{Jn) ===mn for all m, n== 1, 2, ..., where mn is the
Kronecker delta, i.e., mn=== 1 for m==n and mn==O for m::l=n. An ex-
ample is the sequence f{Jn(x) === (2n)-! e inx for n==O, =1= 1, =1=2, . .. in
L 2 (X, #), where X===[O, 2nJ and # is Lebesgue measure. Generally, let
f{Jn (n== 1, 2, . . .) be an orthonormal sequence in the Hilbert space H.
Given the positive integer k and the element fEH, we are interested
in the best approximation of 1 by linear combinations = 1 anf{Jn, i.e.,
we ask for which values of the coefficients aI, . . ., ak the value of
k
11/-  anf{Jnll
1
is minimal. The answer is that the minimal value is obtained for an===-
(I, f{Jn), n== 1, . . " k. Indeed, if aI, ..., ak are arbitrary, and if we de-

Ch.14,62J
ERGODIC TRANSFORMATIONS
437
note (I, qJn) by Cn, then
k
11/-  a n qJnl1 2 == (I, I) -  (I, anqJn) -  (anqJn, I) +   (amqJrn, anqJn)
1
==t/, I)- anCn- anCn+ la n l 2
== (I, I) -  Ic n l2+  (an -Cn) (an -Cn)
k
==11/-  c n qJnl1 2 +  la n -c n I2>11/-  c n qJnl1 2 .
1
Note that the approximation is improved if k is increased. This follows
from
k+l k+l
11/-  c n qJnI1 2 ==11/112-  Ic n l2
1 1
k k
11/112_  Ic n l2== 11/-  c n qJnl1 2 .
1 1
Note also that if aI, a2, . .. is a sequence of complex numbers such
that tk== =1 anqJn satisfies lim Iltkll==O, then a1 ==a2==. . . ==0. Indeed,
if P is a fixed index, then (tk, qJp) ==a p for all k?;;P; on the other hand
(tk, qJp) O as k-+oo in view of I (tk, qJp) 1lltkll'llqJpll== Iltkll. Hence ap==O.
Let now X==[O, 2nJ, # Lebesgue measure, and IEL 2 (X, #). Given
e>O, there is a function g(x), continuous on X, such that 11/-gll<te
(where II/-gil is the L 2 norm of I-g). Furthermore, by the well-known
Weierstrass approximation theorem, there is a trigonometric poly-
nomial
k
P(x) ==  anqJn(X) ,
-k
where qJn(x) == (2n)-! e inx ,
such that Ig(x) -P(x) 1 <ef{2J2n} on X. Then Ilg-PII< ie, so 11/-PII<e.
But since the sequence of all qJn(x) is orthonormal in L2, the best ap-
proximation (in L2) of I(x) by a linear combination of qJo(x), qJ1(X),
qJ-1(X) , . . ., qJk(X), qJ-k(X) is by k cnqJn(x), where C n == (I, qJn). Hence,
writing Sk(X)== k CnqJn(X) , we have II/-skll<e, and then also II/-sjll
<e for all j?;;k (in view of the remark above). In other words, if I(x) ELa

438
ERGODIC THEORY
[Ch. 14,  62
and if
n
S n (x) ==  (I, qJ k) qJ k (x) ,
k=-n
then the sequence sn(x) converges to I(x) with respect to the L 2 norm.
This may be stated in a slightly different way by introducing a new
variable. If X is the set of all complex numbers of absolute value 1,
and if # is 1f2n times Lebesgue measure along the unit circle, then the
sequence of functions qJn(z) ==zn (n==O, =1= 1, =1=2, . . .) is orthonormal in
L 2 (X, #), and for any IEL2 the sequence
n
sn(z) ==  (I, qJk)qJk(Z)
k=-n
satisfies lim 11/-snll==O.
We proceed with the examples.
(3) If X=={z: Izl==l}, if # is 1f2n times Lebesgue measure along X,
if e is a fixed complex number satisfying lei === 1, and Tz==ez, then T is
invertible and measure preserving. We shall prove that T is ergodic if
and only if there is no positive integer n for which e n === 1. Indeed, if
e n == 1 for a positive integer n, then qJn(z) ==zn satisfies qJn(Tz) == (ez) n==
zn==qJn(z) , so qJn(z) is a non-constant invariant function, and this im-
plies that T is not ergodic. Assume now that en -=1-1 for all positive
integers n, and let U be the isometric transformation in L2 induced by
T. Since
U qJk==qJk(Tz) ==qJk(ez) ==ekzk==ekqJk
for all k, the formula
n
lim 11/-  (I, qJk)qJkll==O,
-n
holding for every function IEL2, implies that
n
lim IIUI-  ekt/, qJk)qJkll==O.
-n
Hence, if I is an eigenfunction of U belonging to the eigenvalue A== 1,

Ch. 14,  62J
ERGODIC TRANSFORMATIONS
439
then UI==I, so
n
lim II  { (I, cP k) - e k (I, cP k) }cp k II == O.
-n
But then (I, CPk) -ek(/, CPk) ==0 for k==O, =1= 1, =1=2, ... by the remark
above, so that, in view of e k *l for k*O, we obtain (I, CPk)==O for k*O.
I t follows therefore from lim 11/- n (I, CPk)CPkll ==0 that I(z) == (I, cpo)cpo(z) ,
so I(z) is a constant. Thus every eigenfunction of U belonging to the
eigenvalue A== 1 is seen to be a constant, and this implies that T is
ergodic. The same method shows immediately that the other eigen-
values of U are the numbers en (n== =1= 1, =1=2, . . .) with the constant
multipJes of CPn(z) ==zn as corresponding eigenfunctions.
(4) With X and # as in the preceding example, and TZ==Z2, the
transformation T, although not invertible, is measure preserving (cf.
sec. 58). Writing again CPn(z) ==zn for all integer n, and denoting again
by U the isometric transformation in L2 induced by T, we have UCPk==
CPk(Tz) ==Z2k==CP2k for all k, so
n
lim 11/-  (I, CPk)CPkll==O
-n
implies
n
lim II UI-  (I, CPk)CP2kll==0.
-n
Hence, if UI==AI, then
n n
lim II  (I, cP k) cP 2 k -  A (I, cP k) cP k II == 0,
-n
-n
and it follows easily that (I, CPk) ==0 for k*O, so I(z) is a constant, and
A== 1 if I(z) is not identically zero. This shows that U has A== 1 as a
simple eigenvalue, and there are no other eigenvalues. In particular,
it follows that T is ergodic.
Exercises
A COUNTEREXAMPLE
62.1) Show that the last statement in Theorem 2 is not necessarily
valid if #(X) is infinite.

440
ERGODIC THEORY
[Ch.14,62
ERGODIC TRANSFORMATION IN EUCLIDEAN SPACE
62.2) Show that if X is an open subset of a finite-dimensional Eucli-
dean space such that X is of finite Lebesgue measure, and T is an
ergodic Lebesgue measure preserving transformation of X into itself,
then for almost every XEX the sequence of the points Tnx (n==O, 1,2,
. . .) is dense in X.

CHAPTER 15
NORMED KOTHE SPACES
In this final chapter we resume the investigation (begun in sec. 30) of normed
linear spaces the elements of which are measurable functions. Spaces of this
kind are sometimes called normed Kothe spaces, or also Banach function spaces
if they are norm complete. The spaces Lp are special examples. Several charac-
teristic features of the theory are discussed (Riesz-Fischer property and Fatou
property of the norm; the associate and con jugate spaces; the su bspace of all
functions of absolutely continuous norm; reflexivity conditions). The importance
of some of these features is not always distinctly evident in the Lp case due to,
the special and very simple structure of L p spaces. For a better understanding
of the abstract background we would need several facts from the theory of
normed vector lattices (also called normed Riesz spaces). In the present context,
however, we have avoided any references to this abstract subject. In some of
the sections the discussion is based on the text of a series of notes on Banach
function spaces by W. A. J. LUXEMBURG and the present author [IJ.
63. Function Seminorms and Function Norms; Normed Kothe
Spaces and Banach Function Spaces
In sec. 30 we have defined the general notion of a Banach function
space, and it was shown that Lp spaces (lPoo) are examples of such
spaces. In the present chapter we return to general Banach function
spaces. More precisely, this chapter is devoted to investigating the
properties of certain normed linear spaces the elements of which are
complexvalued measurable functions. If the space is norm complete we
obtain a Banach function space such as in sec. 30. For the particular
case that we have to do with sequence spaces (e.g., when the measure
is the discrete measure on a count ably infinite point set), several of
the notions and results which follow can be traced back already to
work of G. KOTHE and O. TOEPLITZ in the thirties, and for that reason
the normed function spaces which will be introduced in the present
section are sometimes called normed Kothe spaces.

442
NORMED KOTHE SPACES
[Ch. 15,  63
Given the a-finite measure # in the non-empty point set X, we denote
as in earlier sections by M the set of all #-measurable complex functions
on X, and by M+ the subset of M consisting of all fEM such that
f(x)?;:-O on X. Also, just as before, functions in M differing only on a
#-null set are identified. Elements of M will be denoted by f, g, h, ...;
elements of M+ sometimes by u, v, w, . . . . The notation fg will mean
that f and g are real-valued (almost everywhere) and f(x)g(x) almost
everywhere on X.
Assume now that to each 'ltEM+ there corresponds a number p(u)
such that
(a) Op(u)oo (the value +00 is, therefore, admitted),
p(u)==O if u==o (almost everywhere),
p(au) ==ap(u) for every finite constant a?;:-O,
p(u+v)p(u)+p(v) for all u, vEM+.
(b) If U, vEM+ and uv, then p(u)p(v).
The function p(u) thus defined on M+, and mapping M+ into the non-
negative numbers (+00 included, unless p is identically zero), is called a
function seminorm. If the function seminorm p has the additional proper-
ty that p(u) ==0 if and only if u==o (almost everywhere), then p is called a
function norm.
THEOREM 1. If p is a function norm, and UEM+ satisfies p(u)<oo,
then u (x) == 00 holds only on a #-null set.
PROOF. Writing E=={x: u(x)==oo}, we have XEn-1u for n==l, 2,
. . " and so p(XE)n-1p(u) for n== 1, 2, . . . . It follows that p(XE) ==0, so
XE==O almost everywhere.
The set of all function seminorms on M+ is partially ordered in the
natural manner, i.e., P1P2 whenever P1(U)P2(U) for all uEM+. If pI
is a norm and if P1P2, then P2 is a norm. If pn (n== 1, 2, . . .) is a
sequence of function seminorms and an (n== 1, 2, . . .) a sequence of
finite non-negative constants, then p== l anpn, defined by p(u) ==
Ll anpn(u) for every uEM+, is a function seminorm (compare Exer-
cise 30.11). If {p,.} is an arbitrary set of function seminorms, then p==
sup p,., defined by p(u)==SUp,. p,.(u) for every uEM+, is a function semi- ,
norm. Note that p==sup p,. is the least upper bound of the set of all p,.
in the sense of the partial ordering.

Ch. 15,  64J
NORM COMPLETENESS
443
One can extend the domain of definition of any function seminorm
to the whole of M. Since we are mainly interested in norms, we re-
strict ourselves here to function norms. Given the function norm p, we
extend the domain of definition of p to the whole of M by setting p(f) ==
p(lfl) for any fEM, and we denote by L p the set of all fEM satisfying
p(f) <00. In virtue of Theorem 1 any fEL p is finite-valued #-almost
everywhere on X; this guarantees now that L p is a linear subspace of
the linear space of all #-almost everywhere finite functions in M (note
that M itself is no linear space; the addition in M does not obey the
associative law). The function p is a norm on L p , and the norm p has
the additional property that if fEM, gEL p and Ifllgl, then fEL p and
p(f)p(g). Any normed linear space of this kind is sometimes called a
normed Kothe space.
The subset of all non-negative functions in L p will be denoted- by
(L p )+. Evidently, (L p )"+ is the intersection of L p and M+.
64. Norm Completeness
Let L p be a normed Kothe space as defined in the preceding section.
The space L p is not necessarily norm complete, as the following ex-
ample shows.
Example. Let # be the discrete measure in X == {I, 2, . . .} and let,
for every u== (u(l), u(2), . . .) EM+, the function norm p(u) be defined by
00
p(u)==  2- n u(n)+lim sup u(n).
n=l
In order to prove that L p is not norm complete, let VI == (1, 0, 0, . . .),
v2==(I, 1,0,0, .. .), v3==(l, 1, 1,0,0, .. .), and so on. Fork>pwehave
p(Vk-Vp)== +1 2- n -+0 as k, p-+oo, so Vk is a Cauchy sequence. As-
suming the existence of a function fEL p such that p(f-Vk) -+0, we have
p{(f-Vk)XEn}-+O as k-+oo, where XEn is the characteristic function of
the set En consisting of the single point {n}. Hence, since P(XEn)==2- n ,
we have vk(n)-+f(n) as k-+oo for every n. This implies that f(n)== 1 for
every n. It is true that thus obtained f satisfies fEL p , but p{f-Vk)==
r+ 1 2- n + 1 -+ 1 as k-+oo. This contradicts p(f-v k) -+0, and hence L p is
not norm complete.

444
NORMED KOTHE SPACES
[Ch.15,64
We will indicate a condition, necessary and sufficient for norm com-
pleteness of L p .
DEFINITION. The function norm p is said to have the Riesz-Fischer
property if, for any sequence unE(Lp)+ such that 1 p(Un) <00, we have
1 UnELp, i.e., p(l Un) <00.
THEOREM 1. The function norm p has the Riesz-Fischer property if
and only if, for any sequence unE(Lp)+ such that 1 p(U n ) <00, we have
p(l un)l p(u n ).
PROOF. It is evident that if p satisfies the triangle inequality for
infinite sums as stated above, then p has the Riesz-Fischer property.
Conversely, assume that p has the Riesz-Fischer property, but p(l un)
> 1 p(u n ) for some sequence unE(Lp)+ such that 1 p(U n ) <00. Then
there exists a number 8>0 such that p( un) >8+  p(U n ), SO that
after multiplication by appropriate constants we obtain, for every k==
1, 2, . . ., a sequence Unk (n== 1, 2, . . .) in (L p )+ such that
p( Unk) >k+  p(Unk)'
n
n
Subtracting the inequality
p(  Unk)  p(Unk),
n<r n<r
we obtain
p(  Unk) >k+  p(Unk)
nr
nr
for every r== 1, 2, . . ., and for an appropriate r==rk we have nr p(Unk)
<k- 2 . Hence, by reindexing, we obtain for every k== 1, 2, . . . a sequence
Vnk (n== 1, 2, . . .) in (L p )+ such that
 p(Vnk) <k- 2 ,
n
p( Vnk) >k.
n
Let Wn (n== 1, 2, . . .) be the double sequence Vnk (n, k== 1, 2, . . .) ar-
ranged in single order. Then 1 p(wn)< k- 2 <00, so 1 wnE(Lp)+
by the Riesz-Fischer property. But p(l wn»P(n Vnk) >k holds for,
every k== 1, 2, . . " so p(l wn) ==00. Thus a contradiction is obtained.

Ch. 15,  64J
NORM COMPLETENESS
445
THEOREM 2. The normed linear space L p , derived from the function
norm p, is norm comPlete (i.e., L p is a Banach space) if and only it p has
the Riesz-Fischer property.
PROOF. Assume first that p has the Riesz-Fischer property. The
proof that L p is norm complete is similar to earlier proofs (e.g., sec. 30,
Theorem 2). Let fnELp (n== 1,2, ...) and lim p(lm-In) ==0 as m, n-+oo.
There exists a subsequence gn of fn, i.e., gl==ln 1 , g2==ln 2 with n2>n1,
and so on, such that 1 p(gn+1-gn) <00. It follows from the Riesz-
Fischer property that g== 1 Ign+1-gnl satisfies gEL p , and so
Og(x) <00 holds almost everywhere on X. This implies that
1Ign+1(x)-gn(x)1 converges in the classical sense for #-almostevery
XEX, and the same holds then for the series gl(X) + 1 {gn+1(X) -gn(X)}.
Denoting the sum function by I(x), we have I-gp===- :=p {gn+1-gn}
almost everywhere, so p(/-gp); p(gn+1-gn) by the result proved in
the preceding theorem. Hence p(/-gp) -+0 as P-+oo. This shows that
p(/- fn)p(j-gp) +p(gp-fn)-+O
as n (and P) -+00, and so I is the norm limit o{the sequence In.
Conversely, assume that L p is norm complete, and let Un E (L p ) + for
n==1,2, ... and  p(U n ) <00. We have to show that  unEL p . Setting
Sn==U1 +. . . +un for all n, we have p(sm-sn)+l p(Uk) -+0 as m,
n-+oo, and so by the norm completeness there exists a function fEL p
such that p(/-sn) -+0 as n-+oo. Let f===-g+ih (g and h real). Since all
Sn are real, we have I-sn==(g-sn)+ih, so Ihll/-snl for all n. This
implies that p(h) ==0, and so h(x)==O almost everywhere. Hence, we
may assume that f is real. Then f==max(l, O)+min(/, 0)===-1++1-, and
it follows from I/-II/-snl that p(I-)==O, and so f may be assumed
non-negative. For nk we have
ISk-min(/, sk)/==Imin(sn, sk)-min(l, sk)Ilsn-fl,
and since p(sn-f)-+O as n-+oo, it follows that p{sk-min(/, Sk)}==O, so
sk==nlin (I, Sk) holds almost everywhere for k== 1, 2, . .. . But then ISk
holds almost everywhere for k== 1, 2, . . " so j?::- 1 Un. Since p(/) <00,
we conclude that p(l Un) <00, i.e.,  UnELp. This is the desired re-
suit.
The theorems in the present section are due to I. HALPERIN and
W. A. J. LUXEMBURG [lJ.

446
NORMED KOTHE SPACES
[Ch. 15,  65
Exercises
NORM CONVERGENCE AND POINTWISE CONVERGENCE
64.1) Let p be a function norn1 such that L p is norm complete. Show
that if fnELp (n== 1, 2, . . .) and 1 p(f n) <00, then 1 fn(x) converges
pointwise #-almost everywhere on X to a function f(x) EL p , and f==  fn
holds also in L p , i.e., p(f- i f n) -+0 as p-+oo.
65. The Fatou Property
We begin with a definition.
DEFINITION. The function seminorm p is said to have the Fatou proper-
ty whenever it follows from 0U1U2... ju, with all 'UnEM+, that
p(u n ) j p(u).
The function seminorm p is said to have the weak Fatou property when-
ever it follows trom 0U1U2. . . ju, with all unEM+ and lim p(un)
<00, that p(u) <00.
THEOREM 1. If p is a function seminorm with the F atou property,
then p has the weak F atou property.
If p is a function norm with the weak F atou property, then p has the
Riesz-Fischer property, and so L p is then a Banach space (compare sec. 30,
Theorem 2).
PROOF. The first statement is evident. For the proof of the second
statement, assume that p is a function norm with the weak Fatou
property, and let Un EM + for n== 1, 2, . . . and  p( un) < 00. The functions
Sn==U1 + . . . +u n (n== 1, 2, ...) satisfy OSl S2. .. and p(sn)
 p(u n ), so lim p(sn) <00. Hence, since p has the weak Fatou proper-
ty, the function 1 un==lim Sn satisfies p(l un) <00. This shows that
p has the Riesz-Fischer property.
THEOREM 2 (I. AMEMIYA, 1953, [lJ). The function seminorm p has
the weak F atou property if and only if there exists a finite constant k> 1
such that, for every sequence Ounju with all UnEM+ and lim p(un) <00,
we have p(u)k lim p(u n ).

Ch. 15,  65J
THE FATOU PROPERTY
447
PROOF. It is evident that if there exists such a constant k, then p
has the weak Fatou property. Conversely, assume that p has the weak
Fatou property. We will first consider the case that Oun ju with
p(un) ==0 for all n, and we will prove that p(u) ==0 holds in this case (if
p is a function norm, this is evident). Since p has the weak Fatou
property, we have in any case that p(u) <00; assume now that
O<p(u) <00, and set vn==nu n . Then 0V1V2. . . and p(v n ) ==0 for all
n, so the function v==lim Vn satisfies p(v) <00. On the other hand
v?;:-nun?::-pu n for all n?;:-p, so v?;:-pu for every p, and hence p(v)?;:-Pp(u)
for every p== 1, 2, . . " which implies p(v) ==00, and so yields a contra-
diction. Hence, if Ounju with p(u n ) ==0 for all n, then p(u) ==0.
Assume now that there exists no finite constant k?;:-l as asserted.
Then there exists for every p==l, 2, ... a sequence unpju p (as n-+oo)
in M+ such that limn p(u np ) is finite and
p(u p ) >p 3 lim p(unp).
n
(1)
It is impossible that limn p(U np ) ==0 for some p since in view of what
was proved above this would imply p(U p ) ==0, in contradiction to (1).
Hence, multiplying by appropriate constants, we may assume that
lin1n p(u np ) ==P-2, and so p(u p ) >P for every p== 1, 2, . . . . Now, let
Vn==Un1 +U n 2+ . . . +u nn for all n== 1, 2, . .. . Then 0V1 V2. . .,
and since
p(vn)p(Un1) + . . . +p(unn) 1 +2- 2 + . . . +n- 2 ,
we have lim p(vn) n- 2 <00. Hence v==lim V n satisfies p(v) <00. But
v==sup vn?::-supn unp==u p for any fixed p, so p(v)?;:-p(u p ) >P for p==
1, 2, . .. . This yields a contradiction.
The next theorem, by its parallelism to Fatou's lemma for integrals
(of sec. 13, Theorem 8), will show where the name of the Fatou proper-
ty comes from.
THEOREM 3. The function seminorm p has the weak Fatou property if
and only if there exists a finite constant k?;:-l such that, for every sequence
unEM+ with lim inf p(U n ) <00, the function lim inf Un satisfies
p(lim inf un)k lim inf p(u n ).

448
NORMED KOTHE SPACES
[Ch. 15,  6S
T he function seminorm P has the F atou property if and only if this ine-
quality is always satisfied with k == 1.
PROOF. Assume that P has the weak Fatou property, let k be the
finite constant from the preceding theorem, and let UnEM+ (n== 1, 2,
. . .) with lim inf p(un) <00. We set vn==inf(un, U n +1, . . .) for all n==
1, 2, . . . . Then p(vn)P(up) for all p?;:-n, so p(vn)lim inf p(u p ) holds
for every n. Hence, since 0V1V2. . . and lim p(vn)lim inf p(u p ) <00,
the function lim inf un==lim Vn satisfies
p(lim inf un)k lim p(vn)k lim inf p(u n ).
Conversely, if the condition in the present theorem is satisfied, the
particular case of an increasing sequence shows that p has the weak
Fatou property.
If p has the Fatou property, the inequality for increasing sequences
holds with k== 1, and the above proof shows that the inequality for the
more general sequences in the present theorem holds as well with k== 1.
The converse is again evident.
THEOREM 4. If PT' where T runs through an arbitrary index set, is a
set of function seminorms all of which have the Fatou property, then the
function seminorm p==sup PT has the Fatou property.
PROOF. Let Ounju with all unEM+, and let p(un)jex. We have to
show that p(u)==ex. Since ex==lim p(un)P(u), it is sufficient to show
that ex?;:-p(u), Le., ex>{3 for any (3<p(u). Hence, let (3<p(u). Then PT(U)
>(3 for some T, so PT(un) >(3 for this particular T and all n?;:-no, since
PT has the Fatou property. It follows that p(u n ) >(3 for all n?;:-no, and
so ex==lim p(u n ) >{3.
ExamPles. The use of the last theorem is illustrated by the following
examples.
(a) Assume that #(X) ==00, and that # does not possess any atoms;
consequently (by Exercise 10.12) X has plenty of subsets E satisfying
#(E) == 1. For any fEM, let
p(f) ==sup (/ If I d#: #(E) == 1).
E
Evidently p IS a function norm. For any fixed subset EeX with

Ch. 15,  65J
THE FATOU PROPERTY
449
#(E)==l we have that PE(/)/El/ld# is a function seminorm with the
Fatou property, and so p==sup PE has the Fatou property. It follows
tha the corresponding normed linear space L p is norm complete, i.e.,
L p is a Banach space (compare Exercises 30.13-30. 16).
(b) Let # be Lebesgue measure in the closed interval [0, 1J and, for
any IEM, let
h
p(/) ==sup (h- 1 Jill d#: O<h 1).
o
Evidently p is a function norm. For any fixed h such that O<h 1 we
have that Ph (I) ==h- 1 It III d# is a function seminorm with the Fatou
property, and so p==sup Ph has the Fatou property. It follows that the
corresponding normed linear space L p is a Banach space.
All Lp spaces (lPoo) have the Fatou property. For further ex-
amples we refer to the exercises.
Exercises
FUNCTION NORM WITH THE WEAK FATOU PROPERTY BUT WITHOUT THE
FATOU PROPERTY
65.1) Let # be discrete measure in the point set X == {I, 2, . . .} and,
for any uEM+, let
p(U)==sup u(n)+a lim sup u(n),
where a is a positive constant. Show that p is a function norm having
the weak Fatou property but not the Fatou property.
FUNCTION NORM WITH THE RIESZ-FISCHER PROPERTY BUT WITHOUT THE
'\TEAK FATOU PROPERTY
65.2) Let X and # be the same as in the preceding exercise and, for
any uEM+, let p(u)==sup u(n) if u(n)-+O as n-+oo, and p(u)==+oo
otherwise. Show that p is a function norm having the Riesz-Fischer
property but not the weak Fatou property.

450
NORMED KOTHE SPACES
[Ch. 15,  66
66. The Lorentz Seminorm
In the present section it will be shown that, given the function semi-
norm P, there exists a function seminorm PLP such that PL is the
largest seminorm majorized by P and having the Fatou property. The
seminorm PL was introduced by G. G. LORENTZ (unpublished paper)
who proved that PL has the Fatou property. The construction of PL
from P is analogous to the construction of the a-additive part of a
given finitely additive measure (compare Exercise 9.15).
DEFINITION. Let P be a function seminorm. For any uEM+, let
PL(u)==inf (Hm p(un): Oun ju, UnEM+
for n== 1, 2, . . .).
THEOREM 1. PL is a function seminorm such that PLp. If P has the
Fatou property, then PL==p.
PROOF. It is evident that PL(U)p(u) for every uEM+. In order to
show that PL is a function seminorm, it will be sufficient to show that
PL(U+V)PL(U)+PL(V), the other properties being evident. We may as-
sume that PL(U) +PL(V) <00. Then, given e>O, there exist sequences
Oun ju and Ovn j v in M+ such that lim p(u n ) <PL(U) +e and lim p(vn)
<PL(v)+e, so un+vnju+v and limp(un+vn)<PL(u)+PL(v)+2e. It
follows that PL(U+V)PL(U)+PL(V).
Finally, by the definition of the Fatou property, it is evident that
PL==P if P has the Fatou property.
THEOREM 2. PL has the Fatou property, and so (PL)L==PL.
PROOF. We have to show that Ounju in M+ implies that PL(Un)j
PL(U), Since lim PL(Un)PL(U), it will be sufficient to show that PL(U)
lim PL(Un). For that purpose we may assume that lim PL(Un) is finite.
In order to avoid difficulties at the points x where U(x) ==00, we intro-
duce the functions u==min(un, n). Then all u are finite-valued,
Ou ju, and obviously it is sufficient to show that PL(u)lim PL(U),
For every k==l, 2, ... there exists a sequence Ovknjuk (as n-+oo)
in M+ such that
PL(Uk) -k-1p(Vkn)PL(Uk) +k- 1
(1) ,
for all Vkn. Now, since # is a a-finite measure by hypothesis, there ex-

Ch. 15,  66J
THE LORENTZ SEMINORM
451
ists a sequence of sets Xki X with #(X k ) finite for every k. For fixed
k, we have Uk-Vkn! 0 (as n-+oo) on X since Uk is finite-valued, so the
sequence of sets
En=={x: Uk(X)-Vkn(X»k- 1 }nX k
descends to the empty set as n-+oo. This implies that #(En)! 0 since
all En are included in the set X k of finite measure. Hence, for some
n==nk, we have OUk(X) -vkn(x)k-1 on X k except on a subset H k c X k
satisfying #(Hk) <2- k . For brevity, denote V knk by Wk, and note that
the set H==lim sup Hk satisfies #(H) ==0. Given any point xEX-H==
liminf(X-H k ), we have XEXk-H k for all k?::-kx (where kx depends
on x), so
OUk(X) -wk(x)'k-1
for all k?::-kx.
This shows that the sequence Wk converges pointwise almost every-
where (namely, on X-H) to u. By (1) we have lim p(wk)==lim PL(Uk).
Now, for k==l, 2, "', set zk==inf(wk' Wk+1, .. .). Then Ozkiu al-
most everywhere on X, and
p(zk)lim p(Wk+p)== lim PL(U)
for every k.
poo
noo
Hence, by the definitio of PL,
PL(u)lim p(zk)lim PL(U),
This is the desired result, showing that PL has the Fatou property.
Note, for later use, that PL(U) ==lim PL(U), and so OZk i U with p(Zk) i
PL(U),
THEOREM 3. The function seminorm P has the Fatou property if and
only if P==PL.
PROOF. If P has the Fatou property, then P==PL, as observed already
in Theorem 1. Conversely, if P==PL then P has the Fatou property since
PL has the Fatou property.
THEOREM 4. PL is a function norm if and only if P is a function norm.
PROOF. If PL is a norm, it follows immediately from PLP that P is
a norm. Conversely, let P be a norm and let PL(U) ==0 for some uEM+.
As shown in the proof of Theorem 2, there exists a sequence Ozn i U

452
NORMED KOTHE SPACES
[Ch. 15,  66
in M+ such that p(zn) j PL(U), Then p(zn) ==0 for all n, so every Zn is a
null function since P is a norm. It follows then from Ozn ju that u is
a null function. Hence, PL(U) ==0 implies that u is a null function. Then
PL IS a norm.
THEOREM 5. The function seminorm P has the weak Fatou property if
and only if P and PL are equivalent (i.e., for every uEM+ the numbers
p(u) and PL(U) are finite or infinite simultaneously, and their quotient
lies between two fixed positive constants for all UEM+ for which p(u) is
finite). Hence, if P is a function norm, the normed linear spaces L p and
L pL contain the same elements if and only if P has the weak Fatou property.
PROOF. If P has the weak Fatou property, then (by Amemiya's theo-
rem) there exists a finite constant k> 1 such that p(u)k lim p(u n ) for
every sequence Oun ju in M+. It follows that PL(u)p(u)kpL(U), so
p and PL are equivalent.
Conversely, if P and PL are equivalent, then PLpkpL for some finite
constant k, and hence Oun ju in M+ implies that
p(U)kpL(U)==k lim PL(un)k lim p(u n ).
This shows that p has the weak Fatou property.
THEOREM 6 (MAXIMAL PROPERTY OF PL). (a) Given the function semi-
norm P, let A be a function seminorm with the F atou property such that
PLAp. Then A==PL. In other words, the largest seminorm majorized by
P and having the F atou property is pL.
(b) If A is a function seminorm with the weak Fatou property such that
PLAp, then A and PL are equivalent.
PROOF. (a) Since A has the Fatou property, we have A==AL. Further-
more, PLAp implies (PL)LALPL, i.e., AL==PL since (PL)L==PL. Com-
bining the results, we obtain A==AL==PL.
(b) Since A has the weak Fatou property, A and AL are equivalent.
The equality AL==PL holds as above. Hence A and PL are equivalent.

Ch. 15,  67J
SATURATED FUNCTION SEMINORMS
453
Exercises
DETERMINATION OF THE LORENTZ NORM
66.1) Let # be discrete measure in the point set X == {I, 2, . . .} and,
for any uEM+, let
00
P1(U)=== 2-nu(n)+lim sup u(n),
1
P2(U)==SUp u(n)+lim sup u(n),
P3(U) ===sup u(n) if UE (co), and P3(U) ===00 otherwise.
Show that (pl)L(u)===12-nu(n), (P2)L(U)==SUP u(n) and (p3)L(U)==
sup u(n) for any uEM+.
67. Saturated Function Seminorms
If P is a function seminorm such that either p(u)==O for all uEM+
or p(U) ===00 for every uEM+ not identically zero (almost everywhere),
then P is called a trivial function seminorm. Hence, p is nontrivial if
and only if there exists a function uEM+ such that O<p(U) <00. Even
when p is nontrivial, there may exist plenty of sets E e X of positive
measure such that not only p(XE)===OO, but also p(XF)==OO for every
subset FeE satisfying # (F) >0. Any set E of this kind is called a P-
purely infinite set. Evidently, finite or countable unions of p-purely
infinite sets are again p-purely infinite.
THEOREM 1. If #(E) >0, then E is p-purely infinite if and only if any
uEM+ such that p(U) <00 vanishes almost everywhere on E.
PROOF. Let E be p-purely infinite, and let uEM+ with p(U) <00. If
UXE>O on a set of positive measure, there exists a number e>O such
that uXE>e on a set of positive measure, i.e., u>eXF for some FeE
satisfying #(F) >0. But then p(XF) <e- 1 p(u) <00, contradicting the hy-
pothesis that E is p-purely infinite.
Conversely, assume that tXE===O holds for each uEM+ such that
p(U) <00, but E is not p-purely infinite. Then p(XF) <00 for some FeE
satisfying #(F) >0, and so UO===XF satisfies uoEM+ and p(Uo) <00, but

454
NORMED KOTHE SPACES
[Ch.15,67
UOXE==O does not hold. This contradiction shows that E must be p-
purely infinite.
For the particular case that p is a function norm, the theorem shows
that any fEL p vanishes on any p-purely infinite set E; hence, for the
purpose of investigating the normed linear space L p , we may just as
well remove the set E from X. The next theorem will show that the
process of removing p-purely infinite subsets can be accomplished in
one step.
THEOREM 2. There exists a maximal p-purely infinite subset X oo ' i.e.,
Xoo is p-purely infinite and Z ==X -X oo does not possess any p-purely
infinite subsets. The set Xoo is #-uniquely determined. If p is nontrivial,
then Z ==X - Xoo satisfies #(Z) >0.
PROOF. Assume first that #(X) <00, and let
ex==sup (#(E) : E is p-purely infinite).
Then there exists an ascending sequence of p-purely infinite sets En
such that #(En)fex. The set Eoo==r En is p-purely infinite and #(Eoo)
==ex, so that in view of the definition of ex the set X -Eoo does not have
any p-purely infinite subsets. Hence, Eoo is a maximal p-purely infinite
subset of X.
If #(X) ==00, we have X==r X k with all X k disjoint and of finite
measure (since # is a-finite). Each Xk has a maximal p-purely infinite
subset Ek, so Xoo==r Ek is p-purely infinite. If F were a p-purely
infinite subset of X -X oo ' then # (FX k ) >0 for some k, so FX k would
be a p-purely infinite subset of Xk-Xoo==Xk-E k which is impossible.
Hence, Xoo is a maximal p-purely infinite subset of X. The #-uniqueness
of Xoo follows easily.
If p is nontrivial, there exists a function UE M+ such that O<p(u)
<00. Then u>O on a set of positive measure, and since u==o almost
everywhere on Xoo by Tlleorem 1, we must have #(X -X oo ) >0.
DEFINITION. The function seminorm p is said to be saturated if there
do not exist p-purely infinite sets. I n other words, p is saturated if for any ,
set E of positive measure there exists a subset FeE such that F is of
positive measure and p(XF) <00.

Ch. 15,  67J
SATURATED FUNCTION SEMINORMS
455
All examples of function seminorms presented so far are saturated.
It is easy to make (somewhat artificial) examples of non-saturated
semlnorms or norms.
The standard example of the Lp norm shows that even for a satu-
rated function norm p it is not necessarily true that p(xx) <00. We
may ask, however, whether for a saturated function seminorm p there
always exists a sequence X niX such that P(XXn) < 00 for all X n. The
answer is affirmative. For the proof it is convenient to use the termi-
nology of the following definition.
DEFINITION. If X n (n==l, 2, ...) is a sequence of #-measurable sub-
sets of X satisfying Xni X, then the #-measurable set EeX is said to be
{Xn}-bounded if E is included, except for a #-null set, in some Xn.
The following purely measure theoretic theorem holds now.
THEOREM 3 (EXHAUSTION THEOREM). Let Y n (n== 1, 2, . . .) be a se-
quence of subsets of X such that Y niX and #(Y n) <00 for all n. Let (P)
be some property which any {Y n}-bounded subset of X does or does not
possess (it is understood that #-almost equal sets have (P) or do not have
(P) simultaneously). Assume, furthermore, that
(a) if E1 and E2 possess (P), then El +E2 possesses (P),
(b) if E possesses (P), then any measurable subset of E possesses (P),
(c) any {Y n}-bounded set of positive measure has a subset of positive
measure possessing (P).
Then there exists a sequence of measurable sets X n (n== 1, 2, . . .) such
that X n i X and X n e Y n for all n (so #(Xn)<oo for all n), and such
that every {Xn}-bounded set has property (P).
PROOF. Let E be {Y n}-bounded with #(E) >0, and let r be the col-
lection of all FeE possessing property (P). Note that #(E) is finite.
We set cx==sup (#(F) : FEr). There exists a sequence of sets F n E r such
that #(Fn)-+cx, and by assumption (a) we may assume that Fn is as-
cending. The set Foo==l Fn satisfies #(Foo)==cx. If cx<#(E), the set
E-F oo is of positive measure, and so by (c) has a subset FOEr of
positive measure. It follows that Fn+FoErfor all n, and #(Fn+Fo)==
#(F n) + #(F 0) >cx for n sufficiently large. This contradicts the definition

456
NORMED KOTHE SPACES
[Ch. 15,  67
of cx. Hence cx=#(E), i.e.,
sup (#(F) : FEr)=#(E).
I t follows that there exists a sequence of sets X n (n= 1, 2, .. .) such
that XncY n , X n has property (P), and #(Y n -X n )<n- 1 for all n=
1,2, ... . It may be assumed, by (a), that X n is ascending. The set
X'== 1 X n satisfies Y n-X' c Y n-Xn, so #(Y n-X')<n-1 for all n.
But Yn-X'jX-X', so #(X-X')=lim#(Yn-X')==O. Thus the se-
quence X n has the required properties (sets of measure zero are neg-
lected) .
THEOREM 4. The following statements for the function seminorm pare
equivalent.
(a) p is saturated, i.e., for any set E of positive measure there is a
subset F c E of positive measure such that p(XF) is finite.
(b) For any sequence of measurable sets Y niX there exists a sequence
of measurable sets X n i X such that X n c Y nand P(XXn) <00 for all n.
PROOF. From the preceding theorem it follows immediately that
(a) implies (b). Indeed, if we have to prove (b) and the sequence Y niX
is given, we may assume that #(Y n) <00 for all n. For the property
(P) of the set E we choose the property that p(XE) is finite. An appli-
cation of the preceding exhaustion theorem yields the desired sequence
Xn.
It is evident that, conversely, the existence of at least one sequence
X n i X such that P(XXn) <00 for all n already implies that (a) holds.
As observed earlier, if P is a function norm and if it is our purpose
to investigate the corresponding normed linear space L p , we may as-
sume without loss of generality that P is saturated. Theorem 4 guaran-
tees now the existence of a sequence X n i X such that every #-measur-
able function f, bounded and vanishing outside X n , is an element of L p .
One might believe that if #(X) is finite, the choice of the sequence
X n (n== 1, 2, . . .) is arbitrary, subject only to the condition that
X n i X. This is not so, as shown by the example that # is Lebesgue
measure in [0, 1J, and p(u)=fl x- 1 u(x)d#.

Ch. 15,  68J
ASSOCIATE FUNCTION SEMINORMS
457
68. Associate Function Seminorms
In the present section there will be assigned to any function semi-
norm p a sequence of seminorms p(n) (n==O, 1, 2, . . .), the associate
seminorms of p. The method of definition of p(n+1) from p(n) is a well-
known procedure in analysis.
DEFINITION. Given the function seminorm p, let p(O)==p and, for any
uEM+, let
p(n)(u)==sup(fuvd#: p(n-1)(v)1, vEM+)
for n== 1, 2, . . " where the integral denotes integration over the whole set
X. Instead of p(l), p(2), p(3) we shall write p', p", p"'.
THEOREM 1. For n?;:-l, p(n) is a function seminorm having the Fatou
property.
PROOF. It will be sufficient to present the proof for p'; the proof
for p(n), 1>2, follows then by induction. For a fixed vEM+ satisfying
p(v) 1 it is evident that Pv(u) == f uvd# is a function seminorm having
the Fatou property. Hence, by sec. 65, Theorem 4, the same holds for
,
p ==sup Pv.
The function seminorm p(n) is called the n-th associate seminorm of p.
THEOREM 2. (a) (HOLDER INEQUALITY). If p(u) and p'(v) are finite,
then f uvd#p(u)p'(v).
(b) We have p"p and p(n+2)==p(n) for all n>l.
PROOF. (a) For O<p(u)<oo, the inequality f uvd#p(u)p'(v) follows
from the definition of p'(v). Now, let p(u) ==0. If u==o holds almost
everywhere, there is nothing to prove. Assume, therefore, that E ==
{x: u(x»O} is of positive measure, and let En=={x: u(x»n- 1 } for n==
1, 2, . . . . Since En i E, there exists an index no such that #(En) >0
for all n?;:-no. We will show that En is p' -purely infinite for all n?;:-no.
To this end, let n?;:-no and FeE n with #(F) >0. Since u(x) >n- 1 on F,
we have XFnu, so p(XF) ==0. Then p(kXF) ==0 for k== 1, 2, . . ., and
p'(XF)==sup(fwd#: p(w)l, wEM+)
F
 sup k f d#== sup k#(F) ==00.
k= 1,2,... F k= 1,2,...

458
NORMED KOTHE SPACES
[Ch.15,68
This shows that En is p' -purely infinite for n?;:-no. But then E ==
{x: u(x) >O}=== 1 En is also p' -purely infinite. Hence, since p' (v) <00,
we have v==o almost everywhere on E (by sec. 67, Theorem 1), and it
follows that f uvd#==O==p(u)p' (v).
(b) It will be shown first that p"(u)p(u) for every uEM+. We may
assume that p(u) <00. Then
p" (u) ==sup (I tv d#: p' {v) 1 )sup (p(u)p' (v) :p' (v) 1) ==p(u)
by the Holder inequality.
Applying the result to p', we obtain already that p'" p'. On the other
hand, it follows immediately from the definition of the associate semi-
norm that pI P2 implies pi?;:-P2' Hence p" p implies that p"'?:;p'. Com-
bining the thus obtained inequalities, we obtain p'" ==p', and hence
p(n+2) ==p(n) for n?;:-l.
THEOREM 3. If #(E) >0, then E is p-purely infinite if and only if E
is p" -purely infinite.
PROOF. If E is p" -purely infinite, then E is p-purely infinite since
p" p. Assume now that E is p-purely infinite, so u==o on E for any
uEM+ such that p(t)<oo. Then
p'(OO'XE)==sup(/OO'ud#: p(u)l, uEM+) ==0.
E
I t follows that if FeE and #(F) >0, then
p"(XF)==sup(lvd#: p'(v)l, vEM+)?;:- 1 oo'd#==oo,
F F
so E is p" -purely infinite.
THEOREM 4. The following statements are equivalent.
(a) p is saturated,
(b) p' is a norm,
(c) p" is saturated.
PROOF. The preceding theorem shows that p is saturated if and only
if p" is saturated. Assume now that p (and hence p") is saturated, but,
p' fails to be a norm. Then p' (XE) ==0 for some set E of positive measure.
Hence p'(OO'XE)==O since p' has the Fatou property, and it follows as

Ch. 15,  69J ASSOCIATE SPACE OF A NORMED KOTHE SPACE 459
in the proof of the preceding theorem that E is p" -purely infinite,
which is impossible.
Conversely, assume that p' is a norm, but p is not saturated. Then
there exists a p-purely infinite set E (hence u==o almost everywhere on
E for any u_EM+ such that p(U) <00), and so
p'(XE)===sup{lud#: p(u)l, uEM+)==O.
E
Since #(E) >0, this contradicts the hypothesis that p' is a norm.
THEOREM 5 (HOLDER INEQUALITY FOR A SATURATED FUNCTION
NORM). If p is a saturated function norm, then
1 uvd#p(u)p'(v)
for all u, vEM+.
PROOF. Since it was already proved in Theorem 2 that the inequali-
ty holds whenever p(u) and p'(v) are finite, and in view of 0'00===00.0
===0, the only cases left to consider are that p(u) ===0, p' (v) ===00 or p(u)
==00, p'(v)==O. In the first case we have u===o since p is a norm, and
hence f uvd#===O==p(u)p'(v). In the second case we observe that by the
preceding theorem p' is a norm since p is saturated, and hence p' (v) ==0
implies that v==O, and then again f uvd#==O===p(u)p'{v).
69. The Associate Space of a Normed Kothe Space
In the present section, let p be a function norm and L p the corre-
sponding normed linear space. As noted earlier we may assume with-
out loss of generality that p is saturated. It follows from sec. 68, Theo-
rems 4 and 1 that the associate function seminorm p' is now a norm
with -the Fatou property. Hence, the corresponding normed linear
space L p ' is Banach space with respect to the norm p'. Instead of L p '
we will write L;; the space L; is called the associate space of L p . Given
.gEL;, the norm p'(g) is given by
p' (g) ===sup (I Ifgl d#: p(f) 1).
We shall prove that f Ifg I d# may be replaced here by If fg d# I.

460
NORIVIED KOTHE SPACES
[Ch. 15,  69
THEOREM 1. For every gEL; we have
p'(g)==sup(l/fgdfll: p(f)l).
PROOF. If gEL; and p(f)l, then
I Ifgl dflP(f)p' (g)p' (g),
and so the integral f fgdfl exists as a finite number. Evidently
sup (1/fgdfll : p(f)l)sup(/lfgldfl: p(f)l)==p'(g).
For the inverse inequality, let e >0, and choose f1 such that p(f1) 1
and flf1gldfl>p'(g)-e. Since f==lf11/sgng satisfies Ifl==lf11, we have
p(f) ==p(f1) 1 and f fgdfl==f If1g1 dfl. Hence
I/fgdfll== Ilf1gl d fl>p'(g)-e,
and so
sup (II fgdfll : p(f) 1 )p' (g).
The desired result follows.
Given gEL;, the integral f fgdfl exists as a finite number for every
fEL p , and so the given function g defines a linear functional G on L()
by means of G(f) == f fg dfl.
THEOREM 2. If gEL; and G(f)==ffgdfl for every fEL p , then G is a
bounded linear functional on L p (i.e., G is an element of the coniugate
space L;), and IIGII==p'(g).
PROOF. In view of the preceding theorem we have
sup (IG(f)1 : p(f)l)==sup(l/fgdfll: p(f)l)==p'(g).
Hence GEL; and IIGII==p'(g).
It follows that if for any gEL we define GEL; by G(f)==f fgdfl,
there exists a one-one norm preserving linear correspondence between
the elements g of L; and the elements of G of a certain subset of L;.
This subset is a closed linear subspace of L; since L; is a Banach space.
Hence, identifying g and the corresponding G, we nlay say that L; is,
a closed linear subspace of L;. By way of example, if L p is an Lp space
for some p satisfyjng 1 poo, then L; is the space Lq, where q is de-

Ch. 15,  69J ASSOCIATE SPACE OF A NORMED KOTHE SPACE 461
termined by p-1+ q -1== 1 (cf. sec. SO, Lemma y). Hence, L; and L:
coincide for lp<oo, but for p==oo the space L;==L 1 is properly
smaller than L==L: (unless Loo is finite-dimensional).
The elements G of L: belonging to the closed linear subspace L; are
characterized by the existence of a function gEL; such that G (f) ==
f fgd# for all fEL p ; it will be proved that another characterization of
these elements is possible, where it is not assumed in advance that
G(f) can be written as an integral. To this end, we will first prove
several lemmas.
LEMMA ex. The #-measurable function g satisfies gEL; if and only if
fg is #-summable for every fEL p and if in addition G(f)===ffgd# is a
bounded linear functional on L p .
PROOF. We need only prove that if fg is summable for every fEL p
and G(f)===ffgd# is a bounded linear functional on L p , then gEL;.
Given fEL p , we have f1== Ifl/sgn gEL p , and so
f Ifgl d#== f f1gd#===G(f1)IIGllp(f1) === IIGllp(f).
But then p'(g)IIGII<oo by the detil1ition of p', so gEL;.
As on earlier occasions (cf. sec. 50), we denote by Lr) the set of all
realvalued functions in L p . We now recall several results proved in
sec. 50. Given the bounded linear functional G(f) on L p , we write
G(f) ===Gr(f) +iGim(f) for all fELr), where Gr(f) and Gim(f) are the real
and imaginary part, respectively, of G(f). Then Gr(f) and Gim(f) are
real linear bounded functionals on Lr), and on account of sec. 48,
Theorenl1 we have Gr==G:+G; andGim===Gfin+G, whereG:, -G;,
Gfin and -G are non-negative bounded linear functionals on Lr).
We recall that, for OfELr), the number G: (f) is defined by
G: (f) ===sup (G r (f1) : 0f1 f, f1 ELr»).
Similarly for Gfin(f). The functional G:, thus defined on Lr), is extended
on L p by defining G: (f)==G: (g) +iG: (h) for f===g+ih (g and h real) in
L p . Similarly for G;, Gfin and G. Thus we obtain the standard decom-
position G===(G: +G;)+i(Gfin+G) of G into non-negative bounded
linear functionals.

462
NORMED KOTHE SPACES
[Ch. 15,  69
DEFINITION. The bounded linear functional G on L p is called of inte-
gral type (or, briefly, an integral) whenever it follows from OfnELp (n==
1, 2, . . .), f n! 0, that G (f n) -+0.
The next lemma is the same as sec. 50, Lemma €5. The proof pre-
sented in sec. 50 works equally well in the present case where L p is a
normed Kothe space, and not necessarily a Banach function space.
LEMMA fJ. Let the bounded linear functional G on L p be an integral
having the standard decomposition
G== (G: +G;) +i(Giin +G hn ).
Then G:, -G;, Giin and -G hn are integrals. In other words, these
functionals are elementary integrals on Lr).
THEOREM 3. Let G be a bounded linear functional on L p . Then G is
an integral if and only if GEL, i.e., if and only if G(f)==ffgdfl for some
gEL and all fEL p . In this case, G and g determine each other uniquely
(in fact, G and g are identified under the imbedding of L in L:).
PROOF. If G(f)==ffgdfl for some gEL and all fEL p , and if fn!O for
some sequence fnELp (n== 1, 2, . . .), then If1g1 is summable, Ifngllf1gl
for all n, and Ifngl!O holds almost everywhere. Henceflfngldfl!O by
dominated convergence, and so G(f n) -+0 on account of IG(f n) 1==
If fngdfllf Ifngl dfl. This shows that G is an integral.
Assume now, conversely, that GEL: is an integral, and let G==
(G: +G;) +i(Giin +G hn ) be the standard decomposition of G. By the
preceding lemma G: is now an elementary integral on Lr). Let X n
(n== 1, 2, . . .) be a sequence of measurable subsets of X such that
Xni X, and fl(Xn) <00 and XXnELp for every n (such a sequence ex-
ists by sec. 67, Theorem 4), and let Lb be the set of all real bounded
and measurable functions f(x) vanishing outside some X n (in general,
different X n for different f). Evidently Jf==f fdfl is finite for every
fEL b , and Jf is an elementary integral on Lb. Applying the extension
procedure for integrals to this elementary integral, we obtain the same
extended integral as when the extension procedure is applied to the
elementary integral f fdfl, defined initially only on the collection Ls of
all fl-step functions vanishing outside some X n. This is due to the fact '
that Ls c Lb c Li r ), where Li r ) is the set of all real fl-summable functions.

Ch. 15,  69J ASSOCIATE SPACE OF A NORMED KOTHE SPACE 463
It follows that the extended integral referred to is the ordinary ex-
tended Stieltjes-Lebesgue integraloff-===-f fdfl on X. Furthermore, since
XXnELp for every n, we have LbcLr), and so G: is an elementary
integral on Lb (since G: is an elementary integral on Lr»). We shall
denote the corresponding extended integral by . Hence off-===- f fdp,
and f are extensions of elementary integrals, initially defined on Lb.
Since all XXn are elements of Lb (and hence of L p ), we have OXXn==
G: (XxJ < 00 for all n, which shows that the measure induced in X by
 is a-finite. Furthermore, if of(lfl) ==0, then f==O holds fl-almost
everywhere, and so fEL b ; it follows that f===G: (f) -===-G: (0) ==0, which
shows that  is of-absolutely continuous. Hence, the integral version
of the Radon-Nikodym theorem can be applied; consequently, there
exists an of-measurable function gl(X)O such that (f) == f fg1dfl for
all -summable f.
It will be shown now that (/) is finite and equal to G:(f) for all
fEL p ; evidently it will be sufficient to prove this for non-negative
fELr). Writing, in this case, fn==min(f, nx Xn ) where X n is the same
set as above, we have OfnELb and Ofn i f. Hence f-fnELr) for n==
1,2, ... and f-fn!O almost everywhere, which implies G:(f-fn)!O
since G: is an elementary integral on Lr). It follows that G:(fn) iG: (f).
We have also (fn) i (f) by the theorem on integration of monotone
sequences. But (fn)==G:(fn) for all fn since  and G: coincide on
Lb. Hence (f) ==G: (f) for every fEL p , and since G: (f) is finite for all
fEL p , it follows in particular that every fEL p is -summable. Thus we
have obtained the result that G:(f)==(f)-===-ffg1dfl for all fEL p . It re-
mains to prove that gEL;. This follows immediately frorn Lemma ex.
The same argument shows the existence of functions g2, g3, g4 EL ;
such that G;(f)==ffg2dfl, G(f)==ffg3dfl and Ghn(f)==ffg4dfl for all
fEL p , so
g== (gl +g2) +i(g3+g4) EL,
and G(f)===ffgdfl for all fEL p . This completes the proof.
Note that if G is an integral such that G(f)O for every f?;;O in L p ,
then the corresponding function gEL satisfies g(x)?;;O for almost every
x. Similarly, if G is an integral such that G(f) is real for every real f in
L p , then the corresponding gEL; is real-valued for almost every x. The

464
NORMED KOTHE SPACES
[Ch. 15,  70
standard decomposition of such a real integral G is G===G++G-, and
we will show now that if G corresponds with the real function gEL;,
then G+ and G- correspond with g+ and g- respectively.
LEMMA y. If GEL is an integral such that G(f) is real for every real
fEL p , and if G corresponds with (i.e., may be identified with) the real
function gEL;, then G+ and G- correspond with g+ and g- respectively.
PROOF. Given OfELp, we have by definition that
G+(f) ===sup (G(f1) : 0f1 f, f1 EL p )
===SUp(!f1g+ d fl+ !f1g- d fl: 0f1f, f1 EL p).
The least upper bound is evidently attained for f1 === fXE, where E ==
{x: g(x)O}, and hence G+(f)===/ fg+dfl. This holds for every f>O in Lr),
and hence for every fEL p . It follows that G+ corresponds with g+;
consequently, G-===G-G+ corresponds with g-g+===g-.
70. Decomposition of a Bounded Linear Functional into an
Integral and a Singular Functional
As before, we denote by Lr) the real normed linear space of all real
f in L p , and by (Lr»)* the real conjugate space of Lr). The space (Lr»)*
is partially ordered by defining that G1G2 whenever G1(f)G2(f) for
all non-negative f in Lr), Le., whenever G 2 -G 1 is a non-negative
bounded linear functional on Lr). We will denote the null functional
bye. The non-negative bounded linear functionals on Lr) are ex-
actly those satisfying G>e.
Given any GE (Lr»)*, the standard decomposition of G is G===G++G-
with G+>e and G-e. In addition, given any other decomposition
G===G1 +G 2 with G 1 >e and G2e, we have G+G1 (cf. sec. 48, Theo-
rem 1). This shows that, with respect to the partial ordering, G+ is an
upper bound of G and e such that G+ is less than or equal to any other
upper bound of G and e. In other words, G+ is the least upper bound
of G and e; we will express this by "vriting G+===sup(G, e). Similarly,
G- is the greatest lower bound of G and e; we will write G-==inf (G, e).
Given G 1 , G 2 , G 3 , note that sup (G 1 +G 3 , G2+G3) exists jf and only if
sup (G 1 , G 2 ) exists, and in either case
sup (G 1 +G 3 , G 2 +G 3 ) =sup (G 1 , G 2 ) +G 3 .

Ch. 15,  70J
DECOMPOSITION OF A FUNCTIONAL
465
It follows that, for any pair G 1 , G 2 E (Lr»)*, the element sup (G 1 , G 2 )
does indeed exist, and satisfies
sup (G 1 , G 2 ) ==sup(G 1 -G 2 , e) +G2== (G 1 -G 2 )++G 2 .
Similar I Y
inf(G 1 , G 2 )==(G 1 -G 2 )-+G 2 .
Hence
sup (G 1 , G 2 )+inf(G 1 , G2)==(G1-G2)+2G2==G1+G2 (1)
for all G 1 , G 2 E (Lr») *. It follows that
G 1 -inf (G 1 , G2) ==sup (G 1 , G 2 ) -G 2 .
We finally observe that, for all G1, G 2 E (Lr»)*,
(G1 +G2)+G! +Gt.
(2)
(3)
Indeed, GtGl and Gt>G 2 , so Gt+GtG1+G2 and Gt+Gt>e.
It follows that
Gt+Gt>SUp(G 1 +G 2 , e)==(G 1 +G 2 )+.
The first lemma which will be proved now is a decomposition lemma,
and the second lemma mentions a property of the linear subspace of
all integrals (i. e., a property of the linear su bspace L; of L).
LEMMA ex. Given the non-negative bounded linear functionals G, G 1
and G 2 such that eGG1 +G 2 , there exist non-negative bounded linear
functionals Gi and G 2 such that eGiG1, eG2G2 and G==Gi +G 2 .
PROOF. Set Gi ==inf (G, G1) and G 2 ==G-Gi. Then Gi and G 2 are
non-negative, and GiG1 holds evidently. It remains only to prove
that G2G2. By means of (2), this follows from
G 2 ==G-inf (G, G 1 ) ==sup (G, G 1 ) -G1(G1 +G 2 ) -G 1 ==G 2 ,
where we have also used that G 1 +G 2 is an upper bound of G 1 and G.
LEMMA fJ. Given the non-negative bounded linear functionals G and G1
such that eGG1 and G1 is an integral, then G is also an integral. In
other words, if G 1 EL;, then GEL;.

466
NORMED KOTHE SPACES
[Ch. 15,  70
PROOF. By hypothesis we have G 1 (fn)! 0 whenever fn! 0 in L p . Since
0G(fn)G1(fn) for every n, we have also that G(fn)!O whenever fn!O
in L p . This shows that G is an integral.
DEFINITION. The non-negative bounded linear functional G on L p is
called a singular functional on L p whenever it follows from eG1G and
G 1 EL; that G 1 ==e (in other words, every non-negative integral majorized
by G is the null functional). The arbitrary element GEL is called singular
if all the non-negative components in the standard decomposition of G are
singular.
THEOREM 1. The set of all singular functionals on L p is a linear sub-
space of L.
PROOF. We will prove first that the sum of non-negative singular
functionals is singular. Hence, let G 1 , G 2 be non-negative and singular,
and assume that G is an integral satisfying eGG1+G2. We have to
prove that G==e. By Lemma ex there exist non-negative Gi, G 2 such
that eGiG1, eG2G2 and G==Gi +G 2 . Since Gi and G 2 are now
majorized by G, and G is an integral, it follows from Lemma fJ that
Gi and G 2 are integrals. But then Gi==e on account of eGiG1.
Similarly G 2 ==e. Hence G==Gi +G 2 ==e.
Next let G 1 , G2E(Lr»)* be singular, and set G S ==G 1 +G 2 . By defi-
nition, G! and Gt are singular, and by (3) we have eGtG!+Gt,
so Gt is singular by what has already been proved. Since e-G3"
-G 1 -G 2 , similarly as in (3), it follows that -G3" is also singular,
and hence G s is singular.
Finally, given the arbitrary elements G, G' EL, we have G==Gr+iG im
and G'==G;+iG;m, where G r , G im , G; and G;m are real for real f. Since
(G+G')r==Gr+G; and (G+G')im==Gim+G;m, it follows immediately
that G+G' is singular if G and G' are singular.
It remains to prove that if G is singular and a real, then aG is singu-
lar. This is easy, and the desired result follows.
Before proving the main decomposition theorem for elements of L,
we present another lemma.

Ch.15,70J
DECOMPOSITION OF A FUNCTIONAL
467
LEMMA y. If G is a non-negative bounded linear functional on L p , and
G 1 is a non-negative linear functional on L p such that 0G1(f)G(f) for
every to in L p , then G 1 is also bounded.
PROOF. For any fELr) we have
IG 1 (f) I == IG 1 (j+) +G 1 (f-) IG1(f+) +G 1 ( -f-)
==G1(lfl)G(lfl)IIGII' p(f).
If fEL p is arbitrary, then f==g+ih with g and h real. We have Igllfl
and Ihllfl, so p(g)<.p(f) and p(h)<.p(j), and hence
IG 1 (f) 1IG1(g) I + IG 1 (h) 1<.IIGII{p(g) +p(h)}<.21IGII' p(f). ,
This shows that G 1 is a bounded linear functional.
THEOREM 2 (DECOMPOSITION THEOREM). Every GEL: has a unique
decomposition G==G 1 +G2 such that G 1 is an integral and G 2 is singular.
If G is non-negative, then G 1 and G 2 are non-negative.
PROOF. G is an integral or a singular functional if and only if all
its components in the standard decomposition are integrals or singular
functionals respectively. Hence, it will be sufficient to prove the decom-
position theorem only for non-negative G. Furthermore, the only non-
negative G which is an integral and a singular functional simultane-
ously, is the null functional. This proves the uniqueness.
Hence, assume that G is a non-negative bounded linear functional
on L p . For any function u(x»O in L p , let
G 1 (u) ==sup (G' (t) : e<.G' <.G, G' EL).
Evidently, we have 0G1(U)<.G(u), so G 1 (u) is non-negative and finite
for every u>O in L p . Given uO and v>O in L p , it follows immediately
from the definition that G1(U+V)Gl(U)+G1(V). In order to prove the
inverse inequality, let e>O be given, and let the non-negative integrals
G u and G v satisfy G1(t)Gu(u)+e and G1(V)GV(v)+e. By Lemma fJ
it follows from esup (G u , Gv)<.Gu+G v that G' ==sup (G u , G v ) is also
an integral, evidently satisfying G'<.G. Hence
G 1 (u) +G1(V)G' (u) +G' (v) +2e==G' (u+v) +2eG1(u+V) +2e.
This holds for every e>O, and so G 1 (U+v)==G 1 (U)+G 1 (v) for all uO,
vO in L p . It is also evident that G 1 (au)==aG 1 (u) for all uO in L p and

468
NORMED KOTHE SPACES
[Ch.15,70
all constants aO. Defining now G1(f) ==G1(f+) +G 1 (f-) for real fEL p ,
and G 1 (f)==G 1 (g)+iG 1 (h) for f==g+ih (g and h real) in L p , it follows
that G 1 is a linear functional on L p . Lemma y shows that G 1 is bounded.
It will be proved now that G 1 is an integral. To this end, let fI?::-
f2?::-' · · ! 0 in L p . Given e >0, there exists an integral G' such that e
G'G and 0(G1-G')(f1)<e. Then 0(G1-G')(fn)<e for all fn, and
so G 1 (fn)!0 on account of G'(fn)!O.
It remains to prove that G 2 ==G-G 1 is singular. Evidently we have
G 2 =G-G 1 ?::-e, so if G 2 is not singular, there exists a non-negative
integral G", not the null functional and such that eG"G-GI. But
then G 1 +G" is an integral satisfying eG1 +G" G and properly larger
than G 1 in the partial ordering of (Lr»)*. This contradicts the definition
of G 1 .
A rough estimate shows easily that if GEL has the standard decom-
position
G=(G: +G;)+i(Giin+G),
then each of the four components has norm at most equal to 411GII.
Hence, if G' has each of its non-negative components less than or equal
to the corresponding component of G, then (by Lemma y) each of the
components of G' has norm at most equal to 811GII, and hence IIG'II
3211GII. This shows that if G==G 1 +G 2 is the decomposition of GEL
into an integral and a singular functional, then IIG1113211GII and IIG211
3211GII. It follows that (just as the set of all integrals) the set of all
singular linear functionals on L p is a norm closed linear subspace of
L:. Indeed, if IIG-Gnll-+O with all G n singular, and we denote by G s
the singular component of G, then
IIG s - (Gn)sll321IG-Gnll-+0.
But (Gn)s==G n for all n, so we have IIG-Gnll-+O as well as IIGs-GnllO
as n-+oo. This shows that G==G s , i.e., G is singular. It follows that the
linear subspace of all singular linear functionals is norm closed.

Ch. 15,  71 ]
SECOND ASSOCIATE FUNCTION NORM
469
Exercises
ANOTHER DEFINITION FOR THE INTEGRAL COMPONENT
70.1) Let G be a non-negative bounded linear functional on L p . For
any function u>O in L p , set
GL(u)==inf (lim G(u n ): Oun ju, unELp
for n== 1, 2, . . .).
Show that GL is exactly the integral component of G.
PROJECTIONS IN THE SPACE L
70.2) For any GEL, let G==Ge+G s be the decomposition of G into
an integral G e and a singular functional G s , and define the transfor-
mations Pe and Ps of L into L by Pe(G)==G e and Ps(G)==G s . Show
that Pe and Ps are bounded linear transformations of L into L such
that P==Pe and P;==Ps. In other words, Pe and Ps are bounded
projections in L (cf. Exercise 54.6) such that Pe+Ps is the unit
transformation, and the range of Pe is the subspace L; of L.
70.3) Let E be a fixed #-measurable subset of X. For every GEL
and every fEL p , define the number GE(f) by GE(f)==G(fxE). Show that
G E is a bounded linear functional on L p satisfying IIGEIIIIGII. Further-
more, show that (GE)E==G E , so that if we write now GE==PE(G), then
P E is a projection of L; into L with norm IIPEII== 1 if #(E) >0. Show
that for G>e we have eGEG; hence, if G is an integral or a singular
functional respectively, then the same holds for G E . In other words, P E
is a projection in each of the subspaces L; and L:,s separately, where
L:,s is the subspace of all singular functionals.
71. The Second Associate Function Norm
The main result in the present section will be that, for any function
seminorm p, the second associate p" (as defined in sec. 68) is equal to
the Lorentz seminorm PL (as defined in sec. 66). Since P==PL if and only
if P has the Fatou property (sec. 66, Theorem 3) it will follow in par-
ticular that p==p" if and only if p has the Fatou property. The proof
for this particular case will be presented first since the general result
is easily derived from it. The theorem that p"=p whenever p has the

470
NORMED KOTHE SPACES
[Ch. 15,  71
Fatou property is due to G. G. LORENTZ (unpublished) and W. A. J.
LUXEMBURG (1955, [lJ). There exist several proofs, all based on sepa-
ration of a closed convex set and an outside point by a "hyperplane".
In the version presented here (due to I. HALPERIN and W. A. J. LUXEM-
BURG, 1957 [2J) the separation theorem is applied in an L2 space (in fact,
we will use the result of Exercise 29.1, an immediate consequence of
sec. 29, Theorem 6).
THEOREM 1. We have p" ==p il and only il p has the Fatou property.
PROOF. If p" ==p, then p has the Fatou property since p" has the
Fatou property (sec. 68, Theorem 1). Hence, it remains to prove that
if p has the Fatou property, then p" (u) ==p(u) for every uEM+. If p(u) ==0,
then p"(u)p(u)==O, so p"(u) ==0. We may assume, therefore, that p(u)
>0. Let Xoo be the maximal p-purely infinite subset of X. Then, by
sec. 68, Theorem 3, Xoo is also the maximal p" -purely infinite subset of
X. It follows that if u(x) does not vanish almost everywhere on X oo ,
then p(u)==oo==p"(u), and hence we may assume for the remainder of
the proof that u(x) ==0 on Xoo. In other words, we may delete X oo '
which is equivalent to assuming that p and p" are saturated. Hence,
in view of sec. 67, Theorem 4, there exists a sequence X n t X such that
#(X n) <00 and P(XXn) <00 for n== 1, 2, . . . . Writing un==min (u, nX xn )
for n==l, 2, "', we have untu, so p(un)tp(u) as well as p"(un)tp"(u)
since p and p" have the Fatou property. It will be sufficient, therefore,
to prove that p"(Un)==P(u n ) for all.n. In other words, we may restrict
ourselves to the special case that #(X) <00, p(xx) <00, u is a bounded
function in M+ and p(u) >0. Since u is bounded and p(xx) <00, it
follows that p(u) <00, so we may assume in addition that p(u) == 1.
Under these assuITlptions, let U be the set of all #-measurable
functions I satisfying p(l/l) 1, let L 2 ==L 2 (X, #) be the normed linear
space of all #-measurable I satisfying 11/112==(1 1/1 2 d#)!<00, and let U1
be the intersection of U and L2, so U1==UnL2. Obviously, U 1 is
convex (i.e., if 11, 12EU1 and a is a real number such that Oal,
then a/1+(1-a)/2EU1), and U 1 has the property that if IEU1, then
e irp lEU1 for every real cpo We will show that U1 is a closed subset
of the space L2. Indeed, if IEL2 is such that 11/-lnI12-+0 for som
sequence In in U 1 , then Ink-+I pointwise for some subsequence Ink' and
hence Ilnkl-+1/1 point\vise, which implies that p(I/I)lim inf P(l/nkl)l

Ch. 15,  71J
SECOND ASSOCIATE FUNCTION NORM
471
since p has the Fatou property (cf. sec. 65, Theorem 3). This shows
that fE U, so fE U 1. To summarize, U 1 is a norm closed convex subset
of L2 with the property that fE U 1 implies eif{JfE U 1 for every real cpo
Let e>O, and set uo(x)==(l+e)u(x). Then uoEL 2 (since Uo is a
bounded function and fl(X) <00) and p(uo)==l+e, so Uo is not an ele-
ment of the set U 1. Hence, by the result in Exercise 29.1, there exist
a bounded linear functional F on the Hilbert space L 2 and a positive
number c such that IF(f)l<c for all fEU1 and IF(uo)l>c. In view of
the representatioIl theorem for bounded linear functionals on a Hilbert
space this is equivalent to the existence of a function h(x) EL 2 such that
If fh dfl I <c for all fE Uland If uohdfll >c. Setting Ihl ==Vo and observing
that U1 has the property that fEU 1 implies Ifl/sgnhEU1, we obtain
already that f Iflvodfl<c for all fE U 1 and f uovodfl>c.
It will be shown next that
IlflvodflC for all fEU=={f: p(lfl)l}.
To this end, let fE U andfn==min (If I , n) for n== 1, 2, . . . . Thenf Ifnlvodfl
<c for all n since fn E U 1. But fn i If I on X, so f Iflvodflc by the theo-
rem on integration of increasing sequences. It follows that
p' (vo) ==sup (I Iflvodfl: p(lfl) 1 )c.
On the other hand, p' (vo) >0. Indeed, since c is positive, we have
O<c<1 uovodflP(Uo)p' (vo),
so p' (vo) >0. Observe that the Holder inequality may be applied since
p(uo)== 1 +e<oo and p'(vo)c<oo (cf. sec. 68, Theorem 2). Substituting
now Uo== (1 +e)u in J' uovodfl>c, we obtain by (1) that
(1)
(1 +e)1 uvodfl>c>P'(vo) >0,
so
1 u{vo/p'(vo)}dfl> (1 +e)-l.
It follows, therefore, from the definition of p" that p"(u»(l+e)-l.
This holds for every e>O, so p"(u» 1 ==p(u). Since p"(u)p(u) holds
generally, the final result is that p" (u) ==p(u).
THEOREM 2. If p is an arbitrary function seminorm and PL is the
corresponding Lorentz seminorm, then p" ==PL.

472
NORMED KOTHE SPACES
[Ch. 15,  71
PROOF. The function seminorm p" has the Fatou property, and so
(p")L==P" by the definition of (p")L. Hence, it follows from p" p that
p" == (p")LPL. On the other hand, PLP implies that pp', and so
(PL)"p". But PL has the Fatou property by sec. 66, Theorem 2, and
so it follows from the preceding theorem that (PL)"==PL. Hence PL==
(PL)"p". Combining the thus obtained inequalities P"PL and PLP"
we obtain the result that p"==PL.
In view of the properties of PL proved in sec. 66, we have now the
following theorem.
THEOREM 3. (a) The function seminorm p has the Fatou property if
and only if p==p".
(b) p" is a function norm if and only if p is a function norm.
(c) The function seminorm p has the weak Fatou property if and only
if p and p" are equivalent (i.e., for every uEM+ the numbers p(u) and
p" (u) are finite or infinite simultaneously, and their quotient lies between
two fixed positive constants for all UEM+ for which p(u) is finite). Hence,
if p is a function norm, the normed linear spaces L p and L'; contain the
same elements if and only if p has the weak Fatou property.
(d) If p and A are function seminorms such that p" Ap and A has
the Fatou property, then A==p".
In sec. 68, Theorem 4 it was proved that p' is a norm if and only if
p is saturated. We can prove now the same with p and p' interchanged.
THEOREM 4. (a) p is a norm if and only if p' is saturated.
(b) If one of p, p' and p" is a saturated norm, then so are the others.
PROOF. (a) One part is trivial and could have been proved earlier.
If p' is saturated, then p" is a norm by sec. 68, Theorem 4, and so p is
a norm. The nontrivial part is where we assume that p is a norm. Then
p" is a norm by part (b) in Theorenl 3 above, and so p' is saturated by
sec. 68, Theorem 4.
(b) The proof follows easily by combining the just proved result in ,
part (a) and sec. 68, Theorem 4.

Ch. 15,  71 J
SECOND ASSOCIATE FUNCTION NORM
473
Let p be a saturated function norm. Then, as shown in the last theo-
rem, p' and p" are also saturated function norms. Let L p , L; and L'
be the corresponding normed linear spaces. By the results in sec. 67,
in particular Theorem 4 of that section, there exists a sequence of sets
Xni X, each X n measurable and of finite measure such that XXn EL ()
as well as XXn EL; for all n== 1, 2, . . . . Note that this implies that
XXnEL' for all n on account of p"p. Now, if the measurable set E is
{Xn}-bounded, i.e., if E is #-almost included in X n for some n, then
every fEL p is #-summable over E. Indeed, the {Xn}-boundedness of E
implies that p'(XE)<OO, and so fElfld#==flflxEd#p(f)p'(XE)<OO for
every fEL p . Similarly, every function in L; or L' is #-summable over
E. We may say, therefore, that the functions in L p (or in L; or L:
respectively) are locally #-summable.
If L p is an Lp space (for some value of p satisfying lPoo) and
the function f is #-measurable on X, we have by sec. 50, Lemma y,
that fELq (P-1+q-1== 1) if and only if fg is #-summable for every gEL p
(always under the hypothesis that # is a a-finite measure). A similar
theorem holds for general normed linear spaces L p provided the function
norm p has the Riesz-Fischer property. For function norms having the
Fatou property this was first proved by G. G. LORENTZ and D. G.
WERTHEIM (1953, [lJ).
THEOREM 5. Let p be a saturated function norm having the Riesz-
Fischer property (i.e., L p is a Banach space) and let f be a #-measurable
function. Then fEL; if and only if f Ifgld#<oo for every gEL p .
PROOF. If fEL;, then flfgld#<'p'(f)p(g)<oo for every gEL p . Con-
versely, assume that f Ifgl d#<oo for every gEL p , but f not in L;. Then
there exists a sequence gn (n== 1, 2, . . .) in L p such that p(gn) 1 and
f Ifgnl d#>n 3 for n== 1, 2, . . . . It follows that  p(n- 2 Ignl) ==  n- 2 p(gn)
<00 and hence, since p has the Riesz-Fischer property, the function
g==  n- 2 lgnl satisfies gEL p . By hypothesis we have then that f Ifgl dp,
<00. On the other hand,
f Ifgl d#>n- 2 f Ifgnl d#>n
for every n== 1,2, ...,
so f Ifgl d# == 00. This yields a contradiction. It follows that fEL p .

474
NORMED KOTHE SPACES
[Ch. 15,  71
COROLLARY. Let p be a saturated function norm and let f be a #-
measurable function. Then fEL; if and only if f Ifgl d#<oo for every
gEL;.
PROOF. Observe that p' has the Fatou property, and hence the
Riesz-Fischer property.
If L p is not norm complete, it may occur that f Ifgl d#<oo for every
gEL p , and yet p' (f) ==00. By way of example, let # be discrete measure
in X=={1, 2, .. .}, and let, for every u==(u(l), u(2), .. .)EM+, the
function norm p be defined by p(u) ==sup u(n) if the set {n: u(n) >O} is
a finite set, and p(u) ==00 otherwise. Then p' (v) == 1 v(n) for every
v EM+, and p"(u) ==sup u(n) for every UEM+. Let f be defined by f(n) == 1
for all n. Then f Ifgl d#<oo for every gEL p , but p' (f) ==00.
Next, we investigate under which conditions p and p' can be inter-
changed in the last theorem.
THEOREM 6. Let p be a saturated function norm, and let f denote a #-
measurable function. The following statements are equivalent.
(a) fEL p if and only if f Ifgld#<oo for every gEL.
(b) p has the weak F atou property.
PROOF. (a)=>(b). By the last corollary we have that fEL if and
only if f Ifgl d#<oo for every gEL;. Hence, by (a), we obtain that fEL p
if and only if fEL'; Le., the spaces L p and L; contain the same elements.
But then, by Theorem 3 (c), p has the weak Fatou property.
(b)=>(a). If fEL p , then f Ifgld#<oo for every gEL;. Conversely, let
J Ifgl d#<oo for every gEL. Then fEL by the last corollary. But L p
and L'; contain the same elements since p has the weak Fatou property,
and so fEL p .
Once again, let p be a saturated function norm having the Riesz-
Fischer property. Theorem 5 shows that the set (L p ) X of all measurable
f such that fg is summable for every gEL p is exactly the space L.
Theorem 6 shows that the set (L)X of all measurable f such that fg is
summable for every gEL; is again the space L p only provided p has the
weak Fatou property. Hence (Lp)xx==L p if and only if p has the weak,
Fatou property. Spaces satisfying (Lp)xx==L p were called perfect spaces
in the original K6the- Toeplitz theory.

Ch. 15,  72J FUNCTIONS OF ABSOLUTELY CONTINUOUS NORM 475
Exercises
GREATEST LOWER BOUND AND LEAST UPPER BOUND OF FUNCTION
SEMINORMS
71.1) As observed in sec. 63, the set of all function seminorms on
M+ is partially ordered and, given the seminorms PI and P2, the semi-
norm P3==SUP(p1, P2), defined by P3(U) ==max{P1(U) , P2(U)}, is the least
upper bound of PI and p2. Show that p4==inf (PI, P2), defined by
p4(u)==inf (P1(U1)+P2(U2): U1, U2 EM +, U1 +U2==U)
for every uEM+, is the greatest lower bound of PI and p2.
71.2) Show that if p==inf (PI, P2), then p' ==sup (pi, P2).
THE SPACE OF GOULD
71.3) We consider the space of G. G. GOULD, introduced in the
Exercises 30.13-30.16. We recall (cf. also Example (a) in sec. 65) that
this space may be obtained from an atomless a-finite measure # in a
point set X such that #(X) ==00. For any #-measurable I, we set
p(/) ==sup (/ III d#: #(E) == 1).
E
Evidently P is a saturated function norm with the Fatou property, and
hence p" ==p. In order to obtain p', denote the L 1 and Loo norms by PI
and Poo respectively, and observe that p==inf (PI, Poo) by the result in
Exercise 30.16. Show now by means of the preceding exercise that
p' ==sup (Poo, PI). The associate space L; consists, therefore, of all es-
sentially bounded summable functions. This was also proved by G.G.
GOULD [lJ in a different manner.
72. The Subspace of all Functions of Absolutely Continuous
Norm
We assume that p is a saturated function norm. The space L p is the
corresponding normed Kothe space; L; and L  are the first and second
associate spaces respectively. As shown in sec. 70, L; may be considered
as a closed linear subspace of the conjugate space L;; the comple-

476
NORMED KOTHE SP-\CES
[Ch. 15,  72
mentary closed linear subspace of all singular linear functionals will
be denoted by L:,s.
DEFINITION. The function fEL p is said to be of absolutely continuous
norm whenever p(fn)! 0 for every sequence fn (n== 1, 2, . . .) in L p such
that Ifl?::-f1?::-f2?::-. . . ! 0 pointwise almost everywhere on X.
The name is explained by the following lemma.
LEMMA ex. If fEL p is of absolutely continuous norm, there exists for
every e>O a number €5>0 such that #(E)<€5 imPlies p(fXE) <e.
PROOF. Note first that if fEL p is of absolutely continuous norm,
then p(fXEn)! 0 for every sequence En (n== 1, 2, . . .) of measurable sub-
sets of X descending to a set of measure zero. Assuming now that the
condition in the present lemma is not satisfied, there exists a number
eo>O without a corresponding €5>0. It follows that there exists a
sequence En (n== 1, 2, . . .) such that #(En) <2- n and P(fXEn)?::-eo for
all n. Then the sets Fn==En+En+1+'" descend to a set of measure
zero, but P(fXFJ?::-P(fXEn)?::-eo for all n, contradicting the fact noted
above.
If L p is an Lp space for some p satisfying lp<oo, then every
fELp==L p is of absolutely continuous norm. If, however, Lp==Loo and
# is Lebesgue measure on (-00,00), then the only function of abso-
lutely continuous norm is the null function. On the other hand, if L p
is the sequence space loo (i.e., # is discrete measure in the set X ==
{I, 2, . . .} and the norm is the supremum norm), then the functions
of absolutely continuous norm are the sequences converging to zero
(i.e., the functions of absolutely continuous norm are the elements of
the subspace (co)).
THEOREM 1. The function fEL p is of absolutely continuous norm if
and only if p(fXEn)! 0 for every sequence En (n== 1, 2, . . .) of measurable
subsets of X such that En descends to a set of measure zero.
PROOF. If f is of absolutely continuous norm, then f has evidently ,
the property stated in the present theorem. i\.ssume now, conversely,
that f has the property stated in the present theorem, let Ifl?::-f1?::-f2?::-

Ch. 15,  72J FUNCTIONS OF ABSOLUTELY CONTINUOUS NORM 477
. . .!O on X, and let e>O. We recall the existence of a sequence of sets
Xni X such that #(X n ) <00 and O<p(XxJ<oo for all n. Since X-X n
descends to the empty set there exists an index N such that p(fx X - XN )
<ie, and so p(fnXX-XN)<!e for all n. Furthermore, by Lemma ex, there
exists a number €5 >0 such that #(E) <€5 implies p(fXE) < ie. Fix this €5,
set ex===e!{4p(XXN)}' and consider the sets En==={x: fn(x) >ex}nX N for n===
1, 2, . .. . The sets En descend to a set of measure zero, so #(En) <€5
for n):;no. For abbreviation we shall denote the set Eno by Eo. For
n):;no we have fn(x)ex on XN-EO, so
P(fnXXN)P(fnXEo) +p(f nXXN-Eo)
p(fXEo) + exP(XXN) < ie+ ie=== !e.
This shows that p(fn)p(fnxXN)+p(fnxx-XN)<e for n?;:-no, and so
p(fn)!O as n-+oo.
THEOREM 2. The function fEL p is of absolutely continuous norm if
and only if, for every sequence fnELp such that Ifn(x)Ilf(x)1 for all nand
Jim fn(x)===fo(x) on X, we have P(fo-fn)-+O.
PROOF. Let f be of absolutely continuous norm, and let the sequence
/nELp be such that Ifn(x)Ilt(x)1 for all nand lim fn(x)===fo(x) on X.
Then the functions
un(x) ==sup (Ifn+m(x) -fo(x) I: m===O, 1, 2, . . .)
satisfy un(x)2If(x) I and U1>U2>. . . ! 0 on X. Since 21fl is of abso-
lutely continuous norm, we have p(un)!O. But Ifn(x)-fo(x)lun(x), so
P(fo-fn)p(un), and hence P(fo-fn)-+O.
If, conversely, f is in L p and f has the stated dominated convergence
property, then an application to any sequence of the special form
Ifl?;:-f1f2>' · . ! 0 shows that p(fn)! 0 for any sequence of this kind,
and hence f is of absolutely continuous norm.
Let A be a linear subspace of L p with the property that if fEA, g(x)
is measurable and Ig(x)Ilt(x)1 on X, then gEA. Any linear subspace
of this kind is sometimes called an order ideal in L p .

478
NORMED KOTHE SPACES
[Ch.15,72
THEOREM 3. The set L of all fEL p of absolutely continuous norm is
a norm closed order ideal in L p . If O/EL and Ofn i f on X, then
p(fn) i p(f). Finally, p"(f) =p(f) for every fEL.
PROOF. It is evident by Theorem 1 that L is an order ideal in L p ..
I t remains to prove that L  is norm closed. Let fnEL  for n= 1, 2, . . .
and p(f- f n) -+0 as n-+oo. Then, given e >0, there exists an index N
such that p(f-fN) < !e. Now, let En (n= 1,2, . . .) descend to the empty
set. Since
P(fXEJ p{ (I - fN) XEn}+ p(INXEn)  ie+ p(fNXEJ,
and since p(fNXEJ! 0 as n-+oo, we have P(fXEn) <e for n sufficiently
large. This shows that IEL, and so L is norm closed.
Now, let O/EL and Ofnif. Then p(f-fn)!O by the preceding
theorem, and so it follows from p(f)P(fn) +p(f-fn) that p(f)lim p(fn).
Hence p(fn) i p(f). Finally, it follows from the definition of the Lorentz
seminorm pL that PL(f)=p(/) for this particular fEL. Since PL=P", we
have thus that p"(f)=p(f) for O/EL. But fEL implies If I EL, so
p" (f) = p" (I f I) == p (I f I) == p (f)
for every fEL.
Let V be a normed linear space and V* the conjugate space; the
elements of V and V* will be denoted by x and x* respectively, and
the value of x* E V* at the point XE V by <x, x*). Given the subset
A c V, the element y* E V* is said to be orthogonal to A (notation
y* A) whenever <x, y*)==O for all xEA. The set A 1- of all y* A is
clearly a norm closed linear subspace of V*, and A J.. is called the an-
nihilator of A. Given the subset Be V*, the element XE V is said to be
inversely orthogonal to B (notation xT B) whenever <x, y*)==O for all
Y*EB. The set BT of all xBT is clearly a norm closed linear subspace
of V, and BT is called the inverse annihilator of B. It was shown in
Exercise 54.3 that (L1-)T ==L for every closed linear subspace L of V
but if M is a closed linear subspace of V* the inclusion in Me (MT) 1-
may be proper.
It is our purpose to prove now that L is exactly the inverse an-
nihilator of the closed linear subspace L;,s of L. We recall that L;,s. ,
is the set of all singular bounded linear functionals on L p .

Ch. 15,  72J FUNCTIONS OF ABSOLUTELY CONTINUOUS NORM 479
THEOREM 4. We have (L;,s)T =L.
PROOF. Let IEL and GEL;,s' We have to prove that G(/)=O. To
this end there is no loss of generality in assuming that I?;;O and G is
a non-negative functional. Since GEL;,s, the integral component of G
is the null functional, so (cf. Exercise 70.1) we have
inf (lim G(ln) : O/n t I) ==0.
(1)
But IEL, so O/n t I implies by Theorem 2 that p(/-ln) to, and hence
G(ln)G(/). This shows, by (1), that G(/)=O. It has been proved thus
that L  C (L;,s) T.
In order to obtain the inverse inclusion, assume that 1==/1 +i/2 (11
and 12 real) is a function in L p satisfying G(/) =0 for all GEL;,s' Then
G(/)=O for all non-negative G in L;,s, so G(/1)==G(/2)==0 for these G,
and hence G(/1)=G(/2)=0 for all GEL;,s' We have proved thus that
11+i/ 2 E(L;,s)T implies 11, 12E(L;,s)T. Next, let E be a fixed #-measur-
able subset of X. Given GEL, we set GE(/)=G(lxE) for every IEL p .
Then GEEL: and IIGEIIIIGII. Furthermore (GE)E==G E , so if PE(G) is
defined by PE(G)=G E for every GEL, then P E is a projection in the
space L with norm satisfying IIPEII 1. Also, G?;:;e implies eGEG,
so if G is an integral or a singular functional respectively, the same
holds for G E (cf. Exercise 70.3). In other words, if GEL;,s, then GEEL;,s
and so G E (/1) =G E (/2) =0 for the function 1=/1 +i/2 in (L;,s) T which we
consider. For E == {x: 11(X) >O} this yields 0==G E (/1) ==G(/1XE) =G(/t)
for all GEL;.s' Similarly G{/l)=G(lt)=G(/2)==0 for all GEL;,s' Thus
1= 11 +i/2 E (L;,s) T if and only if It,ll' It and 12 are all in (L;,s) T.
Let 1==11 +i/2 be a function in L p which is not in L. Then one at
least of It,ll' It and 12' say It, is not in L. The present proof will be
complete if we can show that I is not in (L;,s) T, and by the preceding
remarks it is sufficient for this purpose to show that G(/t) -=1-0 for some
GEL;,s' We may and will assume, therefore, that I?;;O is in L p but not
in L, and we will prove that G(/) -=1-0 for some GEL;,s' Since I is not
of absolutely continuous norm, there exists a sequence of sets En de-
scending to the empty set such that E1 ==X and p(IXEn)?;:;a for some
a>O and all n. Now, let (c) be the set of all convergent sequences
(C1, C2, . ..) of complex numbers, let Dn==En-En+1 {SO 1 Dn=E1
==X), and let V j be the set of all functions h(x)=l cnl(x)XDn(x) with
(C1, C2, . . .) E (c). If hE Vj, then Ih(x) I{sup Icnl}/(x), so hELp. It follows

480
NORMED KOTHE SPACES
[Ch.15,72
that V j is a linear subspace of L p (we are not interested in V j being
norm closed or not). For any h==l cnfXDnEVj we define G(h)==c oo ,
where coo==lim Cn. Then G(h) is a linear functional on Vj. Since G(h) ===
coo==lim Cn, there exists an index N (depending on h) such that IG(h)1
21Cnl for all n>N. Let hI ==  cnfXDn' Then
00
00
Ih(x) 1lh1(X) I ==  ICnlfXDn!IG(h) I tXDn ==!IG(h) IfxEN'
N J\T
so
p(h)!IG(h) Ip(fxEN»!cxIG(h) I.
This shows that G(h) is a bounded linear functional on the subspace Vj,
so it is possible (by the Hahn-Banach extension theorem) to extend G
as a bounded linear functional onto the whole space L p , Le., the ex-
tended G satisfies GEL. Let G==Ge+G s be the decomposition of G
into an integral and a singular functional, and let h n == fXEn == r=n fXDk
for n== 1, 2, . . . . Then h n E V f and G(h n ) == 1 for all n. Since h n ! 0 on
X, we have Ge(hn)-+O as n-+oo, so G s (h n )-+l. It follows that if G s ==
G1tiG2, where G 1 (g) and G 2 (g) are real for real gEL p , then G 1 EL;,s
and G 1 (h n )-+1 as n-+oo. But then GtEL;,s' and lim Gt(h n »l in view
of Gt(hn)G1(hn)' Since f>fXEn==hnO for all n, it follows finally that
Gt (t) 1. This is the desired result.
THEOREM 5. We have Lp==L if and only if L==L. In other words,
every fEL p is of absolutely continuous norm if and only if every bounded
linear functional on L p is an integral.
PROOF. If L;==L then L;,s consists only of the null functional, and
hence L == (L;,s) T is the whole space L p .
Conversely, if Lp==L, then the inverse annihilator of L;,s is the
whole space L p , which implies that L;,s consists only of the null
functional. But then every bounded linear functional on L p is an inte-
gral, i.e., L==L.
The next problem which arises is under which conditions the an-
nihilator (L)1- is again L;,s. As observed earlier, we have in any case
that L;,s is included in {(L;,s)T}1-==(L)1-. The inclusion may be proper.
Indeed, let # be Lebesgue measure in (-00,00), and Lp==Loo. The
singular functionals form a proper subspace of L: since the subspace of

Ch. 15,  72J FUNCTIONS OF ABSOLUTELY CONTINUOUS NORM 481
the integrals can be identified with L 1 . This shows that L;,s is properly
included in L:. On the other hand, L consists only of the null function,
so (L) J.. ==L:. In order to produce a reasonable solution to the problem
posed above, we first introduce the carrier of an order ideal in L p .
We recall that the linear subspace A of L p is called an order ideal
if it follows from lEA and Igl1/1 with g measurable that gEA. We
will say that the set F c X is disjoint to the order ideal A if every
lEA vanishes almost everywhere on F. Given the set EcX of finite
measure, set
cx==sup(#(F): FcE, F disjoint to A).
There exists an ascending sequence FncE, all Fn disjoint with A,
such that #(F n) i cx. Hence F == 1 F n satisfies # (F) ==cx, and F is still
disjoint with A. Obviously F is a maximal subset of E disjoint with
A, and F is (except for a null set) uniquely determined by this proper-
ty. Now, let Xni X with #(X n ) <00 for all n, let D 1 =X 1 and D n + 1 ==
X n + 1 -X n for n==l, 2, "', and let Fn be the maximal subset of Dn
disjoint with A. Then F=l Fn is the maximal subset of X disjoint
with A. The set C A ==X - F is called the carrier of the order ideal A.
If E is a subset of C A such that #(E) >0, there exists a function lEA
such that I/(x) I >0 on a subset of E of positive measure. Hence, there
exist a number 8>0 and a subset E 1 cE of positive measure such that
I/(x) I>e on E 1 . But then, since III EA and A is an order ideal, we have
XE1EA. Hence, every subset E of C A of positive measure has a subset
E 1 of positive measure such that XE1EA. It follows immediately, by
the exhaustion theorem (sec. 67, Theorem 3) that there exists a se-
quence of measurable sets YniC A such that XYnEA for all n.
THEOREM 6. We have (L)J..==L;,s il and only il the carrier 01 L is
the whole set X.
PROOF. Let the carrier of L be the whole set X, and let Y niX
such that XYnEL for all n. Since L;,s is included in (L)J.., it will be
sufficient to prove that (L) J.. c L;,s. For that purpose, let G E (L) J.. and
let G==Ge+G s be the decomposition of G into an integral G e and a
singular functional G s . Since every singular functional is an element of
(L) J.., we have G e E (L) J... If G e is represented by the function gEL;,
i.e., if Ge(/)==flgd# for every IEL p , it follows from GeE(L).L that

482
NORMED KOTHE SPACES
[Ch. 15,  72
f fg d#== 0 for every fEL. In particular f XEgd#==O for any measurable
set E included in some Y n. Fixing n for the moment, and letting E
run through all measurable subsets of Y n, it follows that g==O on Y n.
But then g==O on X, and so G e is the null functional, i.e., G==G s is
singular. Hence, any GE(L)J.. is singular, which is the desired result.
For the converse part, assume that the carrier of L  is a proper
subset of X in the sense that there exists a set E of positive measure
disjoint to the carrier of L. Hence, since p' is a saturated norm, there
exists a set FeE such that O<p'(XF) <00. It follows that G(f)==
f fXFd# is an integral on L p , not the null functional, and G(f) ==0 for
every fEL, so G E (L) J... This shows that (L) J.. is properly larger than
L;,s. In other words, if (L)J..==L;,s then the carrier of L must be the
set X.
THEOREM 7. The carrier of L is the whole set X if and only if the
conjugate space (L) * can be identified with L; in the sense that every
bounded linear functional G on L  is of the form G(f) == f fg d# for some
gEL; and all fEL, and such that IIGII==p'(g).
PROOF. Assume first that the carrier of L  is X; once more, let
Y n t X such that XY n EL  for all n== 1, 2, . . . . Given OfELp, let fn==
min(f, nXyJ for all n. Then Ofnif pointwise, and so G(fn)-+G(f) for
every integral G on L p . More generally, if f== (f(l) + f(2») +i(f(3) + f(4») is
the standard decomposition of an arbitrary fEL p into non-negative
components f(l), _f(2), f(3) and _f(4), and if fl)==min(f(l), nXYn)
for n== 1, 2, ... and similarly for the other components, then f n==
(fl) + f2») +i(f3) + f4») EL  for all nand f n converges pointwise to f, so
G(fn) -+G(f) for every integral G on L p . In addition, we have Ifn(x) I If(x) I
for all n and x, so P(fn)P(f) for all n. Hence, if e>O is given and fEL p
satisfies p(f)l and IG(f)I>IIGII-e, we have P(fn)l for all nand
IG(fn)I>IIGII-e for n sufficiently large. It follows that
IIGII==sup (IG(f) I : fEL, p(f) 1).
Hence, every integral G(f) == f fg d# on L p is a bounded linear functional
on L with unchanged norm IIGII==p'(g). It remains to prove that every
bounded linear functjonal on L  js of this kind. Given the bounded
linear functional G on L, it may be extended by the Hahn-Banach
theorem onto the whole of L p . Denoting the extension by G ext ' let

Ch. 15,  73J
REFLEXIVITY
483
Gext==Ge+Gs be the decomposition into an integral G e and a singular
functional G s . Since G s vanishes on L and Gext==G on L, we have
G(f)==Ge(f) for all fEL. This is the desired result.
It has been shown in the proof of the preceding theorem that if the
carrier of L is a proper subset of X, then there exists an integral G
on L p , not the null functional, such that G(f)==O for all fEL. Hence,
if it is given that (L) * ==L, the carrier of L  must be the whole set X.
Exercises
THE SPACE OF GOULD
72.1 ) We resume the investigation of the space of Gould considered
in Exercise 71.3. Show that if L p is the space of Gould, then L  con-
sists of all fEL p with the property that for every 8>0 there exists a
set Ae of finite measure such that If(x)I8 outside Ae. Hence L is a
proper subspace of L p , and so L is a proper subspace of L.
72.2) Show that the carrier of L is the whole set X, and hence
(L)*==L and (L)1-==L:,8'
73. Reflexivity
We assume again that p is a saturated function norm.
DEFINITION. If L==Lp, i.e., if every fEL p is of absolutely continuous
norm, we will say that p is an absolutely continuous norm.
It follows from sec. 72, Theorem 3 that if p is an absolutely continu-
ous norm, then OfnifELp implies p(fn)ip(f). Also, p"(f)==p(f) holds
now for every fEL p . Nevertheless it is not true that p has now neces-
sarily the Fatou property, or even the weak Fatou property. Indeed,
it n1ay happen that Ofn i fin M+ with lim P(fn) <00 and p(f) ==00. By
way of example, let # be discrete measure in X == {I, 2, . . .} and, for
every uEM+, let p(u)==sup u(n) if UE(CO) and p(u) ==00 otherwise.
LEMMA ex. If p is absolutely continuous and p has the weak F atou
property, then p has the F atou property.

484
NORMED KOTHE SPACES
[Ch.15,73
PROOF. Let Ouniu in M+. If p(u) is finite, then uELp==L, and
so Op(un) i p(u) as was observed above. If p(u)==oo we must have
lim p(u n ) ==00 in view of the weak Fatou property. Hence p(un) i p(u)
in either case, and so p has the Fatou property.
If p is absolutely continuous, then L;==L; by sec. 72, Theoren15.
Hence, if p and p' are absolutely continuous, then L ;==L; as well as
L';== (L;) *, and so
L" == ( L ' ) * == ( L * ) * == L * *
p p p p .
Explicitly, this means that there exists a one-one linear mapping ff**
of L' onto L;* such that p"(f)==llf**11 and such that G(f)==f**(G) for all
GEL;==L;', i.e., f fgd#==f**(G) for every G represented by a function
gEL;.
Similarly, we will mean by L;* e L'; in the algebraic sense that there
exists a one-one mapping f**f of L;* into L'; (not necessarily onto L;,
and not necessarily norm preserving) such that f**(G) ==G(f) for every
G L ' == L '"
E p p .
THEOREM 1. We have L;* e L' (in the algebraic sense exPlained above)
if and only if p and p' are absolutely continuous, and in that case L';==L;*
holds with norm equality for corresponding elements.
PROOF. It was noted above that L;==L;* holds if p and p' are abso-
lutely continuous. For the converse part, let L;* eL'; and assume that
p is not absolutely continuous, i.e., assume that L; is a proper sub-
space of L;. Then there exists an element f**EL;*, not the null ele-
ment, such that f**(G) ==0 for all GEL;. If f is the element of L' corre-
sponding to f**, then G(f) ==0 for every GEL, i.e., ffgd#==O for every
function gEL;. This, however, implies that f(x) ==0 almost everywhere
on X, and thus contradicts the fact that the corresponding f** EL;* is
not the null element. Hence L;==L; or, in other words, p is absolutely
continuous. But then L;*==(L;)*, and so the hypothesis that L;* eL;
holds is equivalent to (L;) * e (L;)'. Since the inverse inclusion (L;) I e
(L;)* is always satisfied, it follows that (L;)'==(L;)*, and so p' is abso-
lutely continuous.

Ch.15,73J
REFLEXIVITY
485
THEOREM 2. The space L p is reflexive if and only if p and p' are abso-
lutely continuous and p has the weak Fatou property (and hence the Fatou
property) .
PROOF. Assume first that p and p' are absolutely continuous and p
has the weak Fatou property. Then p has the Fatou property by
Lemma ex, so p"==p, and hence Lp==L'. Furthermore, we have L;==L:*
by the preceding theorem. It follows that Lp==L'==L:*, and so L p is
reflexive.
Next assume that L p is reflexive, so Lp==L:*. Since LpcL' is always
satisfied (in the algebraic sense, i.e., not necessarily with preservation
of norm), it follows that L;* c L' (in the algebraic sense). But then, by
the preceding theorem, p and p' are absolutely continuous, and L:* ==L 
holds. It follows that Lp==L:*==L', which shows that L p and L'; con-
tain the same elements. But then p has the weak Fatou property by
sec. 71, Theorem 3.
The last theorem, but restricted to function norms for which it was
assumed in advance that they had the Fatou property, was proved by
I. HALPERIN (1954, [lJ) and independently by W. A. J. LUXEMBURG
(1 955, [1]). I t may be observed that the theorem is a particular case of
a more general theorem for abstract "normed vector lattices" proved
by T. OGASAWARA (1942-44, [lJ, two papers written in Japanese); the
work of Ogasawara on this subject became known outside Japan only
after a number of years.

SOLUTIONS
CHAPTER 1
3.1) There are sets A,B such that Ai=B and d(A,B)==O. Also, if
d(A, B»O, and C==A+B, we have d(A, C) ==0 and d(B, C) ==0,. so
the triangle ineq uali ty fails to hold.
3.2) Take X ==R1, the real line with the usual distance, and let
A n=={x: (n+ 1 )-1 x n-1} for n== 1, 2, ... .
) d ( ) d d(x,z) d(y,z)
3.5 1 x, z + l(Y, z)  1 +d(x, z) +d(y, z) + 1 +d(x, z) +d(y, z)
1 1
= 1 + I/{d(x, z) +d(y, z)}  1 + I/d(x, y) = d1(x, y).
3.6) Let X==R1, and Fn=={x:x>n}.
3.7) Combine Exercises 3.5 and 3.6.
3.10) Let A=={rn} be the set of all rational points, B==R 1 -A its
complement, and assume that A==rrrOn, where all On are open.
Then B== r F n where F n is closed, so R 1 == F n+ {rn} is a countable
union of closed sets. By Baire's category theorem one at least of the
Fn contajns an open interval], so ]eB, a contradiction.
3.11) Let P n, for n== 1, 2, ..., be the subset of R 1 consisting of all
irrational x such that rx>n- 1 , and note that the set of all irrational
points (as a subset of R 1 ) is of the second category. Hence Pn is
of the second category, so that by Baire's theorem one at least of
the sets P n, say for the index no, contains an interval ]. A simple
geometrical argument shows now that all points (x, y) with irrational
XE], 0<y<2/no, are points of B, so any circle C x corresponding to a
rational point XE] has points in common with B.

Ch.1,2]
SOLUTIONS
487
3.12) Let II be an arbitrary closed subinterval of I, and
A k== {x : XEI 1, fk(X) ==O}
for k == 1, 2, .. . .
By Baire's theorem, there is an A k containing an interval J ell. Since
Ik is continuous, we have Ik(X) ==0 at all XEJ for this particular k, so
f(x) ==0 on J by repeated differentiation. Since II was arbitrary, it
follows that I(x) ==0 at all points of a dense set; and since 1 is con-
tinuous, this implies that 1==0 on I.
3.13) Let ra be the radius of convergence of the Taylor series of f
about the point a, Le., l/ra=1im sup I/(n)(a)/n!11/n. Since l/ra<oo for
each aEI, the number Sa==supn I/(n)(a)/n!1 1 /n is finite as well. Let II
be a closed subinterval of I, and
Ak={a: aEl 1 , sa<k}
for k=1,2,....
By Baire's theorem there is an Ak containing an open interval J (so
J ell), so that by the continuity of all/(n) we have I/(n)(x)fn!1 1 /n k
at all XEJ and for all n. It is now easy to show that 1 is analytic on J.
This shows that the set of points of analyticity is dense; since the same
set is open (by the definition of analyticity), its complement is nowhere
dense.
CHAPTER 2
4.1) Consider the class of all a-fields A  To. This class is not empty,
since the collection of all subsets of X is a a-field which includes To.
The intersection of all these a-fields A (i.e., the collection of all subsets
A of X satisfying A EA for all a-fields A  To) is again a a-field AD,
and AD is evidently the smallest a-field including To.
4.3) Let X be the set of all rational numbers in {x: O<x  I}, and
let Tbe the semi-ring of all sets A =={x: a<x b}nX. The total number
of points in X is countable; X == {xo, Xl, X2, .. .}. For each X n *xo there
exists a set AnEr containing Xn but not containing Xo. Then D==
X - r A n is the difference of two a-sets, but D itself, consisting
only of the point Xo, is not a a-set.
A second counterexample is obtained by choosing for X the set
{x: O<xl}, and for the semi-ring Tthe collection of all {x: a<xb}
contained in X. Let A1=={X: l<xi}; A2={X: !<xj} and A 3 ==

488
SOLUTIONS
[Ch. 2
{x: <X}; next, let A4, A5, A6, A7 be the left open middle third
parts of the intervals which remain after the removal of A 1, A 2, A 3,
and so on. Then D ==X - r A n is the difference of two a-sets, and
although D is not empty, D does not contain any subinterval of X.
Hence, D is not a a-set.
5.3) Select a point xoEA. For any EeX, define #(E)==l whenever
xoEE and #(£)==0 otherwise.
7.1) Since fl(5) ==inf #l(A n) ==inf #*(A n) for all possible A n  5
such that all AnEA, and since An5 implies #*(5) #*(An) 
fl*(An), it follows that #*(5) #(5). If fl* IS regular, i.e. if fl==fl*,
then #* is generated by (X, A, #1). If, conversely, #* is generated by
(X, F, #), then Theorem 6 in the present section shows that #* is also
generated by (X, A, fl1), where A is the collection of all #-measurable
sets, and #1 ==#* on A. Hence, #* is regular.
7.2) Let X consist of at least two points; let #*(0)==0, fl*(X)==3
and fl*(5)==2 for 0*5*X. The only measurable sets are 0 and X,
so #(5)==3 for 0*5*X.
7.3) The only property of # which needs a formal proof is the a-
subadditivity. If the sets 5n are arbitrary, and if T is any set included
in 5n and satisfying #*(T) <00, then
#( 5n) ==sup #*(T) ==sup #*( T5n) sup  #*(T5 n ).
But #*(T5 n ) #(5n) for all such T, hence #(5n) #(5n). The
proof that #e-measurability implies #-measurability is trivial, if we
make use of the fact that #(5) ==#*(5) for any 5 satisfying #*(5) <00.
Conversely, let E be #-measurable; we have to prove now that #(5E)
+#(5Ee)#(5) for any 5 satisfying #(5)<00. Then #(5E) and
#(5Ee) are finite; hence, given e >0, there exist sets TIe 5E and
T 2 e SEe such that
#(5E) -e<#*(T 1)#(5E)
and
#(5Ee) -e<#*(T 2)#(5Ee).
For T==T1+T2 we have T e 5, TE==T1, TEe==T2 and #*(T) <00,

Ch. 2J
SOLUTIONS
489
so
#(SE) + #(SEC) <#*(T 1) +#*(T 2) +2e
==#*(TE) +#*(TEe) +2e==#*(T) + 2e#(S) +2e.
Since e>O is arbitrary, the desired result follows.
7.4) Let TncE be a sequence such that #*(Tn)<oo and #e(E)==
lim #*(T n); note that such a sequence exists also when #c(E) ==00.
Furthermore, let On be a a-set of finite measure such that On  T n.
'fhen Fn==EOn is #-measurable, TncFncE, and #(Fn) <00 for all n.
Hence #*(Tn)#(Fn)#c(E), so #e(E)==lim#(Fn).
7.5) If #*(5)==00, then #ie(S)#*(S) holds. If #*(5)<00, and e>O
is given, there exists a set  AnS such that all AnEF and  #(An)
<#*(S)+e, hence
 #ie(An)<#*(S)+e.
It follows that
#ic(S)<#ie( An) #ic(An)<#*(S)+e,
hence #ic(S)#*(S), Similarly #(S)#i(S). If #*(5) and #i(S) are finite,
then
#ic(S)# * (5) == #( S)#i (5) ==#ie(S),
so the inequalities are equalities.
7.6) The only fact which is not immediately evident is that #0-
measurability implies #-measurability. Let E c Y be #o-measurable,
and let 5 c X. Then
Ee==(y -E)+ ye,
so
#*(SEe)#*{S(y -E)}+#*(sye);
hence
#*(SE) +#*(SEc)#*(SE) +#* {S(Y -E)}+ #*(Sye)
==#*(SY) +#*(sye) ==#*(5).
9.1) Let #(X) ==#*(E) +#*(Ee), and let ScX be arbitrary. If T is a
measurable cover of 5, then #*(E) ==#*(ET) +#*(ETe) and #*(Ee) ==

490
SOLUTIONS
[Ch. 2
p*(EeT)+#*(EeTe), so by addition
#(X) ==#*(TE) +#*(TEe) + #*(TeE) +#*(TeEe)
>#(T) +#(Te) =#(X);
hence
#*(TE) +#*(T Ee) +#*(TeE) +#*(TeEe) =#(T) +#(Te).
Subtraction of #*(TeE)+#*(TeEe»#(Te) shows, since all numbers in-
volved are finite, that
#*(TE) + #*(TEe)#(T),
so that finally, since 5E c TE, SEe c TEe and #(T) ==#*(5), we obtain
#*(5E) +#*(5Ee)#*(5).
9.3) If #* is regular, then each 5 has a measurable cover by Theo-
rem 2 (2). Assume, conversely, that each set 5 has a measurable cover.
By Exercise 7.1 the exterior measure #i, generated by (X, A, #1), satis-
fies #*#i; assume now that #*(5) <#i(5) for some 5. If E is a measur-
able cover of 5, then #1(E)==#*(5)<#i(5)#i(E)=#1(E), a contra-
diction.
9.4) Evidently #(5»sup #*(5C) for all a-sets C of finite measure.
Conversely, if Tc5 is such that #*(T)<oo, there exists a a-set CT
of finite measure, and then T c 5C c 5 and #*(5C) <00; hence
sup #*(5C)sup #*(T)=#(5).
It follows already that #(5)=sup #*(5C) for all a-sets C=l Ai
(AiEF) satisfying #(C)<oo. The partial sums Bn= Ai are I-sets,
and #(Bn) i #(C) by sec. 8, Theorem 3, so #*(5B n ) i #*(5C) on account of
#*(5C)#*(5Bn) +#*{5(C -B n )}.
The desired result follows.
9.5) The #l-measure, #2-measure and #-measure of any I-set are
simultaneously finite or infinite; furthermore, if B 1 and B 2 are two
such I-sets of finite measure, then B 1 +B2 is also an I-set of finite
measure. Now, if #(5)<00 and e>O, the preceding exercise shows the,
existence of I-sets B 1 and B 2 of finite measure such that #ie(5) <
pi(5B 1 )+e and #2e(5) <#2(5B 2 )+e. Then the I-set B=B1+B2 is of

Ch. 2J
SOLUTIONS
491
finite measure, and
/-lie (5) + /-l2e(5) </-li(5B) +/-l2(5B) + 2e==/-l*(5B) +2e,
'So /-lie(5)+/-l2e(5)/-l(5), Conversely, if B runs through all I-sets of
finite measure, then
/-l(5)==sup /-l*(5B)sup /-li(5B)+sup /-l2(5B) ==/-lie(5) +/-l2e(S).
9.6) Observe first that any a-set 0 (with respect to r) of finite /-l-
measure is /-ll-measurable, and /-llc(O)==/-l(O); the same holds then for
the limit 0" of any descending sequence of such a-sets. Furthermore,
it follows immediately from the inequality /-lie/-l*, proved in Exer-
.cise 7.5, that any /-l-null set is a /-lIe-null set. Hence, since any /-l-measur-
.able E of finite measure is the difference of a a,,-set 0" and a /-l-null
set, E is /-ll-measurable and /-lle(E)==/-l(E). Next, assume that F is /-l-
measurable, and Fe BEr 1 with /-ll(B) <00 (then /-l(F)may still be +00).
Then
/-le(F) +/-le(B-F) ==/-le(B) ==/-ll(B) <00.
'Since /-le(B-F) is finite, there exists for a given e>O a /-l-measurable
TcB-F such that
/-le(B -F) -e</-l(T)/-le(B -F).
By what has already been proved, this set T is /-ll-measurable, and
/-ll(T) ==/-lle(T) ==/-l(T) >/-le(B-F) -e.
Then B-T is also /-l-measurable as well as /-ll-measurable, and
/-ll(B - T) ==/-ll(B) -/-ll(T) =/-le(B) -/-ll(T) </-le(B) -/-le(B -F) +e
==/-le(F)+e
(all numbers involved are finite). Since FcB-T, it follows that
/-li(F)/-ll(B - T) </-le(F) +e,
hence /-li(F)/-le(F). But then /-li(F)==/-lc(F) by the general relation
Jl/-li. Similarly /-li(B-F) ==/-le(B-F) , so
/-li(F) + /-li(B -F) ==/-ll(B).

492
SOLUTIONS
[Ch. 2
Finally, let E be an arbitrary #-measurable set, and assume that E is
not #l-measurable. Then there exists a set BEI"l of finite #l-measure
such that #l(B) <#i(BE)+#i(B-BE), contradicting what has just
been proved (choose F ==BE). Hence, any #-measurable set is #1-
measurable; the converse is proved similarly. This shows that A==A 1 .
If E EA ==A1, and F is an arbitrary measurable set such that F c E
and #(F) <00, then
#c(E)==sup #(F)==sup #le(F)#lC(E),
and similarly #lc(E)#e(E).
9.8) For the second part, observe that if B 2e is an I-set satisfying
B2eE and v(B 2e -E)<!e, then there exists an I-set CeB2e-E such
that v(Ce)<e. It is not difficult to show that B1e==B2e-Ce is"an I-set
satisfying B 1e cE and B 2e -B 1e c C e , so v(B 2e -B1e)<e.
9.9) The measurability of 0 6 follows from the inclusion theorem.
9.11) In the proof of the first assertion, choose T such that TS}
T is measurable, and v(T)<v*(S)+e.
9.13) For the proof of the assertion corresponding to the assertion
in Exercise 9.6, observe first that any I-set 0 (with respect to r) satis-
fying v(O)<oo is vI-measurable, and V1c(0)==V(0); by the inclusion
theorem in Exercise 9.8 the same holds then for any v-measurable set
E satisfying v(E) <00.
9.14) Evidently #:#i in X, and #a(B)==#f(B) for any I-set B. If
EEAf' and AEF satisfies #(A)<oo, then
#(A)#:(AE) +#:(AEe)#i(AE) + #i(AEe) ==#(A),
hence there is equality, implying that EEAa. It also follows that now
#a(AE)==#f(AE), i.e., #a(E)==#f(E) for any EEAf which is covered by
some A EF of finite measure. The same is then true for any EEAt
which is covered by an I-set of finite measure, that is, for any E EAt
satisfying #f(E) <00. Finally, if E EAf is arbitrary, and B runs through
all I-sets of finite measure, then
#ac(E) ==sup #a(EB) ==sup #f(EB) ==#fe(E).
9.15) Evidently OT(E)v(E) for all EELt, and T(0) ==0. It is suf-,
ficient, therefore, to prove that T is monotone and a-additive. It is easy
to see that T(E1+E2)==T(E1)+T(E2) for disjoint sets El, E 2 of Lt, and

Ch. 2J
SOLUTIONS
493
this shows already that T is monotone. Assume now that EELt, En ELt
(n==1, 2, .. .), where all En are disjoint, and E=='L/f En. Then T(E)
T(l En)==l T(En) for all p, hence T(E)l T(En). For the proof of
the converse we may assume that 1 T(En) is finite. Given e>O, let
FnjELt be such that Fnji En as j-+oo, and limj v (Fnj) <T(En)+ef2n for
n==1,2, .... Writing now Hn==F1n+F2n+...+Fnn for n==I,2,
. . " we have H n iE, and
v(Hn)==v(F1n)+V(F2n) + . . . +v(Fnn)<T(E 1 ) + ... +T(En)+e
00
 T(En)+e
1
for all n, hence T(E)l T(En) +e.
9.16) Observe first that it follows from the parallel for charges of
Exercise 9.7 that (X, F, v) and (X, Lt, v) generate the same collection
A of measurable sets with the same contracted charge Vc. The equalities
A==ATnA p and Ve==Te+Tpe follow from Exercise 9.10 and the parallel
for charges of Exercise 9.5. Finally, Exercise 9.14 shows that Te is a
measure on AT, and so Te is a measure on A.
9.17) Assume first that v is a pure charge on A. Defining T(E) for
all EEA by T(E)==inf{limv(Fn)} for all sequences FniE, FnEA, it
follows from Exercise 9.15 that T is a measure on A satisfying OTV
on A. Hence T(B) ==0 for all BELt, and then also T(E) ==0 for any EEA
satisfying v(E) <00 (since such a set E is covered by some BELt). Given
e>O and such a set E, there exists (in view of T(E)==O) a sequence
F n i E, F n EA , such that lim v(F n) <e. Defining E 1 ==F 1, En==F n- F n-1
for n==2, 3, ..., E==l En is the required decomposition.
Assume, conversely, that for any e>O and any EEA satisfying
v(E)<oo there is such a decomposition, and assume also that # is a
measure on A such that O#(B)v(B) for all BELt. We have to prove
that #(B)==O for all BeLt, and for this purpose we show first that
#(E)v(E) if v(E) <00. Indeed, given r;>0, there exists a set BELt
such that BE and v(B) <v(E) +'Y), hence
#(E)#(B)v(B) <v(E) +'Y),
so #(E)v(E). Now, let BELt and e>O. By hypothesis, B== 1 En
with disjoint EnEA and 1 v(En) <e. Then#(B) == 1 #(En)l v(En)
<e, hence # (B) ==0.

494
SOLUTIONS
[Ch. 2
9.18) Let # be a measure on A satisfying O#Tp on Lt. In view of
the definition of the measure T on Lt there exists, for any BELt and any
e>O, a sequence BnELt such that BniB and limv(Bn)<T(B)+e. It
follows then from O#(Bn)Tp(Bn) ==v(Bn) -T(Bn) and from the a-
additivity of # and T that
O#(B)lim V(Bn)-T(B)<e.
Hence #(B) ==0, i.e. Tpe is a pure charge on A.
9.19) Let e>O and EEA with v(E) <00. By Exercise 9.17 there exist
sequences FniE and FiE such that limTpe(Fn)<e and limTc(F)
<e. It follows that Hn==FnF satisfies HniE, and limTpe(Hn)<e,
lim Tc(Hn)<e. Then
Ilim {Tpe(H n) -T c(H n)}1 <e;
hence (since Tpe==Ve-Te and Tc==Ve-T for the sets Hn) we have
Ilim {T (H n) -Te(H n)}1 <e.
By the a-additivity of Te and.T this implies IT(E)-Te(E)I<e; hence
T (E) ==Te(E) , and then also T c(E) ==Tpe(E).
If E E A is ar bi trary, and B runs through the sets of Lt, then
T(E)==sup T(EB)=sup Te(EB) ==Te(E) ,
and similarly T c(E) ==Tpe(E).
9.20) If A == 1 Ai, where A and the disjoint sets Ai are sets of r,
then
#n(A)== #n(Ai) #(A i )
i i
for all n,
so #(A)i #(A i ). Conversely,
p
 #n(Ai)#n(A)#(A)
i=l
for all nand p,
so f=lll(Ai)#(A) for all p, i.e. 1 #(Ai)#(A).
9.21) All a-sets and a6-sets with respect to rare #-measurable as
well as #n-measurable for all n. It follows then immediately from the
covering theorem in the present section that any #-null set is a #n-
null set for all n. Hence, once more by the covering theorem and ob-

Ch. 2J
SOLUTIONS
495
serving that fl is a-finite, we obtain the result that any fl-measurable
set is fln-measurable for all n.
Next, we prove that fln(E) i fleE) for all fl-measurable E. Since evi-
dently fl1fl2' . . fl for any a-set, the same inequalities hold by the
covering theorem for any fl-measurable set; it is sufficient to prove,
therefore, that fln(E)fl(E). If E is a a-set E== Ai with disjoint
AiEF, we introduce the measures T1==fl1, Tn==fln-fln-1 for n>l. Then
fl(E)== fl(Ai)== {Tn(Ai)}== {Tn(Ai)}== Tn(E)
i inn i n
==lim fln(E).
If E is a a6-set of finite fl-measure, we select a-sets En of finite fl-
measure such that En!E, so E1-E==r (E i -E i + 1 ). Since flnifl for
each Ei-Ei+1, the same holds for E 1 -E, and then also for E. Finally,
observing again that any fl-null set is a fln-null set for all n, we obtain
by means of the covering theorem the desired result for any set of
finite fl-measure, and hence for any fl-measurable set.
Finally, let E be fln-measurable for all n. We have to prove that E
is fl-measurable. By the covering theorem there exist a6-sets F n  E
such that fln(Fn-E) ==0, so F==I1r Fn is a a6- set , and hence fl-
measurable as well as fln-measurable for all n, such that F  E and
fln(F -E)==O for all n. Similarly, by considering the complement Ee==
X-E, we obtain a set GeE, fl-measurable as well as fln-measurable
for all n, such that fln(E-G) ==0 for all n. It follows that F-G is fl-
measurable and fln(F-G)==O for all n, so fl(F-G)==O by what has
already been proved. Then, since GeEeF, the set E-G is a fl-null
set, so E ==G+ (E -G) is fl-measurable.
9.22) If E i is the set consisting of the single point i, and if n>i,
then fln(E i ) ==Pni==Pi, so T(Ei) ==Pi. Furthermore fln(X) ==00 for all n,
so T(X) ==00. But X== Ei, so T( Ei)==oo and  T(Ei)== Pi<OO.
10.2) T and its inverse T-1 transform cells into cells, and fl(AT)==
fl(A) for any cell A. It follows that T and T-1 transform a-sets into
a-sets of the same measure. Furthermore, if the arbitrary set 5 is
covered by the a-set B, then ST is covered by BT, and conversely.
Taking greatest lower bounds, one obtains fl*(ST)==fl*(S),

496
SOLUTIONS
[Ch. 2
If E is measurable and 5 is arbitrary, then
# * (5ET) + #* (5E) == #* {(5 T-IE):P}+ #* {(5 T-IEe)T}
=#*(5 T-IE) +#*(5 T - 1 Ee) =#*(5 T-l) =#*(5),
so ET is measurable. The same argument with T and T-1 interchanged
shows that E is measurable if E T is so.
10.3) Observe that any upper or lower Riemann sum is the measure
of a finite union of two-dimensional cells, and use the inclusion theo-
rem of sec. 9.
10.4) By Exercise 10.2 we may assume that the origin is the centre
of the circle. The result follows then from the preceding exercise; con-
sider first the case that the sector is given in polar coordinates (r, qJ)
by {(r, qJ) : Ora, OqJex} with ex!n, the extension to larger values
of ex is immediate, and the final result for {(r, qJ) : Ora, qJ1qJqJ2}
with qJ2-qJ1 ==ex follows by addition or subtraction.
10.5) Obviously, #*==#1 on Fl. In order to show that #i=# on F,
note that, given A EF and 8>0, there exists a a-set 0, with respect to
F1, such that A cOcB, where BEF and #(B)-#(A)<8.
10.6) We may restrict ourselves to sets E included in a fixed annulus
P given by O<ar<b<oo, since R 2 is the union of the origin and a
sequence of such annuli. The set E c P is n-measurable as well as Le-
besgue measurable if E is a a-set with respect to F1, so also if E is the
limit of a descending sequence of such a-sets. In view of the preceding
exercise it remains only to prove that n(E) ==0 if and only if #(E)=O.
This is derived by observing that #(A)=!(r-ri)(qJ2-qJ1) for any
AEF1, so a#(E)/n(E)b if EcP is a a-set with respect to Fl.
10.7) By Exercise 10.2 we may assume that Fc (0, 1; . . . ; 0, IJ. The
additive group Rk is decomposed into cosets with respect to the sub-
groupd of the rational points in Rk. We choose one point in F from
each coset which has at least one point in common with F, and define
E to be the set of the thus chosen points. Hence E c F c (0, 1; . . . ; 0,
1J. Let {aI, a2, ...} be the set of all rational points in (-1,1; ...;
-1, 1), and En=={x: x=e+a n , eEE}. It is easy to see that the sets En
are mutually disjoint, and that FCl En c (-l, 2; ...; -1,2). As-
suming now that E is measurable and #(E)=ex, we have #(En)=ex for
all n, so that  #(En)=#( En)3k implies #(E)==ex=O. Then 0<
#(F) #(En)=O, a contradiction.

Ch. 2J
SOLUTIONS
497
10.8) There is a a-set O-:::;)E such that #(E»()(,#(O). Let O==An,
where the A n are disj oint cells. Then
 #(EA n) ==#( EA n) =#(E) > ()(,#(O) =()(,  #(A n),
so #(EA n) >()(,#(A n) for at least one value of n, say no. The cell A ==A no
satisfies the required condition.
10.9) Since #(P)=#(Q»!#(A) and PuQcAuB, so #(PuQ)
!#(A), the sets P and Q are not disjoint, i.e. there is a point YEPQ.
Then YEP=EA, and also Y==Y1+X with Y1EP=EA. It follows that
X=Y-Y1EJ(EA) cJ(E).
10.10) If # (G) >0, the difference set J(G) of all points x-Y, xEG,
YEG, contains an open interval 0 around the origin. Since G is a sub-
group, we have 0 c J (G) c G, and it follows that G contains every point
of Rk.
10.11) Since E is not an atom there is a subset F of E such that
#(F) and #(E - F) are positive. One of F and E - F, call it F 1, satisfies
#(F1)!#(E). Since F 1 is not an atom, the procedure may be repeated.
10.12) Consider the collection of all subsets F of E satisfying #(F)()("
identify #-almost equal sets and introduce partial ordering by inclusion
in the collection. Let {F T} be a chain in the collection, and set fJ ==
SUPT #(F T), so fJ()(,. There exists a sequence F n, each F n one of the F T,
such that lim#(Fn)==fJ. Let F==Fn, so #(F)==fJ()(" which shows
that F is in the collection. Take an arbitrary F T from the chain. If
FTC F n for some F n, then FTC F; if F T -:::;) F n for all n, then F T -:::;) F,
and so FT=F on account of #(FT-F)=#(FT)-#(F)fJ-fJ=O. This
shows that F is an upper bound of the chain in the considered col-
lection. But then, by Zorn's lemma, the collection has a maximal ele-
ment E 1 . It follows that #(E 1 )==()(" since otherwise the union of E 1 and
a subset of E-E 1 of sufficiently small positive measure would still be
in the collection, contradicting the maximal property of E 1 .
10.13) Let a and b be irrational numbers satisfying a<b, and let r
be the semi-ring of all rational cells in the real line having no endpoint
in the interval [a, bJ. Let # be Lebesgue measure on r. The semi-ring
r has the stated property, but [a, bJ is an atom for the extended
measure #.
10.14) Assume O<#(E)<oo, and let 0 be a a-set covering E such
that (l-y)#(O)<#(E). We may assume that O== An, where the sets

498
SOLUTIONS
[Ch. 2
An E r are disjoint and #(A n) >0 for all n. To justify the last condition,
neglect the part of E which is covered by the sets An of measure zero,
and hence neglect these sets An. Corresponding to each An we have
BnEr such that Bn cAn and
y#(An) <#(Bn)< (l-y)#(An).
Let E1== EBn. Then E 1 cE and
#(E1)== #(EBn) #(B n )«l-y)  #(An)==(I-y)#(O)<#(E).
We prove that #(E 1 ) >0. If #(E 1 ) ==0, then #(EBn) ==0 for all n, so EAn
is #-almost included in E(An-Bn), which implies that E is #-almost
included in  (An-Bn). But then
#(E) #(An-Bn) < (I-I')  #(A n )== (1-1')#(0) <#(E),
a contradiction. This shows that the arbitrary set E of finite positive
measure is not an atom.
10.16) Any atom FeE is of finite measure since E, and hence F,
is of a-finite measure. Identify #-almost equal atoms. Then different
atoms are disjoint, so the total number of different atoms contained in
E is at most countable, say F 1, F2, . . . . Set E 2 ==  F nand E 1 ==E-E 2 .
Evidently the decomposition is #-almost unique.
11.1) Since F==lim sup Fn==rr;;=l (Fn+Fn+l+" .), we have
F -Fn c (Fn+1 +Fn+2+. . . )-F n
c (F n+1 +F n+2F+1 +F n+3F+2+ . . .) -F n
00
cFn+lF+Fn+2F+1+""  (Fn+p-Fn+p-1) ,
p=l
so
#*(F -F n)<2- n +2-(n+l)+. . . ==2-(n-l).
Also, if xEFn-F, then xEFe, i.e. xEF+p for infinitely many p, so
XE=l (Fn-Fn+p). Hence
00
Fn-Fc  (Fn-Fn+p)==Fn(F+1+F+2+"')
p=l
cFn(F+1+Fn+1F+2+Fn+2F+3+" .)
00
cFnF+1+Fn+1F+2+"'==  (Fn+p-I-Fn+p) ,
p=l

Ch. 2, 3J
SOLUTIONS
499
so #*(F n -F)<2-<n-l). It follows that limp(F, Fn)=O as n-+oo, and
in view of
p(F, En)<.p(F, Fk)+p(Fk' En)
we obtain limp(F, En)=O as n (and k)-+oo.
11.3) v*(5)<.cx follows from
v*(5)<.v*(E n ) +v*(5 -En)<.cx+v*(5 -En).
Note tat A -B is an I-set of positive charge, so that, since the
sequence En (by its definition) invades into each cell contained in A
(and, therefore, invades into A-B), we obtain the result that En-B==
En(A -B) is an ascending sequence of I-sets, non-empty for n):;no.
CHAPTER 3
13.1) Let P(x)EL+ have the exterior ordinate set P, and consider
the intersection PG. This intersection is included in the graph of the
function h==min(l, P), and since J(h) <00, it follows immediately from
the proof of Theorem 5 (2) that the graph of h is of ,a-measure zero.
Hence ,a(PG)=O, and this implies ,a(AG)==O for all A E r. An appeal
to sec. 8, Theorem 1, Corollary (1) con1pletes the proof.
13.2) An immediate application of sec. 6, Theorem 2.
13.3) According to Exercise 9.4, ,ae(E)=sup ,a(EB) for all I-sets B.
Given such an I-set B= r (Pn-Qn), where Pn and Qn are the ex-
terior ordinate sets of PnEL+ and qnEL+ respectively, let p==max(p1,
"', Pr) and q==min(q1, ..., qr). Then A===-p-QE r and BcA, so
,a(EB)<.,a(EA). Hence ,ae(E) <.sup ,a(EA) for all A E r. The inverse ine-
quality is evident. The assertion (b) follows easily from (a), and (c)
follows easily from (b).
13.5) By Exercise 9.8, there exist I-sets B1, B 2 such that v(B 2 -B 1 )
<8 and the ordinate set F of I satisfies B 1 cFcB 2 . Let B1==
r (Pn-Qn), where P n , Qn are the ordinate sets of Pn, qnEL+(Pn):;qn),
and all P n-Qn are disjoint. Then 11 == r (Pn-qn) EL +,/1<'1, and /(/1)
==V(B1)' Similarly, construct 12EL+ such that 12):;1 and /(/2)==v(B2).
13.6) Assume first that /(/) and /(g) are finite. Given 8>0, the
preceding exercise ensures the existence of functions 11,/2, gl, g2 EL +

500
SOLUTIONS
[Ch. 3
such that 11<'1<'/2, gl<.g<.g2, and /(/2-/1)<!e, /(g2-g1)<!e. Then
11 +gl <.1+g<./2+g2,
where 11 +gl, 12+g2 EL + and /(/2+g2)- /(/1 +gl)<e; hence l+gEM+
by the inclusion theorem in Exercise 9.8. It follows then that
/(/2+g2)-/(/+g)<e, and since also
/(/2)- /(/)+ /(g2)- /(g)<e,
we obtain
I/(/+g)- /(1)- /(g)l<e.
Hence /(I+g)==/(/)+/(g).
The proof for the case that one at least of the numbers /(/) and
/(g) is infinite, is the same as in part (d) of the proof of Theorem 10 (1).
13.7) Assume first that v(F -G) <00, where F, G are the exterior
ordinate sets of I, g respectively. Then, given e>O, there exist I-sets
B1, B 2 such that B 1 cF-GcB 2 and v(B 2 -B 1 )<e. Let
r
B1== (P1n-Q1n)
1
and
s
B 2 ==  (P2n -Q2n)
1
with disjoint terms; then
r
b1== (P1n-Q1n)EL+,
1
s
b2== (P2n-Q2n)EL+,
1
b 1 <.I-g<.b 2 , and /(b 2 -b 1 )<e. Hence l-gEM+ by the inclusion theo-'
rem. Next, assume only that /(g) <00, and let PEL+. Then
min(/-g, p)==min(/, p+g)-g for all XEX, and since /{min(/, g+P)}
<00, we have min (I, g+P) -gEM+ by what has just been proved.
Hence min(/-g, P)EM+ for all pEL+, and this implies l-gEM+.
14.2) Let I, gEM+. For PI, P2EL+ we have
J{min (I, P1)}+J{min (g, P2)}==J{min (I, PI) +min (g, P2)}
<.{min (I+g, PI +P2)}<.J e (/+g);
hence f"e(/)+Jc(g)<.Je(/+g). The converse inequality follows from
J{min (/+g, p)}<.{min (I, p)}+J{min (g, P)}<.Je(/) +Jc(g).
The last assertion follows from Exercise 9.7.

Ch. 3 J
SOLUTIONS
501
14.3) The last assertion follows from Exercise 9.13; a direct proof
that the repeated extension yields the same summable functions with
the same integral may be derived from the inclusion of a summable I
between functions 11, 12EL such that /(/2-/1) is arbitrarily small.
14.4) On L the functional /(/) and the Riemann integral are identi-
cal. If I is /-sumn1able, and e>O, there exist 11, 12EL such that 11
1/2 on X and /(/2)-/(/1)<e. The step functions 11 and 12 may be
expressed in terms of the sanle disjoint cells AI, . . " A p (some coef-
ficients may then become zero); let At be a cell satisfying At l An.
Then 11 and 12 may be expressed in terms of disjoint cells B1, . . " Bq
such that At=== f Bn. Since 11 and 12 vanish outside At, the same holds
for I, and since the upper and lower Riemann sums 5 and s of I, corre-
sponding to the decomposition of At into B1, . . " Bq, satisfy / (/1)
s5/(/2), we have 5-s<e. Hence, I is Rien1ann integrable. Finally,
since / (I) and the Riemann integral of I are both included between
/ (11) and / (/2), it follows that / (I) is the Riemann integral of I.
Conversely, if I is Riemann integrable over the cell At and vanishing
outside At, there exists for any e>O a decomposition of At into dis-
joint cells B1, . . ., Bq such that the corresponding upper and lower
Riemann sums 5 and s satisfy 5-s<e. By the definition of 5 and s,
there exist step functions 11 and 12 such that 11//2 on X, and /(/1)
===s, /(/2)==5. Hence, I is /-summable.
16.1) In order to show that L is dense in LIe, let IEL 1e and e>O
be given. Let first I'?:-O. Since fe(/)==sup f{min (I, P)} for all pEL +,
there exists I1EL1 such that 0/11 and f e (/-/1)==f e (/)-f(/1) <!e.
By Theorem 2 there exists 12EL+ such that f(1/1-/2D<!e, i.e.,
f e (1/1-/21)<!e. Hence f e (I/-/21)<le. In the general case, write 1==
1+-(-1-).
For the proof of the last assertion, let g be a function in some equiva-
lence class of LIe. Then, since L is dense in LIe, there is a sequence
InEL such that fe(lg-lnl)--+O. It follows that In is a fundamental
sequence in LIe, hence also in L. But then (II-Inl)--+O for some
IEL1. Since .f(I/-lnl)==fe(I/-lnl), the functions I and g are in the
same equivalence class of LIe.
16.3) In order to show that, for the classical Riemann integral in
R1, the metric space is not complete, we refer to the example in Exer-

502
SOLUTIONS
[Ch. 3, 4
cise 11.3. The cell (0, 1 J is denoted by A, and Ene A is the increasing
sequence of f-sets defined in that exercise. If f n= XEn' then cf (f n) i ()("
where O<()(,< 1. Assume now that cf(lf-fnl) -+0 for some cf-summable
f. Then the negative part f- of f is a cf-null function, since If-I
If-Inl for all n. Hence, we may assume that rO for all X. Similarly,
we may assume that f 1 for all x, f= 1 on every En, and f==O outside
the cell A. Furthermore, cf(f)()(, on account of
cf (f)cf (fn) + cf(lf- Inl)()(,+ cf (If- fnl).
Hence, there is a step function f2?;;f such that cf (f2) < 1. Again, we
may assume that 0<f2< 1, 12= 1 on each En, and f2==0 outside A.
Then xA-/2 is a non-negative step function l anXBn' zero on each En,
and such that cf(XA-f2) >0. This implies that one at least of the cells
Bn is disjoint to all En, contradicting the fact that the sequence En
invades into each cell.
CHAPTER 4
17.1) For the case that #(X)<oo and Of(x)M, introduce
5(f; P)=  lk+1#(E k ), and observe that s(f; P)J(f)5(f; P) and
5(f; P)-s(f; P)(P)#(X). Of course, Lebesgue could not assume
beforehand the existence of J(f), so that he had to prove first that if
the partition P 2 is a refinement of the partition PI, then
s(f; P1)s(f; P2)5(f; P2)5(f; PI),
which implies by a well-known argument that sup s(f; P)inf 5(f; P);
hence, it follows then from 5(/; P)-s(f; P)<(P)#(X) that sup s(f; P)
==inf 5 (f; P). The last assertion in the exercise follows from Theorem 4.
17.2) Follows from Theorem 4.
17.3) Let E be a non-measurable subset of (0, 1J; let f(x) = 1 on E,
f(x)==-l on (0, 1J-E, and f(x) ==0 outside (0, 1J.
17.4) For any fEL+, the function min(f, 1) is either a function in
L + or the uniform limit of a sequence of functions in L +.
17.5) If f, gEL, then 4fg= If+gI 2 -lf-gI 2 holds for all XEX, since I,
and g do not assume the values :l:oo. Hence, fgEM. If f and g are
summable, there exist sequences fnEL, gnEL such that lim fn==f,

Ch. 4J
SOLUTIONS
503
lim gn==g, except on a null set. Then fngnEM, and lim Ingn==fg except
on a null set, so fgEM.
Let IEL +. Since 1 2 EM+ by hypothesis, the function gn==min (f2, nf)
is summable for n==l, 2, ..., so that gnlEM+ by the just established
result. Since gn/i 1 3 , it follows that f3EM+. By induction, fnEM+ for
n==4, 5, . . . .
17.6) Evidently Pn(y»O for all n. By induction, Pn(Y) 1 and Pn+1(Y)
?:-Pn(Y) on [0, 1J for n==O, 1,2, . . . . Hence, Pn(Y) i l(y) on [0, 1J, and
Ol(y) 1. It follows from the defining relation that 2l(y) == 1 _y+l2(y),
so l-l(y)==yi, and this implies that qn(Y)== 1-Pn(Y) !yi. By Dini's
theorem the convergence is uniform on [0, 1 J, and so Q n (y) ==
qn(Y) -qn(O) converges to yi uniformly on [0, 1 J.
On account of this result there exist polynomials P n(y), n== 1, 2, . . "
such that Pn(O)==O for all nand IPn(y)-yil<n-1 on [0, nJ. Then
lim P n(y) ==yi on [0, 00).
Let I be finite-valued on X and fnEM+ for n==1,2, .... Then
lim Pn{/(x)}==/i(x) on X, and since the functions Pn{f(x)} are measur-
able by hypothesis (we use here that Pn has no constant term), it
follows that fiEM+. Note now that fn+iEM+ for n== 1,2, . . " since
lim ynPn(y)==yn+i on [0,00). Hence, the proof may be repeated for
fi instead of I, and it follows by induction that 1 1 , It, . . . EM+.
17.7) We haveg n ==min(f,/ 2 - n )EM+forn==1, 2, .. .,andgn!min(/, 1)
on X. Hence, min(f, l)EM+.
17.8) For the proof of (c), observe that for any Je-summable f>O
there exists a sequence of non-negative ve-step functions fn such that
Je(1 n) i ofe(/). Hence, th repeated procedure yields the same value
Je(f) for all Je-summable 1>0. It follows easily that the repeated
procedure yields the same value ofe(/) for any of-measurable 1>0. For
(d), observe that
ve(E) ==Je(XE) ==sup J{min(XE' P)}
for all PEL+. By the representation theorem we may use for L+ the
collection of all non-negative v-step functions, hence
ve(E)==sup of {min (XE, XA)}==SUP v(EA)
for all A of finite v-measure, and in view of sec. 7, Theorem 4, this is
the desired result.

504
SOLUTIONS
[Ch. 4
17.9) The only fact not immediately evident is that E, FEA implies
E-FEA, since this does not follow immediately from Exercise 13.7.
We may assume that FeE. Then
min(XE-F, p)==min(XE, p)-min(XF, P)EM+
for all PEL+,
since the result in the exercise referred to is now applicable. Hence
XE-FEM+.
17.10) Since min(xx, s)EL: for any sEL:, the desired result follows
from Exercise 13.4.
17.12) Observe first that if h(x»O satisfies h(x)k(x) for some J-
summable k(x), then h fails to be J-summable if and only if h fails to
be J -measurable. Hence, if this is so, there exists e >0 such that
J(h2) - J(h 1 »e for any pair of J-summable h 1 (x) and h 2 (x) satis-
fying h1hh2. We shall say then, for the purposes of the present
proof, that h is unmeasurable of degree e. Now, let 0<AO<A1 < . . . <
Am+1 <00 be any finite sequence of positive numbers, and let
gn=={min(f, An)-min(f, A n -1)}f(An- A n -1)
for n==l, 2, ...,m+1. Then gn is J-summable, gn(x)==O for f(x)
An-I, O<gn(x) < 1 for A n -1 <f(x) <An, and gn(x) == 1 for f(X»An. De-
noting the characteristic function of {x: f(x) >A} by XA' we have
gm+1XAmgm' . · XA1gl'
Fix AO, Al and e>O (observe that gl(X) is then fixed), and select the
remaining values of An from among the values of A for which XA is
unmeasurable of degree e (if possible). Then J(gn) - J(gn+1»e for
n==2, . . ., m; hence
m
(m-1 )e  {J(gn) - J(gn+1)}= J(g2) - J(gm+1)J(g2)J(gl)'
n=2
This shows that the number m of values A>A1 for which XA is unmeasur-
able of degree e is bounded by the fixed constant 1 +e- 1 J(gl). Letting
e!O, it follows that the number of values A>A1 for which XA fails to be
J-summable is at most countable. Finally, letting A1!0, the result
follows.
17.13) By hypothesis, there exists a J -summable set E  {x: f(x) >O}.
If f(x)M on X, and e>O is given, let 0<Ao<A1 < . . . <An==M be a

Ch. 4J
SOLUTIONS
505
partition of [0, MJ such that Ao<e{v(E)}-l, Ai-Ai-1 <e{v(E)}-l for i==
1, . . " n, and F i == {x: f>Ai} is J -summable for i==O, 1, . . ., n. Then
F ==E -Fo is also J-summable. Let gi==XFt-l-XFt' and set g==  Ai-1gi
and h==AoXF+  Aigi. Then g<.f<.h on X, and
n
J(h-g)==AoJ(XF)+ (Ai- A i-1)J(gi)
1
n
<e{v(E)}-l{J(XF) +  J(gi)}==e.
1
Hence, 1 is J -summable by the inclusion theorem. Since J (g) == f g dv
and J(h)==f hdv, the function 1 is also summable with respect to the
integralfldv; finally, since J(/) andffdv are both included between
J(g) and J(h), we have J(f)==ffdv.
17.14) If J (I) is a Stieltjes-Riemann integral, then XXEM+ by Exer-
cise 17.10. Conversely, let xxEM+. It is trivial that any v-step function
s(x»O is J-summable, and J(s)==f sdv. Furthermore, it follows from
the last two exercises that any IEL + is summable with respect to
ffdv, and J(f)==fldv. But then J(/)==ffdv holds generally in virtue
of Exercise 7.12.
17.15) Note that, given the J-summable 1>0, there exists a v-step
function s such that I<.s on X; hence, 1 is bounded and vanishing
outside a J -summable set.
17.16) Let EEA 1 satisfy v1(E)==J(XE) <00. Given e>O, there exist
v-step functions g, h, assuming their values on sets of A, such that O
g<'XE<.h and J(h-g)<e. But then there also exist sets E1, E 2 EA such
that E 1 cEcE 2 and v(E 2 -E 1 ) ==J(XE 2 -XE 1 ) <e. Hence, EEAv, and
it also follows easily that V(E)==V1(E). Now, let EEA1, where v1(E)==
J(XE) is not necessarily finite. Consider the intersection of E and an
arbitrary AEA satisfying V(A)==V1(A)<00. Then EAEAv by what has
already been proved, so E EAv.
Conversely, let EEAv satisfy v(E) <00. Given e>O, there exist sets
E1, E 2 EA such that E 1 cEcE 2 and v(E 2 -E 1 )<e. Then XE 1 <'XE<'XE 2 ,
where J(XE 2 -XEJ==v(E 2 -E 1 )<e. It follows that XE is J-summable,
and so EEA 1 . We also have v1(E)==v(E). Finally, let EEAv, where v(E)
is not necessarily finite. We have to prove that XEEM+. Given pEL +
(i.e., p is a v-step function), there exists a set A EA such that P(x) ==0

506
SOLUTIONS
[Ch. 4
outside A. Since AE EAv and 1'(AE) <00, we have AE EA 1 by what has
already been proved, so min(XE' p)==min(XAE, P)EM+. This implies
XE E M +.
Finally, since 1'l(E)<oo or 1'(E) <00 implies 1'l(E)==1'(E), it follows
that 1'1 ==1' holds generally on Al ==Av.
18.4) Consider the Lebesgue integral over [0, 1J of the function equal
to kjlog k on k-1<X(k-1 )-1 for k==2, 3, . . . .
18.5) Let I(x) be equal to k- 1 on kx<k+1 for k==l, 2, "', and
let In(x)==/(x) on [1, n) and In(x) ==0 on [n, 00). Then In converges to
f uniformly on X == [1, 00), but I is not Lebesgue summable over X.
A second example, with Lebesgue integration over [0, 00), is obtained
by setting In(x)==n- 1 on [0, nJ and In(x) ==0 on (n, 00). Then the limit
function 1==0 is summable, but J(ln)== 1 for all nand J(/)==O. For
the last assertion, note first that I is summable by the theorem on
dominated convergence; since lim I/-In/==O and 11-lnl2g almost
everywhere on X, a second application of the same theorem shows
that lim J(I/-Inl) ==0.
18.6) Note that, by sec. 16, Theorem 1, Corollary, In contains a
subsequence, converging almost everywhere on X to I.
18.7) For the last assertion, let # be Lebesgue measure in X == [0, 1J,
and choose gn(x)==n on [0, n- 1 J, gn(x) ==0 on (n- 1 , 1J for n==2, 3, ... .
18.8) Observe that 1(ln+gn)-(/+g)la implies one at least of the
inequalities I/n-/I!a and Ign-glla.
18.9) Writing En=={x: I/n-/la}, and observing that the sequence
Dp is descending, we have lim D p == IT Dp==lim sup En. Furthermore
p(lim sup En) ==0, since all In and I are finite almost everywhere, and
lim In==1 almost everywhere. On the set where g!a we have Iin-/l
la, and this shows that all sets En are included in the set of finite
measure where g>!a. It follows that #(Dp) <00 for all p, so
lim #(Dp)==#(lim Dp)==#(lim sup En) ==0.
18.12) If
00
D(P, q)=  {x: Iln-/lq-1}
n=p
for p, q== 1, 2, . . "
then
lim #{D(P, q)}==O
as poo.

Ch. 4J
SOLUTIONS
507
Hence
#{D(pq, q)}<e/2 q
for some p==pq,
and
00
#(De) <e
for
De==  D(pq, q).
q=l
Furthermore, if XEX -De, then XEX -D(pq, q) for all q, so I/n(x) -/(x) I
<q-1 for n?::-pq. This shows that In converges to I uniformly on X-De.
18.14) Assume that In converges in measure to I on every subset of
finite measure of the set X of a-finite measure. The same is then true
for every subsequence, so that it is sufficient to prove first that In
contains a subsequence converging almost everywhere on the fixed
subset S of finite measure to I. By means of the well-known diagonal
method it is then easy to obtain a subsequence which converges almost
everywhere on X to I. We restrict ourselves, therefore, to points XES.
In view of the convergence in measure there exists, if the integer k>O
is given, an index nk such that
#{x: Iln(x)-/(x)lk-1}<2-k
for n?::-nk,
and we may assume that n1<n2<... . Hence, if Ek=={X: I/nk(x)-/(x)1
k-1}, we have #(Ek) <2- k , so
00
#(lim sup Ek)  #(Ek) <2- n
k=n+l
for all n,
that is, #(lim sup E k) ===0. This is the desired result, since lim I nk(x) == I(x)
on the set S-lim sup Ek.
Assume, conversely, that every subsequence of In contains a subse-
quence which converges to I almost everywhere on X, and let the
subset S of finite measure and e>O be given. If En is the subset of S
on which Iln(x)-/(x)le, we have to prove that lim#(En)==O. If this
fails to hold, we have #(EnJ>€5 for some €5>0 and some subsequence
nk. But (selecting a suitable subsequence of nk if necessary) we may
assume that Ink converges to I almost everywhere on S, so # (lim sup E nk )
===0. Hence O==#(lim sup EnJ?::-lim sup #(E n J?::-€5>O, and this is a
con tradiction.
18.15) Let all In and I be summable. If pn==/xl/-Inld# tends to
zero, and a>O is arbitrary, it follows from #{x: I/n-/la}<Pna-1 that

508
SOLUTIONS
[Ch. 4
In converges to I in measure. If # is Lebesgue measure in X == [0, 1J,
and In(x)==n on [0, n- 1 J and zero elsewhere, then In converges in
measure to 1==0, but Ix I/n -II d#== 1 for all n. If Ilnlg almost every-
where on X for a summable g, and In converges in measure to I, then
(since the set on which gi=O is of a-finite measure) we may conclude
from the preceding exercise that every subsequence of In contains a
subsequence which converges pointwise to I almost everywhere on X.
Hence, if Pnk==lxllnk-/ld#>>O for some subsequence nk, we may
assume that Ink converges to I almost everywhere, and then lim Pn1C==O
by the theorem on dominated convergence. It follows that Pnk>>O
is impossible.
18.16) If tn!to, then Yn=={x: -oo</tn} descends to Yo=={x:
- oo</to}, so #(Y n) ! #(Y 0) on account of #(X) < 00.
18.17) The result holds if E is a cell, so it holds also if E is a a-set,
and also if E is the limit of a descending sequence of a-sets. If #p(E) ==0,
consider a a-set of arbitrary small measure covering E.
18.18) The first assertion follows by observing that F+ and F- corre-
spond to G+ and G- respectively, and by making use of approximation
by step functions. For the second assertion, observe that F(t) ==t is
always #p-measurable, and that the corresponding G(x) is I(x).
18.21) Prove first that lim Pn(t)==P(t), except at most at the values
of t for which #{x: I(x) ==t} >0. These values of t are the discontinuity
points of P(t), and their number is at most countable. Let #{x: I(x) ==t}
==0; hence, if E=={x: I(x)t} and F=={x: I(x) <t}, then #(E)==#(F).
Writing En=={x: In(x)t}, it is seen easily that
Fclim inf Enclim sup En cE ,
so lim #(En)==#(E), i.e., lim Pn(t)==p(t).
Next, prove that lim 1(Y)==I*(y), except at most at the disconti-
nuity points of I*(y). Let 1* be continuous at y, and write I*(y)==to.
Then P(t) >y for t>to and P(t)<y for t<to. If t is fixed such that
to<t<to+e and lim Pn(t) ==P(t), then Pn(t) >y for n?;:-nt; hence
1:(y)t<to+e==I*(y) +e
for n?;:-nt.
Also, if t' is fixed such that to-e<t' <to and lim Pn(t') ==P(t'), then
Pn(t')<y for nnt'; hence 1:(y)t'>I*(y)-e for nnt', It follows that
lim I(y) ==I*(y).

Ch. 4J
SOLUTIONS
509
18.22) Check that the result holds if I is a step function; for the
general case, consider a sequence of step functions In such that I:}; i t+
and I;; !/-. Then lim Pn(t) ==P(t) almost everywhere on R 1 by the proof
of the preceding exercise; more specifically, Pn(t) i P(t) almost every-
where on (-00, OJ, and Pn(t)!p(t) almost everywhere on [0,00). The
desired result follows, therefore, by the theorem on integration of
monotone sequences.
20.1) We may assume that I(x) is non-negative. If the condition is
satisfied, then IXE (as a limit of measurable functions) is measurable
for any E EA of finite measure; hence, min (I, s) is measurable for any
s(x) EL, and this shows that I is measurable.
Conversely, if t?:-O is measurable, and E EA is of finite measure,
then IXE is measurable. Let In==min (IXE, n) for n== 1,2, . . . . Then all
In are summable, and In i IXE' Since L is dense in the metric space of
all real summable functions, there exist step functions sn(x) EL such
that! Iln-snld#<2-n. Then lltn(x)-sn(x)1 is summable, and hence
finite almost everywhere. It follows that I n(x) -sn(x) -+0 almost every-
where, so sn(x) -+1 (x) XE (x) almost everywhere.
21.1) We may assume that I is real, and that Lt is a cell. Let first I
be continuous almost everywhere on Lt, and let  be a sequence of
nets of the kind introduced in the proof of Theorem 1. Adhering to the
notations of that proof, we have m(x)==/(x)==M(x) for any x at which
j is continuous, i.e., almost everywhere. It follows that
lim  mi#(Lt i ) == (L)f m(x) d#== (L)f I(x) d#=== (L)f M(x) d#
  
==lim  Mi#(Lt i ),
and since this holds for any sequence of the required kind, I is Riemann
integrable.
Let now, conversely, I be Riemann integrable, and let, again,  be
a sequence of nets as introduced in the proof of Theorem 1. Then, if
Xo is a point of Lt not lying on any face of any of the cells of any net
, and if in addition m(xo) ==M (xo) , the set of all such points Xo has
the same measure as Lt. It is now easy to derive that I is continuous at
all points Xo of this kind.

510
SOLUTIONS
[Ch. 4
21.2) Note that the integral of It I over [kn, (k+ 1 )nJ (k== 1, 2, . . .)
exceeds {(k+ l)n}-l times the integral of Isin xl over the same interval.
It follows that /00 f+dx== +00 and /00 f-dx== -00.
21.3) Note that
If(x) 12x-1Icos x- 2 1-2x>x- 1 -2x
on each of the intervals
{(2n+ !)n }-tx {(2n - !)n }-!.
21.4) Note that, for O<x<l,
00
x P - 1 (1 +xQ)-1===XP-1(I-xQ+x2Q-x3q+.. .)==  fn(x),
n=O
where fn(x)==(1-xQ)xp-1+2nQ>0; the result follows therefore from the
theorem on integration of increasing sequences.
21.6) Since sin nx==-sin n(2n-x), it is sufficient to prove that
sn(x) is uniformly bounded on [0, nJ. We have
!X
sn(x) == -!x+ f {sin (2n+ 1 )tjsin t}dt,
o
and by the remark concerning uniform convergence in the preceding
exercise there exists an index no such that
!X tX
If {sin (2n+ 1 )tjsin t}dt- f {sin (2n+ 1 )tjt}dtl < 1
o 0
for all nno and all x in [0, nJ. Furthermore, there exists a constant
A such that
tX !(2n+ l)x
If {sin (2n+ 1 )tjt}dtl == I f {sin tjt}dtlA
o 0
for all n and x, so
ISn(x)IA+ 1 +!lxlA+ 1 +!n
for nno and Oxn.
The formula for 1 n- 2 cos nx follows then by the theorem on domi-
nated convergence. Setting x==n, we obtain
 (-I)nn-2=== n- 2 -1n 2 ,

Ch. 4J
SOLUTIONS
511
and the desired results follow by observing that
00
2- 2 +4- 2 +6- 2 +. . . ==!  n- 2 ,
1
so that
00
1 +3- 2 +5- 2 + . . . ==1  n- 2 .
1
21.9) By addition,
in in
f sin 2x f
2.f == log 2 dx == log sin 2x dx - in log 2
o 0
n
=! f log sin tdt-in log 2=.F -in log 2.
o
21.11) Obviously,
x
dF f
- == 2e- x2 e- t2 dt
dx .
o
Denoting by f(x, t) the integrand in the integral for G(x), the partial
derivative
of/ox== _2xe- x2 (1+t 2 )
exists for all x>O, and lof/oxl2xe-X2A for some constant A and all
(x, t) such that x>O and Ot 1. Then
f(x', t) - f(x, t)
A
x'-x
for all pairs x, x' and all t satisfying Ot 1, so that, since g(t) ==A is
summable over [0, IJ, we have
1 1 x
dG == f !L dt == -2xe-X2 f e-x2t2 dt == -2e- X2 f e- t2 dt
dx OX
000
by sec. 18, Theorem 5. Hence dF/dx+dG/dx==O for x>O. Since F(O) ==0
and G(O)==!n, we obtain F(x)+G(x)==!n. Finally, since G(x)!ne-X2,
it follows that lim G(x) ==0 as x-+oo, so lim F(x) ==!n as x-+oo.

512
SOLUTIONS
[Ch. 4
21.13) Keep b >0 fixed, and consider /(x, a) ==log (a 2 cos 2 x+b 2 sin 2 x)
for 0<a1<a<a2<00. Then o/joa exists and does not exceed the
summable function 2a2 cos 2 xj(ai cos 2 x+b 2 sin 2 x). Hence, for a-=l-b,
in 00
of f 2a cos 2 X f dt
----a;:; = a 2 cos 2 x+b 2 sin 2 X dx = 2a (a2+b2t2)(1 +t 2 )
o 0
00
- 2a f ( b2 _ 1 ) dt - __
- b 2 -a 2 a 2 +b 2 t 2 1 +t 2 - a+b '
o
and for a==b the same result follows trivially. Then F(a, b)==
n log (a+b)+G(b). Setting a==b, we obtain n log b==n log 2b+G(b).
21.14) F(r)==F(-r) is evident. Then
n
2F(r) ==F(r) +F( -r) == flog{(1-2r cosx+r2) (1 +2r cos x+r2)}dx
o
n 2n
== flog (1-2r 2 cos 2x+r 4 ) dx==tflog (1-2r 2 cos y+r 4 ) dy==F(r 2 ).
o 0
If O<r< 1, then (l-r)2< 1-2r cos x+r2«1 +r)2, so F(r) tends to zero
as r to. This implies, by the relation F(r)==!F(r 2 ), that F(r)==O for
r 2 <1. Setting r==l in F(r 2 )==2F(r), we obtain F(l)==O. Finally, set
t==r- 1 for r>1 (so F(t) ==0). Then
n n
F(r) = f log ( 1 -  cosx + * )dX = - f logt 2 dx
o 0
== -nlogt 2 == nlogr 2 .
21.16) The differential equations for F and G are obtained by
partial integration of the integrals for dFjdb and dGjdb respectively;
note that dGjdb+2a- 1 bG(b)==a- 1 implies d(eb2/aG)jdb==a-1eb2/a.
21.17) Since (1-y)1/ Y <e- 1 for O<y<l, we have, for O<x<n, that
(l-xjn)n<e- x . Hence, if /n(x)==xq(log x)r(l-xjn)n on [0, n], and /n(x)
==0 on (n, 00), then l/n(x)l<xqllog xlre- x on (0,00) for all n. Also
1 1 1
f xqllog xl r e- 1 dx<f xqllog xl r e-xdx<f xqllog xl r dx,
000

Ch. 4J
SOLUTIONS
513
so the integral in the middle is finite. Furthermore, there exists a
positive constant k such that xq(log x)r e-ix<k for x> 1, hence
xq(log x)re-x<ke- tx on [1,00), which shows that xq(log x)re- X is
summable over [1,00). It follows that xq(log x)re- X is summable over
[0, 00), and since lim I n(x) ===xq(log x)r e- X for all x, the desired result is
a consequence of the theorem on dominated convergence.
21.18) By the preceding exercise the integral is the limit of  n as
n-+oo, where
n 1
.F n = f (logx)(l -x/n)ndx=n f (logn + logy)(l -y)ndy
o 0
1
=== n l ogn+n f tnlog(l-t)dt,
n+l
o
and the last term is
1 1
n f n f _ tn+1-1
log (1 - t)d(t n + 1 - 1) == - dt.
n+ 1 n+ 1 t-1
o 0
21.19) We have
b b b
(R) f gn(x) ===n f I(x+ 1 In) -n f I(x)
a a a
b+ l/n a+ l/n
===n f I(x) -n f I(x),
b a
and this obviously tends to I(b)-I(a) as n-+oo. If I'(X) exists and is
bounded on [a, bJ, we first modify I(x) on [b, b+ 1J, making I(x) linear
with derivative I' (b) on this interval. For the modified function we
have lim g n(x) === I' (x) on [a, b J, and g n(x) === I' () for some  satisfying
x<<x+ 1/n. Hence, j'(X) is Lebesgue summable over [a, bJ, and
b b b
lim (R) f gn(x) ==lim f gn(x) dx=== f I' (x) dx
a a a
by the theorem on dominated convergence.

514
SOLUTIONS
[Ch. 4
21.20) The proof that J(/)==/(xo) is linear and non-negative on L is
trivial. If v is the corresponding measure, we consider first a bounded
closed interval A not containing Xo. If IEL + is chosen then such that
1?;:-1 on A and I(xo) ==0, we have v(A)fldv==/(xo)==O, and it follows
readily from this result that v(Rk-{xo})==O. This shows already that
R k - {xo}, and hence any of its subsets, is a v-null set. Since Rk and
Rk-{XO} are measurable, the set {xo} is measurable, and it remains
to show only that v ({xo}) == 1, i.e. V(Rk) == 1. This is derived by se-
lecting a sequence In E L + satisfying OI n 1, In t XRk' and In (xo) == 1 for
all n.
21.21) Note that VI is a measure in Rk by the remark following
sec. 18, Theorem 3, and the measure VI satisfies the hypotheses of the
second part of the Riesz representation theorem.
21.22) If fly is a Stieltjes-Lebesgue measure, generated by g(x), then
/(/)==fldfly is a non-negative linear functional on L, as observed in
the last paragraph of this section. It follows from sec. 17, Theorem 9
that the measure v, induced in R 1 by /(/), satisfies V===fly, and it is
sufficient, therefore, to prove that the restriction /L(/) of /(1) to the
collection L yields upon extension again /(/). This can be done, simi-
larly as in the proof of Theorem 2, by showing that for any cell A the
function XA becomes /L-summable with /L(XA) == / (XA).
Conversely, if J(/) is a non-negative linear functional on L, then J{/}
induces (after extension) a measure v in R1, such that all cells are v-
measurable and of finite v-measure. The restriction fly of v to the semi-
ring r of all cells is, therefore, a Stieltjes-Lebesgue measure generated
by some function g(x). It remains to prove that the extension pro-
cedure for measures, applied to (R1, r, fly), yields again v. It is suf-
ficient, for this purpose, to show that the restriction Jy(/) of J(/) to
the collection of all step functions, with respect to r and fly, yields
upon extension again J(/). This is easy; given IEL+, there exists a
sequence Sn of such step functions such that Osn t I on R 1 .
21.23) Let InEC, O/n!O on Rk, and assume that J(tn) !ex>O. We
may assume without loss of generality that 11 (x) >0 at all xER k ; if
necessary, replace 11 by max(/1, 1). Let g=l (In-(3/1)+, where (3=
exj{2J(/1)}' Given xOERk, there exists N such that (IN-(3/1) (xo) <0,
hence (In-(3/1) + (x) ==0 for all n?;:-N and all x in a neighbourhood of xo.
This shows that g is continuous at Xo. It follows that J(g) is finite. On

Ch. 4, 5J
SOLUTIONS
515
the other hand J(g)1 J(fn-{3f1)==1 {J(fn)-lcx}>lkcx for k== 1,
2, . . . . This is a contradiction.
21.24) Let B(r) denote the closed sphere of radius r around the
origin, and A (r1, r2) the closure of B(r2)-B(r1) for r2>r1. Assume that
there exists no set Eo with the desired property. Then there is a set F
disjoint to B(l) such that #(F) >0. It follows that there exists also a
set A (r1, r2) of positive measure CX1 with r2>r1> 1. Similarly, there
exists a set A (r3, r4) of positive measure CX2 such that r4>r3>r2, and
so on. Let f be a non-negative continuous function such that f==cx 1 1
on A (r1, r2), f==cx 2 1 on A (r3, r4), and so on. Then J(f» =1 cxncx:;l=k
for k== 1, 2, . . . ; a contradiction.
CHAPTER 5
23.2) The functions f and g are continuous at each interior point of
their respective domains of integration. Hence, for each real number a,
the sets where f>a and g>a respectively, are open. This shows that
f and g are (two-dimensionally) Lebesgue measurable. Since f and g
are non-negative, the interchange of the order of integration is per-
mitted by Theorem 4, Corollary. In order to compute I fdt, observe
that, for x>O, y>O, x*y,
1 1 { X2 Y2 }
(1 +x 2 t 2 )(1 +y 2 t 2 ) = x 2 _y2 1+x 2 t 2 - 1+y 2 t 2 ·
23.4) Since It [/ If I dyJ dx<oo, the order of integration may be
interchanged by Theorem 5. Note that It fdx==(1 +y4)-1+g(b, y),
where Ig(b,y)dy tends to zero as b-+oo. It follows that In!A=
I (1 +y4)-ldy. 1'he last integral may be computed in an ele-
mentary manner by observing (y=t- 1 ) that P=/ (1 +y4)-ldy==
I y2(1 +y4)-ldy, so
00 00
p==![! (y2+y 2 + 1 )-ldy+ ! (y2-y2+ 1 )-ldyJ.
o 0
It follows that P== l6 n2+ f6 n2=!n2.
23.8) Any cell in Rk is vI-measurable and of finite VI-measure by the
Riesz representation theorem, and similarly, any cell in Rz is V2-measur-

516
SOLUTIONS
[Ch. 5
able and of finite V2-measure. Hence, any cell in Rk+l is (VI X V2)-
measurable. Since any open set in Rk+l is a countable union of cells,
it follows that any open set in Rk+l is (VI X v2)-measurable. For any
real number a, the set on which h(x, y) >a is open, and this implies
that h is (VI X v2)-measurable.
23.9) Let cf(/) be another non-negative linear functional on C such
that cf{/(x+a)}==cf{/(x)}, and select the fixed function gEC+ such
that g*(x)==g( -x) satisfies J(g*) >0. Let IEC+. Then, if VI and V2 are
the measures induced in Rk by J and cf respectively, the non-negative
functions I(x+y)g(y) and I(y)g(y-x) are (VI X v2)-measurable by the
preceding exercise. Hence,
cf (g)J (I) == cfyJxg(y)/(x) == cfyJxg(y)/(x+y)
== J xcfyg(y) I(x + y) == J xcfyg(y - x) I(y)
== cfyJxg* (x-y)/(y) == cfyJxg* (x)/(y) ==J (g*) cf (I),
so
cf (I) == cJ (I)
for
c== cf (g) jJ(g*).
23.13) We have to show that IjEL, Ijto implies Jljto, Le., if Ij is
a decreasing sequence of functions of L, and O<eJlj for all f, we
have to show that there exists a point (aI, a2, . . .) such that Ij(a1, a2,
. . .) does not tend to zero. We may assume that all Ij 1, so all Jlj 1,
and all J(n)/j 1. If Fj== {Xl: J(l)/j>e(l-!)}, then the set Fj is #1-
measurable, and it follows from
Jlj== f f(l)/jd#l + f J(l)/ j d#l
Fj XI-Fj
that eJlj#l(Fj) +e(l-!), so #l(Fj»!e for all f. Hence, there ex-
ists a point a1EX1, contained in all Fj (since Fj is a descending
sequence, and #l(Fj) does not tend to zero), and this implies that
J(1)/j(a1, X2, X3, .. .»e(l-!)
f or all j.
This argument may be repeated, and we obtain a point a2EX2 such
that
J(2) l(a1, a2, X3, . . .) >e( 1- i-i)
for all j.
Continuing, we obtain a sequence aI, a2, . . . such that anEXn and
J(n) Ij(al, a2, . . . , an, xn+ 1, . . .) >!e
for all n and all f.

Ch. 5J
SOLUTIONS
517
Assume now that Ij(a1, a2, . . . )<le for some index j. Since IjEL,
there exists an index k==k(j) such that
Ij(a1' . . " ak, Xk+1, · · .) ==lj(a1, · · ., ak, ak+1, . · .)
for all (Xk+1' Xk+2, . . .) EXk),
so Ij(a1' . . ., ak, Xk+1, . . .) < Ie. But then J(k)/j(a1, . . " ak, Xk+1, . . .)
< !e, a contradiction. Hence Ij(a1' a2, . . . )!e for all j.
23.14) If F n ==E 1 x... xEnxxn), then Fn is a descending se-
quence of v-measurable sets such that E ==lim F nand v(F n) ==
rrf= 1 #i(E i ).
23.16) We have Jlgnlkn==(J1" .In)(n)kn==(J1'' .In)III==
Jill; the last equality is due to the fact that the function I/I is a
constant. The proof for h n is similar.
23.17) We obtain Jlgn-J1.. .Jnll<e and Jlhn-J(n)ll<e, i.e.,
Jlgn-ll<e and Ihn-ll<e for n sufficiently large, where l de-
notes the constant function whose value is Jl. It follows that Jign-gi
Jlgn-ll+Jlg-ll==Jlgn-Jll+JIJI-ll<2e and Jlhn-/I
Jlh n -ll +I/-li <2e for n sufficiently large. The extension to complex
I is immediate.
23.21) Assume that v is separable. Then the metric space of all J-
summable functions is separable, and it follows that the collection L,
regarded as a subset of that metric space, is separable; let the sequence
of functions In be dense in L. Each In is essentially a function of a
finite number of the variables x,., so there exists at least one index 1'0
such that all Indo not depend on X,.o' Let A,.o c X,.o satisfy #,.o(A,.o) == ex,
where O<ex<l; let Co==A,.oXXbTO), and lo==xc o ' Then 10EL and, for
any In, I/n-/ol! either for all x,.oEA,.o or for all X,.oEX,.o-A,.o' so
Jl/n-/ol! min(ex, I-ex) for all n, contradicting the hypothesis that
In is dense in L.
23.22) Let the sequence An be dense in the metric space of all sets
of finite measure with distance function p1(E, F) ==#(E-F)+#(F -E),
and let Y ==  An. Then Y is of a-finite measure, and assuming that
#(X - Y) >0, there exists a set Fe X - Y of finite positive measure.
Hence p1(F, An) <!#(F) for some An since An is dense, i.e.,
#(F)+#(An)<!#(F), a contradiction.

518
SOLUTIONS
[Ch. 5
24.1) If such a number €5 >0 does not exist, there is a sequence En
such that #(E n )<2- n and v(En)==IEnlfld#?;;e for all n. Then the set
E==lim sup En satisfies
00
#(E)#(En+En+1+" .)  2-p==2-(n-1)
p=n
for all n,
so #(E) ==0, and therefore also v(E) ==0. On the other hand, the finite
measure v satisfies
v(E) ==v(lim sup En) ?;;lim 'sup v(E n) ?;;e,
a contradiction.
24.2) Let # be generated by g(x) ; the function g(x) is non-decreasing
and continuous on R1 (the left continuity is implied by the fact that
#({x})=O for every point x). Select a point XOE [a, bJ. It follows from
the continuity of g at Xo that, given €5>0, there exists a number ho>O
such that Ig(x) -g(xo) I <€5 for Ix-xolho. In other words, #([x, xoJ) <
or #([xo, xJ)<€5 for these values of x. The continuity of F(x) at Xo
follows then from the preceding exercise.
24.3) Set Q(x) ==Q(a) for all xa, and Q(x) ==Q(b) for all x?;;b, and
introduce the function Q*(x)==Q(x+). Note that Q*(x)==Q(a) for all
x<a, but not necessarily for x==a. The function Q*(x) is non-decreasing
and right continuous, and Q*(a-)==Q(a). If 1: is the Stieltjes-Lebesgue
measure generated by Q*(x), and T(x)==1:([a, xJ) as in Theorem 1, then
T(x) ==Q*(x) -Q*(a-) ==Q*(x) -Q(a)
for a<.x<.b.
Let f==f++f- (f+?;;O,I-<.O), and let v be the measure in [a, bJ, de-
fined by v(E)==IEI+d# for any #-measurable set Ec [a, bJ. Then N(x)
==1: f+d# is continuous by the preceding exercise, so N(x)==N(x-),
and it follows by the Theorems 1 and 2 that
b b b
fTI+d#== fTdv==N(b)T(b)- fNd1:.
a a a
A similar result holds for 1-, so F(x) ==1: fd# satisfies
b b
fTfd#==F(b)T(b) - f F die
a a
Observe now that I: Fd1:==exi([a, bJ)==exT(b), where ex IS a number

Ch. 5J
SOLUTIONS
519
lying between the bounds of F(x) on [a, bJ. Since F is continuous,
there exists a number  in [a, bJ satisfying ex==F(). Hence
b
J Tjdfl==F(b) T(b) -F()T(b),
a
so
b b
f Q*fdfl-Q(a) f fdfl= {F(b) -F()}Q*(b) -F(b)Q(a) + F()Q(a).
a a
In view of f: jdfl==F(b), Q*(b)==Q(b), and fQ*fdfl==fQfdfl (since the
set where Q*Q* is of fl-measure zero), we obtain the desired result.
25.3) Let XE (a, b); choose a closed subinterval [c, dJ containing x
in its interior, and let If(t)IC for all tE[C, dJ. Next, given e>O, choose
the positive integer n such that 2C /n<e, and then the number h
(either positive or negative) such that x-nh and x+nh are in [c, dJ.
Since
2f(x)f(x+h) +f(x-h),
it follows that
f(x) -f(x-h)f(x+h) -f(x),
hence
f(x) -j(x-nh)n{f(x+h) -f(x)}f(x+nh) -f(x).
Then
If(x+h)-f(x) 12C/n<e,
i.e., f is continuous at the point x.
Let now Xl and X2 be points in (a, b). It follows easily from the
hypothesis that
f{ exX.1 + (1-<X)X2}exf(X1) + (1-ex)f(x2)
for any ex==k/2n;
n== 1, 2, .. " k==O, 1, 2, . . . . Since the set of these numbers ex is lying
dense in [0, 1J, the continuity of f implies that the same inequality
holds for any ex in [0, 1 J.
25.4) Given Xl, X2 in (a, b) such that Xl <X2, choose X3 between a
and Xl, and let €5=={f(X1) -f(X3)}/(X1-X3). Then €5{f(x) -f(X1)}/(X-X1)
for all x satisfying Xl <XX2, so f(x»f(X1) + (X-X1)€5. Hence f(x»f(X1)
if €50, and f(x)f(X1)+(X2-X1)€5 if €5<O. This shows that f is bounded

520
SOLUTIONS
[Ch. 5
from below in [Xl, X2J. Since j(x)max {f(X1), j(X2)} in [Xl, X2J, f is also
bounded from above.
25.5) If either P2==ff 2 d# or q2==-fg2d# equals zero, then f==-O or
g==O almost everywhere, so the inequality holds. Let, therefore, P-=lO
and q-=lO. Substitute a==j(x)fP and b==-g(x)fq in ab!a2+!b2, and inte-
grate over E.
25.6) The first assertion follows from Exercise 25.4. Since it follows
already from Exercise 25.1 that F(x) is continuous on (0, 00), it will be
sufficient by Exercise 25.3 to show that u(x)==log F(x) satisfies 2u(x)
u(x+h) +u(x-h), i.e., that {r(x)}2F(x+h)r(x-h). This is proved by
means of the Schwarz-Buniakowski inequality:
00
{r (x) }2== {f e-tttHx+h-l) e-itt!(x-h-l) dt}2
o
00 00
<1 e- t t x + h - 1 dt. f e- t t X - h - 1 dt==F(x+h)r(x-h).
o 0
The proof for B(x, y) is similar.
25.7) For any j, satisfying the conditions, we have
j(x+n)==-(x+n-1)(x+n-2). . .xj(x),
f(n) == (n-1) !.
Choose now a fixed value of X such that O<x< 1, and let n2 be
integer. Then
log j(x+n) -log j(n)
log f(n) -log f(n-1)< < log f(n+ 1) -log f(n)
X
by Exercise 25.2, so
(n-1 )X(n-1) !<j(x+n)<nX(n-1) !.
This implies
( n - 1 ) x ( n - 1 ) ! n X ( n - 1 ) !
<f)< .
x(x+1).. .(x+n-1) x(x+1).. .(x+n-1)
Changing n into n+ 1 in the inequality on the left, we obtain
n! n! x+n
 j(x) 
x(x+ 1). . . (x+n) x(x+ 1). . . (x+n) n

Ch. 5J
SOLUTIONS
521
so
n nXn!
f(x)  f(x).
n+x x(x+ 1). . . (x+n)
Letting noo, the result follows for O<x 1. Denoting the expression
in the middle by F n(x), we have F n(x+ 1) ==xnF n(x) I (x+n+ 1), hence
lim Fn(x+1)==xf(x)==f(x+1) for O<xl, Le., f(x)==lim Fn(x) for
1 <x2. By induction, f(x) ==lim F n(x) for all x >0.
25.8) Observe that f, as a product of logarithmically convex
functions, is logarithmically convex.
25.10) The function f(x)==n-!2 x - 1 F(ix)F{i(x+ I)} satisfies the con-
ditions (a), (b) and (c) of Exercise 25.7. Observe that log 2 x - 1 ==
(x-I) log 2 is a convex function of x.
25.12) Denoting the maximum of Ig(x) I in [0, 1J by M, we have
Ig(x) Ii-M +i-M ==iM for all x in [0, 1J, in particular M iM, which
implies M ==0. Hence g(x) ==0 for all x, so log cp(x)==ax+b. By the peri-
odicity of log cp(x) we obtain log cp(x) ==b, so cp(x) is constant. Obvi-
ously, it follows from the given functional equation that cp(x) ==c for
all x.
25.13) The equality cp(x+ 1) ==cp(x) for non-integer x follows by
means of F(l-x)==-xr(-x) and sin n(x+1)==-sin nx. In order to
show that cp(x) >0 for all x it is sufficient, therefore, to observe that
cp(x) >0 for Oxl. For XE[ -1, 1J, x*o, we have
cp(x)==x- 1 F(x+ l)F(l-x) sin nx,
hence
( n3x2 n 5 x 4 )
cp(x)==F(x+1)F(I-x) n- + - ... .
3! 5!
Obviously, denoting the function on the right by 1p(x) , we have also
cp(O) ==n==1p(O); hence cp==1p in [-1, 1J. Since 1p has derivatives of all
orders in this interval, also at the point x==O, the same is true of cp(x);
hence, by the periodicity, cp has derivatives of all orders for all x.
Multiplying now, for any non-integer x, the relations
cp(ix)==r(ix)F(1-1x) sin lnx,
cp{i(x+ l)}==F{l(x+ l)}F{i(l-x)} cos inx,

522
SOLUTIONS
[Ch. 5, 6
and observing that by Legendre's relation
n t
F(!x)F{i(x+ I)} == 2 x - 1 F(x),
n t
F{!(l-x)}F(I-!x) == 2- x F(I-x),
we obtain qJ(!x)cp{!(x+ l)}==nF(x)F(l-x) sin nx==ncp(x). By conti-
nuity, the same holds if x is an integer.
25.14) If x is an integer, the assertion is obviously true. If x is not
an integer, we have by Gauss' representation
1
F(x)
1
F(l-x)
= lim :x (1 + T) (1 +  )... (1 + : ).
-X;(-X) =lim nx (1-  )(1-  )...(1- : ).
and the result follows by multiplication.
25.15) By setting x==t/(l-t), it is easily seen that the integral is
B(ex, 1 -ex) =r (ex)r (1 -ex) ::;=n/sin nex.
25.16) Let ex==n+{3, where n is integer and.O{3<l, and let h(x)==
/l(X) -/(x). Then J(Xh(x) ==0, so n+1h(x) ==J1-p(Xh(x) ==0 for all x, Le.
f Jnh(t)dt=O for all x in [0, b]. Hence Jnh(x) ==0 almost everywhere
on [0, b], by sec. 18, Theorem 4. If n==O, the proof is complete; if n>O,
then nh(x)=O for all x in [0, b], since Jnh(x) is continuous in that
case (cf. Exercise 24.2). It follows then similarly that I n - 1 h(x)==0
almost everywhere; by induction, h(x) ==0 almost everywhere.
CHAPTER 6
26.1) For the proof of the last assertion, let D be the non-empty
collection of all linearly independent subsets x== {x(X}. The collection D
is partially ordered by inclusion; X1-<X2 whenever any X(XEXI satisfies
also X(X EX 2. If C=={xp} is a chain in D, then Up xp is also a linearly
independent subset of V; hence, it is an upper bound of the chain C.
It follows by Zorn's lemma that D contains a maximal element xm.
Then x m is a Hamel basis of V. Indeed, if some XOE V is not a finite

'Ch. 6J
SOLUTIONS
523
linear combination of the elements in x nb the subset x m U Xo would
still be linearly independent, and it would be properly larger than xm.
26.2) Let {XlX} be a Hamel basis of V, and let Xn, n== 1, 2, . . ., be a
countably infinite subset of {XlX}. It is no restriction of the generality
to assume that Ilxnll == 1 for n== 1, 2, . . . . Let Y *0 be an element of W,
and define TXn==ny for n== 1, 2, . . ., TXlX==O for the other XlX in the
Hamel basis. Furthermore, if XE V has the representation X==lXlXl +
· · · +nxC(n' define TX==lTxlXl + . . . +nTxC(n' Then T is linear, and
.1I Tx nll==nIIYII== (nIIYII) Ilxnll, so T is unbounded.
27.1) It is evident that Ilxll* is a norm. If X n is a fundamental
sequence with respect to Ilxll*, then the sequence of its j-th coordinate
(j== 1, . . " k) is a fundamental sequence of complex numbers, having
a limit. The element xo, having these limits as coordinates, is then the
limit element of the sequence Xn. Hence, V is a Banach space with
respect to Ilxll*. Let Ilxll be another norm in V. If M ==max {lfe111, ...,
'1Iekll} and x==  jej, then IlxllM Ijl ==Mllxll*. In order to prove an
inequality in the converse direction, assume that there exists no C >0
satisfying Ilxll*Cllxll for all XEV. Then Ilxnll*>nllxnlf for some sequence
X n ; hence, for Yn==xn/llxnll*, we have IfYnll-+O and IIYnll*== 1, Le.,
=1 1171 == 1 for Yn==  17ej. By a repeated application of the Bolzano-
Weierstrass theorem it follows that Yn contains a subsequence Zn==
f= 1 Cej such that C -+Cj as n-+oo for j == 1, . . ., k. Hence, if we set
z==  Cjej, then Ilzll*== 1, Le., z*O. On the other hand
k
IIz-znll  ICj-CI'llejll-+O
1=1
as n-+oo,
so Ilzllllznll+llz-znll-+O, Le., z==o. This yields the desired contra-
diction. The remaining statements are now evident.
27.2) Evidently, each Fj is linear; furthermore IFj(x)Ilfxll*, where
11xll* is the same as in the preceding exercise. Since Ilxl[*Cllxll by the
preceding exercise, it follows that Fj is bounded. Finally, if a1 F 1 + . . .
+akF k==O in V*, then
a1 F l(X) + . . . +ak F k(X) ==0
for all XE V,
in particular for x==ej. This shows that aj==O for all j, i.e., F 1, . . " F k
are linearly independent. Finally, if F is an arbitrary linear functiona]

524
SOLUTIONS
[Ch. 6
on V, and F(ej) ==aj for f== 1, . . ., k, then x==  jej implies F(x) ==
 jar==  ajFj(x), i.e., F ==  ajFj.
27.3) Let el, . . " ek be k linearly independent elements of V, and
let V k be the k-dimensional subspace consisting of all linear combi-
nations of e1, "', eke The functionals fj(x), defined for f==l, ..., k on
V k by fj(x) ==j for x== 1 jej, are bounded, linear and linearly inde-
pendent. Extending each of these to the whole space V, we obtain
linearly independent bounded linear functionals F 1, . . . , F k.
28.2) Define the sublinear functional P(f) on V by P(f) ==
inf M(f; aI, ..., an), where
n
M(f; aI, . . ., an) ==lim SUp n- 1  f(x+ai)'
xoo i= 1
28.4) Follows immediately from Theorem 1.
30.1) Let, for any complex a, the number sgn a be defined by
sgn a==a/lal for 0< lal <00, and sgn a== 1 for a==O or lal ==00. Then
h(x) == l/sgn g(x) ELoo, and Ilhlloo== 1. Hence, for p==oo, setting f(x) ==
h(x), we have Iffgdfll==flgldfl==llgI11 (and, therefore, also flfgldfl==
IlgI11)' For 1 <P<oo, we have
k(x) == Ig(x) Iq-1/ sgn g(x) ELp
on account of IkI P == Iklq/(q-l)== Iglq, and Ilkll p == Ilgll/p. Hence
f(x) ==k(x) /lIgll/P ELp,
and Ilflfp== 1. Furthermore,
If fgdfll == f Iglq dfl/"gll/P== (I/gllq)q-q/p== IIgllq.
30.2) Let the additional condition be satisfied. Then, given e>O,
the set {x: Ig(x) I> Ilglloo -e} contains a subset E of finite positive
measure, so that, if we define f(x) == {fl(E) sgn g(x) }-1 on E and f(x) ==0
on X -E, the function f(x) satisfies IIfl11 == 1. Furthermore,
Iffgdfll== flgldfl/fl(E)l/glloo-e.
E
If X contains a set F of infinite measure without a subset of finite
positive measure, and if g(x) ==XF(X) , then Ilglloo==l, but ffgdfl==O for
any fELl since f==O almost everywhere on F.

Ch. 6J
SOLUTIONS
525
30.3) On account of 1 rp<oo, we have I/lr 1 + I/IP on X; hence,
if IELp, then IELr. Furthermore, writing r==pa with O<al) we have
by Holder's inequality
/ I/lrd#== / (1/IP)exd#(/ I/IPd#)ex(/ d#)l-ex
== (/ I/I P d#)ex{#(X)}l-ex,
so
/ I/lr d#/#(X){/ I/IP d#/#(X)}ex,
which implies Mr(/)M p(/).
30.4) It is no restriction of the generality to assume that 11/11p== 1.
Then Ilnll for all n, so  I/nlr I/nI P ==l. Hence 1I/IIrl==ll/llp.
30.5) It will be sufficient to consider the case that #(X)==oo, and
to show the existence of the sets X r for r== 1, 2, . . " since for any non-
integer r> 1 we may set X r equal to X[T]' where [rJ is the greatest
integer contained in r. Let An t X, where all #(An) <00. We may as-
sume, passing if necessary to a subsequence, that #(A 1) > 1, and
#(An+1)-#(An) >1 for all n. If #(A 1 )==a1, and [a1J is the greatest
integer in aI, we set X 1 ==. . . ==X[ad==0; if #(A 2 )==a2, we set X[ad+1 ==
. . . ==X[a2]==A 1 , and so on.
Let now I(x) be #-measurable on X. For 11/1100==0, the assertion con-
cerning 11/1100 is true; for 11/1100>0, choose O<M <11/1100, and let E be a
set of finite positive measure on which I/(x)I>M, chosen such that
E c X r for sufficiently large values of r. Then
#r== { / I/lr d#/ # (X r ) }l/TM {#(E) }l / T{#(X r )} -l/r
Xr
>M r -l/r {# (E) } 1 IT,
so lim inf #rM. This holds for all M <11/1100, hence lim inf #r>ll/lloo' On
the other hand #rll/lloo for all r, so lim sup #rll/lloo'
30.8) Note that Ig-lngn==(/-ln)g+/(g-gn) + (In-I) (g-gn).
30.9) If # is Lebesgue measure in the real line R1, then Loo(R1' #)
does not possess an absolutely continuous norm. As regards the second
assertion, it follows immediately from Ilnlg and 1/1g that InELp
and IEL p . For any given index n and any XEX, let un(x)==
sUPml/n+m(x) -/(x) I for m==O, 1, 2, . . . . Then U1(X)U2(X», . . ! 0 and
un(x)2g(x) for all n andx, so unEL p . Hence Ilunll! 0 by the absolute conti-
nuityofthenorm, and since I/n(x) -/(x) Iun(x), we obtain lim 111- Inll==O.

526
SOLUTIONS
[Ch. 6
30.13) Let I be summable over each set of finite measure, and let
An=={x: n2<I/(x)I(n+1)2}
for n== 1, 2, . .. .
Assume first that ;;=no n 2 #(An) ==00 for all no. If nJ#(A n)?;:; 1 for some
infinitesubsequencenj, then there exist sets BjCAni such that nJ#(Bj)
== 1. The integral of III over  Bj is then infinite, and #( Bj) < 00;
this is a contradiction. If n 2 #(An) < 1 for all n?;:;N, then, similarly, the
integral of III over  An is infinite (since  n 2 #(An) ==00), but
#( An) < 00; this is also a contradiction. Hence, there exists an
index no such that :n2#(An)<00, and A==:An is. the required
set of finite measure. Writing 11==lxA and 12 == Ixx-A , we have 1=
11 + 12 with 11 ELI and 12 EL oo. Conversely, if I is a measurable function
such that 1=/1+/2 with I1EL1 and 12ELoo, and if E is an arbitrary set
of finite measure, then
f III d#f 1/11 d#+ f 1/21 d#1[/1111 +#(E) 11/21100<00.
E x E
30.14) Proof of the last assertion: If IEL p , then I is summable over
each set of finite measure in view of
f III d#{#(F) + 1 }p(/) ;
F
hence 1==11 +12 with 11 ELI and 12ELoo by the preceding exercise. Con-
versely, if 1==/1+/2 with I1EL1 and 12ELoo, and if #(E)==l, then
f III d#f 1/11 d#+ f 1/21 d#11/1111 + 11/21100;
E X E
hence 1I/1111/1111 + 11/21100 <00.
30.15) Let, for any A?;:;O, the set FA be defined by FA=={X: I/(x)I?;:;A}.
Then Fo==X, so #(Fo)===oo; furthermore, FA is descending as A in-
creases, and #(FA)!O as Ajoo (in view of 1==/1+/2 with I1EL1 and
12 EL oo). If #(F Ao )== 1 for some AO, then Ef==FAo is the required Ef.
Otherwise, there exists a number AO such that #(F A o ) > 1 and #(F A) < 1
for all A>AO. Select a sequence An!AO; if #( FAJ==l, then Ef== FAn
is the required Ef; if #( F A J<l, then it is sufficient to take the
union of  FAn and an appropriate subset of {x: I/(x) I ==Ao} in order to
obtain the required E f .

Ch. 6J
SOLUTIONS
527
30.16) Let 1==/1 +12 be an arbitrary decomposition of IEL p into
I1ELl and 12ELoo. For any set E satisfying #(E)== 1 we have
f III d#11/1111 + 11/21100'
E
so
1I/1I/1/1111 + 11/21100'
This shows that
11/11inf (11/1111 + 11/21100)'
It remains to prove the existence of at least one decomposition 1==
11 + 12 such that 11/11== 1111111 + 11/21100' Assume first that IO on X, and let
E f and AO be defined as in Exercise 30.15. Set 12(x)==min(/(x), Ao) and
11==1-/2' Then, obviously, 12 EL oo, 12(x)==-Ao on Ef and 12(x)==-I(x) on
X-E f , whereas 11(x)==/(x)-Ao on E f and Il(x)==O on X-E f . Hence
/xl1 d #==/E f l 1d #==/E,{/(x)-Ao}d#, and it follows that I1EL1 since I is
summable over Ef. In addition,
11/1/== f (11 +12) d#== f 11d#+Ao#(Ef)== f Ild#+Ao== 11/1/11 + 11/21100'
Ef Ef X
If 1 is complex, we apply the obtained result to III, and then multiply
by sgn I(x).
30.17) The collection of all continuous functions possessing a bound-
ed carrier may be chosen as the initial domain of definition of the Le-
besgue integral. The existence of a continuous g(x), possessing a bounded
carrier and satisfying 11/(x) -g(x) I/p< ie, follows therefore from Theo-
rem 6. Then 11/(x+h)-g(x+h)llp<le for every h; furthermore,
Ilg(x+h)-g(x)llp<!e for h sufficiently near the origin on account of
uniform continuity.
30.18) Note that
IF(x+h) -F(x) 1II/(x-t+h) -/(x-t) Ilpllg(t) Ilq,
and use the result of the preceding exercise for lp<oo. For p==oo,
observe that F(x) may also be written as F(x) == / I(t)g(x-t) dt.
30.19) It is no restriction of the generality to assume that I and g
are non-negative. Let In be a sequence of bou1!ded measurable functions
on Rk, such that O/n i I. Then
F:(x)== f In(x-t)g(t) dt

528
SOLUTIONS
[Ch. 6
sa tisfies
IF:(x)IllgI11{SUp Ifn(t) I}
t
for every x
(i.e., F:(x) is bounded), and F(x) i F(x) on Rk. In addition, F:(x) is
continuous by the preceding exercise; it follows that F(x) is measur-
able, so F(x) is measurable as well. Set F n(x) ==F:(x) for XE [ -n, n;
...; -n, n] and Fn(x) ==0 for all other x. Then Fn(x)ELp, and Fn(x)i
F(x) on Rk. By the Exercises 30.1 and 30.2 we have IIF nllp==
sup f IF n(x)h(x) I dx for all hELq satisfying Ilhllq 1. Herein we may
restrict ourselves to non-negative h(x), and for any such h we have
J F n(x)h(x) dxJ {J f(x-t)g(t) dt}h(x) dx
== J {J f(x-t)h(x) dx}g(t) dt
llfllpllhllqJ g(t) dtllgI111Ifllp,
Hence IIF nllpllgIl11[fllp, It follows from F n i F that IIFllpllg[llllfllp,
30.20) Let fELp(X, #) for some value of p satisfying 1 pCX). De-
fine f(x) ==0 for x outside X==[O, b], and define g(u)==u lX - 1 for Oub,
and g(u) ==0 outside [0, b]. Then
x
F(a)oflXf(x) == J (x-t)lX-1f(t) dt== J g(x-t)f(t) dt
o Rl
for Oxb.
The preceding exercises show that lXf(x) is again in Lp for all a>O
and, for 1 <pCX), the function oflXf(x) is continuous on [0, bJ whenever
gELq (where 1/P+1fq==I), i.e., whenever (a-1)q>-1. This condition
is equivalent to a> IIp. For p== 1, the function oflXf(x) is continuous
whenever gELoo, i.e., whenever a> 1.
30.21) If Ilfn-fll does not tend to zero, then Ilgk-fll>e for some
e>O and some subsequence gk==fnk' By Exercise 18.14 there is a
subsequence hj==gkj such that hj-+f pointwise, and hence Ilhj-fll-+ O .
This contradicts Ilgk-fll>e.
31.2) Note first that, given the element V*EV*, the non-negative
number c and e>O, there exists an element VEV such that Ilvll -=- c and
cllv*ll-e<v*, v>cllv*ll.

Ch. 6]
SOLUTIONS
529
This follows from the fact that cllv*11 ===sup I <v*, v> I for all v E V satisfying
Ilvl!===c. Insertion of a suitable factor eiq; (qJ real) ensures that we may
write <v*, v> instead of I<v*, v>l. Let now t'(x)==1 V7XE i (X) be a step
function on X into V*; we may assume that the sets E i are disjoint.
The proof of Exercise 30.1 shows the existence of a non-negative step
function c(x) == 1 CiXEi(X) satisfying
pp{c(x)}== 1
and
I c(x) lit' (x) II dfl===Pq(llt'II).
Let e>O be given, and set 'Y)===e/{1 fl(E i )}. Determining, for i== 1, . . .,
k, the elements ViE V such that Ilvill===Ci and
cillv:II-'Y) <v;, Vi>Cillv:ll,
we obtain a step function t(x) === 1 ViXE/X) on X into V, satisfying
pp(llt(x)II)===pp{c(x)}== 1, such that I <t'(x), t(x»dfl===1 <v7, vi>fl(E i ) lies
between I c(x) lit' (x) II dfl-e and I c(x) lit' (x) II dfl, i.e., between pq(llt'll)-e
and pq(llt'II).
Finally, letg(x) EBq(X, V*, fl). Since lq<oo, there exists a sequence
of step functions t(x) on X into V* such that pq(llg-tll) -+0; hence
pq(llgll)==lim pq(lltll). To each t we determine, by the method above,
a step function tn(x) on X into V such that pp(lltnll)=== 1 and
pq(lltll) -n-l1 <t, t n > dflPq(lltl\) ;
hence, the integral in the middle tends to pq(llgll) as n-+oo. Since
II <g, tn>dfl- I <t, tn>dfllPp(lltnll)Pq(llg-tI\)-+O,
the numbers II <g, tn>dfll tend to pq(lIgll) as well, and this is equivalent
to the desired result.
31.3) Let the additional condition be satisfied. Then, given e >0,
the set {x: Ilg(x)ll>poo(llgll)-e} contains a subset E 1 of finite positive
measure, and there exists a sequence of step functions t(x) on X into
V*, vanishing outside E1, and tending to g(X)XEl(X). By Egoroff's theo-
rem (cf. Exercise 18.12; the proof for vectorvalued functions is almost
the same as for numerical functions) the set E 1 contains a subset E of
positive measure such that the convergence is uniform on E; we may
assume, therefore, that Poo(llgXE-tll) <e, and hence that
Ilt(x)ll>poo(llgl\)-2e on E.

530
SOLUTIONS
[Ch. 6
Setting c(x) == {fl(E) }-l XE (x), we have
PI {c(x)}== 1 and J c(x) I It;" (x) II dfl >Poo(llgll) -28
for all t;". Now we determine, similarly as in the preceding exercise,
step functions tn(x) on X into V such that Iltn(x) II==c(x) (hence PI (1Itnll)
== 1), and such that
I <t;", t n ) dfl > I c(x) I It;" (x) II dfl-8,
1.e.,
I <t;", tn)dfl>Poo(llgll) -38.
Since
II <g, tn)dfl- I <t;", tn)dfllP1(lltnll)Poo(llgXE-t;"II)<8,
it follows that If <g, t n ) dfll >Poo(llgll) -48, and this is equivalent to the
desired result.
31.4) Note that, given the element VEV, the non-negative number
c and 8>0, there exists an element V*EV* such that IIv*ll==c and
<v*, v)==clivil. This follows from the Hahn-Banach extension theorem
in sec. 27; cf. in particular the proof of sec. 27, Theorem 2.
31.5) Assume first that lp<oo. If t(X)==ViXEt(X) is a step
function on X into V, with disjoint sets Ei, then
I IITt(x) IIPdfl==  IITviIIPfl(Ei)IITIIP  IlviIIPfl(E i )
== IITIIP J Ilt(x) liP dfl;
hence, pp(lrTtll)IITllpp(lltll). Similarly, if f(x) EBp(X, V; fl) and tn(x) is
a sequence of step functions such that lirn pp(llf-tnll) ==0 and lim tn(x)
== f(x) almost everywhere, then lim Ttn(x) == Tf(x) almost everywhere,
and
pp(IITt m - Ttnll) IITllpp(lltm -tnll) -+0,
so there exists a function g(x) EBp(X, W; fl) satisfying lirn pp(llg- Ttnll)
==0. Since, for a subsequence of Ttn(x), there is pointwise convergence
almost everywhere to g(x) and Tf(x) simultaneously, we obtain g(x) ==
Tf(x). Hence
Tf(x) ELp(X, W; fl)
It follows that
and
pp(IITtll) ==lim pp(I[Ttnll).
pp(11 Tfll)  II Tllpp(llfll).

Ch. 6J
SOLUTIONS
531
Finally, given e>O, select VEV such that l[vll==l and IITvll>IITII-e.
Then the step function t(x) ==VXE(X) satisfies
pp(lltlD == {#(E) }l/ P
and
pp(IITtID == IITvll{#(E)}l/ P ,
and this shows that the quotient pp(IITtll)fpp(lltll) is included between
IITII-e and IITII.
If p==oo, and E is any set of finite measure, then there exists a
sequence of step functions tn(x) satisfying limll/(x)-tn(x)II==O almost
everywhere on E. Then
lim IITI(x) - Ttn(x) 11==0
almost everywhere on E,
and this shows that TI(x) is measurable. The inequality IITI(x) II
IITII'II/(x)ll, holding for every XEX, shows then that
Poo (1ITIII)  II Tilpoo (11/11).
Finally, if VEV is such that l[vl/==l and IITvll>IITII-e, the function
I(x) ==Vxx(x) satisfies
Poo(II/ID == 1
and
Poo(IITIII) == IITvll > IITII-e.
31.6) Note that I(x)g(x) is measurable by remark (h).
31.7) The proof for uniform continuity is along the usual lines. It
follows that, given the continuous g(x) possessing the bounded carrier
E, there exists a sequence of step functions vanishing outside E and
converging uniformly to g(x) on E. This implies readily that g(x) is
Bochner integrable.
Next, let lp<oo, and let t(X)==VXE(X) be a step function con-
taining one single term VXE(X) such that the set E is bounded. De-
termine a numerical continuous g'(x), possessing a bounded carrier,
such that pp(IXE-g'l)<efl/vll. Then, if g(x)==vg'(x), we have pp(llt-gID
<c. It follows that, for any step function t(x) vanishing outside a
bounded set, there exists a continuous g(x) with a bounded carrier such
thatpp(llt-gll)<e. Let now I(X)EBp(Rk, V; #), and set In(x)==/(x) for
xELt n ==[ -n,n; . . . ; -n, nJ and/n(x)==OoutsideLt n . Thenlimpp(II/-lnll)
==0 as n-+oo by dominated convergence; hence pp(ll/- INII) <!e for
some N. Corresponding to this IN, there exists a step function t(x),
vanishing outside Lt N , such that pp(I[/N-tll) < !e, and finally there ex-
ists a continuous g(x), possessing a bounded carrier, such that pp(llt-gID

532
SOLUTIONS
[Ch. 6, 7
< le. Hence pp(ll/-gll) <e. The proof for
lim pp(llf(x+h)-f(x)II)==O
as h-+O
is the same as in Exercise 30.17.
31.8) Proof as in Exercise 30.18.
31.9) Proof as in the preceding exercise, by means of the result in
Exercise 31.6.
31.10) By Exercise 30.19 the function F * (x) == f Ilg(t) 11'II/(x-t) II dt is
in L p ; hence F*(x) <00 for almost every x. Furthermore, it is easy to
see that, for any x, the function <g(t), f(x-t» is a measurable function
of t. It follows then that F(x)==f <g(t), I(x-t»dt exists for all x at
which F*(x) <00, i.e., almost everywhere, and
IF(x) If I <g(t), f(x-t»1 dt<.F *(x)
at these points x.
It remains only to prove that F(x) is measurable, since pp(IFI)<,pp(F*)
<'p1(lIgll)pp(ll/ll) follows then again by Exercise 30.19.
Set I n(t) == f(t) at all t where Ilf(t) II<.n, and I n(t) ==0 elsewhere. Then
Ilfn(t)II<.II/(t) II and lim fn(t) ==f(t) , so
lim <g(t), In(x-t»== <g(t), I(x-t» for all x and all t.
Since
I <g(t), fn(x-t»Illg(t)II'llf(x-t)ll,
and since the function on the right is summable (with respect to t) for
almost every x, the dominated convergence theorem shows now that
lim f <g(t), fn(x-t» dt== f <g(t) , f(x-t» dt==F(x)
\
for almost every x; it is sufficient, therefore, to show that all functions
Fn(x)==f<g(t),fn(x-t»dt are measurable. This is so, since by Exer-
cise 31.8 each F n(x) is continuous.
31.11) Proof similarly as in the preceding exercise.
CHAPTER 7
32.1) For the proof that cf f is an elementary integral on .L, note
that fnEL, In!O implies fnlo!O of-almost everywhere, so cffn==of(fnfo)
! 0 by dominated convergence. Obviously, extending cf, the relation

Ch. 7J
SOLUTIONS
533
/1==.f(llo) holds for any a-function I, and hence for any a6-function
I which is /-summable. Next, let I be a non-negative /-null function.
Then, by sec. 13, Lemma ex, there exists a a6-function g1 such that
/ g==O. Hence .f(g/o) ==0, and so .f(llo) ==0 on account of Ollog/o.
It follows that /1==.f(llo)==O holds for any /-null function I. Once
more by sec. 13, Lemma ex, the relation /1==.f(llo) is then seen to be
valid for any / -summable I.
In order to show that / is .f -absolutely continuous, and / xx< 00,
we prove first that /1==.f(llo) holds for any a6-function IO, whether
/ -summable or not. Let k be a a-function such that k?::-I, and let
knEL+, knik. Then the functions Pn==min(kn,/) are /-summable
(since k n is /-summable and I is /-measurable), so /Pn==.f(Pnlo).
Hence /1==.f(llo) in view of Pn i I.
Assume now that I?::-O is an .f-null function. Then there exists a
a6-function h?::-I such that .fh==O, so .f(hlo)==O as well. But /h==
.f(hlo) since h is a a6-function, so /h==O. On account of O/h it
follows that /1==0, and this shows that / is .f-absolutely continuous.
In order to show that / xx<oo, it is not permitted to put immedi-
ately I==xx in /1==.f(llo) , since we do not know yet that Xx is /-
summable. For this reason we prove first that /1==.f(llo) holds for
any .f-summable I?::-O (both sides may be +00). Observing that .f-
measurability implies /-measurability on account of the .f-absolute
continuity of /, the proof is the same as above for a a6- f unction.
Since by hypothesis xx== 1 XAn' where each XA n is .f-summable, it
is now easy to see that / xx==.f(xxlo) ==.f/ o , and this shows that
/xx<oo on account of the .f-summability of 10.
Finally, by the Radon-Nikodym theorem there exists a function
lo?::-O such that /1==.f(llo) for all IEL 1 (/), in particular
.f(xElo)==/XE==.f(XElo) for any .f-measurable subset E of X.
It follows easily that 10 and 10 are .f-almost equal.
32.2) Note first that the given integral .fl is still the extension of
the elementary integral .fl on L. Since .fl induces a a-finite measure
in each X n, the conclusions in the preceding exercise hold for each X n
separately, and the desired result is easily obtained by addition.
32.3) In view of the result in Exercise 32.1 and the Radon-Nikodym
/!o
theorem, the function I is / -summable if and only if 110 is .f -summable,

534
SOLUTIONS
[Ch. 7
 
and JI==J(llo) for any such I. In particular, JI=J(llo)=JI for any

IEL. Similarly, JI==JI for any IEL. This shows, by sec. 17, Lemma ex,

that J and J are identical.
32.4) If 10 is not bounded J-almost everywhere, and we select for L
the collection of all finite-valued J-summable functions, then J(llo) is
not necessarily finite for all IEL +.
32.5) We prove first that :Ytl induces in X a a-finite measure. In-
deed, since J and J are Stieltjes-Lebesgue integrals, the function XX
is J-measurable as well as J-measurable; hence XX is :Yt-measurable,
so :Ytl is a Stieltjes-Lebesgue integral. Furthermore, since J and J
induce a-finite measures, there exist sequences An t X and Bn t X such
that JXAn and JXBn are finite for all n, so :Yt*XAnBn<oo for all n in
view of :Yt*I=J*I+ J*I for any I?;:-O. It follows that the ordinate set
of XX is the union of a countable number of sets of finite exterior
measure (the exterior measure generated by :Yt in X X Rt), and the
desired result is easily obtained. The integrals JI and JI are :Yt-abso-
lutely continuous; hence, by the Radon-Nikodym theorem, there exist
:Yt-measurable (and hence J-measurable as well as J-measurable)
functions 11 and 12 such that 0$;.11  1, 0<'/2 1, J 1==:Yt (111) for all
IEL 1 (of) and JI=:Yt(112) for all IEL1(J). As indicated, we let F=
{x: 11(X»0} and G==X-F=={x: 11(X)==0}, and we set 10=XF/2 and
10==xGI2' Then 10 and 10 are :Yt-measurable, and 10+10==/2. Setting
now J1/==:Yt(llo) and J2/=:Yt(llo) on L, it follows by Exercise 32.2
that J11 and J21 are elementary integrals on L and that, after ex-
tension, the relations J1/=:Yt(llo) and J2/=:Yt(llo) are preserved. We
wish to show now that JI=J1/+J21 holds not merely for aII/EL,
but also for all IEL1(J). Assuming (as clearly permitted) that 0<.
IEL1(J), the function Ilo==XFI12 is :Yt-measurable since XF and 112 are
:Yt-measurable; in addition we have 110<.112 EL1(:Yt) , so 110 is $'-
summable. But :Yt-summability of 110 is equivalent to J1-summability
of I, and J1/=:Yt(llo) in this case. Similarly, J21 exists and J2/=
:Yt(llo). It follows that J1/+J2/==:Yt(112)=JI.
It remains to prove that J11 is J-absolutely continuous, and J2.f.
In order to prove the first assertion, let J XE=O. Then :Yt(xEl1) ==0,
i.e., E nF is a :Yt-null set. Since 10==XF/2, it follows that E n{x: lo-(x) *O}
is a :Yt-null set, so :Yt(xElo) ==0, i.e., J1XE=0. The orthogonality of .f

Ch. 7J
SOLUTIONS
535
and /2 follows from JXG==:Yt(XGf1) ==0 and
. /2XF==:Yt(XFfo) ==:Yt(xFXGf2) ==0.
32.6) Let / ==/3+ /4, where /3 is J-absolutely continuous, and
/4J. The sets F and G, and the integrals /1 and /2, are defined as
in the preceding exercise; hence J XG==O and /2XF==0. Conversely, if
J XE==O for some set E c X, then E is included in G except for a /-
null set. Indeed, if JXE==O, then :Yt(xEf1) ==0, so EnF is a :Yt-null set,
and hence a/-null set. Since /4 and J are orthogonal, there exists a
decomposition X==F 1 +G 1 into disjoint sets such that JXG1==0 and
/4XF 1 ==0. It follows by the above remark that G 1 is /-almost included
in G, i.e., F is /-almost included in Fl. Then F is /4-almost included
in F1, so /4XF==0. For fEL 1 (/), we write f==fxF+fxG, and then
/1 (fxG) ==/3 (fxG) ==0 since /1 and /3 are J-absolutely continuous,
and /2 (fXF) == /4 (fXF) ==0. Hence
/If- /3f== (/1- /3) (fXF) == (/4 - /2) (fXF) ==0.
32.8) Write F 1 ==x -G 1 and F 2==X -G 2 . Then J 1 XF 1 ==:Yt(XF 1 g1) ==0,
so J 2 XF 1 ==0 since J 2 is J 1 -absolutely continuous. This shows that
(XFlg2)==0, so g2(X) ==0 :Yt-almost everywhere on Fl. Hence F 1 nG 2
is a :Yt-null set, i.e., :YtXF 1 -F 2 ==0. Similarly :YtXF 2 -F 1 ==0. It follows that.
G 1 and G 2 are :Yt -almost equal. Next, since :Yt =:. :Yt', there exist non-
negative :Yt-measurable functions hand h' such that :Ytf==:Yt' (fh) and
:Yt'f==:Yt(fh'). Then :Ytf==:Yt(fhh') for all f EL 1(:Yt) , so hh'==1 holds:Yt-
almost everywhere. In particular, h>O and h'>O :Yt-almost every-
where. It follows now from :Yt'(fg')==J1f==:Yt(fg1)==:Yt'(fg1h) that g'==
glh holds :Yt'-almost everywhere, i.e., :Yt-almost everywhere. Similarly,
gl == g' h' :Yt -almost everywhere. Hence, G 1 and G' are :Yt -almost equal.
By Exercise 32.2 we have that Jf==:Yt(fxG) is a :Yt-absolutely conti-
nuous integral. In order to show that J =:. e.11, note that each of the
following assertions is equivalent to the next one: J 1 XE==0<=>:Yt(XEg1) ==
==O<=>EnG is a :Yt-null set <=>:Yt(XEXG)==O<=>JXE==O.
32.9) Any [JJ,. is represented by a :Yt-measurable subset G,. eX
satisfying :Yt(txaJ==f,.fE[JJ,., and the representation is unique if :Yt-
almost equal sets are identified. Assume, for simplicity, first that :Yt
induces in X a finite measure #; hence #(X) ==:Yt xx< 00. Form all
finite unions of the sets G,., evaluate # for each such finite union, and

536
SOLUTIONS
[Ch. 7
let cx be the least upper bound of the numbers so obtained. If Sn is a
sequence of such finite unions satisfying lim fl(Sn) ==cx, and if S == r Sn,
then the integral .ff==:Yt(fxs) represents the least upper bound [.fJ of
[.fJ,.. Indeed, it follows from the definition of cx that fl(G,.-S) ==0 for
all G,., hence [.fJ,.-<[.fJ. If, on the other hand, [/J satisfies [.fJ,.-<
[/J for all i, and [/J is represented by the :Yt -measurable set T c X,
then G,. c T (except for a -null set) for all i, so Sn c T, and hence
ScT (except for a :Yt-null set), i.e., [.fJ-<[/J. The extension to the
case that  induces a a-finite measure is immediate.
32.10) Form the least upper bound of the collection of all integrals
[/J(1 satisfying [/J(1-<[.fJ,. for all i. Note that the collection [/J(1 is
non-empty, since [OJ, represented by the integral which is identically
zero, is such a [/J (1.
32.12) In order to prove formula (1), let fl== +&. Then  -<fl,
&-<fl, and
f {(d.fjd:Yt)(d/jd)}!dflK== f {(d.fjd)(d/jd:Yt)}!(djdfl) dflQ==
== f {(d.f jd) (d jdfl)}! {(d / jd) (d:Yt jdfl)}! dflQ
== f {(d.fjdfl) (d/jdfl)}ldflQ.
Similarly
f {(d.f jd&) (d / jd&)}l dflP== f {(d.f jdfl) (d / jdfl)}! dflQ,
"
and formula (1) follows.
Given.f and /, the integral  ==.f + / satisfies.f -< and / -<.ff.
The inequality p(.f, /){.f Xx' / xx}l follows by Schwarz's inequality.
Finally, if .f and / are orthogonal, there exists a decomposition of X
into disjoint sets F and G such that .fXG==O and /XF==O. Then
djd ==0 holds -almost everywhere on G, and d/jd ==0 holds:Yt-
almost everywhere on F, so p(.f, /) ==0. Conversely, if p(.f, /) ==0,
then the non-negative function (d.fjd)(d/jd) is -almost every-
where zero. Hence, if F=={x: d.fjd>O} and G==X-F, then .fXG==O
and d/jd==O -almost everywhere on F, so /XF==O.
32.13) For any a>O and any fEL+, the set E=={x:f(x»a} is-
measurable. Hence, the collection of all .f-measurable sets is a a-field
containing all such sets E. It follows that the smallest a-field A con-
taining all such E is included in the a-field of the .f-measurable sets.

Ch. 7J
SOLUTIONS
537
In other words, any Borel set is J-measurable. Evidently, off is an
elementary integral on the collection of the Borel step functions, and
it follows from sec. 17, Theorem 4 that, given fEL+, there exists a
sequence of Borel step functions fn such that fn i f, so offn i Jf. This
shows that, starting from the elementary integral off on the Borel
step functions, the integral Jf on L + is reobtained, hence (cf. sec. 17,
Lemma ex) the given integral Jf is reobtained. Denoting by fl the
measure induced in X by Jf, it follows now from sec. 17, Theorem 9
that, applying the extension procedure for measures to (X, A, fl), we
obtain the same measure fl in X. Hence, given the J-summable set E
(that is, fl(E) <00), there exists by sec. 9, Theorem 2 (3) a a6-set 0 6
(with respect to A) such that E c 0 6 and fl(06-E) ==0. But then 06EA
since A is a a-field, hence E is included in and J-almost equal to a
Borel set. It is evident from this result that any fl-step function is J-
almost equal to a Borel step function, and the proof of the last as-
sertion follows then immediately.
32.14) By the preceding exercise any fli-measurable set in Xi is fli-
almost equal to a Borel set. The present assertion follows from this
result by proving step by step that if Jf is taken in the restricted
sense, then any set C ===A 1 X . . . X A n, where Ai C Xi is fli-measurable
but not necessarily a Borel set, becomes J-measurable with Jxc===
rr fli(A i)'
32.15) By the preceding exercise the integrals off, eff, and also the
auxiliary majorant :Ytf, are extensions of elementary integrals on the
initial domain of definition L, where L consists of all step functions
f(x)==="'2/i CkXCk with Ck===A 1k X ... xA nk such that Aik is a Borel set
with respect to L(i)' Setting p=== rr dofild$i, it follows then immedi-
ately that Jf===:Yt(fP) holds for all fEL, and Exercise 32.1 shows that
the relation off===:Yt(fP) is preserved after extension of Jf. Similarly,
eff==:Yt(fq) for q=== rr defild$i and all ef-summable f. It follows that
p(of, ef) === rr p(J i , efi)' Evidently, p(J, ef) ==0 if and only if p(J i , eft)
==0 for at least one value of i, that is, J ef if and only if Jieft for
at least one value of i.
If eft-<.of i for all i, we may choose $i==J i for all i, so :Yt ===of. The
rela tion
ef f == :Yt(fq)
with
n deft
q==rr
1 d$i

538
SOLUTIONS
[Ch. 7
becomes then
/f == J(fq)
with
d/ n dJi
dJ =q= I{ dJi '
Assume, conversely, that / -<.J. In order to show, e.g., that /1 -<.J 1 ,
let A 1 c X 1 be an J 1 -null set. Then P==A 1 xX 2 X... xX n satisfies
JXP==O, so /Xp==O. But then P is /-measurable, so /XP==(/lXA 1 )'
(/2XX 2 ). . . (/nXxJ, and since JiXXt>O, this shows that /lXA 1 ==0.
32.16) If Ji-<.f i for all i, it is evident from the preceding exercise
that fl n ==/l' . ./n-<.f1' . · In==f!JJ n for all n, and 1pn(x)==(dfl n ldf!JJ n )t===-
(ITi=1 dJildJi)t is an element of the Hilbert space L 2 (J) such that
n n
II1pnl/ 2 ===-J(dfl n ldf!JJ n) == IT fi(dJildJ i ) == IT JiXXt== 1.
i= 1 i= 1
For m>n we have then
II1pm-1pnI1 2 ==2-2(1pm,1pn)
n m
==2-2J 1 . . · Jm{IT (dJildf i ) IT (dJildJi)l}
1 n+1
m m
==2-2 IT f i {(dJildJ i )!}==2-2 IT p(fi, Ji).
n+1 n+1
Hence, if
00
IT p(fi, Ji) >0,
1
we have
m
lim n p(J i , Ji) == 1
n+1
as m, noo,
.and so limll1pm-1pnll==O. Let 1p(x)EL2(J) be the limit element of the
fundamental sequence 1pn(X) , and let A==A 1 x... xAnx IT;;+1 Xi,
with Ai (i== 1, . . ., n) a Borel set (with respect to L(i») in Xi. For m?;:-n
we have then
n n
/XA== IT JiXAt===- IT Ji{XA/dJildf i )}
i= 1 i= 1
n m
==J{XA IT (dJildJi)}==J{XA IT (dJildJ i )}==II1pmXAI1 2 ,
i=1 i=1

Ch. 7J
SOLUTIONS
539
and for m-+oo this tends to II1pXAI1 2 ==J(1p2 XA ). But cf XA is independent
of m, so cfxA==J(1p2 XA )' It follows that cff==J(f1p2) holds for all cf-
summable f. This shows that cf -<.J, and dcfldJ==1p2. Hence
p(J, cf)==J1p==1im J1pn==lim J 1 . · · J n 1pn
n n 00
==lim n Ji{(d/dJi)t}==lim 11 p(J i , )== 11 p(J i , ).
111
If III p(J i , )==O and c>O is given, there exists an index k such
that 111 p(J i , )<c. Let
k
B=={x: 11 d/dJi>l}.
1
Note that B is of the form A X Ilr+l Xi with A C Xl X . . . X X k . Then
k k k
JXBJ{xB(Il d/dJi)!}f1" .Jk{Il (d/dfi)!}==Il p(J i , )<c,
1 1 1
and
k k
cf XXw-B==J{XXw-B 11 (d/dfi) }J{XXw-B 11 (d/dfi)t}
1 1
k k
J1. . .fk{Il (d/dfi)t}== 11 p(J i , )<c.
1 1
If, now, cn==2- n , and Bn is such that JXBn<cn and cfXxw-Bn<cn,
then B==lim sup Bn satisfies J XB==O and cf XXw-B==O. Hence cf -.iJ, so
00
p(J, cf) ==0== 11 p(J i , ).
1
32.17) Evidently, xx==l, and J-<. as well as cf-<.. Since
(l-ex)(dJld)(x)+ex(dcfld)(x)==l for -almost every x, we have
(dJld)(x)(l-ex)-l for -almost every x. It follows immediately
that
p(J, cf)=={(dJld)t(dcfld)!}
(l-ex)-t {(dcf Id)!}== (l-ex)-!p(cf, ).
For the proof of p(J, ) I-ex, let A == {x: dJld> I} and B==X-A

540
SOLUTIONS
[Ch. 7
==={x: d.fjd%l}. Then
p(.f, %) ===% {(d.f jd%)t}?;;-% XA +% {XB(d.f jd%)}
==% XA +.f XB== (l-ex).f XA +ex/ XA + (l-ex).f XB+ex.f XB
=== (l-ex).f xx+ex(/ XA +.f XB) == 1-ex+ex(/ XA +.f XB)
 I-ex.
32.18) Since 1-exip(.fi, ) 1 by the preceding exercise, and
III (l- ex i) >0 by Exercise 11.4, we have III p(.fi' ) >0.
In the case that / -<.%, set
CPn=={d(.f 1 .. ..f n )jd(%l" .%n)}t
and
1pn=={d(/l' · ./n)fd($].. . .%n)}!.
Then cpn and 1pn converge in the Hilbert space L 2 (%) to cP== (d.fjd%)!
and 1p== (d/ jd%)t respectively, so p(.f, /) === (cp, 1p) ===lim (CPn, 1pn) ===
lim Il p(.fi' /i).
In the case that / %, it follows from .f -<.% that / .f, so
p(.f, /)===0, and it remains to prove that III p(.fi' /i) ==0. Observe
first that (l-exi)-t (1 + 2exi)t, so 11 1 (l- ex i)-l is finite by Exercise 11.4.
Hence, since III p(/i, )===O on account of / %, and p(.fi' /i)
(l- ex i)-t p (/i, ) by the preceding exercise, it follows that III p(.fi' /i)
===0.
The proof that / -<..f implies /i-<..fi for all values of i is the same
as in Exercise 32.15. In order to obtain an example where neither
.f / nor / -<..f or .f -<./ holds, let Xi be the interval [0, 1J for all
values of i with .f i / C;lnd /it the Lebesgue integral for i==2, 3, . .. .
For i=== 1, however, let .fIt be the Lebesgue integral of itXA and /1/
the Lebesgue integral of itXB, where A ==[0, iJ and B==[!, 1J. Then
p(.f1' /1) ===! and p(.fi, /i) === 1 for i2, so p(.f, /) ===! which shows al-
ready that .f / does not hold. In addition, we have neither /-<..f
nor .f -<./, since neither /1 -<..f 1 nor .f 1 -<'/1,
32.19) The first assertion is derived by observing that Pn 1 +  ak.
For the second assertion it remains to show that q>O implies the con-
vergence of 1 an. If q>O, then the sequence q;;l converges, ap.d it
follows then from 1 +ak(l-ak)-l that Pn== II (1 +ak) converges, so
1 an<oo.

Ch. 7J
SOLUTIONS
541
32.20) Writing ri== 1-p(J i , efi), it follows from Exercise 32.16 that
J and ef are equivalent or orthogonal according as qn=== Ilf= 1 (1 -r i)
converges to a number q>O or to zero; hence, by the preceding exer-
cise, according as 1 ri converges or diverges. But ri==!;Yi, and so the
first assertion follows.
If a(Xil-a and a{:Jil-a for a fixed a>O, then
I (Xi - {:J i I I (Xi - {:J i I !! I (Xi - {:J i I I (Xi - {:J i I
2 < !+P! - = l<Xi-Pil = !+p! <; 2! -,
(X. . (X.. a
   
and similar inequalities hold for I (I-(Xi)! - (l-{:Ji)!I. Hence, the con-
vergence of  Yi is then equivalent to the convergence of  ((Xi-{:Ji) 2.
32.21) If the points of R2 are denoted by (x, y), then lo(x, y)===
(x2+y2)-!.
32.22) By Exercise 9.21 the set E e X is fl-measurable if and only
if E is fln-measurable for all n, and fln(E) i fl(E) for any such E. Simi-
larly for v. All fln are, therefore, fl-absolutely continuous, and the
functions gn(x), satisfying fln(E)===fXEgndfl for all fln-measurable E
(i.e., fln(E)===fEgndfl for all fl-measurable E), are such that gni 1 holds
p-almost everywhere on X. Also, if E is a fl-null set, then fln(E)===O
for all n, so vn(E) ==0 for all n, and this shows that E is v-measurable,
and v(E)==lim vn(E) ==0. Hence, v is fl-absolutely continuous, i.e., v(E)
== f XEldfl for some non-negative fl-measurable I and all v-measurable E.
For E any v-measurable set, we have then
vn(E) === f XElndfln=== f XElngndfl if XEldfl===V(E) ,
in particular fElngndflifEldfl for any fl-measurable E, so Ingnil holds
p-almost everywhere on X. It follows that I===lim In holds fl-almost
everywhere on X.
34.1) Let fl have the finite subset property, and assume that the
..?-measurable function 1>0 satisfiesJI>O. Then the set E =={x: I(x) >O}
is J-measurable, and fl(E) >0. Hence, E contains a measurable subset
F of finite positive measure. Let g==min (I, XF).
Assume, conversely, that for any 1>0 with JI>O there exists a
function g(x) such that Og1 and O<..?g<oo, and let E eX be fl-
measurable with fl(E) >0. For I(x) == XE(X) , there exists by hypothesis
a function g(x) such that Og(X)XE(X) and O<Jg<oo. It follows

542
SOLUTIONS
[Ch. 7
that for some a >0 the set F == {x: g(x) >a} is of finite positive measure.
Since, obviously, FeE, this shows that fl has the finite subset property.
34.2) The equivalence of (a) and (b) follows from the preceding
exercise, and the remaining equivalence proofs are similar to the proof
of Theorem 2.
34.4) Note that the sets E for which flc(E) ==0 are exactly the local
J-null sets. The desired result follows then from part (b) in sec. 33,
Lemma ex.
35.1) Let {X lX } be the same maximal collection as used before in
this section, and set PlXP(X) == XXa(x)fp(x). Then s==supp P{3 exists by
sec. 34, Theorem 1, and f* == sup s exists since fl is localizable (proof
as in part (a) of Theorem 2). Hence f*==SUPlX,p P{3' Also, 1;==suplX PfJ
for any (3, so f*==sup l*p.
35.2) By the definition of fle, there exists a sequence of fl-summable
sets F neE such that fl(F n) i fle(E) , and we may assume that FIe F 2
c... . Then F=='L/l Fn satisfies fl(F) ==fle(E) , so fle(E-F) ==0. Since
fle-null sets and local fl-null sets are identical, this shows that E - F
is a local fl-null set. If F ' is another fl-summable set such that F' c E
and fl(F')==fle(E), then F-F' and F'-F are local fl-null sets, and
hence fl-null sets (since F and F ' are of finite fl-measure). This shows
that F and F ' are fl-almost equal.
35.4) Assume that fl is localizable, and let {F p} be a collection of sets
F p of finite fle-measure. We have to show that sup F*p* exists. By
Exercise 35.2 each F p has a fl-unique subset E p such that fl(E p )==
fle(F p ). Note that E*p*==F*p*. By hypothesis F*==sup E*p exists, and it
follows from F*>E; that F**>E*p*, so F** is an upper bound of {E;*}.
Assume that Q** is another upper bound, so Q**?;;E;* for all {3. Then
Ep-Q is a fle-null set, and hence a fl-null set (since E p is of finite fl-
measure). This shows that Q*>E*p for all {J, so Q*>sup E*p==F*, im-
plying that Q**>F**. It follows that
F**==sup E-p*==sup F;*.
Since fle has the finite subset property, it follows from Theorem 2
that fle is localizable if and only if any cross section <1;*> is determined
by a function f defined on X, and by the preceding exercise this is so
if and only if any cross section <I;> is determined by a function f de-

Ch. 7, 8J
SOLUTIONS
543
fined on X. This shows that (a) and (b) are equivalent. Finally by
sec. 33, Theorem 5, (b) and (c) are equivalent.
35.5) The proof of the first statement follows by means of the result
in Exercise 35.2.
35.6) The proof that # is not localizable may be given as follows.
Let X(X be the horizontal segment {(x, y): Oxl, y===cx}, and assume
that P* ===sup X exists. If P is an arbitrary set in the equivalence
class P*, then P has infinitely many points on each horizontal segment;
hence, selecting the set Q c P such that Q has one point on each hori-
zontal segment, (P-Q)* is still an upper bound of {X}. But #(Q) ==00,
so this contradicts P* ===sup X.
The localizability of #e follows by observing that X is the direct sum
of the horizontal segments X(X.
CHAPTER 8
36.1) Follows from the Lebesgue decomposition theorem for inte-
grals in Exercises 32.5 and 32.6.
36.2) Since VI is #-absolutely continuous, and V1(X) <00, there exists
by the Radon-Nikodym theorem a non-negative #-summable function
to such that VI (E) === / XEto d# for any vI-measurable set E. In particular
v1(E)==/Etod# for any #-measurable set E, since #-measurability im-
plies vI-measurability. It follows therefore that DV1 (x) === to (x) holds #-
almost everywhere, hence v1(E)==/XE(X)Dv1(X)d# for any VI-measur-
able E, and in particular for any v-measurable E.
The sets F and Hare v2-measurable, so
1'2 (E) =1'2 (EF) +v2(EH) ===v2(EH)
for any v-measurable E. Also, since #(H) ==0 implies 1'1 (H) ===0, we have
v1(EH) ===0. This shows that EH, being vI-measurable as well as V2-
measurable, is v-measurable, and
v (EH) ===v1(EH) +v2(EH) =v2(EH) ==v2(E).
Hence
. v(E) ==v2(E) +v1(E) ===v(EH) + f XE(x)Dv1(X) d#
for any v-measurable set E.

544
SOLUTIONS
[Ch. 8
36.3) Assume that  is regular with respect to #+v, and let E be
a subset of (0, 1J such that #(E) >0. Then, given 8>0, there exists a
sequence of sets 1 nEJVO==U1 /Vk such that
(#+v)(E- 1 n )==0
and
(#+v)( 1n)(#+v)(E)+8==#(E)+8,
so
#(E- 1 n )==0
and
2#( 1 n)#(E) +8.
But #(E-  1 n)==O implies #(E)#( 1 n), so we obtain 2#(E)#(E)+
+8. This holds for any 8>0, a contradiction.
36.4) Note first that any (#+v)-measurable set is #-measurable, v-
measurable, vI-measurable and v2-measurable; this holds in particular
for any 1 EJVO== Ul' Let E be (#+v)-measurable, and #(E) <00.
Then, given 8>0, there exists a sequence of sets 1 n from JVO such that
(#+v) (E -  1 n)==O and (#+v) ( 1 n-E)8. It follows that#(E -  1 n)
==0, so#{ (1 n nE)}==#(E), and alsov2( 1 n -E)8, SOV2( 1 n)v2(E) +
+8. The same method as in the proof of Theorem 1 shows now that if
D V2(x)?;:;a on E, then v2(E)?;:;a#(E), and it is easily derived from this
result that if V2(P) ==0 for the (#+v)-measurable set P, then DV2(X) ==0
holds #-almost everywhere on P. The last result is applied to the set
F in the decomposition X==F+H with v2(F)==#(H) ==0. The set F is
}t-measurable, so vI-measurable; F is also v2-measurable, so F is v-
measurable. It follows that F is (#+v)-measurable; the equality v2(F)
==0 implies therefore that DV2(X) ==0 holds #-almost everywhere on F,
i.e., #-almost everywhere on X. Hence, by Exercise 36.2, Dv(x) ==
DV1(X) exists as a finite number #-almost everywhere on X, and the
desired result follows.
36.5) It will be sufficient to prove that the regularity condition is
satisfied for any bounded closed interval; note that all open and closed
sets are (#+v)-measurable. With only small changes the proof is the
same as in Lemma 5.
37.1) Combine the final result in the present section and the results
in the Exercises 37.2, 37.4 and 37.5 in the preceding section.
37.2) If g is right continuous, and g(oo) -g( -00) is finite, consider
the Stieltjes-Lebesgue measure v, defined by v(A)==g(b)-g(a) for any
cell A==(a, bJ, and apply the result in the preceding exercise. The re-

Ch. 8J
SOLUTIONS
545
striction that g(oo) -g( -00) is finite may be removed easily by con-
sidering auxiliary functions equal to g on a bounded interval and
constant to the left and the right of this interval. Finally, assume that
g is not right continuous. Then G(x) ==g(x+) ==lim g(x+h) for h! 0 is
non-decreasing and right continuous, and G==g except on a set H of
Lebesgue measure zero. Let x be any point in R 1 - H where G' (x) exists
as a finite number, and let x1-=1-x. Then
G(X1) ==g(X1 + »g(X1)G(X1-),
and since {G(X1)-G(X)}j(X1-X) as well as {G(X1-)-G(X)}j(X1-X)
tend to G' (x) as Xl --+x by the preceding exercise, it follows that
{g(X1) -g(x) }j(X1-X) tends to G' (x). Hence, g' (x) exists almost every-
where as a finite number.
37.3) It may be useful to observe first that all measures Vii may be
regarded as extensions of measures initially defined on A; this follows
from the proof of the Lebesgue decomposition theorem in Exercise 32.5.
It follows from the Lebesgue decompositions Vn==Vn1 +Vn2 and Vn+1-Vn
==Tn1 +Tn2, by observing that any vn+1-measurable set is vn-measur-
able as well as (vn+1-vn)-measurable, that
'Vn+1 == (Vn1 +Tn1) + (Vn2+ T n2)
for every vn+1-measurable set.
Also, it is evident that Vn1 +Tn1 is #-absolutely continuous, and Vn2+Tn2
is orthogonal to #. Hence, by the uniqueness of the decomposition,
Vn1 +Tn1 ==Vn+1,1 and Vn2+ T n2==Vn+1,2. This shows that the sequences
Vn1 and Vn2 are increasing on the sets of A, where all these measures
are certainly defined. Since VnlV and Vn2V on A, the sequences Vn1
and Vn2 converge to finite-valued set functions v! and V2 on A, and by
Exercise 9.20, vi and V2 are measures on A. It follows then from Exer-
cise 9.21, that Vn1 t vi for all v! -measurable sets, and Vn2 t V2 for all V2-
measurable sets.
Let E be an arbitrary v-measurable set. Then E is vn-measurable
for all n by Exercise 9.21, so E is measurable with respect to all Vn1
and all Vn2. Once again by Exercise 9.21, it follows now that E is
measurable with respect to vi and V2, so
(Vn1 +Vn2) (E) t (vi +V2) (E),
1. e.,
v(E) ==vi(E) +v2(E).
Next, we prove that vi is #-absolutely continuous. Indeed, #(E)==O

546
SOLUTIONS
[Ch. 8, 9
implies vn1(E) ==0 for all n, so that, by Exercise 9.21, E is vi-measur-
able and vi(E) ==0. Also V2#, for if F nand H n are disjoint sets such
that Fn+Hn==X and #(Hn) ==vn2(F n ) ==0, then H==Hn satisfies
#(H)==O, and F==X-H satisfies FeF n for all n, so vn2(F) ==0 for all
n, and this implies that F is v2-measurable and v2(F) ==0. Hence, by
the uniqueness, v==vi +V2 is the Lebesgue decomposition of v, i.e.,
vi ==V01 and V2==V02, so Vn1 t VOl and Vn2 t V02. It follows then from Exer-
cise 32.22 that fn t f holds #-almost everywhere on X.
37.4) Combine the results in the Exercises 37.1 and 37.3.
37.5) Since
hk(X) ===g(x) -snk(x)g(b) -snk(b) <2- k
for all k and all xE[a, b], the series 1 hk(x) converges to a finite
sumfunction, and it follows by the argument already used before that
1 hk(x) converges almost everywhere to a finite sumfunction, so
hkO holds almost everywhere.
37.6) Given xER 1 -E and s>O, choose a rational point p satisfying
If(x)-pl<!s. Then, assuming for simplicity that h>O, we have
h h h
J If(x+t) -f(x) I dtJ If(x+t) -pi dt+ J Ip-f(x) I dt;
000
the first integral on the right, divided by h, tends to If(x) -pi < is as
h-+O, and the second integral on the right does not exceed ish. Hence,
for h sufficiently small, f If(x+t) -f(x) I dtsh.
CHAPTER 9
38.1) In order to show that the measure v, generated by qJ(x), is #-
absolutely continuous, use Exercise 24.1 or, equivalently, sec. 36,
Lemma 3.
40.1) Let F e 01 be compact, and for each xEF with image point
y==h(x) let %(x) and %(y) be neighbourhoods of x and y respectively
which are in one-one correspondence. The set F is covered by a, finite
number of the %(x), say JVi, . . ., JVp. Denote the corresponding %(y)
by %(Y1), "', %(Yp). Then, if f(y»O is measurable on 02, we have

Ch. 9J
SOLUTIONS
547
by the proof of Theorem 1 that
ff{h(x)}](x)dfl= f f(y)dfl
JV i ."Y'(Yi)
for i == 1, .. " p,
and the equality continues to hold if Jtii and JV(Yi) are replaced by
any fl-measurable EicJtii and its image h(Ei) cJV(Yi) respectively.
Apply this to E1=JVinF, E 2 ==(JV2nF)-E 1 , .", Ep==(JVpnF)-
(E1 + . . . +E p - 1 ), and sum the so obtained equalities. The sum on the
left is!Ff{h(x)}](x)dfl; the sum on the right is!h(F)NF{y)f(y)dfl, where
for any YEh(F) we denote by N F(Y) the number of different XEF
satisfying h(x) =y. Note that in view of h(F) == f= 1 h(Ei) the function
N F(y) assumes only a finite number of values on h(F), each of these
values on a measurable set.
For all xER k , let d(x) denote the distance of x to the complement
O of 0 1 . Then d(x) is continuous, and d(x) >0 if and only if XE01. It
follows that Dn={x: d(x)n-1} is a closed subset of 01 for n= 1, 2,
. . " so F n==DnBn is compact where Bn is the closed sphere of radius
n around the origin. Evidently F n t 0 1 , so h(F n) t O 2 , and for each
YE02 we have N Fn(Y) t N(y), so N(y) is measurable. Since
f f{h (x)}] (x) dfl= f N Fn(Y)Xh(Fn)(Y)f(Y) dfl,
Fn 02
we obtain (n-+oo)
f f{h (x)}] (x) dfl= f N(y)f(y) dfl.
01 02
40.2) Assume, e.g., that the k-th row of the matrix of the tip(a) is
a linear combination of the other rows, i.e.,
k-l
ikp(a) =  (Xiiip(a)
i=l
for p == 1, .. " k.
Write now
k-l
l(y) == (Y1, · . · , Yk-1, Yk-  (XiYi)
1
f or all Y == (y 1, · · ., Y k) .
Then l(y) is a linear transformation on Rk onto Rk with its Jacobian
equal to 1; we set H(x)==l{h(x)} for all XE01. Since fl(E)=fl{l(E)} for
every measurable set E, we obtain
fl{h(A) } ==fl [l{h (A )}J == fl{H (A)}

548
SOLUTIONS
[Ch. 9
for every compact set A c 0 1 (note that h(A) is also compact, and
hence measurable). Assume now that C 1 is a closed cube of centre a,
contained in 0 1 . Then there exists a number M >0 such that liip(X) I
M for all XEC 1 and all i, p. Furthermore, 8H k (a)j8x p ==0 for p== 1,
..., k, so there exists a closed cube U(a) cC 1 of centre a such that
18H k(X) j8x p l ej[(2k).k Mk-1 ]
for all XE U(a) and p== 1, . . ., k. This U(a) is the desired cube around a.
Indeed, if C is a closed cube such that C c U(a) and if x, x' EC, then
k
Hi(X) -Hi(X') ==  (xP-X)iiP(b(i»)
p=l
for all i, where b(i) is some point on the segment from x to x'. Hence,
if #(C) =()(,k, we obtain
IHi(X) -Hi(X') lkM ()(,
for i == 1, . . " k -1,
and
IH k(X) -H k(X') lke()(,j[(2k)kMk-1].
It follows that the set H(C) is included in an interval of measure
(2kM ()(,)k-l. 2k()(,ej[(2k)kMk-1] ==e()(,k==e#(C).
Hence
#{h(C) }==#{H (C) }e#(C).
Similarly, if C is not a cube, but an arbitrary closed interval contained
in U(a).
40.3) Given 'Y»O, cover E by a a-set  Bn such that  #(Bn)<
#(E) +'Y). If C n is the intersection of U(a) and the closed interval with
the same vertices as Bn, then E is covered by  C n, so h(E) is covered
by h( Cn)== h(C n ). Hence
#*{h(E)}#*{ h(Cn)} #{h(Cn)}e  #(Cn)e  #(Bn)
<e{#(E) +'Y)}.
This holds for every 'Y»O, so #*{h(E)}e#(E).
40.4) Let B be the subset of 0 1 where J(x)=O. Evidently! B is
closed. The desired result will follow from the result in Exercise 40.1
if we can prove that #{h(B)}=O. If Fn is a sequence of compact sets

Ch. 9J
SOLUTIONS
549
such that F n j 0 1 , then all F n nB are compact, and F n nBj B. We may
assume for the proof, therefore, that B is compact. For each point
aEB, we let U(a) be the closed cube around a of the preceding exer-
cise, so #*{h(E)}e#(E) for every measurable Ec U(a). The interiors of
all U(a), aEB, cover B, and so B is already covered by a finite number
of these. Hence, B is covered by a finite number of the U(a), say UI,
. . . , Up, Now, ) U i may be written as a union i Vi of closed inter-
vals VI, . . " V q having only boundary points in common. Hence, if
Pi==BV i , then #(B)==  #(P i ). Since B is covered by i Vi, the image
h(B) is covered by h(i Pi), so
q q
#*{h(B)} #{h(Pi)}e  #(Pi)==e#(B).
1 1
This holds for all e >0, i.e., #{h(B) }==O.
The proof presented here is essentially due to J. MARiK (1956, [lJ).
40.5) If J(x) ==0 at the point x, then the system of linear equations
k
 iip(X)exp==O,
p=l
i==l, ...,k,
has a non-zero solution, and then   iiP(X)exiexp=='O for this solution.
Hence J(x) -=1-0 at every x.
Assume now that a-=l-b and h(a)==h(b), i.e., hi(a)==hi(b) for i== 1, . . "
k. Let cp(x) == = 1 hi(x)exi, where exi==bi-ai. Then cp(b) -cp(a) ==0, so by
the mean value theorem and the fact that 0 is convex we obtain
k
0==  {ocp(c) joxp}exp,
p=l
where c is some point on the segment from a to b. But
ocp(c) jox p ==  iip(C)exi,
i
1. e.,
0==   iiP(C)exiexp,
i p
and this contradicts the hypotheses.
40.6) By
u(x, y)==ax+by+c, v(x, y)==alx+b 1 Y+C1
we map 0 onto the open set 0 1 =={(u, v): lul P + Ivl q < I} in the (u, v)-
plane. The mapping is one-one, and J(x, y)==ILtI, so J*(u, v)==ILtI-1.

550
SOLUTIONS
[Ch. 9
The measure of 0 is therefore
1 (l-U P )I/(l 1
II f du f dv= II f (l-u p ) ljQ du = - P'I B(+, ++ 1)=
o 0 0
4
PqlLtI
r(ljp)r(ljq)
r(1/p+1jq+l)
4
(p+q) ILtI
r(ljp)r(ljq)
r(ljp+ 1jq) .
40.7) Note the following facts: The boundary of 0 1 consists of
the horizontal line segment {(u, v) : Oua, v== I},
the vertical line segment {(u, v): u==a, (1-a2)tv1}
and the circular arc{(u, v): U 2 +v 2 == 1, Oua, v>O}.
It is not true, however, that the boundary of 0 1 is the image of the
boundary of O. Since horizontal segments in 0 are mapped onto hori-
zontal segments in 0 1 , it follows easily that the mapping is one-one.
Furthermore,
J(x, y)=={sin 2 ylsin xl}jcos 2 x==u 2 /sin xl,
and u 2 cos 2 x==1-v 2 implies sin 2 x==(u 2 +v 2 -1)ju 2 , so J(x, y)==
u(u 2 +v 2 -1 )t. The measure of 0 is therefore
f ... 1
dudv.
J u(u 2 +v 2 -1)!
01
Integration of (u 2 +v 2 -1 )-t with respect to v between v== (1-u 2 )! and
v== 1 is the same as integration of (t 2 -1)-! between t== 1 and t==
(l-u 2 )-t; hence, writing t- 1 ==sin ex, we obtain for this integral
tn
f dex
. == [log tg lexJ: == -log tg lexo,
SIn ex
eta
where exo is the number between 0 and In determined by sin exo==
(1-u 2 )t. It follows that log tg !exo==! log{(l-u)j(l +u)}; the measure
of 0 is therefore
a a
1 f 1 1 +u f a3 a 5 -
- -log du == (1+!u 2 +t u4 +.. .)du==a+ - + _ 2 +...
2 u 1 -u 3 2 5
o 0

Ch. 9J
SOLUTIONS
551
by the theorem on integration of increasing sequences. For a== 1 this
gives the well-known value in 2 .
40.8) For the indicated mapping we have J(X)==Xn-1, so
D== /1(x1+... +xn-1)xf1-1.. ,x1+Pn-1xn-1-1(1-xn)Pn-1dx1" .dxn
01
==B(Pn-l, Pn)/ l(x1 +. · · +Xn_1)xf1-1. . .xi+Pn-1dx1. . . dX n-1,
where B(Pn-1, Pn) is the Beta function, and the last integral is the
(n-1 )-dimensional analogue of D (with only Pn-1 replaced by Pn-1 +
+Pn). The desired result is easily obtained by repeating the argument.
40.10) Consider, for OA 1, the matrix with elements Cij(A) ==Aaij+
+ (l-A)eij, where eij is the unit matrix, so eij== 1 for i==i and eij==O
for i*i. The matrix Cij(A) is symmetric and positive definite for all
these values of A, and hence its determinant cannot vanish (if the
determinant vanishes, say for Ao, then the system of linear equations
n
 Cij(AO)CXj== 0 ;
1=1
i == 1, ..., n,
has a non-null solution, and hence ii=1 Cij(Ao)CXiCXj would vanish for
this solution). Since the determinant is continuous in A, and since it is
positive for A==O, it must be positive for A== 1.
In order to compute the integral cf, we use the fact that there exists
another rectangular coordinate system in Rn such that ij aijXiXj==
i AiY;, where Yi (i== 1, "', n) are the coordinates with respect to this
new system, and where Ai (i== 1, . . ., n) are the roots of the character-
istic equation det (A -AE) ==0. By the positive definiteness all Ai are
positive, and on account of det (A -AE)== ITi=1 (Ai-A) it follows that
ITi=1 Ai==det(A). The transformation determinant from one rectangu-
lar coordinate system to another one is of absolute value one, and so
00
cf = f e-A'Y" dp, = il ( f e-A'Yi' d Yi )
Rn -00
n nt ntn ntn
_ IT - -
- i = 1 A  - (IT Ai) t - { det (A)}t ·

552
SOLUTIONS
[Ch. 9, 10
40.11 ) We may assume without loss of generality that O<A< 1. Set
P==A- 1 and q== (l-A)-l, so 1 <P<oo and p- 1 +q-1== 1. Finally, set
e-)''2:.aijXiXj == f(x),
e - (1-)')'2:.bijXiXj == g(x).
Then
n tn == J e-'2:.CijXiXj dfl == J fg dfl 
{det (C)}t 
< (J fp dp, YIP (f gq dp, Ylq = (f e-'E.aux,xi dp, Y (f e-'E.biix,xi dp, Y-A,
which yields the desired result.
40.12) Introduce new variables Yi in  aijXiXj by setting Xi== -Yi
for i== 1, . . " k and Xi==Yi for i==k+ 1, ..., n. Then ii=1 aijXiXj be-
comes
k n
( + 
ij= 1 ij=k+ 1
k n
- 
i=1 j=k+1
n k
  )(aijYiYj)==
i=k+1 j=1
==Sl (y) +S2(Y) -S3(Y) -S4(Y),
and hence, introducing these new variables in the integral cf, we obtain
cf == 1 e-'2:.aijXtXj dfl== 1 e-(Sl+s2-sa-s4) dfl.
It follows by addition that
2cf == 1 (e- S1 - S2 ){P(x)+P-1(x)}dfl
with P(x)==e- sa - s4 >0. Since P(x)+P-1(x»2 for all x, this yields
2cf?;:,2(1 e- S1 dX1. . . dXk) (I e- S2 dXk+1. . . dXn),
from which the desired result follows easily. The inequality det (A)
rr aii is now derived by induction.
CHAPTER 10
41.1) Assume first that g is right continuous. If v is the mesure
generated by g(x), there exists by Exercise 37.1 a set H .of Lebesgue
measure zero such that v(E) ==v(EH) + f lEg' dfl for any bounded v-

Ch. 10J
SOLUTIONS
553
measurable set E. In particular
b
g(b) -g(a) ==v{(a, bJ nH}+ f g' (x) dfl
a
for a<b, so that, defining gl(X) and g2(X) by
gl(O) ==0,
b
gl(b) -gl(a) == f g' (x) dfl,
a
g2(X) ==g(x) -gl (x),
the desired result follows. If g is not right continuous, apply the ob-
tained result to G(x)==g(x+)-g(O+), so G==G 1 +G 2 with G 1 (x)==
It G'(t)dfl, and set gl==G 1 , g2==g-G 1 .
41.2) Select the positive numbers An (n==O, 1, 2, ...) such that
00
 2nAn== I-ex,
n=O
e.g., An== (1-ex)j3 n + 1 . Make the length of the open interval which is
removed at the first step equal to Ao; generally, make the length of
each of the 2 n open intervals which are removed at the (n+ 1 )-th step
equal to An.
41.3) Let II, 1 2 , 1 3 , . .. be the closed intervals [0, tJ, [t, IJ, [0, 1J,
[1, tJ, [t, !J, [!, 1J, [0, 1J, [1, 1J, ..., and let gn(x) be the Cantor
function of In. Then g(x) == 1 gn(x)j2 n converges uniformly, so g(x)
is continuous. Evidently g(x) is non-decreasing, and since the length
of the maximal subinterval of [0, 1J on which the partial sum sn(x) is
constant tends to zero as n-+oo, the function g(x) is strictly increasing
on [0, 1J. Finally, by Exercise 37.5, the derivative g'(x) is zero almost
everywhere.
41.4) The absolute continuity of I follows by observing that
f I' dfl== 1 ==/(1) -/(0).
D
Since E 1 == I(E) is a subset of C 1, we have fl(E 1) ==0, so g==XEl is measur-
able. Since
{x: g{/(x)}> 1 }==E,
the function g{/(x)} is not measurable. The last statement follows by
observing that I' (x) ==0 almost everywhere on C.

554
SOLUTIONS
[Ch. 10
41.6) The set of all rational numbers in X is dense and of measure
zero. The set of all irrational numbers in X is a boundary set and of
measure one. The set of all rational numbers is of measure zero, but
not nowhere dense. Given ex such that Oex< 1, there exists a Cantor
set C of measure ex, and C is nowhere dense. If C n is a sequence of
Cantor sets with #(C n) t 1, then E ==  C n is of the first category and
p(E) == 1. The complement F ==X - E is of the second category, and
p(F) ==0.
41.7) For k>O, and 0<h<h 1 , all these numbers being sufficiently
small, we have
q(x-k, x)q(x, x+h)q(x, X+h1).
Hence, q(x, x+h) is decreasing as h!O, and bounded from below by
q(x-k, x). It follows that D+f(x) exists as a finite number. Similarly
for D-f(x). It is also immediately evident that D-f(x)D+f(x). Note
that the continuity of f, proved earlier in Exercise 25.4, follows now
easily. In order to show that D-f and D+f are increasing, assume that
Xl <X3 and choose X2 such that Xl <X2<X3. Then
D+f(X1)q(X1, X2)q(X2, x3)D-f(X3),
and so D-f(X1)D-f(X3) as well as D+f(X1)D+f(X3)'
41.8) Let x be a point of continuity of D-f. For h>O we have
D+f(x)D-f(x+h), so (h!O) we obtain D+f(x)D-f(x). It follows that
D+f(x) == D-f(x).
41.9) Let [Xl, X2J c (a, b). Then D-f and D+f are bounded on [Xl, X2J;
let M be a common upper bound for ID-f(x) I and ID+f(x) I. If [X3, X4J
is a closed subinterval of [Xl, X2J, then
D+f(X3) q(X3, X4) D-f(X4),
so If(x4)-f(x3)IM(x4-x3)' It follows immediately that f is absolutely
continuous on [Xl, X2J.
41.11) We have
Xs X2
ex J gdtex(X3-Xl)g(X3) == (I-ex) (x2-x3)g(x3)(1-ex) J gdt.
Xl
Xs
Hence, f(x) == fx g(t) dt is convex, so f(x) == fx f' (t) dt by the preceding
exercise. This holds for all x, and so f'==g almost everywhere in (a, b),

Ch. 1 OJ
SOLUTIONS
555
and in particular at every point where both I' and g are continuous.
It follows that I'==g except on an at most countable set.
42.1 ) It will be sufficient to indicate the completeness proof for
BV o . Let InEBVO for n==l, 2, ..., and Illn-lmll-+O as m, n-+oo. For
any xE[a, bJ, we have
I/n(x) -1m (x) IV(ln - Im)llln -Imll,
so In converges unifqrmly on [a, bJ to a function lo(x), with 10(a)==0.
Given e>O, we select N such that Illn-Imlle for m, n>N, and we let
f!JJ be an arbitrary partition of [a, bJ. Then
S&J(IN-Im)II/N-Imlle
for m>N,
so (m-+oo) we obtain S&J(IN-Io)<.e. This holds for all partitions f!JJ, and
it follows that IN-IoEBVo, so 10EBV o . Since, similarly, S&J(ln-Io)e
for all f!JJ and all n>N, we obtain Il/n-Iolle for n>N, so 10 is the limit
.of In in BVo.
42.2) Let Il/n-Ioll-+O as n-+oo. This implies uniform convergence
-on [a, bJ of In to 10; hence, if all In are continuous, then 10 is continuous.
I t follows that C B V 0 is closed.
Assume now that InEACo for all n, and Il/n-Ioll-+O. Given e>O, we
select N such that IIIN-Ioll<e. Since INEAC o , there exists €5>0 such
that if  (bi-ai) <€5 for the finite union  (ai, biJ of disjoint cells,
then
k
 IIN(b i ) -IN(ai) 1 <e.
1
It follows easily that
k
 I/o(b i ) - 10(ai) I <2e,
1
i.e., 10EAC o .
42.3) Given IEC and e>O, there exists in view of the uniform conti-
nuity of I a continuous function 10 such that 11/-loll<e, and the graph
-of 10 consists of a finite number of straight line segments. Then 10EAC,
.and this shows that A C is dense in C.
In order to prove that CBV is of the first category in C, consider
first the set An of all IECBV satisfying V':)<.n, where n is a positive

556
SOLUTIONS
[Ch. 10
integer. This set is closed in C. Indeed, if Vjin (i== 1, 2, . . .) and
IIfi-fllO, then Sfin for any partition f!}J of [a, bJ and all i. Hence
for ioo, we obtain Sfn, i.e., Vbjn. Evidently, CBV == 1 An.
Assuming now that CBV is of the second category in C, one at least
of the An, say A no ' contains a sphere {f: Ilf-foll<p} in C. Hence, in
particular, Vbjono for the centre fo of the sphere. Let g(x) be a conti-
nuous function whose graph consists of straight line segments, such
that Ig(x)l<p for all x and Vg>2no. The existence of such a function
g is obvious. Then f==fo+g is contained in the phere {f: Ilf-foll<p},
and hence in A no ' but Vf Vg- Vfo>2no-no==no. Hence, CBV is
of the first category in C. It follows by Baire's category theorem that
CBV is a boundary set in C. The set AlX=={f: Vbj(X} is then also a
boundary set, and since A lX is closed, the set A lX is nowhere dense in C.
42.4) Determine the partition fl such that
Sflf> V-!e,
and let N be the number of partition points in fl. Then determine
€5>0 such that Ix-x'I<€5 implies If(x)-f(x') I <e/4N. Let f!}J be an arbi-
trary partition
f!}J=={a==xo<x1 < . . . <xn==b}
satisfying maxi (Xi-Xi-1) <€5, and denote the sum of the terms in Sf
corresponding to intervals having no partition points of fl in their
interior by Sf. Finally, let f!}J1 be the common refinement of f!}J and
fl. Then SltSf, but the difference is at most 2N times e/4N, so
5 fS &ilf - !e>S flf - !e.
It follows that S&if>Sflf-!e, so Sf> Vbj-e. It is evident from
If(Xi) -f(Xi-1) IQ:-lf that Sff= 1 Q:-lf. Finally, since there exists a
partition f!}J2, refining f!}J and having as additional partition points in
[Xi-I, XiJ the points where f attains its maximum and minimum, we
have
n
 Q:_lfS&i2fVbj.
i=l
Note that if f(x) ==0 on [0, 1J, except at x==! where f==l, then VAf==2,
and for all f!}J not having x==! as a partition point we have Sf==O and
 Q:-lf== 1.

Ch. 10J
SOLUTIONS
557
42.5) Consider the sequence of partitions f!JJ n of [a, bJ, where fJJ n
divides [a, bJ into 2 n equal subintervals. Then, denoting the sum
i Q:-lf for f!JJ n by Qnf, we have lim Qnf== Vf by the preceding exer-
cise. Let [Xi-I, XiJ be the i-th subinterval of f!JJ n , and let mi and M i be
the minimum and maximum of f on this subinterval. The function
Li(Y) is defined by Li(Y)== 1 on [mi, MiJ and Li(Y) ==0 at all other real
y. Then
Q:_lf==Mi-mi== f Li(Y) dy,
and it follows that if N n(Y) ==L1(Y) + . . . +L 2n (y), then Qnf== f N n(Y) dYe
It is easy to see that, for each y, N n(Y) does not decrease as n increases,
1. e.,
N n(Y) i N*(y)
and
f N*(y) dy==lim Qnf== V':J.
Since Nn(y)Nf(Y) for all nand y, we have N*(y)Nf(Y)' In order to
show that Nf(y)N*(y), let y be given and select the finite non-nega-
tive integer p such that PN(y). Then there exist p different points
in [a, bJ where f assumes the value y. For n sufficiently large these
points lie in different subintervals of f!JJ n, so N n(y)?::-P for n large. It
follows that N*(y)?::-p, and so N*(y)Nf(Y)'
42.7) Let
x
v+(x) == f d 1 (t) dt+r1(x)
a
and
x
-v-(x) == f d 2 (t) dt+r2(X)
a
be the Lebesgue decompositions of v+ and -v-, so d 1 (x) and d 2 (x) are
non-negative. Since f(x) - f(a) ==v+(x) -{ -v-(x)}, we obtain the de-
sired representation. At almost every X, one at least of d1(X) and d 2 (x)
vanishes. Indeed, if h==min (d 1 , d 2 ), then
x
W1(X) == f (d 1 -h) dt+r1(x)
a
and
x
W2(X) == f (d 2 -h) dt+r2(x)
a
are non-decreasing, WI (a) ==w2(a) ===0 and f(x) - f(a) ==W1 (x) -W2(X); it

558
SOLUTIONS
[Ch. 1 0
follows from Theorem 3 in the present section that W1(X»V+(x), and
so h(x)==O almost everywhere. This implies that d 1 (x)+d 2 (x)==ld(x)l,
where we had already that d(x)==d 1 (x)-d 2 (x), and it follows by ad-
dition of the expressions for v+ and -v- that
x
v (x) == f Id(t) I dt+{r1(x) +r2(x)}.
a
Hence, v' (x) == Id(x) I almost everywhere.
42.8) Follows immediately from
max (5&,g, 5&,h)5&'/5&,g+5&,h,
holding for all f!lJ.
42.9) If I==g+ih, g and h real, and Vg==Vg(t), Vh==Vh(t), then
v(t') -v(t){Vg(t') -vg(t)}+{Vh(t') -Vh(t)}
for all t'>t, so v is absolutely continuous since v g and Vk are so. Hence
t'
v(t') -v(t) == f v' (u) du
t
for all t't.
Let f!lJ be an arbitrary partition of [a, bJ. Then
n n ti n b
5&'1==  I/(ti) -/(ti-1) I ==  I J I' (u) dul f II' (u) I du== f II' (u) I du,
i= 1 i= 1 ti-l 1 a
so
b
v(b) -v(a) == V':Jf II' (u) I duo
a
Similarly
t'
v(t') -v(t)f 1/'(u) I du
t
for all t' >t;
hence, v' (t) II' (t) I at almost every t. Conversely, since
v(t') -v(t»I/(t') -/(t) I
for all t' >t,
we obtain v'(t»lf'(t) I for almost every t.

Ch. 10]
SOLUTIONS
559
42.10) Let 1==11 + 12 be the Lebesgue decomposition of I; hence,
t
I(t) == J I' (u) du+ 12(t)
a
with 12(t)==0 at almost every t. In the particular case that 1'(t)==O
almost everywhere, we have l(t)==/2(t). Hence, if I==g+ih, the real
functions g and h of finite variation satisfy g'(t)==h'(t)=O almost
everywhere, so by Exercise 42.7 the derivatives of their total vari-
ations vg(t) and Vh(t) vanish almost everywhere. Since
V(t') -v(t){Vg(t') -Vg(t) }+{Vh(t') -Vh(t)}
for all t'>t, we obtain then that V'(t) ==0 almost everywhere.
In the general case, if f!JJ is an arbitrary partition,
b
S/S&J/1 +S&J/2J !/'(u) I du+ V2,
a
so
b
v(b)-v(a)JI/'(u)ldu+ V2'
a
Similarly
t'
v(t') -v(t)J II' (u) I du+ Vf/2,
t
and so v'(t)I/'(t)1 at almost every t, since the derivative of the total
variation of 12 is zero almost everywhere by what has already been
proved. The inverse inequality v' (t) II' (t) I follows again from v(t') -v(t)
 I/(t') - I(t) I for all t' >t.
42.11) The curve is rectifiable if and only if I is of finite variation,
and by Exercise 42.8 this is so if and only if g and h are of finite vari-
ation. In this case
v' (t) == II' (t) ! =={ (g') 2 + (h') 2}i
at almost every t by the preceding exercise, hence
t t
v(t) f v' (u) du=== J {(g')2+ (h')2}i du
a a
for all t in view of the Lebesgue decomposition of v. Evidently, there

560
SOLUTIONS
[Ch. 10
is equality for all t if and only if v(t) is absolutely continuous, that is
(cf. Exercise 42.9), if and only if f(t) is absolutely continuous.
42.13) For ex==O the statement is trivially true; we may assume
therefore that ex>O. Let first g' (x) ==0 almost everywhere, and write
f(x)==x+ig(x). Since Lg== Vf by definition, we have
Lg== V{x+ig(x)}1 + Vg== 1 +ex.
In order to prove an inequality in the converse direction, we select a
sequence hn!O and a number e>O. If
A =={x: g' (x) ==O}
and
An=={x: g(x+hn)-g(x)ehn},
then #(A)== 1 where # is Lebesgue measure. Also, xEA implies xEA n
for all nnx, so A c lim inf A n. Then # (lim inf An) == 1, and since
p(liminfAn)liminf#(An), we have also liminf#(A n )==I. Hence
lim #(A n) == 1. Now select the index p and the integer N such that
p(Ap»l-e and h p N>I, and let e1==ejN. Take X1EAp such that
xl-inf (x: x EA p) <e1,
and let Y1==X1+hp. Next, take X2EAp such that
0x2-inf(x: xEA p n[Y1, 1J)<e1,
and let Y2==X2+hp. Continuing with this procedure, we obtain at most
N closed intervals [Xi, YiJ with disjoint interiors, and
#([0, 1J- [Xi, YiJ)#([O, 1J-Ap)+Ne1<e+e==2e,
so #( [Xi, YiJ) > 1-2e. Now consider the partition f!JJ of [0, 1J having
as partition points, besides the endpoints of [0, 1J, all Xi and all Yi.
The part i If(Yi)-f(Xi) I of S&Jf is at least 1-2e, and since
 {g(Yi)-g(xi)}e  (Yi-xi)e,
i i
whereas g( 1) -g(O) ==ex, the remaining terms of 5 f yield at least ex-c.
Hence
S&Jt 1-2e+ex-e== 1 +ex-3e,
and it follows that Lg= Vf 1 +ex.

Ch. 10, 11J
SOLUTIONS
561
Assume now, conversely, that Lg== 1 +cx, and let g==gl +g2 be the
Lebesgue decomposition of g, hence gl(X) ==10 g' (t) dt and g2(X) ==0 al-
most everywhere. Evidently, since gl and g2 are non-decreasing, Vg==
Vgl + Vg2. Then
1 +cx== Vf== V(X+ig1 +ig2)V(x+ig1)+ Vg2
 1 + Vgl + Vg2== 1 + Vg== 1 +cx,
so V(X+ig1) == 1 + Vgl. It follows by Exercise 42.11 that
1 1
j{1+(gi)2}!dx== j(l+gi)dx.
o 0
But
[1 +{gi(x)}2J! 1 +gi(x)
for almost every x.
Hence, these functions are equal almost everywhere, and so gi (x) == 0
almost everywhere. Since g' (x) ==gi(x) holds almost everywhere, we
arrive at the desired result that g' (x) ==0 almost everywhere. The
second half of this proof is due to W. A. J. Luxemburg.
CHAPTER 11
43.1) The (-v-)-measure of E+==lim inf Dn satisfies
-v-(E+) ==lim inf {-v-(Dn) }==O.
The set E-==E-E+==lim sup (E-Dn) satisfies E-c =k (E-Dn) for
all k, so
00
v+(E-)  2- n
n=k
for all k,
i.e., v+(E-) ==0.
43.2) Since
IT(D) I Ifll (D) + Ivl (D)
for all DEAE,
the set function T is finite on A E . The same inequality shows that if
D==  Dn with disjoint Dn, then  T(Dn) is absolutely convergent.
Hence, if FncDn for all n, then F== FncD and
 {fl(F n) +v(Dn- F n)}==fl(F) +v(D-F)T(D),

562
SOLUTIONS
[Ch. 11
so  T(Dn)T(D). Conversely, if FeD, then F== Fn for Fn==FDn, so
/-l(F)+v(D-F)== {/-l(Fn)+v(Dn-Fn)} T(Dn),
and it follows that T(D) T(Dn). This shows that T is a signed measure
on A E . The choice F==D in T(D)):;/-l(F)+v(D-F) for all FeD implies
that T):;/-l, and the choice F==0 implies T);:V. Finally, assume that a is
a signed measure on A E such that a):;/-l and a):;v. We have to prove
that a):;T. This is easy; for all DEA E and all FeD we have
a(D) ==a(F) +a(D-F));:/-l(F) +v(D-F),
so a(D)T(D).
For the proof of the last statement, let D==F++F- be a Hahn
decomposition of D with respect to /-l-V, so (/-l-v)-(F+)==(/-l-v)+(F-)
==0, which implies that
l/-l-vl (F+) == (/-l-v)+(F+) == (/-l-v) (F+)
and
l/-l- v l (F-) == - (/-l-v)-(F-) == - (/-l-v) (F-).
Denote (/-l+v)+I/-l-vl by w. Then
w(F+) =={(/-l+v) + (/-l-v)}(F+) ==2/-l(F+)
and
w(F-) =={(/-l+v) - (/-l-v)}(F-) ==2v(F-) ,
so
w(D) ==w(F+) +w(F-) ==2{/-l(F+) +v(F-)}2T(D),
i.e., tWT=/-lVV on A E . On the other hand, since l/-l-vl);:/-l-v and
l/-l-vl):;v-/-l, we have !w):;/-l and twv. Hence tW=/-lVv.
43.3) Write
f max (d/-l/dA, dv/dA) dA==a(D)
D
for all D EA E . Then a is a signed measure on A E such that a):;/-l and
av. Assume that a1 is another signed measure on A E satisfying 1/-l
and a1v. We have to show that aa1' For this purpose, let Al be a
finite measure on A E such that A and la11 are AI-absolutely continuous.

Ch. 11 ]
SOLUTIONS
563
Then
a(D) == f{ max ( d d  '  )} dAl = f max (   ) dA1
A dA dA1 dA1' dA1
D D
since dA/dA1>0, and a1(D) == In ( da 1/ dA 1) dA1. Now, alfl implies that
da1/dA1dfl/dA1 holds AI-almost everywhere (since da1/dA1 <dfl/dA1 on
a set D of positive AI-measure would imply that a1(D)<fl(D)). Simi-
larly, a1v implies that da1/dA1dv/dA1' Hence
da1/dA1>max (dfl/dA1' dv/ dA 1) ,
and this shows that a1(D»a(D) for all DEA E . The result is that a==
flYv. The proof for flAv follows by observing that -(flAv)==(-fl)Y(-v).
43.4) If, again, A denotes a finite measure such that Ifll and Ivl are
A-absolutely continuous, then IflIAlvl==O is equivalent to min(dlfll/dA,
dlvl/dA) ==0 holding A-almost everywhere. Hence, if F =={x: dlfll/dA>O}
and G==E-F, then Ifll (G) ==0 and Ivl(F)==O, so Ifll..llvl in the sense of
Exercise 32.5. The converse is also evident.
Since dlfll/dA== Idfl/dAI and dlvl/dA== Idv/dAI in view of the manner
these functions are introduced by means of the Radon-Nikodym theo-
rem, it is now evident that fl..lv if and only if (dfl/dA) (dv/dA) ==0 holds
A-almost everywhere, i.e., if and only if
dfl dv dfl dv
dA + di dA - di
holds A-almost everywhere, that is, if and only if Ifl+vl==lfl-vl. The
last statement follows by observing that, for f==dfl/dA and g==dv/dA,
we have fg==O if and only if If+gl==max(lfl, Igl).
43.5) Lg> V{x-g(x)} follows from
[(X2- X 1)2+{g(X2) -g(X1)}2Ji> 1 (X2- X 1) -{g(X2) -g(X1) }I,
holding for 0X1 X2 1.
43.6) Let Vo be one of the measures VlX, and let TlX=='VO YVlX for all ex.
Any upper bound of {VlX} is an upper bound of {TlX} and conversely, and
VOTlXfl for all ex. The existence of sup TlX is evidently equivalent to
the existence of sup (TlX-VO); hence, we may assume from the beginning
on that OvlXfl for all ex. Since all VlX are now fl-absolutely continuous
measures, the existence of sup VlX follows from sec. 34, Theorem 1.

564
SOLUTIONS
[Ch. 11
44.1) Given e>O, let E== Dn be a decomposition of E into a
countable number of sets such that
 Iv(Dn)I>lvl(E)-e,
and let
p
 Iv(Dn) I> Ivl (E) -2e.
1
Then, if En==Dn for n== 1, ..., p, and Ep+1==E-) Dn, we have E==
f+1 En and
p+1
 Iv(En)I>lvl(E)-2e.
1
44.2) If
&=={a==XO<X1 < . . . <xn==b}
is an arbitrary partition of [a, b], and D i == (Xi-I, Xi], then
n n
S&Jf==  If(Xi) -f(Xi-1) I ==  Iv(Di) I Ivl (E),
1 1
so Vlvl(E).
In order to prove the converse inequality, let e>O be given, and let
E===) En be a decomposition of E into disjoint sets EnEAE such that
p
 Iv(En)I>lvl(E)-e.
1
A decomposition of this kind exists by the preceding exercise. For each
En there exists a countable union of cells OnEn such that Ivl(On-En)
<efp2. This implies that Iv(On) -v(En) I <efp2 for n== 1, . . " p. Further-
more, if F n is a suitable finite union of the cells which constitute On,
then Iv(On)-v(Fn)l<efp2. Hence Iv(En)-v(Fn) I <2efp2, which implies
that certainly
p p
 Iv(F n) I >  Iv(En) 1-2e> Ivl (E) -3e.
1 1
The sets F n (n=== 1, . . ., P) may overlap, but
F1==F1(F2+'" +Fp)+{F1-(F2+'" +Fp)}==Fi+F, '
and since Ivl(FiFj)lvl(OiOj)<2efp2, we have Ivl(Fi)<2efp, so Iv(Fi)l<

Ch. 11 J
SOLUTIONS
565
2e/p, and hence
Iv(F) I> Iv(F 1) 1- 2e/p.
Similarly for F;, . . ., F;, and this yields
p
 Iv(F) I> Ivl (E) -5e.
t
The sets F are disjoint, and each F is a finite union of cells; hence,
if f!JJ is the corresponding partition of E===[a, bJ, then
p
S/> Iv(F)I>lvl(E)-5e.
t
It follows that V>lvl(E).
45.3) The set {x: In(x»O} is of measure 1/2n+1; hence, the set
E =={x: In(x) ==0 for all n}
is of measure not less than !. Then !Elndx==O for all n, and fEdx>!.
45.4) The case A== -00 offers no difficulties; we may assume, there-
fore, that A>-oo. Lebesgue measure will be denoted by #. Let
-oo<ex<A and
E a==={X: x E [0, 1 J, cp(x»ex}.
Then #(Ea)==fJ for some fJ such that O<fJl. If
Ena=={x: XE[O, 1J, cp(nx»ex}
for the same ex and n== 1, 2, . . ., then #(Ena) ==fJ with the same fJ as
above. Let Sa===lim supn Ena and let I == [a, bJ be an arbitrary sub-
interval of [0, 1J. Since it is evident that #(EnanI) -+fJ(b-a) as n-+oo,
we have
#(SaI»lim sup #(Ena nI) >!fJ(b-a).
Assume now that lim sup In <A on a subset F 1 c [0, 1J of positive
measure. Then there exists a finite number ex<A such that lim sup In
<ex on a subset Fe F 1 with #(F) >0. Let fJ be the number corresponding
to this ex according to the preceding paragraph, and choose y such that
O<y<l and !fJ+y>l. Since #(F) >0, there exists an interval 1==
[a, bJc[O, 1J such that #(FI»y(b-a); cf. Exercise 10.8. On FI we

566
SOLUTIONS
[Ch. 11, 12
have lim sup fn<a and on SaI we have lim sup fn==lim sup cp(nx»a;
hence, FI and SaI are disjoint. On the other hand their measures
exceed y#(I) and tfJ#(I) respectively, with y+tfJ> 1. This yields a
con tradiction.
45.6) Set h1==!n(g+f) and h 2 ==!n(g-f) on [0, 2nJ; note that hI and
h 2 are non-negative. The sequence
f: (x) ===h 1 (x) (Isin nxl +sin nx)
satisfies fEf dx-+(2/n) fEh1dx; similarly,
-f;(x) ===h2(X) (Isin nxl-sin nx)
satisfies IE (-f;) dx-+(2/n) fE h 2 dx . The functions f: and -f;; are non-
negative, and at each point XE[O, 2nJ one at least of them is zero.
Hence, if we define fn===f+f;, then Ifnl===f:-f;. It follows that
J fn dx -+(2/n)f (h 1 -h 2 ) dx=== J fdx
E E E
and
J Ifni dx-+(2/n) f (hI +h 2 ) dx=== f g dx.
E E E
CHAPTER 12
47.2) In view of Exercise 32.13 the result holds if E is v-summable.
Hence, since any v-measurable E is a countable union of v-summable
sets, the result holds for every v-measurable E.
47.3) In Exercise 32.13 it was proved that J(f) is an elementary
integral on the Borel step functions, such that after application of the
extension procedure for integrals the initial J(f) is reobtained. The
desired result follows,- therefore, from sec. 17, Theorem 9.
48.1) It was proved in Theorem 1 that IF+(f)IF+(lfl) for all fEV.
Similarly IF-(f)I-F-(lfl). Hence
IF(f) 1IF+(f) I + IF-(f) I(F+-F-)(Ifl) = IFI (Ifl).
The proof of IIFI(f)IIFI(lfl) is similar. Then
IIIFIII=== sup IIFI (f) I sup IFI (f)IIIFIII,
I!flll fO,lIflll
and (b) follows immediately.

Ch. 12J
SOLUTIONS
567
For the proof of (c), let O<'/EV. If gn (Ogn<'/) is a sequence such
that F(gn)-+F+(/), then F(/-gn)-+F-(/). Denoting by (3 the number
supF(/1-/2) for all decompositions 1=/1+/2, 0<./1 E V, 0<.f2EV, it
follows immediately that
IFI (I) ==F+(f) -F-(f) =lim{F(gn) -F(f-gn)}<.{3.
On the other hand, if 1=/1 +12 is such a decomposition, then
F(/1-/2)<.IF(f1) I + IF(/2) IIFI (11 +/2) == IFI (f)
by (a), and so ,8<. IFI (f).
For the proof of (d), let O<./E V and denote by y the number sup IF(g) I
for all gEV satisfying Igl<.l. Since IF(g)IIFI(lgl)<'IFI(f), we have y<.
IFI(f). On the other hand, for any e>O, there is a decomposition 1=
11 +f2, 0<./1 E V, 0/2E V, such that F(/1-/2)IFI (I) -e. Hence, writing
Il-f2==g, we have Igl1 and IF(g)I=F(g)IFI(/)-e, so yIFI(f)-e.
Finally, in order to prove (e), observe that IIFIIIIIFIII follows from
IF(/)I<.IFI(I/I). The converse inequality IIIFIII<'IIFII follows by com-
bining (b) and (d).
48.2) If OfEV, and gEV satisfies Igl<'f, then IF(g)IIIF"'llgll
IIFII'llfll, and so IFI(/) is well defined and 5atisfies O<.IFI(/)<.IIFII'II/II.
The proof that IFI is a non-negative linear functional on V follows now
very closely the pattern of the proof in Theorem 1 that F+ is linear.
If f E V is real, then
IIFI (I) I IFI (f+) + IFI (-1-) == IFI (lfl)<.IIFII'llfll.
Hence, if lEV is complex and IFI(f)==re iqJ (r>O, cp real), then fe- iqJ ==
11 ==gl +ih1 satisfies
IIFI (f) I ==r== IFI (f1) == IFI (gl)IIFII'lIg111IIFII'11/111= IIFII'llfll,
and this shows that IIIFIII<'IIFII. The converse follows from
IIFII== sup IF(f)l= sup (sup IF(g)l)== sup IFI(/)IIIFIII.
Ilflll IIflll,fO Iglf IIflll,fO
48.3) We prove only that F+==FYO, where 0 denotes the null
functional. It is evident that F+>F and F+>O, so it is sufficient to
prove that G?;:-F and G>O implies G?;:-F+. For O<.g<.1 (I, gEV) we have
G(g)F(g), and so G+(f»F+(f). But G+==G since G>O, so G(f»F+(f)
for O<'fEV, i.e., G>F+.

568
SOLUTIONS
[Ch. 12
48.4) Let first n==2, so 0FF1+F2. Let G 1 ==FAF 1 and G 2 ==
F -G 1 . Then G 1 and G 2 are non-negative and F==G 1 +G 2 . Evidently
G1F1, and from F+F1== (FAF 1 ) + (FVF 1 ) it follows that
G 2 ==F - (FAF 1) == (FVF 1) -F 1 == (F -F 1)VOF2.
The general result is now derived by a simple induction argument.
48.5) By Exercise 48.3 we have
F1AF2==F2+ (F1-F2)-==F1-(F1-F2)+,
so 2(F1AF2)==F1+F2-IF1-F21. Once more by Exercise 48.3 we have
(F 1 VF 2 ) + (F 1 AF 2 ) ==F 1 +F 2,
and so 2(F 1 VF 2) ==F 1 +F 2+ IF 1 -F 2 1. The triangle inequality IF 1 +F 21
IF11+IF21 follows immediately from the definition of IFI. For the
proof of the last assertion, observe that FG and -FG (for real
functionals F and G) implies FV( -F)G, so IFIG.
49.1) Let FE C: and let F 1 be the restriction on C K of F. Then
there exists a unique complex measure v on Ab such that F 1 (f)==lfdv
for all fEC K . The integral I fdv exists for every fEC oo ' and evidently
F 2 (f)==lfdv is a bounded linear functional on Coo which extends Fl.
Since CK is dense in Coo by Exercise 47.1, the extensions F and F 2 of
F1 must be identical, and IIFII==IIF 1 11==lvl(R k ).
49.2) Note that E is v-measurable if and only if E is {L-measurable,
where {L== t=l fli. Note, furthermore, that any {L-measurable set is {L-
almost equal to a Borel set by Exercise 47.2.
49.3) If F and G are bounded linear functionals on C) corresponding
to the signed measures v and a respectively, then FG if and only if
va. Since the correspondence is one-one, it follows then that F 1 VF 2
and F 1AF 2 correspond to VI VV2 and V1Av2 respectively. This yields a
second proof of the fact that F+==FVO and F-==FAO correspond to
vVO==v+ and vAO===-v- respectively. Then IFI ==F+-F- corresponds to
v+ -v-== Ivl. The assertions on orthogonality are now immediate conse-
quences of the corresponding assertions for signed measures in Exer-
cise 43.4.
49.4) We have to prove that, for every rO in C K , it is true that
I fdlvl==sup II gdvl for all gEC K satisfying Iglf, and evidently it will

Ch. 12J
SOLUTIONS
569
be sufficient to show that, given e>O, there exists gEC K such that
Igl1 and Ifgdvl>ftdlvl-e. On account of the uniform continuity of
1 there exists a step function
p
Sl (x) ==  cnXAn (x)
n=l
with positive coefficients C n and disjoint cells An, such that
OSl <I
and
f sl d l v l > f Idlvl-e.
Corresponding to each cell A n there is a closed interval B n cAn (in
the "right upper corner" of A n) such that S2(X) ==) cnXBn (x) still
sa tisfi es
f s2 d l v l > f Idlvl-e.
Since the closed intervals B1, . . " Bp are now disjoint, and since s2<f
on each Bn, there exist disjoint open intervals C n  Bn for n== 1, . . " P
such that s=) cnXCn still satisfies Os<l; of course, we have also
fsdlvlfs2dlvl> fldlvl- e .
Evidently, it will be sufficient, therefore, to prove that for any open
interval C we have Ivl(C)==suplfgdvl for all gECK satisfying Iglxc.
Given e>O, let B be a closed interval such that BcC and Ivl(B}>
Ivl(C)-e, and let B== En with disjoint Borel sets En such that
 Iv(En)I>l v l(B)-e>l v l(C)-2e.
If v(En) == Iv(En) I ei<pn, then k(x) ==  e-i<Pn XEn(x) is a Borel function satis-
fying
fkdv= Iv(En)l.
Note that Ik(x) 1== Ion Band k(x) ==0 outside B. Writing V==(fl1-fl2)+
i(fl3-fl4) and fl=t=l fli, there exists a function g*EC K such that
f Ig* -kl dfl<e, and similarly as in the proof of the Riesz theorem it
may be assumed that Ig*(x)ll for all x. It may happen, however,
that g*(x) *0 for some x outside C. Let h(x) EC K , Ohl, h==O outside
C and h== 1 on B. Then g==hg*EC K , Iglxc and Ig-kllg*-kl, so
f Ig-kl dfl<e. It follows then as in the proof of the Riesz theorem that
IJ'gdvllf kdvl-e==  Iv(En) I-e> Ivl (C) -3e.

570
SOLUTIONS
[Ch. 12
If IF1IAIF21==0, then IF11+IF21==IIF11-IF211 by Exercise 48.5. The
triangle inequality in the same exercise shows that
IIF11-IF211IF 1-F21IF 11 + IF21
and
IIF11-IF211== IIF11-I-F211IF1 +F21IF11+ IF 2 1;
hence, all these functionals are now equal, in particular IF1+F21==
JF 1 - F 21. It follows then also from
(IF 11A IF 21) + (IF 11 V I F 21) == IF 11 + IF 21 == IF 1 +F21
that IF1+F21==IF1IVIF21. Conversely, if F 1 is not the null functional
and F2==iF1, then IF1 +F21== IF 1 -F 2 1 but IF1IAIF21== IF 11 *0. Also, if
F 3 ==( -!+!iJ3)F1'
then IF1IVIF31==IF1+F31 but IF1IAIF31==IF11*0.
50.1) Let # have the finite subset property. By Holder's inequality
J I/gl d#pq(1/1) for Ilgllp 1, and so it is sufficient to show that f I/gnl d#
tends to pq(1/1) for some sequence gn satisfying Ilgnllp 1. Assume first
that 1 <q<oo, and let IT==IXx for all possible X T of a-finite measure.
If
b==sup pq(I/TI),
T
there exists an ascending sequence X n C X of sets of a-finite measure
such that the corresponding In==IXxn satisfy pq(l/nl) i b. We shall prove
that b==pq(ltl). If b==oo, this is evident since pq(I/I»pq(l/nl) for all n.
If b<oo, we write X o == 1 X n , and observe that Xo is of a-finite
measure. Assuming that 1*0 on a subset of X -X o of positive measure,
we have (by the finite subset property) that 1*0 on a subset X T C
X -X o of finite positive measure, and so pq(I/IXx+xo) >b. This is a
contradiction. It follows that 1==0 almost everywhere outside X o , i.e.,
f==lxx o almost everywhere. This shows that
pq(l/l) ==pq(l/lxx 0) ==lim pq(11 nl) ==b.
By Lemma y we have pq(llnl)==sup f I/gl d# for all gEL p vanishing
outside X n and satisfying Ilgllp 1, and so it follows from pq(l/nl) i pq(1/1)

Ch. 12J
SOLUTIONS
571
that f I/gnl dfl-+Pq(l/l) for an appropriately chosen sequence gnELp,
Ilgnllpl.
Next, let q==oo, and assume (with the same notations) that
Poo(1/1) >b==sup Poo(I/,.I).
T
Then I/(x) I >b on a set of positive measure, and hence on a set of finite
positive measure, which immediately yields a contradiction. Hence,
similarly as above,
Poo(l/l) ==sup Poo(l/nl) ==lim J I/gnl dfl
.for an appropriate sequence gn EL 1, IlgnI111.
If fl does not have the finite subset property, there exists a set E of
infinite measure, containing no subset of finite positive measure. Then,
if 1 <q< 00, the function XE satisfies Pq(XE) ==00, but f IXEgl dfl==O for
every gEL p . If q==oo, we have Poo(XE) == 1,. but f IXEgl dfl==O for every
gELl'
50.2) Let fl have the finite subset property, and assume that I is
measurable and Ig is summable for every gEL p . Furthermore, let 1,.==
IXXT for all possible X,. of a-finite measure. On account of Lemma y we
have I,.ELq for all T, and in the proof of the preceding exercise it was
shown that
pq(1/1) ==sup 11/,.llq.
T
We have to show, therefore, that
b==sup 11/,.llq<oo.
T
As in the proof of the preceding exercise, there is an increasing sequence
X n c X, each X n of a-finite measure, such that In== IXxn satisfies Il/nllq i b.
Writing Xo==r X n , the set Xo is of a-finite measure, so 10==lxxoELq,
once more by Lemma y. Furthermore, since I/nl i 1/01, we have
Il/ollq==lim Il/nllq==b.
Hence b<oo.
If fl does not have the finite subset property, there exists a set E of
infinite measure containing no subset of finite positive measure. Then,
if 1 <q<oo, the function XE is not in Lq, but f IXEgl dfl==O for every
gEL p . If q==oo, the function 1==00' XE is not in Loo, but f I/gl dfl==O for
every gEL1.

572
SOLUTIONS
[Ch. 12
50.3) It will be sufficient to prove that if FE (a-L oo ) * satisfies F(gn)
-+0 whenever gnEa-Loo, gn!O, then F(g)==flgdfl for some tELl and
all gEa-Loo. If X does not contain any set of a-finite measure, this is
obviously true. If X'T C X is of a-finite measure, there exists by Theo-
rem 2 a unique l'TEL1, vanishing outside X'T' such that F(g)==f l'Tgdfl
for all gEa-L oo vanishing outside X'T' and "1'T111IIFII. For different sets
X'Tl and X'T2 of a-finite measure, the uniqueness of l'Tl and 1'T2 implies
that l'Tl==I'T2 on the intersection of X'Tl and X'T2' Let b==suplll'T111, so
bIIFII<oo. There exists an ascending sequence XncX, each X n of
a-finite measure, such that the corresponding functions InEL1, repre-
senting F on X n , satisfy IItnl11 jb, and it follows as in the proof of Theo-
rem 3 that In is a fundamental sequence, and its limit function IEL1
satisfies F(g) == f Igdfl for all gEa-L oo '
50.4) If In is the interval (n+ 1)-1<xn-1, consider the function
gl(X) such that gl(X) == 1 for Ixi Ell +13+ . . . and gl(X) ==0 for Ixi E12+
14+..., and also the function g2(x)==1-g1(X), Then
F(gl) ==F(g2) ==F(gl +g2) == 1.
A similar construction by means of 11+14+17+.. . and 12+15+18+
. . . shows that F 1 is not linear.
50.5) By Exercise 28.2 there exists a non-negative linear functional
F(/) on C B such that F(/) ==limxoo I(x) whenever the ordinary limit on
the right exists. For n== 1, 2, . . ., let In be bounded and continuous,
In(x)==O for xn, In(x)==1 for xn+1, and O/n(x)l for n<x<
n+l. Then Into on R1, and F(ln)==l for all n. Hence, F is not an
elementary integral on C B.
50.6) In order to show that F 1 VF 2 and F 1 AF 2 correspond to
max (/1,/2) and min (/1,/2) respectively, it will be sufficient to prove
that F+==FVO corresponds to I+==max (I, 0). For this purpose, let O
gEL p . Then
F+(g)==sup F(gl)==SUP flg1dfl
for all glELp satisfying Oglg. Obviously, this least upper bound is
attained for gl ==gXE, where E =={x: IO}, and the maximal value is
then f I+gdfl. This shows that F+ corresponds to 1+. Similarly, f-==
FAO corresponds to 1-==min(/, 0), so IFI==F+-F- corresponds to
1+-1-== III.

Ch. 12J
SOLUTIONS
573
50.7) Let OgELp. Then
IPI (g) ===sup IP(gl) I ===sup If /gl dfll
for all glELp satisfying Ig11g. This least upper bound is attained for
gl ==gfsgn I, and the maximal value is then f I/lgdfl.
50.8) If gEl1, then G(/) ==}2 /ngn is obviously a bounded linear
functional on (co), and IIGIIllgI11 by Holder's inequality. The choice
fn=== 1fsgn gn for n== 1, ..., N, and /n===O for n>N shows, for N -+cx>,
that IIGII===llgI11' In order to show now that every GE(CO)* is of this
particular form, we introduce the elements en E (co) for n=== 1, 2, . . . ,
where en is defined as having its n-th coordinate equal to one and all
other coordinates zero. Given GE (co)*, let G(e n ) ===gn. If
/ === (11, /2, · · .) E (c 0 )
is arbitrary, and
k
Sk===}2 /n e n==(/1,/2, "',/k,O,O, ...),
n=l
it is obvious that II/-skll oo -+0, and so
k 00
G(/) ==lim G(Sk) ===lim }2 /ngn==}2 Ingn.
1 1
The choice /n===lfsgngn for n==l, ..., N, and /n===O for n>N shows
that }2f IgnlIIGII for all N, and so }21lgnl <cx>, i.e., g== (gl, g2, . . .) Ell.
50.9) If
g== (gO, gl, g2, . · .) Ell,
then G(/)==}2'; /ngn is obviously a bounded linear functional on (c),
and IIGIIllgI11 by Holder's inequality. The choice In== 1fsgn gn for n==
1, . . . , N, and 1 n=== 1fsgn go for n > N shows, for N -+cx>, that IIGII== Ilg111.
In order to show now that every GE(C)* is of this particular form, we
observe that if /==(/1,/2, ...)E(C) with lo===liml n , then /-/oeoE(co),
where eo is the element in (c) having all its coordinates one. The re-
striction of G on (co) is a bounded linear functional on (co), hence there
exists
g== (0, gl, g2, · . .) Ell
such that
00
G(/-Ioeo) ==}2 (In -IO)gn.
1

574
SOLUTIONS
[Ch. 12
Setting G(eo) ==go, we obtain
00
00
00
00
G(/)==logo+  (In-Io)gn==lo(go-  gn)+  Ingn ==Iogo +  Ingn,
1 1 1 1
and evidently g== (gO, gl, g2, . . .) Ell.
51.2) By Lemma ex in the preceding section there exist functions
gnELp such that
Ilgnllp1
and
If Ingndfl I >lllnllq-n- 1 .
Hence, defining the bounded linear functionals F n on Lp by F n(g) ==
flngdfl, we have lim sup IFn(gn) 1==00. It follows then from the co-
rollary of the Banach-Steinhaus theorem that there is a sub collection
Z c Lp of the second category such that
lim sup If Ingdfll ==00
for every gEZ.
52.1) Let 10 be the characteristic function of a bounded open inter-
val, and let In be a sequence of non-negative functions in C K satisfying
In t 10 for all xER k . Then, if (/)==f Idfl is any non-negative linear
functional on C K , we have (In) t .f(/o) , and this implies that lim F(ln)
= f lodv for any bounded linear functional F(/) == f Idv on Coo (cf. Exer-
cise 49. 1 ). It follows that In is a weak fundamental sequence in Coo. If
there exists a weak limit gOEC oo , we have f lodv== f godv for every v;
in particular (choosing for v the measure corresponding to a unit mass
at the point x; cf. Exercise 21.20) we have lo(x)==go(x) at every point
x, contrary to the assumption that go(x) is continuous and lo(x) is not.
52.2) Assume that Xo is not in VI. Since VIis closed, there exists.
now, in view of the Hahn-Banach extension theorem, a bounded linear
functional F on V such that F ==0 on VIand F(xo) == 1. Hence 1 ==
F(xo) ==lim F(x n ) ==0, a contradiction.
52.3) Let X==l X n with all X n disjoint and of positive measure,
and let VI be the closed linear subspace of Loo consisting of all IELex>
which are constant on each X n. It will be sufficient to prove that VI
is not weakly complete, and for this purpose we may as well assume
that VIis the sequence space loo. By the same argument, it will be
sufficient to prove that the closed linear subspace (c) of loo (cf. Exer-

Ch. 12J
SOLUTIONS
575
cise 50.9) is not weakly complete. Let I(n)' for n== 1, 2, . . ., be the ele-
ment of (c) having its first n coordinates equal to zero, and all other
coordinates equal to one. Then it is easy to check, since we know all
bounded linear functionals on (c), that I(n) is a weak fundamental
sequence having no weak limit.
52.4) We have
I Inrkbklll/nr(kr-1, k r ) 11-lllnr(O, k r - 1 ) 11-lllnr(k r , 00) II
k
............1 1 1 _ 1
"""-::;- 2 8 -- 8 -- 8 --- 8
 5 5 10'
52.5) For 1 <poo, the hypotheses made imply that fElndfl-+
JElodfl for any set E of finite measure. The same argument as in
sec. 45 shows then that lim inf In/olim sup In holds almost every-
where on E. The same is then true on any set of a-finite measure, and
this concludes the proof for 1 <p<oo. For p===oo it follows that the
set on which the two inequalities do not hold simultaneously is a local
null set, and hence a null set since fl has the finite subset property.
53.1) By Exercise 52.1 the space Coo is not weakly complete.
53.2) If xo===O, there is nothing to prove; we may assume, therefore,
that Ilxoll=== 1 and all xn*O. Since X n converges weakly to Xo and Ilxnll-+
Ilxoll=== 1, the sequence xn==xn/llxnli converges weakly to Xo, and Ilxnll=== 1
for all n. Then xn+XO converges weakly to 2xo; hencel12xoll
lim infllxn+xoll by Theorem 2, so that, given 8>0, there exists N 1
such that
Il x n+ x oll>2-e
for nN 1.
On the other hand, it is evident that
Ilxn+xoll2
for all n.
It follows that Ilxn+xoll-+2, so Ilxn-xoll-+O by the uniform convexity,
and hence Ilxn-xoll-+O.
53.5) We may assume that 0<8< 1. Let F 1 be the closed unit circle
in the complex plane, and F 2 the open circle of centre 1 and radius 8.
Then z may run through r1-F2. Let ex be the point on the circumfer-
ence of r 2 such that arg (ex-I) ===!:n;, and let r 3 be the open circle having
the origin as centre and lexl as radius. For ZE (r 1 -r 2 ) -F3 there is a

576
SOLUTIONS
[Ch. 12
constant fJ(e) such that
O<fJ(e) < 1
and
11 + z 1'fJ ( e) (1 + I z I ) ,
hence
l+z P ( l+lzl ) p 1+lzlP
 fJ(e)  fJ(e) .
2 2 2
For zEF3 we have Izla==lcxl< 1, and hence, by the preceding exercise,
l+z P ( l+lzl ) p { _ ( )} l+lzlP
  1 a
222
for all ZEr 3 . Setting 1-y(e) ==max {fJ(e) , l-(a)}, we obtain the de-
sired result. Observe that a depends on e, so (a) depends on e.
53.6) On the complement Ee==X -E we have
If-glcx(lfl + Igl) ==2cx{!(lfl + \gl)},
so
It-gIP2cx{!(lfl + Igl)}pcx(lfIP+ IgIP),
and it follows that
f If-gIPdflcx f (lfI P + IgIP) dfl==2cx==!e.
EC
Hence, since Ilt-gIPdfl?::-e, we have IE If-gIPdfl>!e. Furthermore,
Since
If-gllfl + Igi ==2{l(lfl + Igl)},
we have
f If-glpdfl2P f {!(Ifl + Igl)}Pdfl2P-1 f {lfI P + IgIP}dfl.
E E E
It follows already that
f {lfI P + IgIP}dfl?::-ej2 P .
E
Observing now that
{!(IfIP+ IgIP)}-{!lt+gl}pO on X,

Ch. 12J
SOLUTIONS
577
we have
1 - f I+g P d === j a { I/IP+lglp _ I+g P } d
2 # 2 2 #
 f { I/IP; Iglp - I;g P} dp,  iy(<x) f {I/IP+ Iglp}dp,
E E
by the preceding exercise. Hence
1- J 1!(/+g) IPd#'Y((X)eI2p+1=='Y(!e)eI2p+1==q(e).
This proof is essentially due to E. J. MCSHANE (1950, [2J).
53.7) Assume that III nllp== IIgnllp== 1 for n== 1, 2, . · " III n +gnllp--+2
and Il/n-gnllp does not tend to zero. There is no loss of generality in
assuming that I I/n-gnlpd#e for some e>O and all n. Then, by the
preceding exercise, Ili(ln+gn)llp--+l is not satisfied. This is a contra-
diction.
53.8) Let
In(x)== 1-cos nx
for n== 1, 2, . . . .
Then In converges weakly to 10, where 10 (x) === 1 for all n. In addition,
Il/nll===lltoll==2n for all n. But Il/o-lnll==4 for all n.
53.9) Assume (a) and (c) satisfied, but not II/-Inll--+O, where the
norm denotes the L1 norm. Similarly as in Exercise 30.21, it is easy to
see, by passing to a subsequence, that we may assume as well that
t n --+1 pointwise. Furthermore, X can be assumed to be of a-finite
measure. Let Xki X, where all X k are of finite measure. Given e>O,
there exists, in view of the weak convergence and the Hahn-Saks theo-
rem, an index ko such that the integrals of III and alll/ni over X-X ko
do not exceed e. Hence, it may be assumed as well that X is of finite
measure. There exists >O such that #(E)< implies IE III d#<e as
well as IE I/nl d#<e for all n. Hence, writing X ==E +F, where #(E) <
and In converges uniformly to I on F (Egoroff's theorem), it follows
now easily that II/-Inll--+O.
For l<P<oo, take X===[O, 1J, # Lebesgue measure, En==[O, n- p ]
.
for n== 1,2, . . ., In===nXEn' Then In converges pointwise and weakly
to zero, but III nllp=== 1 for all n.

578
SOLUTIONS
[Ch. 12, 13
53.10) Follows immediately from Theorem 1 (b).
53.11) By the result of the preceding exercise there exists for each
XEV a number Mx>O such that IIT,.xllMx for all T. Hence, by the
Banach-Steinhaus theorem, the transformations T,. are uniformly
bounded.
CHAPTER 13
54.1) It is sufficient to show that Tx==O if and only if (x, T*y)==O
for all YEH, and this is an immediate consequence of the definition
of T*.
54.2) The properties in (a) are evident. As regards (b), we have
IIT*TIIIIT*II'IITII=IITII2
and
IIT11 2 ==sup (Tx, Tx) Illxl1 2 ==sup (x, T*Tx) IlIxI1 2 <IIT*TII
by Schwarz's inequality. Next,
Ilxll 2 + IITxI1 2 == (x, x) + (T*Tx, x)II(E + T*T)xll '1Ixll,
so II(E+T*T)xll>llxll, showing that (E+T*T)x==O if and only if x==o.
The inverse (E + T*T)-l exists therefore, with domain of definition the
range of E+T*T. Since
(E + T*T) * ==E + T*T,
the null space of (E + T*T) * consists only of the null element, so that
by the preceding exercise the range of E + T*T is dense in H. Given
any YoEH, there exists then a sequence X n such that (E+T*T)x n con-
verges to Yo, so that, since (E + T*T)-l is norm-decreasing, Xn is a
fundamental sequence and converges to some xoEH. It follows that
(E + T*T)xo==yo, showing that the range of E + T*T is the whole of
H. Hence, (E+T*T)-l is a bounded linear transformation on H of
norm not exceeding one.
54.3) Evidently, L c (L1.) T whenever L is a closed linear subspace
of V. If the inclusion is proper, there exists an element Xo E (L 1.) T not
belonging to L. Since Xo is not in L, an application of the H:ahn-
Banach theorem shows the existence of Y*EV* such that <x, y*)==O

Ch. 13J
SOLUTIONS
579
for all xEL, and <xo, y*)*O. Then Y*ELJ.. since <x, y*)==O for all xEL,
and so <xo, y*)==O since XOE (L..L) T. This is a contradiction.
54.4) If <x, x*)== <Tx, y*) for some fixed y*, then x* is clearly a
linear functional on VI, and x* is bounded since
I <x, x*)IIIY*lf '1ITII'lfxll.
The transformation x*==T*y* is then linear, and IIx*IIIITII'IIY*/1 implies
that IrT*IIIITII. Given e>O, choose XOE VI such that Ilxoll== 1 and IITxol1
>IITII-e. Then, if yo==Txo, there exists (by the Hahn-Banach theo-
rem) an element y* such that
IfY*II== 1
and
<Yo, y*)== IIYolf.
It follows that
<xo, T*y*)==<Txo, Y*)==<Yo, Y*)==/lYoll==IITxoll>(IITlf-e)llxoll,
showing that
IIT*Y*II > IITII-e== (1ITII-e) Ily*ll.
Hence IIT*II>IITII.
Finally, if Y*EV2, we have T*y*==O if and only if <x, T*y*)==O for
all x, that is, if and only if <Tx, y*)==O for all x. Similarly, if XEV 1 ,
then Tx==O if and only if <Tx, y*)==O for all y* (by sec. 27, Theorem 2),
that is, if and only if <x, T*y*)==O for all y*.
54.5) We indicate the proof of the last statement. Let xER(P i ).
Then
IlxI12>IIPxI12== (Px, x) ==  (PjX, x) ==  IIPjxI12>IIPiXI12== Ilxl/ 2 ,
so Pjx==O for all i*i. It follows that PjPi is the null transformation
for i *i.
54.7) It follows immediately from the definitions that (P*)2==P*, so
P* is a projection. By Exercise 54.4 we have
N(P*) =={R(P)}..L
and
R(P*) c {N(P)}J..,
and hence it is sufficient to show that R(P*) is actually equal to
{N(P)}J... In the present situation this follows easily by observing that
the adjoint of the projection E-P is the projection E*-P* (where E*

580
SOLUTIONS
[Ch. 13
is the unit transformation in V*), so
R(P*) ==N(E* - P*) == {R(E - P)}-L = {N(P)}-L.
54.8) It is trivial that (PI + P 2 ) 2==P 1 + P 2 if P1 P 2 and P 2 Pl are
eq ual to the null transformation. Assume now con versel y that (PI + P 2) 2
=P1+P2, so P 1 P 2 X+P 2 P 1 X==0 for every XEV. If xEN(P 1 ), then
P2P1X==0 and so P 1 P 2 X==0 as well. If xER(P 1 ), then P 1 P 2 X+P 2 X==0,
so that, writing P 2 x==y+z with YER(P 1 ) and zEN(P 1 ), we obtain
0==P 1 (y+z) +y+z=2y+z.
But then y==z==O since R(P 1 ) and N(P 1 ) have only the null element in
common. This shows that P2X==0, and so P1P2X==P2P1X==0. The final
result is that P1P2X==P2P1X=0 for every XE V.
56.1) First of all it is evident that cp(O) ==0; indeed, take s==O and
IEL2 arbitrary in the first of the given relations. Next, let t>O and f
the characteristic function of the interval [0, tJ. Then the corresponding
g satisfies
s t st s
f g(x) dx = f cpX) dx = f CP) du = f cpv) dv
o 0 0 0
for all s, and so g(x)==cp(tx)jx almost everywhere. It follows then from
the second of the given relations that
00
f cp(sx)cp(tx) dx
x 2
-00
equals min (s, t) or 0 according as s>O or s<O. A similar argument
works for t<O. In particular (s==t== 1), we have cp(x)jxEL 2 .
56.2) We have g(s) = Isl-1/(s-1) and I(s) == Isl-lg(s-l) for almost eve-
ry s.
56.3) We have
S-1
g(s) == (sgn s) J I(x) dx-l s l- 1 /(s-1),
o
and the same formula with 1 and g interchanged.

Ch. 13J
SOLUTIONS
581
56.4) We have
00
g(s) ==s-l f {f(x) fx}dx+sf(s-l)
8- 1
for s >0,
and
8- 1
g(s) ==s-l f {f(x) fx}dx-sf(s-l)
for s<O,
-00
and the same formulas with f and g interchanged.
56.5) Evidently
x+a y+a
IF(x, Y)I f du f If(u, v)ldv,
x-a y-a
so
00 x+a 00 y+a
f IF(x, y) I dy f du f dy f If(u, v) I dv
-00 x-a -00 y-a
x+a 00 v+a x+a 00
== f du f dv f If(u, v) I dy==2a f du f If(u, v) I dv.
x-a -00 v-a x-a -00
It follows easily that
f IF(x, y) I d(x, y)4a2 f If(u, v) I d(u, v).
R2 R2
Obviously,
IF(x, y) I f If(u, v) I d(u, v)
R2
for all (x, y),
so F is bounded. The continuity of F is also evident.
56.6) It was shown in the solution of the preceding exercise that
foo IF(x, y) I dy exists, as a finite number, for every x. Hence,
foo F(x, y) dy exists for every x, and
00 x+a 00 y+ Va 2 -(u-x)2
f F(x, y) dy== f du f dy f f(u, v) dv
-00 x-a -00 y- V a 2 _(u-x)2
x+a 00 v+ Va!_(u-x)2
== f du fdv f f(u,v)dy
x-a -00 v- Va2_(u-x)2
x+a 00
== f du{2-J a 2 - (U-X)2} f f(u, v) dv==O.
x-a -00

582
SOLUTIONS
[Ch. 13
56.7) If x' and y' are introduced as stated, then
00
f F e ipx ' dy' ==0
-00
for any x' and any real p,
so that, integrating over x' and returning to the variables (x, y), we
obtain
f F(x, y) eip(xcosa-ysina) d(x, y) ==0,
R2
.
1. e.,
f F(x, y) e-'i(sx+ty) d(x, y) ==0
R2
for every point (s, t). It follows then from Plancherel's theorem and
the continuity of F that F(x, y)==O for all (x, y). Finally, the theorem
on differentiation of indefinite integrals shows that I(x, y) ==0 holds
almost everywhere in R 2 . There exist several solutions of the problem
set in the present exercise; the present solution is due to C. Visser.
.56.11) Set f fAI(x, y)d(x, y)==M. It is immediately evident that if
the lines of support of A parallel to the y-axis are x==a and x=b,
then M =k(b-a), so b-a==Mk- 1 . Similarly, the width of A in every
direction is w==Mk- 1 , so A is of constant width W. Since A is non-
empty and open, the width w is positive, and so M #-0.
Now consider the centre of gravity Z(xz, yz) of the "mass distri-
bution with density I", i.e.,
XZ==M-1 f f xl(x, y) d(x, y) ;
A
YZ==M-1 f f yj(x, y) d(x, y),
A
so
b
xz== [k(b-a)J-1[kf xdxJ==l(a+b).
a
It follows that every line of support has a distance lw from the fixed
point Z, and it is easily derived now that A is the interior of a circle
(cf. Exercise 29.2). This proof, much simpler than the original proof,
is essentially due to N. G. de Bruyn.
56.12) Let (x, y, 11(x, y) +8) be an interior point of A, and B a
closed sphere around this point such that Be A. Consider the, cone
tangent to B and having (x, y, 11(x, y)) as vertex; consider also the
part of the cone below the vertex.

Ch. 13, 14J
SOLUTIONS
583
56.13) It follows from Exercise 56.11 that any plane intersection of
A is a circle. Consider all plane intersections parallel to a fixed plane n.
Assume now that three of these circles have centres M i (i== 1, 2, 3)
not on one line. Then consider the plane n1 through M i (i== 1,2,3),
and observe that n1 intersects A in a circle. This yields a contra-
diction. Hence, all intersections parallel to n have their centres on a
line l. Observing that any plane through l intersects A in a circle, it
follows that l is perpendicular to n, and so A is the interior of a sphere.
56.14) Assume the existence of such a function I. If 5 is a segment
of B of height h, then the integral of lover 5 is hk, so the integral of
III over 5 is at least hlkl. The spherical surface of 5 is 2nh; so for
small h there are at least
1 4n 1
---- == -
2 2nh h
non-overlapping segments of height h, and their total contribution to
the integral of III is then at least Ikl. On the other hand, the contri-
bution of the point set between radius 1 and radius 1-h tends to zero
as h-+O. Contradiction. This simple proof is due to N. G. de Bruyn.
CHAPTER 14
59.1) The first statement is a reformulation of the recurrence theo-
rem. In order to prove the second statement, let
Ek=={X: I(x) >k- 1 }
for k == 1, 2, . . . .
For almost every xEE k the points Tnx are in Ek for infinitely many
values of n, so  I(Tnx) diverges for these points x. Since E ==  E k,
the result follows.
59.3) Consider the countable collection of all open spheres with
rational centre and radius and having non-empty intersection with X.
If these intersections are B1, B2, . . ., then  Bp=X. Almost every
point of B p is recurrent in the strong sense; denoting the exceptional
subset of B p by Bp, we have therefore that #(Bp) ==0. Every point of
X -  Bp has now the required property. Indeed, if 0 is an arbitrary
open neighbourhood of such a point x, there exists a set Bp such that

584 '
SOLUTIONS
[Ch. 14
XEBpcO. Since xEBp-B, the point x is recurrent with respect to
Bp, i.e., there are infinitely many Tnx in Bp and therefore in O.
61.1) Let X ==R1, fl Lebesgue measure, and Tx==x-1. For f(x) we
take the characteristic function of (0, 1J. Then f(Tk x ) is the character-
istic function of (k, k+ 1J, so an(f; x)==n- 1 on (0, nJ and zero elsewhere.
It follows that f*(x)==O on X. Then ff*dfl==O, whereas ffdfl==l.
Furthermore,
Jlf*-a n ldfl==l
for all n.
61.2) Since fl(X) <00, it follows from fEL p that fELl, so that the
convergence to a limit function f* is guaranteed. Furthermore, writing
fk(X) ==f(Tkx) for brevity, we have
n-l n-l
l/a n ll p ==n- 1 11  fkllpn-l  Ilfkllp== Ilfllp,
o 0
so
Ilf*llplim inf Ilanllpllfllp
by Fatou's lemma. The proof for lim 11/* -anllp==O is the same as for
p== 1.
61.3) For any fO we have
t t
J {J If(T u X ) I dfl}du== J IIfl11 du==tllflll,
o X 0
so that by sec. 23, Theorem 4 (Corollary) and Theorem 5 the function
f(T u X ) is v-summable over X X [0, tJ. It follows by Fubini's theorem
that, for every t>O, the function fJ f(T uX) du is defined (almost every-
where) on X and fl-summable over X. In particular, if
1
g(x) == J f(T u X ) du,
o
we have
1 1
J g(x) dfl== J {J f(T u X ) du}dfl== J {J f(T u X ) dfl}du
x x 0 0 X
1
== J {J f(x) dfl}du== J f(x) dfl.
o x x

Ch. 14]
SOLUTIONS
585
Denoting T1 simply by T, we have Tk==Tk, so
1 k+l
gk(X) ==g(Tk x ) == f I(T U+kX) du== f I(T u x ) du,
o k
and this implies that
n-l n
an(x) ==n- 1  gk(X) ==n- 1 f I(T u x ) du
o 0
for n== 1, 2, . . . . Application of the individual ergodic theorem to
T==T1 shows, therefore, that an(x) converges almost everywhere on
X to a fl-summable limit function I*(x). If t is a number between the
integers nand n+ 1, then
t n+ 1 1
If I(T u x ) dul f It(T u x ) I du== f I/(T n+u X ) I du
n n 0
1
== f I/(T u Tnx ) I du==h(Tnx)
o
for h(x) == JlI/(T ux) I duo Since, by the ergodic theorem applied to h(x),
the quotient n- 1 h(T n x) tends to zero almost everywhere on X as n-+oo,
the same is true for t- 1 J I(T ux) du as t-+oo (where n is the largest
integer t). Hence, at(x) converges to I*(x) almost everywhere on X. If
fl(X) <00, then
fl*dfl== fgdfl== fldfl,
and the formula limJ 1t*-atldfl==O follows immediately from
lim f 1/* -ani dfl==O
and
t n+l
t- 1 f If I(T u x ) dul dflt-1 f f I/(T u x ) 1 dfldu==t- 1 1111/1 O
x n n X
as t-+oo (where, once more, n is the largest nteger t). This completes
the proof.
62.1) Let X be the set of all integers with discrete measure, and
let Tx==x+2. Then T is not ergodic, but for every summable I(x) the
time mean I*(x) is identically zero. This follows from the fact that the

586
SOLUTIONS
[Ch. 14, 15
constant zero is the only function which is summable as well as in-
variant.
62.2) Denote by {S} the countable set of all open spheres Si (i==
1, 2, . . .) with rational centre and rational positive radius, and by
{P} the countable set of those intersections Pi=SinX which are non-
empty. Evidently # (Pi) >0 for all PiE {P}. The sequence Tnx fails to
be dense in X if and only if there exists a set PiE{P} such that
00
X E n {X-T-n(Pi)}==Qi'
n=O
Since
00
T-1(Qi)== IT {X-T-n(P i )},
n=l
we have T-1(Qi) Qi' Also, #{T-1(Qi)}==#(Qi) is finite (since #(X) is
finite), and these facts taken together show that Qi is invariant. Since
the invariant set Qi is disjoint to Pi, and # (Pi) >0, the ergodicity of
T implies that #(Qi) =0. Hence, for all x not contained in the null set
i:l Qi, the sequence Tnx is dense in X.
CHAPTER 15
65.1 ) It is evident that
p(u)(a+ 1) Ilulloo,
when Ilulloo==sup u(n) is the Zoo norm of u. Let v1==(1, 0, 0, .. .), V2==
(1,1,0,0, .. .), v3==(I, 1, 1,0,0, .. .), and so on. Then p(vk)==1 for
every k, and Vk t v with v(n) == 1 for every n. Hence p(Vk) t 1, but v=
limvk satisfies p(v)==I+a>1. It follows that p does not have the
Fatou property.
If OUk t u, then Ilukll oo t Ilull oo , so
p(u)(a+ 1) Ilull oo = (a+ 1) lim Iluklloo(a+ 1) lim p(Uk)'
This shows that p has the weak Fatou property. The particular
sequence V n (n== 1, 2, . . .) above shows that k==a+ 1 is the best possi-
ble constant in the inequality p(u)k lim p(Uk)'
65.2) L p is isomorphic (algebraically and isometrically) with the
subspace (co) of Zoo, i.e., L p consists of all sequences converging to zero

Ch. 15J
SOLUTIONS
587
provided with the Zoo norm. Hence L p is norm complete (cf. Exercise
50.8), Le., L p has the Riesz-Fischer property. The same sequence V n
(n== 1, 2, .. .) as in the preceding exercise shows that L p does not have
the weak Fatou property.
70.1) It is not difficult to prove that GL(U+v)==GL(U)+GL(v) for
u>O, v>O in L p and GL(au)==aGL(u) for u>O in L p and all constants
a>O. The restriction of G to the set of all u>O in L p is a function
seminorm if we add the definition that G(u)==oo for every fl-measur-
able u(x»O satisfying p(u) ==00. The corresponding Lorentz seminorm
is exactly G L . But then G L has the Fatou property, and so Oun tu
in L p implies that GL(un) tGL(u). This is equivalent to GL(ln)!O when-
ever In!O in L p . In other words, G L is an integral. Hence, denoting by
G e the integral component of G, we have eGLGeG since G e is
the largest integral majorized by G. It follows from GeG that (Ge)L
GL. On the other hand, the definition of G L shows that (GC)L==G c .
Hence Ge==(Ge)LGL' This, together with GLGc, shows that GL==G e .
71.2) It follows from PP1 and PP2 that p'>pi and p'>P2, so p'>
sup (pi, P2). In order to prove that p'(v){sup (pi, P2)}(V) for every
vEM+, we may assume that pi(v) and P2(V) are finite. Let uEM+,
p(u)I, and U==U1+U2 with U1, U2EM+ such that P1(U1) and P2(U2)
are finite. Then
J uvdfl== J u1 vd fl+ J u2vdflP1(U1)pi(v) +P2(U2)P2(V)
{P1(U1) +P2(U2)}' max {pi (v) , P2(V)},
so
J uvdflmax{pi(v), P2(V)}.
This holds for every uEM+ satisfying p(u)l, and so
p'(v)max{pi(v), P2(V)}.
72.1) Assume that IEL p has the property that for every 8>0 there
exists a set At of finite measure such that
I/(x) 1 8
outside At.
Let 1/1>/1>/2)::;. . . ! O. On account of PPoo (where Poo is the Zoo norm)

588
SOLUTIONS
[Ch. 15
we have P(IXx-AJe, so
P(lnXX-AJe for all n.
Since 1 is summable over Ae, it follows from 1/1>ln!O that fAElnd#!O.
Hence, on account of P<'P1 (where PI is the L 1 norm), we have
p(lnXAJ!O. Combining the results, we obtain that p(ln)<2e for n
sufficiently large, i.e., p(ln)! o. This shows that IEL.
For the converse, assume that IEL p and that, for some e>O and
every set A of finite measure, the subset B of X-A on which I/(x) I >e
holds is of positive measure. Then we must have #(B) ==00; indeed, if
#(B) <00 for some such pair A, B, we would have I/(x)I8 outside the
set A +B of finite measure, contrary to our assumption. Hence, I/(x) I
>8 holds on a set of infinite measure. Since # has no atoms, it follows
easily that there exists a sequence of disjoint sets En (n== 1, 2, . . .)
such that #(En)==l for all nand I/(x) 1 >8 on every En. Set
5n==En+ E n+1 +. · ·
for n== 1, 2, . . "
and In==I/IXsn' Then 1/1>tn!O and
p(ln» fln d #== fl/ld#>e
En En
for all n.
This shows that t is not an element of L.

BIBLIOGRAPHICAL REFERENCES
(1 n most of the following references the section where the
reference is used is indicated)
I. AMEMIYA
[IJ A generalization of Riesz-Fischer's theorem, Journal Math. Soc. of Japan
5 (1953), 353-354 [sec. 65].
E. SPARRE ANDERSEN and B. JESSEN
[1 J Some Jimit theorems on integrals in an abstract set, Danske Vide Selsk.
Math.Fys. Medd. 22 (1946), Nor. 14 [sec. 23].
E. ARTIN
[ 1 J Einfiihrung in die Theorie der Gammafunktion, Leipzig (1 931 ) [sec. 25J.
G.AUMANN
[IJ Reelle Funktionen, Berlin-Gottingen-Heidelberg (1954).
R. BAIRE
[IJ Sur les fonctions de variables reelles, AnnaH Mat. Pura e Appl. (3) 3
(1899),1-122 [sec. 3J.
S. BANACH
[ 1 J Theorie des operations lineaires, W arsa w (1 932) .
[2J Sur Ie probleme de la mesure, Fundamenta Math. 4 (1923), 7-33 [sec. 28J.
[3J Sur les lignes rectifiables et les surfaces dont l'aire est finie, Fundamenta
Math. 7 (1925),225-236 [sec. 42J.
[4J Sur les fonctionnelles lineaires I, II, Studia Math. 1 (1929), 211-216 and
223-239 [sec. 27].
[5J The Lebesgue integral in abstract saces, Note in S. SAKS, Theory of the
integral, Warsaw (1937), 320-330 [sec. 13J.
S. BANACH and H. STEINHAUS
[1 J Sur Ie principe de la condensation de singularites, Fundamenta Math. 9
(1927), 50-61 [sec. 51].
G. BAXTER
[1J An ergodic theorem with weighted averages, Journal Math. and Mech. 13
(1964), 481-488 [sec. 61].
S. K. BERBERIAN
[IJ Measure and integration, New York (1965).
G. D. BIRKHOFF
[1 J Proof of the ergodic theorem, Proc. N at. Acad. Sci. U.S.A. 1 7 (1931),
656-660 [sec. 61J.

590
REFERENCES
S. BOCHNER
[IJ Inversion formulae and unitary transformations, Annals of Maths 35
(1934), 111-115 [sec. 55].
[2J Integration von Funktionen deren Werte die Elemente eines Vektor-
raumes sind, Fundamenta Math. 20 (1935), 262-276 [sec. 31].
H. F. BOHNENBLUST and A. SOBCZYK
[IJ Extension of functionals on complex linear spaces, Bull. Am. Math. Soc.
44 (1938), 91-93 [sec. 27].
E. BOREL
[IJ Le<;ons sur la theorie des fonctions, Paris (1st ed. 1898, 2nd ed. 1914, 3rd
ed. 1928) [sec. 1 OJ.
N. BOURBAKI
[IJ Integration (Actualites sci. et indo 1175; 1244; 1281; 1306), Paris (1952,
2nd ed. 1965; 1956; 1959; 1963).
J. C. BURKILL
[IJ The Lebesgue integral, Cambridge (1953).
C. CARATHEODORY
[IJ Vorlesungen TIber reelle Funktionen, Leipzig-Berlin (1st ed. 1918, 2nd ed.
1927) [sec. 7].
[2J Masz und Integral und ihre Algebraisierung, Basel (1956).
[3J Uber den Wiederkehrsatz von Poincare, Sitzungsberichte der Preussi-
schen Akad. der Wiss. 32 (1919), 580-584 [sec. 59].
P. J. DANIELL
[IJ A general form of integral, Annals of Maths 19 (1917-18), 279-294 [sec.
12, 13J.
[2J Integrals in an infinite number of dimensions, Annals of Maths 20 (1918-
19),281-288 [sec. 23J.
[3J Further properties of the genera] integral, Annals of Maths 21 (1919-20),
203-220 [sec. 12, 13].
[4J The derivative of the general integral, Annals of Maths 25 (1923-24), 193-
204 [sec. 32J.
N. DUNFORD
[IJ Integration in general analysis, Transactions Am. Iath. Soc. 37 (1935),
441-453 and 38 (1935), 600-601 [sec. 31].
N. DUNFORD and J. T. SCHWARTZ
[IJ Linear operators, New York; Part I: General theory (1958); Part II:
Spectral theory (1 963) .
D. T. EGOROFF
[IJ Sur les suites des fonctions mesurables. Comptes Rendus Acad. Sci. Paris
152 (1 911 ), 244-246 [sec. 18].
P. FATOU
[IJ Series trigonometriques et series de Taylor, Acta Math. O (1906), 335-
400 [sec. 13J.

REFERENCES
591
M. FRECHET
[IJ Sur les ensembles de fonctions et les operations lineaires, Comptes Rendus
Acad. Sci. Paris 144 (1907), 1414-1416 [sec. 30,50].
[2J Sur l'inttgrale d'une fonctionnelle etendue a un ensemble abstrait, Bull.
Soc. Math. de France 43 (1915), 249-267 [sec. 10].
G. FUBINI
[IJ Sugli integrali multipli, Rend. Accad. Lincei Roma 16 (1907), 608-614
[sec. 23J.
[2J Sulla derivazione per serie, Rend. Accad. Lincei Roma 24 (1915), 204-206
[sec. 37J. .
A. M. GARSIA
[IJ A simple proof of E. Hopf's maximal ergodic theorem, Journal Math. and
Mech. 14 (1965), 381-382 [sec. 61].
C. GOFFMAN
[IJ Real functions, New York (1953).
H. H. GOLDSTINE
[IJ Linear functionals and integrals in abstract spaces, Bull. Am. Math. Soc.
47 (1941), 615-621 [sec. 13].
G. G. GOULD
[1 J On a class of integration spaces, Journal London Math. Soc. 34 (1959),
161-172 [sec. 30].
L. M. GRAVES
[IJ The theory of functions of real variables, New York (1st ed. 1946, 2nd
ed. 1956).
J. W. GREEN
[IJ On the determination of a function in the plane by its integrals over
straight lines, Proc. Am. Math. Soc. 9 (1958), 758-762 [sec. 56].
H. HAHN
[IJ Uber Folgen linearer Operationen, Monatshefte fur Math. und Physik
32 (1922), 1-88 [sec. 45].
[2J Uber line are Gleichungen in linearen Raumen, Journal fur reine und an-
gew. Math. 157 (1927), 214-229 [sec. 27].
H. HAHN and A. ROSENTHAL
[IJ Set functions, Albuquerque (1948).
P. R. HALMOS
[IJ Measure theory, New York (1950).
[2J Lectures on ergodic theory, Tokyo (1956) [sec. 62].
1. HALPERIN
[IJ Reflexivity in the LA function spaces, Duke Math. Journal 21 (1954),
205-208 [sec. 73J.
1. HALPERIN and W. A. J. LUXEMBURG
[1 J The Riesz- Fischer completeness theorem for function spaces and vector
lattices, Transactions Royal Soc. of Canada 50 (1956), 33-39 [sec. 64J.

592
REFERENCES
[2J Reflexivity of the length function, Proc. Am. Math. Soc. 8 (1957), 496-
499 [sec. 71].
O. HAUPT and G. AUMANN
[IJ Differential- und Integralrechnung III (with the cooperation of y. PAUC),
Berlin (1955).
F. HAUSDORFF
[IJ Grundzuge der Mengenlehre, Leipzig (1st ed. 1914) [sec. 3].
E. HEWITT
[IJ Integration by parts for Stieltjes integrals, Am. Math. Monthly 67 (1960),
419-423 [sec. 24].
E. HEWITT and K. STROMBERG
[IJ Real and abstract analysis, Berlin-Heidelberg-New York (1965).
D. HILBERT
[IJ Grundzuge einer allgemeinen Theorie der linearen Integralgleichungen,
Leipzig (1912) [sec. 29J.
T. H. HILDEBRANDT
[IJ Introduction to the theory of integration, New York (1963).
E. W. HOBSON
[IJ The theory of functions of a real variable and the theory of Fourier's
series, Cambridge (1st ed. 1907, 2nd and 3rd ed. 1921-1927).
R. L. JEFFERY
[IJ The theory of functions of a real variable, Toronto (1951).
B. JESSEN
[IJ The theory of integration in a space of an infinite number of dimensions,
Acta Math. 63 (1934), 249-323 [sec. 23].
P. JORDAN and J. VON NEUMANN
[IJ On inner products in linear metric spaces, Annals of Maths 36 (1935),
719-723 [sec. 29].
S. KAKUTANI
[IJ Notes on infinite product measure spaces I, Proc. Imp. Acad. Tokyo 19
(1943), 148-151 [sec. 23].
[2J On equivalence of infinite product measures, Annals of Maths 49 (1948),
214-224 [sec. 32].
E. KAMKE
[IJ Das Lebesgue-Stieltjesintegral, Leipzig (1956).
H. KESTELMAN
[1 J Modern theories of integration, Oxford (1937).
A. N. KOLMOGOROFF and S. V. FOMIN
[IJ Measure, Lebesgue integrals and Hilbert space, New York (1961),
translated from the Russian.
B. O. KOOPMAN
[IJ Hamiltonian systems and transformations in Hilbert space, Proc. Nat.
Acad. Sci. U.S.A. 17 (1931), 315-318 [sec. 60J.

REFERENCES
593
S. KUREPA
[IJ Note on the difference set of two measurable sets in En, Glasnik Mat.-Fiz.
Astr. II 15 (1960), 99-105 [sec. 37].
Kv FAN
[IJ On a theorem of Weyl concerning eigenvalues of linear transformations I,
II, Proc. N at. Acad. Sci. U.S.A. 35 (1949), 652-655 and 36 (1950), 31-35
[sec. 40J. .
H. LEBESGUE
[IJ Le<;ons sur l'integration et la recherche des fonctions primitives, Paris
(1st ed. 1904, 2nd ed. 1928) [sec. 9, 17, 32, 37, 41].
[2J Integrale, longueur, aire, These, Paris (1902).
[3J Integrale, longueur, aire, Annali Mat. Pura e Appl. (3) 7 (1902), 231-359
[sec. 10, 12, 23].
[4J Sur l'integration des fonctions discontinues, Annales Ecole Norm. Sup.
(3) 27 (1910), 361-450 [sec. 14,37].
B. LEVI
[IJ Sopra l'integrazione delle serie, Rend. Inst. Lombardo di Sci. e Lett.
(2) 39 (1906), 775-780 [sec. 13].
L. H. LOOMIS
[IJ Linear functionals and content, Am. Journal of Maths 76 (1954), 168-182
[ sec. 1 7J.
G. G. LORENTZ and D. G. WERTHEIM
[IJ Representation of linear functionals on Kothe spaces, Can. Journal of
Maths 5 (1953), 568-575 [sec. 71].
W. A. J. LUXEMBURG
[IJ Banach function spaces, Assen, Netherlands (1955).
W. A. J. LUXEMBURG and A. C. ZAANEN
[IJ Notes on Banach function spaces, Proc. Acad. Sci. Amsterdam 66 (1963);
Note I: 135-147; Note II: 148-153; Note III: 239-250; Note IV: 251-
263 ; Note V: 496-504 (Indagationes Math. 25) [Ch. 15].
J. MARIK
[IJ Transformation of nt-dimensional Lebesgue integrals, Czechoslovak Math.
Journal 6 (1956), 212-216 [sec. 40].
S. MAZUR
[IJ 0 metodach sumowalnosci, supplement to Annales de la Soc. Polonaise
de Math. (1929), 102-107 [sec. 28J.
E. J. MCSHANE
[IJ Integration, Princeton (1947).
[2J Linear functionals on certain Banach spaces, Proc. Am. Math. Soc. 1
(1950), 402-408 [sec. 53J.
E. J. MCSHANE and T. A. BOTTS
[IJ Real analysis, Princeton (1959).

594
REFERENCES
M. E. MUNROE
[IJ Introduction to measure and integration, Cambridge, Massachusetts
( 1 953) .
L. NACHBIN
[1 J The Haar integral, Princeton (1965).
I. P. NATANSON
[ 1 J Theorie der Funktionen einer reellen V erander lichen, Berlin ( 1 954) ,
translated from the 1950 Russian edition.
[2J Theory of functions of a real variable, New York (VoL 1, 1955 ; VoL 2,
1960), English translation of the same book.
J. VON NEUMANN
[1 J Functional operators I (Annals of Maths Studies 21), Princeton (1950)
[sec. 4].
[2J Allgemeine Eigenwerttheorie Hermitescher Funktionaloperationen, Math.
Annalen 102 (1929), 49-131 [sec. 29].
[3J Proof of the quasi-ergodic hypothesis, Proc. Nat. Acad. Sci. U.S.A. 18
1 932), 70-82 [sec. 60J.
[4J On rings of operators III, Annals of Maths 41 (1940),94-161 [sec. 32].
O. M. NIKODYM
[IJ Sur une generalisation des integrales de M. Radon, Fundamenta Math. 15
(1930),131-179 [sec. 32J.
[2J Sur les suites convergentes de fonctions parfaitement additives d'ensem-
ble abstrait, Monatshefte fur Math. und Physik 40 (1933), 427-432 [sec.
45J.
T. OGASAWARA
[IJ Theory of vector lattices, Journal Sci., Hiroshima Univ.; Part I: A 12
(1942),37-100; Part II: A 13 (1944),41-161 [sec. 73J.
C. H. VAN OS
[IJ Moderne integraalrekening, Groningen (1925).
W. F. OSGOOD
[IJ Zweite Note uber analytische Funktionen mehrerer Veranderlichen, Math.
Annalen 53 (1 900), 461-463 [sec. 3J.
H. R. PITT
[IJ Integration, measure and probability, Edinburgh-London (1963).
M. PLANCHEREL
[1 J Contribution a l' etude de la representation d'une fonction arbitraire par
des integrales definies, Rend. Circ. Math. Palermo 30 (1910), 289-335
[sec. 56].
H. POINCARE
[IJ Sur Ie probleme des trois corps et les equations de la dynamique, Acta
Math. 13 (1890), 1-270 [sec. 59].
R. DE POSSEL
[IJ Sur la derivation abstraite des fonctions d'ensembles, Comptes Rendus
Acad. Sci. Paris 201 (1935), 579-581 [sec. 36].

REFERENCES
595
J. RADON
[IJ Theorie und Anwendungen der absolut additiven Mengenfunktionen, Sit-
zungsberichte Akad. der Wiss. Wien 122 (.1913), IIa, 1295-1438 [sec. 10,
32].
F. RIESZ
[IJ Sur les operations fonctionnelles lineaires, Comptes Rendus Acad. Sci.
Paris 149 (1909), 974-977 [sec. 21].
[2J Untersuchungen iiber Systeme integrierbarer Funktionen, Math. Annalen
69 (1910), 449-497 [sec. 50J.
[3J Zur Theorie des Hilbertschen Raumes, Acta Sci. Math. Szeged 7 (1934),
34-38 [sec. 29J.
F. RIESZ and B. SZ.-NAGY
[IJ Le<;ons d'analyse fonctionnelle, Budapest (1st ed. 1952, 2nd ed. 1953, 3rd
ed. 1955, 4th ed. 1965).
[2J Uber Kontraktionen des Hilbertschen Raumes, Acta Sci. Math. Szeged
10 (1943), 202-205 [sec. 60J.
W. W. ROGOSINSKI
[IJ Volume and integral, Edinburgh-London (1952).
H. L. ROYDEN
[IJ Real analysis, New York (1963).
S. SAKS
[IJ Theory of the integral, Warsaw (1937) [sec. 17].
[2J Addition to the note on some functionals, Transactions Am. Math. Soc.
35 (1933), 965-970 [sec. 45J.
J. J. SCHAFFER
[IJ Function spaces with translations, Math. Annalen 137 (1959), 209-262
[sec. 30J.
J . SCHWARTZ
[IJ The formula for change in variables in a multiple integral, Am. Math.
Monthly 61 (1954),81-85 [sec. 40].
I. E. SEGAL
[1 J Equivalence of measure spaces, Am. Journal of Maths 73 (1951), 275-313
[sec. 35].
G. A. SOUKHOMLINOV
[IJ Uber Fortsetzung von linearen Funktionalen in linearen komplexen
Raumen, Mat. Sbornik N.S. 3, 45 (1938), 353-358, in Russian with Ger-
man summary [sec. 27].
H. STEINHAUS
[1 J Additive und stetige Funktionaloperationen, Math. Zeitschrift 5 (1918),
186-221 [sec. 50J.
M. H. STONE
[1 ] Notes on integration, Proc. N at. Acad. Sci. U.S.A.; N oteI: 34 (1948),
336-342; Note II: 447-455; Note III: 483-490; Note IV: 35 (1949), 50-58
[sec. 13, 17, 32].

596
REFERENCES
B.C. STRYDOM
[IJ Abstract Riemann integration, Assen, Netherlands (1959) [sec. 9, 13].
A. E. TAYLOR
[IJ General theory of functions and integration, New York-Toronto-London
(1965) .
E. C. TITCHMARSH
[1 J The theory of functions, Oxford (1 st ed. 1932, 2nd ed. 1939).
C. J. DE LA VALLEE POUSSIN
[IJ Inhgrales de Lebesgue, fonctions d'ensemble, classes de Baire, Paris (1st
ed. 1916, 2nd ed. 1934).
[2J Sur l'integrale de Lebesgue, Transactions Am. Math. Soc. 16 (1915), 435-
50 1 [sec. 37].
G. VITALI
[IJ SuI problema della misura dei gruppi di punta di una retta, Bologna
( 1905) [sec. 10].
G. N. WATSON
[1 J General transforms, Proc. London Math. Soc. (2) 35 (1933), 156-199
[sec. 56].
J. H. WILLIAMSON
[IJ Lebesgue integration, New York (1962).
E. WITT
[IJ Sobre el teorema de Zorn, Riv. Mat. Hisp. Amer. (1950),3-6 [sec. 2].
M. A. WOODBURY
[1 J A decomposition theorem for finitely additive set functions (preliminary
report), Bull. Am. Math. Soc. 56 (1950), 171 [sec. 9J.
K. Y OSIDA and E. HEWITT
[1 J Finitely additive measures, Transactions Am. Math. Soc. 72 (1952), 46-66
[sec. 9].
A.C. ZAANEN
[IJ Linear analysis, Amsterdam-Groningen (1953).
E. ZERMELO
[IJ Beweis, dasz jede Menge wohlgeordnet werden kann, Math. Annalen 59
( 1 904), 514-516 [sec. 2J.

INDEX
Abel, A.'s integral equation 182
absolute, a. value of a bounded linear
functional 348
absolutely continuous, a.c. function
302; a.c. integral 230; a.c. measure
237; a.c. signed measure 328; a.c.
complex measure 328; a.c. norm
214, 483; function of a.c. norm 476
abstract, a. Riemann integral 101
accumulation, point of a. 17
additive, finitely a. integral 94; finitely
a. measure 45, 195; a-a. measure 30
adjoint, a. space 188; a. transformation
388
almost, a. equal functions 103; a.
equal sets 47; a. everywhere 47; a.
every point of a set 47; locally a.
equal functions 249
analytic, a. function 23
annihilator, 392, 478; inverse a. 392,
478
anti-symmetric, a.-s. relation 7
ascending, a. sequence of sets 2
associate, a. seminorm 457
atom, 67, 372
axiom, a. of choice 8
Baire, B.'s category theorem 19
Banach, B. space 185; B. function
space 207; B. indicatrix of a func-
tion 315; B.-Mazur limit 198; B.-
Steinhaus theorem 369
basis, Hamel b. 189
Beta, B. function 1 76
Bochner, B. integral 219; B. integrable
function 220
Bohnen blust, B. -So bczy k -Soukhomli-
nov extension theorem 1 92
Bolzano, B.-Weierstrass theorem 20
Bp, B p space 225
Borel, B. function 343; B. step func-
tion (with respect to an integral)
244; B. set 342; B. set (with respect
to a collection of functions) 244
boundary, b. set 18
bounded, b. functional 188; b. set in a
metric space 14; b. transformation
186
(c), space (c) 368
(co), space (co) 368
Cantor, C.'s ternary set 304
carrier, c. of a continuous function 77;
c. of an order ideal 481
Cartsian, C. product of sets 80
category, sets of the first and second c.
18; Baire's c. theorem 19
Cauchy, C. sequence of points 17; weak
C. sequence 371
cell, 31; cubic c. 59
chain, c. in a partially ordered set 7;
maximal c. 8
characteristic, c. function of a set 5
charge, 45, 195; exterior c. 46; Jordan
c. 71; pure c. 57
choice, c. function 4

598
closed, c. set 15
closure, c. of a set 16
collection, linear c. of functions 74
comparable, c. elements in a partially
ordered set 7
complement, c. of a set 1; orthogonal
c. of a linear su bspace 389
complete, c. metric space 17; weakly
sequentially c. space 371
complex, c. function 4; c. measurable
function 134; c. summable function
134; c. null function 135; c. linear
functional 135, 188; c. measure 326;
absolutely continuous c. measure
328; c. linear vector space 184
conjugate, c. space 188; c. sequences of
nets 278
content, 45, 195
continuous, c. curve 317; c. trans-
formation 186
continuum, c. hypothesis 13
contracted, c. exterior measure 44; c.
in tegral 93
contraction, 415; positive (strong) c.
417
convergence, c. in measure 131; weak
c. 371
convergent, c. sequence of points 17;
c. sequence of sets 2
convex, c. function 180; logarithmical-
ly c. function 180; c. set 202; uni-
formly c. Banach space 384
convolution, 216; c. for vectorvalued
functions 229
cross section, 250
cubic, c. cell 59
curve, continuous c. 317; rectifiable c.
317
Daniell, D. integral 79
decomposition, Hahn d. of a measur-
able set 320; Jordan d. of a real
bounded linear functional 347; J or-
dan d. of a signed measure 320; Le-
INDEX
besgue d. of an integral 242; Le-
besgue d. of a measure 274
definite, positive d. matrix 299
degenerate, d. interval 59
dense, d. set 18
density, d. theorem 271, 280
derivative, (upper and lower) d. of a
set function (with respect to a
measure and a sequence of nets) 268;
(upper and lower) regular d. 277;
(upper and lower) symmetric d. 276
descending, d. sequence of sets 2
diameter, d. of a set in metric space 14
difference, d. of two sets 1; d. set of a
set 66; d. set of two sets 283
differentiation, d. of an indefinite
integral 272; d. of an indefinite
Lebesgue integral 280
dimension, d. of a linear vector space
184
Dini, D.'s theorem 78
direct, d. sum of sets 250
discrete, d. measure 62
disjoint, d. sets 2
distance, d. in a metric space 14
distribution, d. functions 133
domain, d. of a function 4
domina.ted, d. convergence theorem 98
Egoroff, E.'s theorem 132
eigenelement 433
eigenfunction 433
eigenspace 433
eigenvalue 433; simple e. 433
element, maximal e. in a partially
ordered set 8
elementary, e. integral 78
empty, e. set 1
equidistributed, e. functions 134
equivalent, e. norms 193
ergodic, mean e. theorem 415; maxi-
mal e. theorem 418; individual e.
theorem 421, 425, 428; e. transfor-
mation 431

-essential, e. upper bound of a function
207
Euclidean, general E. space 200
-extension, Bohnenblust-Sobczyk-
Soukhomlinov e. theorem 192;
Hahn-Banach e. theorem 191
.exterior, e. charge 46; e. measure 39; e.
measure generated by a measure 35;
contracted e. measure 44; regular e.
measure 44; relative e. measure 45;
e. ordinate set 80
j, f -set 45
F atou, F. 's lemma 86; F. property
446; weak F. property 446
field, f. of sets 27; a-f. of sets 27
fine, indefinitely f. (of a sequence of
nets) 274
finite, f. measure 51; a-f. measure 51;
f. subset property (of a measure)
257; function of f. variation 309
finitely, f. additive integral 94; f.
additive measure 45, 195
Fourier, F. transformation 401
fractional, f. integral 1 78; f. integral
for vectorvalued functions 229
Fubini, F. 's theorem 157
function, 3; absolutely continuous f.
302; f. of absolutely continuous
norm 476; almost equal f. 103;
anal ytic f. 23; Banach f. space 207;
Beta f. 176; Bochner integrable f.
220 ; Borel f. 343 ; Borel step f.
(with respect to an integral) 244;
characteristic f. of a set 5; choice f.
4; complex f. 4; complex measur-
able f. 134; complex null f. 135;
complex summable f. 134; convex
f. 180; distribution f. 133; equi-
distributed f. 134; f. of finite
(bounded) variation 309; Gamma f.
176; inverse f. 4; local null f. 248;
locally almost equal f. 249; loga-
rithmically convex f. 180; non-
INDEX
599
negative measurable f. 82; null f.
102; real f. 4; real integrable f. 95;
real measurable f. 95; real sum-
mable f. 95; a-f. 87; a6-f. 87; step f.
73; vectorvalued f. 217; vector-
valued measurable f. 222; f. norm
442; f. seminorm 442
functional, 188; bounded f. 188; com-
plex linear f. 135, 188; linear f. 78,
188; non-negative linear f. 78;
singular f. 466; subadditive f. 190;
sublinear f. 190
fundamental, f. sequence of points 17;
weakly f. sequence 371
Gamma, G. function 176
Gauss, G.'s product representation for
the Gamma function 181
Gould, space of G. 475
graph, g. of a function 84
Hahn, H.-Banach extension theorem
191; H. decomposition of a measur-
able set 320; H.-Saks theorem 332
Hamel, H. basis 189
Hausdorff, H.-Kuratowski principle 9
Heine, H.-Borel-Lebesgue covering
theorem 20
Hilbert, H. space 201
Holder, H.'s inequality 208, 457, 459
J, J-absolutely continuous integral
230; oF-integrable set 112; J-
measurable set 112; J-summable
set 112
ideal, order i. 477
image, i. of a point 4; i. of a set 4,
408; inverse i. of a set 6, 408
incomparable, i. elements in a partially
ordered set 7
indefinitely, i. fine (of a sequence of
nets) 274
indicatrix, Banach i. of a function 315

600
individual, i. ergodic theorem 421, 425,
428
induced, i. measure 112
infinite, purely i. set 453
inner, i. product 199; i. product space
200
integrable, Bochner i. function 220;
real i. function 95; i. set 112
integral, 462; absolutely continuous i.
230; abstract Riemann i. 101; Boch-
ner i. 219; i. with respect to a
complex measure 327; contracted
i. 93; Daniell i. 79; elementary i. 78;
finitely additive i. 94; fractional i.
178; fractional i. for vectorvalued
functions 229; i. of Laplace 148 ;
Lebesgue i. 79; (infinite) product i.
245; i. with respect to a signed
measure 323; Stieltjes-Lebesgue-
Radon i. 79; vectorvalued i. 217
integration, partial i. 173
interior, i. point of a set 15; i. ordinate
set of a function 80
intersection, i. of sets 1
inverse, i. annihilator 392, 478; i.
function 4; i. image of a set 6, 408
invertible, i. transformation 409
isometric, i. linear transformation 387
Jessen, J.' s theorem 169
Jordan, J. charge 71; J. decomposition
of a real bounded linear functional
347; J. decomposition of a signed
measure 320
Kakutani, K.'s theorem 246
kernel, Poisson k. 147
Kothe, normed K. space 443
Kronecker, K. delta 436
L1, L1 norm 107
Lp, Lp space 209
Laplace, integral of L. 148
least, 1. upper bound of a subset of a
INDEX
partially ordered set 8
Lebesgue, L. decomposition of an
integral 242; L. decomposition of a
measure 274; L. integral 79; L.
measure 58; L. measurable set 58;
L. points 282
Legendre, L.'s relation 181
length, 1. of a curve 317
limit, 1. of a sequence of points 17;
(upper and lo\ver) 1. of a sequence of
sets 2; Banach-Mazur 1. 198
line, 1. of support of a convex set 406
linear, 1. collection of functions 74; L
functional 78, 188; non-negative l.
functional 78; complex 1. functional
135, 188; normed 1. space 185; 1.
ordering of a set 7; 1. subspace 188;
1. transformation 186; 1. variety 202;
1. vectorspace 184
local, 1. null set 248; 1. null function
248
localizable, 1. measure 262
locally, 1. almost equal functions 249
logarithmically, 1. convex function 180
Lorentz, L. seminorm 450
lower, 1. derivative (of a set function
, with respect to a measure and a
sequence of nets) 268; regular 1.
derivative 277; symmetric 1. de-
rivative 276; 1. limit of a sequence of
sets 2
mapping, 3
maximal, m. chain in a partiall y
ordered set 8; m. element in a
partially ordered set 8; m. ergodic
theorem 418
meager, m. set 18
mean, m. ergodic theorem 415; m.
value theorem 126; second m. value
theorem 1 75
measurable, complex m. function 134;
non-negative m. function 82; real m.
function 95; vectorvalued m. func-

tion 222; m. set (with respect to a
charge) 46; m. set (with respect to a
measure) 39; oF-me set 112; Lebes-
gue m. set 58; Stieltjes-Lebesgue m.
set 64; m. transformation 409
measure, 30; absolutely continuous m.
237; absolutely continuous signed
m. 328; absolutely continuous com-
plex m. 328; complex m. 326; con-
vergence in m. 13 1; discrete m. 62;
exterior m. 39; exterior m. gener-
ated by a measure 35; contracted
exterior m. 44; regular exterior m.
44; relative exterior m. 45; finite m.
51; a-finite m. 51; finitely additive
m. 45, 195; induced m. 112; Le-
besgue m. 58; localizable m. 262;
monotone m. 30; orthogonal m. 242;
polar m. 65; m. preserving trans-
formation 409; product m. 161, 167,
170; a-additive m. 30; separable m.
69; signed m. 319; singular m. (with
respect to another measure) 242 ;
Stieltjes-Lebesgue m. 64
metric, m. space 14
metrically, m. transitive transfor-
mation 431
monotone, m. exterior measure 36; m.
measure 30; m. sequence of nets
268; m. sequence of sets 2
negative, n. part of a function 95; n.
variation of a function 310; n.
variation of a signed measure 319
neighbourhood, 15
net, 268
non-negative, n.-n. linear functional
78; n.-n. measurable function 82
norm, n. of an element in a normed
linear space 185; n. of a linear
transformation 187; absolutely con-
tinuous n. 214, 483; equivalent n.
193; function n. 442; Ll-n. 107;
uniform n. 185
INDEX
601
normed, n. Kothe space 443; n. linear
space 185
nowhere dense, n.d. set 18
null, n. function 102; complex n.
function 135; local n. function 248;
n. set 46, 103; local n. set 248; n.
space 390, 392; n. transformation
187
open, O. set 15
order, o. ideal 477
ordering, linear O. of a set 7; partial o.
of a set 7; simple O. of a set 7
ordinate, exterior and in terior o. sets
of a function 80
orthogonal, o. complement of a linear
subspace 389; o. elements in a Hil-
bert space 201; o. measures 242;
o. projection in Hilbert space 390;
o. system of elements in Hilbert
space 436; o. transformation 388
orthogonality, O. of signed measures
325
oscillation, O. of a function 315
parallelogram, p . law 201
parameter, p. of regularity 277
partial, p. integration 173; p. ordering
7
,
perfect, p. space 474
Plancherel, P. 's theorem 399
Poincare, P.' s recurrence theorem 412
point, p. of a set 1; p. of accumulation
17; interior p. 15; Lebesgue p. 282
Poisson, P. kernel 147
polar, p. measure 65
positive, p. (strong) contraction 417; p.
definite matrix 299; p. part of a
function 95; p. variation of a func-
tion 310; p. variation of a signed
measure 31 9
preserving, measure p. transformation
409
product, Cartesian p. of sets 80; inner

602
p. 199 ; (infinite) p. integral 245; p.
measure 161, 167, 170
projection, p. in Banach space 394;
orthogonal p. in Hilbert space 390
property, Fatou p. 446; weak Fatou p.
446; finite subset p. 257; Radon-
Nikodym p. 253; Riesz-Fischer p.
444
pure, p. charge 57
purely, p. infinite set 453
Pythagorean, P. theorem 201
radius, I. of a spherical neighbourhood
15
Radon, R.-Nikodym property 253; R.-
Nikodym theorem 236, 237; R.-
Nikodym theorem (non-a-finite ver-
sion) 252
range, r. of a function 4
real, r. function 4; r. integrable
function 95; r. measurable function
95; r. summable function 95; r.
linear vectorspace 184
rectifiable, r. curve 31 7
recurrence, Poincare's r. theorem 412
reflexive, r. relation 7; r. normed
linear space 379
regular, r. (upper and lower) derivative
277; r. exterior measure 44; r.
sequence of nets 269
regularity, parameter of r. 277
relation, anti-symmetric r. 7; re-
flexive r. 7; transitive r. 7
relative, r. exterior measure 45
representation, Gauss' product r. 181;
Riesz r. theorem 142, 350; r.
theorem for the Stieltjes-Lebesgue
integral 118
Riemann, abstract R. integral 101; R.-
Lebesgue theorem 336
Riesz, R.-Fischer property 444; R.
representation theorem 142, 350
ring, r. of sets 26; a- r. of sets 26
INDEX
0', a-additive measure 30; a-field of
sets 27 ; a-finite measure 51; a-
function 87; a6-function 87; a-ring
of sets 26; a-set 25; a6-set 51 ; set of
a-finite measure 51; a-subadditive
exterior measure 36
saturated, s. seminorm 454
Schwarz, S. 's inequality 200
second, s. mean value theorem 175
section, cross s. 250
Segal, S. 's theorem 265
seminorm, associate s. 457; function s.
442; Lorentz s. 450; saturated s. 454
semi-ring, s.-r. of sets 25
separable, s. measure 69; s. metric
space 20
separation, s. theorem 204
sequence, ascending and descending s.
of sets 2; Cauchy s. of points 17;
conjugate s. of nets 278; convergent
s. of points 17; convergent s. of sets
2; fundamental s. of points 17;
monotone s. of nets 268; monotone
s. of sets 2; s. space 212
sequential, s. covering of a set by a
semi-ring 35
sequentially, weakly s. coplete space
371
set, almost equal s. 47; Borel s. 342;
Borel s. (with respect to a collection
of functions) 244; boundary s. 18;
Cantor's ternary s. 304; s. of the
first and second category 18; closed
s. 15; convex s. 202; dense s. 18;
difference s. of a set 66; difference s.
of two sets 283; disjoint s. 2; empty
s. 1; f-s. 45 ; oF-integrable s. 112; oF-
measurable s. 112; oF -summable s.
112; meager s. 18; measurable s.
(with respect to a charge) 46;
measurable s. (with respect to a
measure) 39; Lebesgue ,measurable
s. 58; nowhere dense s. 18; null s.
46, 103; local null s. 248; open s. 15;

INDEX
exterior and interior ordinate s. of a
function 80; purely infinite s. 453;
a-so 25; a6-S. 51; s. of a-finite
measure 51; Stieltjes-Lebesgue
measurable s. 64
sheaf, 250
signed, s. measure 319; absolutely con-
tinuous s. measure 328
simple, s. eigenvalue 433; s. ordering of
a set 7
singular, s. functional 466; s. measure
(with respect to another measure)
242
space, adjoint s. 188; Banach s. 185;
Banach function s. 207; B p-s. 225;
s. (c) 368; s. (co) 368; weakly
sequentially complete s. 371; con-
jugate s. 188; general Euclidean s.
200; s. of Gould 475; Hilbert s. 201 ;
inner product s. 200; normed Kothe
s. 443; Lp-s. 209; normed linear s.
185; metric s. 14; null s. 390, 392;
perfects. 474; sequences. 212; torus
s. 168
standard, s. representation of a step
function 73
step, s. function 73; Borel s. function
(with respect to an integral) 244
Stieltjes, S.-Lebesgue-Radon integral
79; S.-Lebesgue measure 64
strong, positive (s.) contraction 417
subadditive, s. functional 190; a-so
exterior measure 36
sublinear, s. functional 190
subspace, linear s. 188
sum, direct s. of sets 250
summable, complex s. function 134;
real s. function 95; J-s. set 112
support, line of s. of a convex set 406
symmetric, (upper and lower) s.
derivative 276
ternary, Cantor's t. set 304
theorem, Banach-Steinhaus t. 369;
603
Bohnen blust-So bczy k -Soukhomli-
nov extension t. 192; dominated
convergence t. 98; Egoroff's t. 132;
individual ergodic t. 421, 425, 428;
maximal ergodic t. 418; mean
ergodic t. 415; Fubini's t. 157;
Hahn-Saks t. 332; Hahn-Banach
extension t. 191; t. on integration
of increasing sequences 85; Jessen's
t. 169; Kakutani's t. 246; mean
value t. 126; second mean value t.
175; Plancherel's t. 399; Poincare's
recurrence t. 412; Radon-Nikodym
t. 236, 237; Radon-Nikodym t.
(non-a-finite version) 252; repre-
sentation t. for the Stieltjes-Lebes-
gue integral 118; Riemann-Lebes-
gue t. 336; Riesz representation t.
142, 350; Segal's t. 265; Zermelo's
t. 13
torus, t. space 168
total, t. variation (of a function) 308;
t. variation (of a complex measure)
326; t. variation (of a signed
measure) 319
transform, Watson's general t. 398
transformation, 186; adjoint t. 388;
bounded t. 186; continuous t. 18b;
ergodic t. 431; Fourier t. 401;
invertible t. 409; linear t. 186; iso-
metric linear t. 387; measurable t.
409; null t. 187; orthogonal t. 388;
measure preserving t. 409; metri-
cally transitive t. 431 ; unitary t. 388
transitive, t. relation 7; metrically t.
transformation 431
triangle, t. inequality in a metric space
14
unbounded, u. interval 172
uniform, u. norm 185
uniformly, u. convex Banach space 384
union, u. of sets 1
unitary, u. transformation 388

604
upper, u. bound and least u. bound of a
subset of a partially ordered set 8;
u. derivative (of a set function with
respect to a measure and a sequence
of nets) 268; regular u. derivative
277; symmetric u. derivative 276;
u. limit of a sequence of sets 2
value, v. of a function 4
variation, positive and negative v. of a
function 3 10; total v. of a function
308; total v. of a complex measure
326; positive and negative v. of a
signed measure 319; total v. of a
signed measure 319; function of finite
(bounded) v. 309
variety, linear v. 202
INDEX
vectorspace, linear v. 184
vectorvalued, v. function 21 7 ; v.
measurable function 222; v. integral
217
Watson, W.'s general transform 398
weak, w. Cauchy (fundamental) se-
quence 371; \v. convergence 371;
w. Fatou property 446
weakly, w. sequentially complete space
371
well ordering, 13
width, w. of a convex set 406
Zermelo, z. '8 theorem 13
Zorn, Z.'s lemma 9