/
Text
MEASURE AND INTEGRAL
An Introduction to Real Analysis
Richard L. Wheeden
Department of Mathematics
Rutgers, the State University
of New Jersey
New Brunswick, New Jersey
Antoni Zygmund
Department of Mathematics
University of Chicago
Chicago, Illinois
MARCEL DEKKER, INC. NEW YORK AND BASEL
PURE AND APPLIED MATHEMATICS
A Program of Monographs, Textbooks, and Lecture Notes
Executive Editors—Monographs, Textbooks, and Lecture Notes
Earl J. Taft
Rutgers University
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University of Washington
Seattle, Washington
Chairman of the Editorial Board
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University of California, Berkeley
Berkeley, California
Editorial Board
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MONOGRAPHS AND TEXTBOOKS IN
PURE AND APPLIED MATHEMATICS
1. K. Yano, Integral Formulas in Riemannian Geometry (1970) (out of print}
2. S. Kobayashi, Hyperbolic Manifolds and Holomorphic Mappings
(1970) (out of print}
3. V. S. Vladimirov, Equations of Mathematical Physics (A. Jeffrey, editor;
A. Littlewood, translator) (1970) (out of print}
4. B. N. Pshenichnyi, Necessary Conditions for an Extremum (L. Neustadt,
translation editor; K. Makowski, translator) (1971)
5. L. Narici, E. Beckenstein, and G. Bachman, Functional Analysis and
Valuation Theory (1971)
6. D. S. Passman, Infinite Group Rings (1971)
7. L. Dornhoff, Group Representation Theory (in two parts). Part A:
Ordinary Representation Theory. Part B: Modular Representation Theory
(1971,1972)
8. W. Boothby and G: L. Weiss (edsj, Symmetric Spaces: Short Courses
Presented at Washington University (1972) (out of print}
9. Y. Matsushima, Differentiable Manifolds (E. T. Kobayashi, translator)
(1972)
10. L. E. Ward, Jr., Topology: An Outline for a First Course (1972) (outofpti.
11. A. Babakhanian, Cohomological Methods in Group Theory (1972)
12. R. Gilmer, Multiplicative Ideal Theory (1972) (out of print}
13. J. Yeh, Stochastic Processes and the Wiener Integral (1973) (out of print}
14. J. Barros-Neto, Introduction to the Theory of Distributions (1973) (out of
15. R. Larsen, Functional Analysis: An Introduction (1973) (out of print}
16. K. Yano and S. Ishihara, Tangent and Cotangent Bundles: Differential
Geometry (1973) (out of print}
17. C. Procesi, Rings with Polynomial Identities (1973)
18. R. Hermann, Geometry, Physics, and Systems (1973) (out of print}
19. N. R. Wallach, Harmonic Analysis on Homogeneous Spaces (1973) (out о
20. J. Dieudonne, Introduction to the Theory of Formal Groups (1973)
21. I. Vaisman, Cohomology and Differential Forms (1973) %
22. B.-Y. Chen, Geometry of Submanifolds (1973) (out of print}
23. M. Marcus, Finite Dimensional Multilinear Algebra (in two parts)
(1973, 1975)
24. R. Larsen, Banach Algebras: An Introduction (1973)
25. R. 0. Kujala and A. L. Vitter (edsj, Value Distribution Theory: Part A;
Part B. Deficit and Bezout Estimates by Wilhelm Stoll (1973)
26. К. B. Stolarsky, Algebraic Numbers and Diophantine Approximation (1974)
27. A. R. Magid, The Separable Galois Theory of Commutative Rings (1974)
28. B. R. McDonald, Finite Rings with Identity (1974)
29. J. Satake, Linear Algebra (S. Koh, T. Akiba, and S. Ihara, translators)
(1975)
30. J. S. Golan, Localization of Noncommutative Rings (1975)
31. G Klambauer, Mathematical Analysis (1975)
32. M. K. Agoston, Algebraic Topology: A First Course (1976)
33. K. R. Goodearl, Ring Theory: Nonsingular Rings and Modules (1976)
34. L. E. Mansfield, Linear Algebra with Geometric Applications (1976)
35. N. J. Pullman, Matrix Theory and its Applications: Selected Topics (1976)
36. B. R. McDonald, Geometric Algebra Over Local Rings (1976)
37. C W. Groetsch, Generalized Inverses of Linear Operators: Representation . _
and Approximation (1977)
38. J. E. Kuczkowski and J. L. Gersting, Abstract Algebra: A First Look
(1977)
39. С. O. Christenson and IV. L. Voxman, Aspects of Topology (1977)
40. M. Nagata, Field Theory (1977)
41. R. L. Long, Algebraic Number Theory (1977)
42. W. F. Pfeffer, Integrals and Measures (1977)
43. R. L. Wheeden and A. Zygmund, Measure and Integral: An Introduction to
Real Analysis (1977)
Library of Congress Cataloging in Publication Data
Wheeden, Richard L.
Measure and integral.
(Monographs and textbooks in pure and applied
mathematics ; 43)
Includes index.
1. Measure theory. 2. Integrals, Generalized.
Copyright © 1977 by Marcel Dekker, Inc. All rights reserved.
Neither this book nor any part may by reproduced or transmitted in any form or
by any means, electronic or mechanical, including photocopying, microfilming, and
recording, or by any information storage and retrieval system, without permission
in writing from the publisher.
Marcel Dekker, Inc.
270 Madison Avenue, New York, New York 10016
PRINTED IN THE UNITED STATES OF AMERICA
To our families
Introduction
The modern theory of measure and integration was created, primarily
through the work of Lebesgue, at the turn of this century. Although the
basic ideas are by now well established, there are ever widening applications
which have made the theory one of the central parts of mathematical analysis.
However, different applications require different emphasis on various aspects
of the theory. For example, certain facts are of primary interest for real and
bomplex analysis, others for functional analysis, and still others for prob-
ability and statistics. This text is written from the point of view of real
variables, and treats the theory primarily as a modern calculus.
The book presupposes that the reader has a feeling for rigor and some
knowledge of elementary facts from calculus. Some material which is no
doubt familiar to many readers has been included; its inclusion seemed
desirable in order to make the presentation clear and self-contained.
The approach of the book is to develop the theory of measure and
integration first in the simple setting of Euclidean space. In this case, there
is a rich theory having a close relation to familiar facts from calculus and
generalizing those facts. Later on, we introduce a more general treatment
based on abstract notions characterized by axioms and with less geometric
content. We have chosen this approach purposely, even though it leads to
some repetition, since considering a special case first usually helps in de-
veloping a better understanding of the general situation. Anyway, we all
“learn by repetition.”
The outline of the book is as follows. Chapter 1 is primarily a collection
of various background information, including elementary definitions and
results that will be taken for granted later in the book; the reader should
already be familiar with most of this material. Very few proofs are given in
Chapter 1. Actual presentation of the theory begins in Chapter 2, which
treats notions associated with functions of bounded variation, such as the
Riemann-Stieltjes integral. Strictly speaking, a reading of Chapter 2 could
be postponed until Chapter 5, where we use the Riemann-Stieltjes integral
as a way of representing the Lebesgue integral.
Chapter 3 deals with Lebesgue measure in Euclidean space, via the notion
of outer measure. Chapter 4 gives the theory of measurable functions, and
Chapter 5 considers the Lebesgue integral, again in Euclidean space. In
Chapter 6, we study repeated integration, the central result being Fubini’s
theorem. Chapter 7 treats the process which is the inverse of integration,
namely, differentiation. Here, we consider the differentiation of integrals
treated as set functions, as well as the differentiation of real-valued functions
of a single variable, such as the differentiability of monotone functions.
Chapters 3-7 complete the treatment of the general theory of integration in
Euclidean spaces.
In Chapters 8 and 9 we consider special classes of functions, like L2 and
Lp, and special results for these classes, such as the behavior of convolution
operators, the Hardy-Littlewood maximal function, and the integral of
Marcinkiewicz.
In Chapters 10 and 11 we give an abstract treatment of Lebesgue measure
and integration. Here, there are several possible approaches. We have chosen
to start with an abstract definition of measure and develop the theory of
integration following the pattern of earlier chapters. This is done in Chapter
10. It is natural to ask how such abstract measures actually arise. This ques-
tion is answered to some extent in Chapter 11, where we use the notion of
abstract outer measure to construct some specific examples of measures.
Chapter 12 plays a special role and can be read immediately after Chapter
9. It deals with an application of the Lebesgue integral to a specific branch
of analysis—harmonic analysis. This is a very broad field, and we consider
only a few problems indicative of the role that Lebesgue integration plays
in applications. Harmonic analysis also happens to be a field whose develop-
ment had a great impact on the theory of integration.
At the end of each chapter, we list a number of problems as exercises,
sometimes with parenthetical hints at solutions. Some relatively important
results are given in the exercises, but as a rule, the text does not require facts
which have appeared earlier only as exercises.
We would like to express our thanks to the Departments of Mathematics
of Rutgers University and the University of Chicago, and in particular to
Professor William H. Meyer for the friendly help he gave us during the
preparation of the manuscript of the book. Special thanks also go to Joanne
Darken and Dr. Edward Lotkowski, both of whom proofread almost the
entire manuscript and offered many helpful comments, and to Michele
Ginouves for her help with the cover design. Finally, thanks to Mrs. Annette
Roselli, our typist, for a really excellent job.
Richard L. Wheeden and Antoni Zygmund
Contents
Introduction
Chapter 1 Preliminaries
1. Points and Sets in Rn 2. Rn as a Metric Space 3. Open and Closed Sets in R"; Special Sets 4. Compact Sets; the Heine-Borel Theorem 5. Functions 6. Continuous Functions and Transformations I1 7. The Riertiann Integral 1 Exercises 1
Chapter 2 Functions of Bounded Variation; the Riemann- Stieltjes Integral 1 1. Functions of Bounded Variation 1 2. Rectifiable Curves • 2 3. The Riemann-Stieltjes Integral 2 4. Further Results About Riemann-Stieltjes Integrals 2 Exercises 3
Chapter/ 3 / Lebesgue Measure and Outer Measure 3
1. Lebesgue Outer Measure; the Cantor Set 3 2. Lebesgue Measurable Sets 3 3. Two Properties of Lebesgue Measure 4 4. Characterizations of Measurability 4 5. Lipschitz Transformations of Rn 4 6. A Nonmeasurable Set 4 Exercises 4
viii
bonumis
Chapter ) Lebesgue Measurable Functions 50
1. Elementary Properties of Measurable Functions 51
2. Semicontinuous Functions 55
3. Properties of Measurable Functions: Egorov’s
Theorem and Lusin’s Theorem 56
4. Convergence in Measure 59
Exercises 61
Chapterf5\ The Lebesgue Integral 64 64 66 71 76 83 85
1. 2. 3. 4. 5. Definition of the Integral of a Nonnegative Function Properties of the Integral The Integral of an Arbitrary Measurable f A Relation Between Riemann-Stieltjes and Lebesgue Integrals; the Lp Spaces, 0 < p < oo Riemann and Lebesgue Integrals Exercises
Chapter\6 j Repeated Integration 87
1. Fubini’s Theorem 87
2. Tonelli’s Theorem 91
3. Applications of Fubini’s Theorem 93
Exercises 96
Chapter 7 Differentiation 98
1. The Indefinite Integral 98
2. Lebesgue’s Differentiation Theorem 100
3. The Vitali Covering Lemma 109
4. Differentiation of Monotone Functions 111
5. Absolutely Continuous and Singular Functions 115
6. Convex Functions 118
Exercises 123
Chaptek 8 J If Classes 125
1. Definition of Lp 125
2. Holder’s Inequality; Minkowski’s Inequality 127
3. Classes lp 130
4. Banach and Metric Space Properties 131
Content
/ 5. The Space L2; Orthogonality 135
6. Fourier Series; Parseval’s Formula 137
7. Hilbert Spaces 141
Exercises 143
Chapter 9 Approximations of the Identity; Maximal Functions 14f
1. Convolutions 145
2. Approximations of the Identity 14£
3. The Hardy-Littlewood Maximal Function 155
4. The Marcinkiewicz Integral 15S
Exercises 15S
Chapter 10 Abstract Integration 16J
1. Additive Set Functions; Measures 16’
2. Measurable Functions; Integration 16'
3. Absolutely Continuous and Singular Set
Functions and Measures 17z
4. The Dual Space of Lp 18:
5. Relative Differentiation of Measures 18:
Exercises 19(
Chapter 11 Outer Measure; Measure 19:
1. Constructing Measures from Outer Measures 19.'
2. Metric Outer Measures 19(
3. Lebesgue-Stieltjes Measure 19'
4. Hausdorff Measure 20
5. The Caratheodory-Hahn Extension Theorem 20^
Exercises 20<
Chapter 12 A Few Facts From Harmonic Analysis 21
1. Trigonometric Fourier Series 21
2. Theorems about Fourier Coefficients 21
3. Convergence of 5[/] and S[/] 22
4. Divergence of Fourier Series 22
5. Summability of Sequences and Series 22
6. Summability of S[/] and S[/] by the
Method of the Arithmetic Mean 23
X i к ut;i i ta
7. Summability of 5[/] by Abel Means 246
8. Existence of J 250
9. Properties of J for f g Lp, 1 < p < oo 256
10. Application of Conjugate Functions to Partial
Sums of S[f] 259
Exercises 260
Notation 265
Index 267
Chapter 1
Preliminaries
This book is devoted to Lebesgue integration and related topics, a basic pai
of modern analysis. There are classical and abstract approaches to the intc
gral, and we have chosen the classical one, postponing a more abstract trea
ment until later in the book. The classical approach is based on the theory c
measure (while in some modern treatments the integral is introduced as
linear functional). Measure can be defined and studied in various spaces, bi
we will primarily consider л-dimensional Euclidean space, R". A prerequisite
undertaken in this chapter, is a review of elementary notions about R". W
have not attempted to present these in a thorough manner, but only to Ih
some of the definitions and notation that will be used throughout the bool
and state some background facts that a reader should know. We assume
knowledge of various properties of the real line R1 and of functions define
on R1, and leave as exercises the proofs of many facts which are either simila
to or derivable from their one-dimensional analogues.
1. Points and Sets in R”
Let n be a positive integer. By n-dimensional Euclidean space R", we mean th
collection of all «-tuples x = (xn . . . , xn) of real numbers xk, — co < xk <
+ oo, Л = 1, ...,«. If x = (xj,. . ., xn) and у = (y1?..., y„) are points c
R", we say that x = у if xk = yk for 1 < к < n. R" is a vector space over th
reals if for x == (x1? . . ., хи), у = (yls. . . , y„), and a e R1 we define x + у =
(Xi + У1, . . . , xn + y„) and ax — (axn . . ., axn). The point each of whos
coordinates is zero is called the origin and denoted 0 = (0,. . . , 0). By th<
vector emanating from x and terminating at y, we mean the line segmen
connecting x and y, directed from x to y. The points of this segment are о
the form (1 — + /у, 0 < t < 1. We will identify vectors that have equa
length and direction. We will also identify x with the vector emanating fror
0 and terminating at x.
If E is a set of points of R”, we use the notation CE = R” — E for th'
2
Preliminaries
(1.1)
complement of E. The complement of R” is the empty set 0. If = {E} is a
family of subsets of R", the union and intersection of the sets E in & are
defined respectively by
U E = {x : x e E for some E e
Ee&
P| E = {x : x e E for all E e
Ee&
(Here, and systematically below, we use the notation {x : . . to denote the
set of points x which satisfy. . . .)
If & is countable (i.e., finite or countably infinite), it will be called a
sequence of sets, and denoted & = {Ek: к = 1,2, . . .}. The corresponding
union and intersection will be written Ek and Pfc Ek. A sequence {Ek} of
sets is said to increase to Ek if Ek cz Efc+1 for all k, and to decrease to
P* Ek if Ek z> Ek+1 for all k', we use the notations Ek / Ek and Ek \
P* Ek to denote these two possibilities. If {Ek}f^ j is a sequence of sets, we
define
(1.1) limsup Ek = Q f Q e\ liminf Ek = Q ( Q Ek
J=l\k = j J J=l\k = J
noting that the sets Uj — ; Ek and Vj = P*LEk satisfy Uj \ limsup Ek
and Vj / liminf Ek. We leave it as a simple exercise to verify that limsup Ek
consists of those points of R” which belong to infinitely many Ek, and liminf Ek
of those which belong to all Ek for к > k0 (where k0 may vary from point to
point). Thus, liminf Ek c limsup Ek.
If Et and E2 axe two sets, we define E{ — E2 by Et — E2 = Er n CE2,
and call it the difference of E{ and E2, or the relative complement of E2 in E{.
We will often have occasion to use the De Morgan laws, which govern rela-
tions between complements, unions, and intersections; these state that
C( U £) = Cl СЕ, C( Г) £) = (J CE, .
Ee& Ee& EeP EeS?
and are easily verified. The set-theoretic notions discussed above are not
confined to R1 and hold for subsets of an arbitrary set S.
2. Rn as a Metric Space
r
R" also has, of course, a metric space structure. If x — (x1?. . ., xn) and у =
(Ti, • • • > К), we define their inner (dot) product by
n
x-y = E Xkyk.
k= 1
We have x-y = y-x, ax-y — a(x-y) for real a, and x-(y + z) = x-y +
re иь a ivieinc ораие
х • z. Noting that х • х > О, we define the absolute value of x, or the length of x.
by
/ n \ 1/2
|x| == (x-x)1/2 = ( £ xk\ .
\k=l /
We will use this notation regardless of the dimension n, Thus, if x e R1 (ii) (iii)
|x| means the usual one-dimensional absolute value of x. Then in any dimen-
sion |x| has the following properties:
(i) |x| > 0 and |x| — 0 if and only if x = 0
(ii) |ocx| = |a||x| for a e R1
(iii) |x 4- y| < |x| + |y| (the triangle inequality)
To verify (iii), observe that if we square both sides of the inequality, (iii) i<
equivalent to showing that (x + y)-(x + У) lxl2 + 2|x||y| + |y|2. Since
(x + y)-(x + y) = |x|2 + 2(x*y) + |y|2, the problem reduces to showing
that x-y < |x||y|; i.e., that
(1.2) Ё xkyk < ( £ xf) ( Ё ) •
This important inequality is called the Schwarz (or Cauchy-Schwarz) in
equality and can be proved as follows. For a,/? e R1, the fact that (a — fi)2 > (
gives a/J < la2 + I/?2. Therefore, 3=i xkyk < Qxj2 + |j2) =
|(|x|2 + ly|2)- Inequality (1.2) follows immediately if ]x| = |y| = 1, since
then ^к=1хкУк < i(l + 1) = 1 = |x||y|. Moreover, (1.2) is obvious i
either |x| = 0 or |y| = 0 since then both sides must be zero. Finally, i
|x| > 0 and |y| > 0, let xk = xfc/|x|, X = Л/1УI, x' = W, •••>•<) = x/lx
and y' = (ji,. . . 9y'„) = y/|y|. Then |x'| = |y'| = 1, so that by the case al
ready proved, ^=1 хкУк 15 that is, ££=1 xkyk < |x||y|, as claimed.
If we now define the distance between two points x and у by d(x9y) -
|x — y|, we immediately obtain the characteristic metric space properties:
(i) d(x9y) = J(y,x)
(ii) d(x,y) > 0, and d(x9y) = 0 if and only if x = у
(iii) d(x,y) < d(x9z) + J(z,y)
We have used the symbol xk to denote the £th coordinate of x. When nc
confusion should arise, we will also use {xfc} to denote a sequence of points о
R”. If x g R", we say that a sequence {xk} converges to x, or that x is thi
limit point of {xk}9 if |x — xk\ -> 0 as к -> co. We denote this by writing eithe
x = lim^oo xk or xk -> x as к co. A point x g R” is called a limit point of i
set E if it is the limit point of a sequence of distinct points of E. A point x g у
is called an isolated point of E if it is not the limit of any sequence in E (ex
eluding the trivial sequence {xj where xk ~ x for all k). It follows that:
4
Preliminaries
(1.3)
is isolated if and only if there is a <5 > 0 such that |x — y| > <5 for every
У G £, у / X.
For sequences {xk} in R1, we will write limk_>00 xk = 4- oo, or xk -> + co
as к -> oo, if given M > 0, there is an integer К such that xk> M whenever
к > К. A similar definition holds for lim^^ xk = — oo.
A sequence {xfc} in R” is called a Cauchy sequence if given e > 0, there is
an integer К such that |xfc — x;| < e for all kJ > K. We leave it as an exercise
to prove that R" is a complete metric space, that is, that every Cauchy sequence
in Rn converges to a point of R".
A set E c Et is said to be dense in Er if for every x2 e Et and s > 0,
there is a point x e E such that 0 < |x — xj < e. Thus, E is dense in Er if
every point of Ev is a limit point of E. If E = Ev, we say E is dense in itself.
As an example, the set of points of Rn each of whose coordinates is a rational
number is dense in R”. Since this set is also countable, it follows that R” is
separable, by which we mean that Rn has a countable dense subset.
For subsets E of R1, we use the standard notations sup E and inf E for
the supremum (least upper bound) and infimum (greatest lower bound) of E.
In case sup E belongs to E, it will be called max E; similarly, inf E will be
called min E if it belongs to E.
If {б7к}^°=1 is a sequence of points in R1, let bj — supk^j ak and Cj =
inf^j ak, j — 1, 2, ... . Then — oo < Cj < bj < 4-oo, and {bj} and {Cj} are
monotone decreasing and increasing, respectively; that is, bj > bj+l and
Cj < cj+1. Define limsup^^ ak and liminf^>o0 ak by
limsup ak = lim bj = lim {sup ak},
(j fc-►oo j-*oo j—►oo k^j
liminf ak = lim Cj = lim {inf ak}.
k-+co j—> <x> k^>j
We leave it as an exercise to show that — oo < liminfk_>00 ak < limsup^ ak
< + oo, and that the following characterizations hold.
(1.4) Theorem
(a) L = limsupfc„>00 ak if and only if (i) there is a subsequence {akj} of
{ak} which converges to L, and (ii) if Lf > L, there is an integer К
such that ak < L' for к > K.
(b) / = liminf^ ю ak if and only if (i) there is a subsequence {akj} of
{ak} which converges to I, and (ii) if Г < I, there is an integer К
such that ak > Г for к > К.
Thus, when they are finite, limsupfc„4(Zi ak and liminf^^ ak are the largest
and smallest limit points of {ak}, respectively. We leave it as a problem to
show that {ak} converges to a, — co < a < +oo, if and only if limsup^^ ak
— liminf^ ak ~ a.
v I .и;
upcii dfkU LrlUbCU ouu HI n , Opeciai OGlb
We can also use the metric on Rn to define the diameter of a set E by
letting
5(E) = diam E = sup {|x — y| : x,y e E}.
If the diameter of E is finite, E is said to be bounded. Equivalently, E r
bounded if there is a finite constant M such that |x| < M for all x e E. If E
and E2 are two sets, the distance between Er and E2 is defined by
d(Ei,Ez) = inf {|x - y| : x e Ех, у e E2}.
3. Open and Closed Sets in R”; Special Sets
For x e R" and b > 0, the set
_B(x;<5) = {y : |x - y| < <5}
is called the open ball with center x and radius b. A point x of a set E is callee
an interior point of E if there exists b > 0 such that B(x;<5) cz E. The collectioi
of all interior points of E is called the interior of E, and denoted Ё. A set E i
said to be open if E — Ё; that is, E is open if for each x e E there exists <5 > (
such that j£(x;<5) с E. The empty set 0 is open by convention. The whol
space R" is clearly open, and we leave it as an exercise to prove that B(x;<5) i
open. We will generally denote open sets by the letter G.
A set E is called closed if CE is open. Note that 0 and R” are closed
Closed sets will generally be denoted by the letter F. The union of a set E am
all its limit points is called the closure of E, and written E. We leave it to th
reader to prove the following facts.
(1.5) Theorem
(i) B(x,b) = {y : |x - y| < 5}.
(ii) E is closed if and only if E = E; that is, E is closed if and only i
it contains all its limit points.
(iii) E is closed, and E is the smallest closed set containing E; that и
if F is closed and E c F, then Ё a F.
By the boundary of E, we mean the set E — Ё.
The open subsets of R” satisfy the conditions listed in the next theorem
(1.6) Theorem
(i) The union of any number of open sets is open.
(ii) The intersection of a finite number of open sets is open.
Verification is left to the reader. Using De Morgan’s laws, we obtain th
following equivalent statements.
6
rreummanes
U •'/
(1.7) Theorem
(i) The intersection of any number of closed sets is closed.
(ii) The union of a finite number of closed sets is closed.
A subset Er of E is said to be relatively open with respect to E if it can be
written Er ~ E n G for some open set G. Similarly, Er is relatively closed
with respect to E if Er = E n F for some closed F. Note that the relative
complement of a relatively open set is relatively closed. A useful alternate
characterization of relatively closed is as follows.
(1.8) Theorem A set Et cz E is relatively closed with respect to E if and only
if Er = E n Et, that is, if and only if every limit point of E{ which lies
in E is in Et.
The proof is left as an exercise.
Consider a collection {A} of sets A. Then a set is said to be of type Ad
if it can be written as a countable intersection of sets A, and to be of type Aa
if it can be written as a countable union of sets A. Thus, “5” stands for inter-
section and “a” for union. The most common uses of this notation are Gd
and Fc, where {G} denotes the open sets in R” and {F} the closed sets. Hence,
H is of type Gd if
н = 0 Gk, Gk open,
к
and H is of type Fa if
H = {J Fk, Fk closed.
к
The complement of a Gd set is an Fa set, and vice versa. A Gd (Fff) set is of
course not generally open (closed); in fact, any closed (open) set in R1 is of
type Gd (Ffp. see exercise l(j). These two special types of sets will be very use-
ful later in the measure approximation of general sets.
Another special type of set which we will have occasion to use is a perfect
set, by which we mean a closed set C each of whose points is a limit point of C.
Thus, a perfect set is a closed set which is dense in itself. One particular
property of perfect sets we will use is stated in the following theorem. The
proof is postponed until Section 4 of this chapter.
(1.9) Theorem A perfect set is uncountable.
Other special sets which will be important are ^-dimensional intervals.
When n = 1 and a < b, we will use the usual notations [a,b\ — {x : a <
x < b}, (a,b) = {x : a < x < b}, = {x : a < x < b} and (a,b\ =
{x : a < x < b} for closed, open, and partly open intervals. Whenever we use
just the word interval, we generally mean closed interval. An n-dimensional
interval I is a subset of R" of the form I = {x = (xn . . ., xn): ak < xk < bk.
к — 1, . . ., n}, where ak < bk, к = 1, . . . , n. An interval is thus closed,
and we say it has edges parallel to the coordinate axes. If the edge length
bk — ak are all equal, I will be called an n-dimensional cube with edges paralie
to the coordinate axes. Cubes will usually be denoted by the letter Q. Twc
intervals Ц and I2 are said to be nonoverlapping if their interiors are disjoint
i.e., if the most they have in common is some part of their boundaries. A se-
equal to an interval minus some part of its boundary will be called a partly
open interval. By definition, the volume v(J) of the interval I = {(xb . . . , xn)
ak < xk < bk, к = 1,. . . , n} is
v(J) = fl (bk - ak)-
k=l
Somewhat more generally, if {efc}^= t is any given set of n vectors emanat-
ing from a point in R", we will consider the closed parallelepiped
p = {x : X = £ 0 < tk < 1}.
1
Note that the edges of P are parallel translates of the efc. Thus, P is an interval
if the efc are parallel to the coordinate axes. The volume n(P) of P is by defini-
tion the absolute value of the n x n determinant having e1? . . . , e„ as rows/
In case P is an interval, this definition agrees with the one given above. A
linear transformation T of R! transforms a parallelpiped P into a parallel-
piped P( with volume v(P') = |det T\v(P)fi In particular, a rotation of axej
in IV (which is an orthogonal linear transformation) does not change the
volume of a parallelepiped. We will assume simple facts about volume: e.g.
N N . N
if I C U Ik,N finite, then К (7) < S V(Ik), and if \Ik f are nonoverlappin^
1 1 n 1
intervals contained in a parallelepiped P, then S V(Ik) < V(P).
We shall use the notion of interval to obtain a basic decomposition o]
open sets in R". We consider first the case n = 1, which is somewhat simplei
than n > 1.
(1.10) Theorem Every open set in R1 can be written as a countable union oj
disjoint open intervals.
Proof. Let G be an open set in R1. For x e G, let Ix denote the maxima!
open interval containing x which is in (7; that is, Ix is the union of all open
intervals which contain x and which lie in G. If x,x' e G and x x', then Ix
and Ix> must either be disjoint or identical, since if they intersect, their union
is an open interval containing x and xf. Clearly, G — (JxeG 7X. Since each Ix
contains a rational number, the number of distinct Ix must be countable,
and the theorem follows.
*See, e.g., G. Birkhoff and S. Mac Lane, A Survey of Modern Algebra, 3d ed., Macmillan.
Ne& York, 1965, Theorem 8, p. 290.
Mbid.. Theorem 9. o. 290.
8
Preliminaries
(LH)
The construction used in this proof fails in Rn if n > 1, since the union of
(overlapping) intervals is not generally an interval. The theorem itself fails
when n > 1, as is easily seen by considering any open ball. As a substitute,
we have the following useful result.
(1.11) Theorem Every open set in Rn, n > 1, can be written as a countable
union of nonoverlapping (closed) cubes.
Proof. Consider the lattice of points of R" with integral coordinates and
the corresponding net KQ of cubes with edge length 1 and vertices at these
lattice points. Bisecting each edge of a cube in KQ, we obtain from it T sub-
cubes of edge length |. The total collection of these subcubes for every cube
in Kq forms a net Kx of cubes. If we continue bisecting, we obtain finer and
finer nets Kj of cubes such that each cube in Kj has edge length 2~7 and is the
union of 2" nonoverlapping cubes in Kj+i.
Now let G be any open set in R”. Let So be the collection of all cubes in
KQ which lie entirely in G. Let Si be those cubes in Kx which lie in G but
which are not subcubes of any cube in So. More generally, for j > 1, let Sj
be the cubes in Kj which lie in G but which are not subcubes of any cube in
So, ..., Sj_1. If 5 denotes the total collection of cubes from all the Sj, then
5 is countable since each Kj is countable, and the cubes in S are nonover-
lapping by construction. Moreover, since G is open and the cubes in Kj
become arbitrarily small as j oo, each point of G will eventually be caught
in a cube in some Sj. Hence, G = (JQeS an<^ Pro°f is complete.
The collection {Q : Qe Kj, j = 1, 2, . . .} constructed above is called a
family of dyadic cubes. In general, by dyadic cubes we mean the family of
cubes obtained from repeated bisection of any initial net of cubes in R".
It follows from (1.10) that any closed set in R1 can be constructed by
deleting a countable number of open disjoint intervals from R1. A perfect set
results by removing the intervals in such a way as to create no isolated points;
thus, we would not remove any two open intervals with a common endpoint.
4. Compact Sets; the Heine-Borel Theorem
By a cover of a set E, we mean a family of sets A such that E c (JAe& A.
A subcover & x of a cover & is a cover with the property that A j e J* when-
ever Ax g x. A cover & is called an open cover if each set in is open.
We say E is compact if every open cover of E has a finite subcover. Two
equivalent statements, whose proofs are left as exercises, are as follows.
(1.12) Theorem
(i) (The Heine-Borel theorem) A set E c R” is compact if and only if
it is closed and bounded.
(ii) A set E c R" is compact if and only if every sequence of points of E
has a subsequence which converges to a point of E.
We leave it as an exercise to show that the distance between nonempty,
compact, disjoint sets is positive, and that the intersection of a countable
sequence of decreasing, nonempty, compact sets is nonempty. Thus, a nested
sequence of closed intervals has a nonempty intersection.
With these facts, we can now prove (1.9).
Proof of theorem (7.9). Let C be a perfect set in R", and suppose that C is
countable: C = {cfc}/=1. Let Ck = C — {cj, к > 1. Given xt g C15 let Qr
be a (closed) cube with center Xj such that Cj ф Qr. Then n C is compact
(closed and bounded) and not empty. Since e C and C is perfect, Xj is a
limit point of C, and so also of C2. It follows that C2 n 6i is not empty.
Let x2 g C2 n 6i and choose a cube Q2 with center x2 such that Q2 c
and c2 ф Q2. Then Q2 n C is a compact, nonempty subset of n C. Con-
tinuing in this way, we obtain a decreasing sequence Qk n C of compact,
nonempty sets such that cfc ф Qk. It follows that (Qk n C) is a nonempty
subset of C which contains no ck. This contradiction proves that C must be
uncountable and establishes the theorem.
5. Functions
By a function f = f(x) defined for x in a set E c R", we will always mean a
real-valued function, unless explicitly stated otherwise. We allow f to take the
values ± oo; if |/(x)| < + oo for all xgE, we say f is finite (or finite-valued)
on E. A finite function /is said to be bounded on E if there is a finite number
M such that |/(x)| < M for x g E\ that is,/is bounded on E if supxe£ |/(x)| is
finite. A sequence {fk} of functions is said to be uniformly bounded on E if
there is a finite M such that |/^(x)| M for x g 2Г and all k.
By the support of / we mean the closure of the set where / is not zero.
Thus, the support of a function is always closed. It follows that a function
defined in R" has compact support if and only if it vanishes outside some
bounded set.
A function / defined on an interval I in R1 is called monotone increasing
(decreasing) if f(x) < f(y) [f(x) > /(y)] whenever x < у and x,y g I. By
strictly monotone increasing (decreasing), we mean that f(x) < f(y) [/(x) >
/(y)] if x < У and Х>У g I.
Let/be defined on E c R" and let x0 be a limit point of E. Let B'(x0, 5) —
jB(x0;<5) — {x0} denote the punctured ball with center x0 and radius <5, and let
M(x0/) - sup /(x), m(xQfi) = inf /(x).
Хб1Г(х0;<5) r\E хеВ'(хо;д)пЕ
As <5 0, Л/(х0 ;<5) decreases and m(x0 ;b) increases, and we define
10
Preliminaries
(1.13)
limsup /(x) = lim M(x0;<5),
x-*x0;xeE <5->0
liminf /(x) = lim m(x0;<5).
x->xo:xeE <5->0
We leave it as an exercise to show that the following characterizations are
valid.
(1.14) Theorem
(a) M — limsupx_>xo;xe£/(x) if and only if (i) there exists {xk} in E
such that xk -> x0 and/(xk) -> M, and (ii) if M’ > M, there exists
3 > 0 such that /(x) < M' for x e B'(x0;3) n E.
(b) m = liminfx^Xo;xeE/(x) if and only if (i) there exists {xk} in E
such that xk -> x0 and f(xk) -> m, and (ii) if m' < m, there exists
3 > 0 such that /(x) > m' for x g B'(xq ;<5) n E.
We also define limsup^i-^.^/^) and liminf|x|^;xe£/(x). For example,
M = limsupjxi^.^/Xx) means (i) there exist {xk} in E such that |xk| -> oo
and /(xk) -+ M, and (ii) if M' > M, there exists N such that /(x) < M' if
|x| > N and x g E. These notions should not be confused with lim sup^^
fk(x) and liminfk_>00/k(x), which denote the limsup and liminf of the se-
quence {/t(x)}.
6. Continuous Functions and Transformations
A function f defined in a neighborhood of x0 is said to be continuous at x0
if/(x0) is finite and Нт^^Дх) = /(x0). If/is not continuous at x0, it follows
that unless /(x0) is infinite, either limx^Xo/(x) does not exist or is different
from/(x0).
For functions on R1, we will use the notation
/Oo + ) = I™ f(x) and /(x0-) = lim f(x)
x-*xo;x>xo x-*xo;x<xo
for the right and left-hand limits of f at x0, when they exist. If/(x0 + )/(x0-),
and/(x0) exist and are finite, but/is not continuous at x0, then either/(x0 + )
ф /(x0-) or Л*о + ) /(x0 —) Л*о)- In the first case, x0 is called a jump
discontinuity of / and in the second, a removable discontinuity of /(since by
changing the value of / at x0> we can make it continuous there). Such dis-
continuities are said to be of the first kind, as distinguished from those of the
second kind, for which either /(x0+) or /(x0 — ) does not exist or for which
Л*о+),Л*о“), or/(x0) is infinite.
If/is defined only in a set E containing x0, E c Rn, then/is said to be con-
tinuous at x0 relative to E if/(x0) is finite and either x0 is an isolated point of E
or x0 is a limit point of E and linix-^.^/fx) = Дх0). If Et с E, a function
(1.18)
The Kiemann integral
i i
is said to be continuous in Er relative to E if it is continuous relative to E at
every point of Ex. The proofs of the following basic facts are left as exercises.
(1.15) Theorem Let Ebe a compact set in R" andf be continuous in Erelative
to E. Then
(i) f is bounded on E; that is, supxeE |/(x)| < oo.
(ii) /attains its supremum and infimum on E; i.e., there exist хх,х2 g E
such thatf(xj = supxeE/(x),/(x2) = infxeE/(x).
(iii) f is uniformly continuous on E relative to E; i.e., given 8 > 0, there
exists b > 0 such that |/(x) — /(y)| < 8 if\x — y| < <5 andx,y g E.
A sequence of functions {fk} defined on E is said to converge uniformly on
E to a finite f if given e > 0, there exists К such that |/fc(x) — /(x)| < e for
к > К and x g E. We will use the following fact, whose proof is again left to
the reader.
(1.16) Theorem Let {fk} be a sequence of functions defined on E which are
continuous in E relative to E and which converge uniformly on E to a
finite f. Then f is continuous in E relative to E.
A transformation T of a set E cz R" into R" is a mapping у = Tx which
carries points x g E into points у g R". If у = (j^, . . . , y^, then T can be
identified with the collection of coordinate functions yk =fk(x), к = 1,... ,n,
which are induced by T. The image of E under T is the set {у : у = Tx for
some xeE}. T is continuous at x0 g E relative to E (by which we mean
limx_+xo;xe£ Tx = Txf) if and only if each Д is continuous at x0 relative to E.
We will use the following result in Chapter 3.
(1.17) Theorem Let у = Tx be a transformation ofW which is continuous in
E relative to E. If E is compact, then so is its image ТЕ.
7. The Riemann Integral
We shall see that the Lebesgue integral is more general than the Riemann
integral, in the sense that whenever the Riemann integral of a function exists
then so does its Lebesgue integral, and the two are equal [theorem (5.52)].
The Riemann integral is nonetheless useful, its significance being simplicity
and computability.
If f is defined and bounded on an interval I = {x : x = (x1? . . ., xn),
ak < xk < bk, к = 1, . . . , n} in Rn, its Riemann integral will be denoted by
Г&1 rbn r
(1.18) (R)\ /(x,,. . ., x„) dxi • • • dx„ or (A) /(x)r/x,
J at J an /. JI
12
Preliminaries
(1.19)
and is defined as follows. Partition I into a finite collection Г of nonoverlap-
ping intervals, Г = {/&}*= 15 and define the norm |Г| of Г by |Г| = max*
(diam Zfe). Select a point £k in Ik for к > 1, and let
лг = яг(£1;..., = £/(4Ш),
(1-19) N t=1
Ur = Z [sup/(x)]p(4), Lr = Z [ inf/(x)№).
k~l xelk k~l xelk
We then define the Riemann integral by saying that A = (R) Jz/(x) dx if
and only if lim|rp0 Rr exists and equals A; i.e., if and only if given £ > 0,
there exists d > 0 such that }A — Rr| < £ for any Г and any chosen {£J,
provided only that |F[ < <5. This definition is actually equivalent to the
statement that
(1.20) inf UT = supTr = A.
г г
The integral of course exists if f is continuous on I. Proofs of these facts are
left as exercises; the treatment given in Chapter 2 for Riemann-Stieltjes
integrals should serve as a review for many facts about Riemann integrals.
See also Theorem 5.54.
Exercises
1. Prove the following facts, which were left as exercises above.
(a) For a sequence of sets limsup consists of those points which belong
to infinitely many Ek, and liminf Ek consists of those points which belong
to all Ek from some к on.
(b) The De Morgan laws.
(c) Every Cauchy sequence in R" converges to a point of R”. (This can be
deduced from its analogue in R1 by noting that the entries in a given
coordinate position of the points in a Cauchy sequence in R" form a Cauchy
sequence in R1.)
(d) Theorem 1.4.
(e) A sequence {ak} in R1 converges to a, — oo < a < +oo, if and only if
limsup* ak = liminf^ ak = a.
(f) jf?(x; <5) is open.
(g) Theorem 1.5.
(h) Theorems 1.6 and 1.7.
(i) Theorem 1.8. J
(j) Any closed (open) set in R" is of type G6(Fa). [If F is closed, consider the
sets {x; dist(x,F) < (1/A:)}, к = 1, 2, . . .]
(к) Theorem 1.12.
Exercises
(1) The distance between two nonempty, compact, disjoint sets in R” is positive.
(m) The intersection of a countable sequence of decreasing, nonempty, compact
sets is nonempty.
(n) Theorem 1.14.
(o) Theorem 1.15.
(p) Theorem 1.16.
(q) Theorem 1.17.
(r) The Riemann integral A = (R) J\/(x) dx of a bounded f over an interval/
exists if and only if infr Ur = suprLr = A.
(s) If/is continuous on an interval 7, then (R)$ //(x) dx exists.
2. Find limsup Ek and liminf Ek if Ek = [—(1/Zr),l] for к odd and Ek = [— 1 ,(1/Лг)]
for к even.
3. (a) Show that C(limsup Ek) — liminf CEk.
(b) Show that if Ek / E or Ek \ E, then limsup Ek = liminf Ek ~ E.
4. (a) Show that limsup^oo (—ак) = —liminffc-.a, ak.
(b) Show that Пш8ирк_ад (ak 4- bk) < limsup^oo ak 4- limsup^oo bk, provided
that the expression on the right does not have the form oo 4- (—oo) or — oo
4- oo.
(c) If {ak} and {bk} are nonnegative, bounded sequences, show that limsup*^
(akbk) < (limsup^oo ^(limsup^ bk).
(d) Give examples for which the inequalities in parts (b) and (c) are not equali-
ties. Show that if either {afc} or {/>*} converges, equality holds in (b) and (c).
5. Find analogues of the statements in exercise 4 for limsup*->Xo;xeE/(x).
6. Compare limsupk^oo ak and limsup (—oo,#k).
7. Show that A n E2 = (£\ n E2)°, and Д u Ё2 c (Ei u E^0, Give an
example when Ex u Ё2 (Et u E2) °.
8. Let E be relatively open with respect to an interval 7. Show that E can be
written as a countable union of nonoverlapping intervals.
9. Prove that any closed subset of a compact set is compact.
10. Let {xk} be a bounded infinite sequence in R". Show that {xfc} has a limit point.
(This is the Bolzano-Weierstrass theorem in R”.)
11. Give an example of a decreasing sequence of nonempty closed sets in R" whose
intersection is empty.
12. Give an example of two disjoint closed sets Fx and F2 in R" for which dist(Fl}F2)
- 0.
13. If/is defined and uniformly continuous on F, show there is a function /defined
and continuous on E such that / = / on E.
14. If / is defined and uniformly continuous on a bounded set F, show that / is
bounded on F.
15. Show that a bounded / is Riemann integrable on 7 if and only if given & > 0,
there is a partition Г of I such that 0 < Ur — Lr < e.
16. If {fk} is a sequence of bounded, Riemann integrable functions on an interval I
14
Exercises
which converges uniformly on I to f, show that f is Riemann integrable on I
and that
(Я)£Л(х)Л^(Я)£/(х)</х.
17. Let /be a finite function on R" and define
ш{ё) = sup {|/(x) —/(y)|: |x - y| < J},
<5 > 0, to be the modulus of continuity of /. Show that co(<5) decreases as d
decreases to 0 and that / is uniformly continuous if and only if w(<5) —> 0 as
<5 —> 0.
18. Let Fbe a closed subset of (-00,4-00), and let /be continuous relative to F.
Show that there is a continuous function g on (—00,4-00) which equals/in F.
If l/WI < M for хе/ show that g can be chosen so that |#(x)| < M for
— 00 < x < 4-00. (This is the Tietze extension theorem for the real line.)
Chapter 2
Functions of Bounded Variation;
the Riemann-Stieltjes Integral
In the chapters ahead, we will study the Lebesgue integral. In this chapter
we introduce the Riemann-Stieltjes integral and, as a natural preliminary
step, study functions of bounded variation. The justification for doing sc
is that Lebesgue integration is intimately connected with Riemann-Stieltje;
integration, although this is not apparent from the definitions. We shall set
in (5.43) that Lebesgue integrals can be represented as Riemann-Stieltjej
integrals.
1. Functions of Bounded Variation
Let f(x) be a real-valued function which is defined and finite for all x in i
closed bounded interval a < x < b. Let
Г = {x0, , Xm}
be a partition of [#,/>]; that is, Г is a collection of points xh i = 0, 1, . . . , m
satisfying x0 = a, xm — b, and r < xt for i = 1, . . ., m. With each parti
tion Г, we associate the sum
m
Sr = W; a,b] = X l/(x<) - Лхг_,)|.
i=l
The variation of f over [a,b\ is defined as
V = K[/; a,b\ = sup Sr,
where the supremum is taken over all partitions Г of [a,b]. Since 0 < Sr <
+ oo, we have 0 < F < + co. If К < + oo,/is said to be of bounded variatio,
on [a,b]; if V = Too,/is of unbounded variation on
We list several simple examples.
15
16
Functions of Bounded Variation
Example 1. Suppose f is monotone in [a,b\. Then, clearly, each 5r equals
|/(ft) -/(n)|, and therefore V = |/(ft) ~/(n)|.
Example 2. Suppose the graph of f can be split into a finite number of
monotone arcs; that is, suppose [a, ft] = i \.a»ai +1] and /is monotone in
each [ahai + J. Then V = |/(ni+J —/(nf)|. To see this, we use the result
of Example 1 and the fact, to be proved in Theorem 2.2, that V — F[a,b] =
£t=i V[ahai+1],
Example 3. Let /be defined by /(x) = 0 when x / 0 and /(0) = 1, and let
[a,b\ be any interval containing 0 in its interior. Then 5r is either 2 or 0,
depending on whether or not x = 0 is a partitioning point of Г. Thus,
V[a,b] = 2.
If Г = {x0, xn . . . , xm} is a partition of [n,ft], let |F|, the norm of Г,
be defined as the length of a longest subinterval of Г:
- |T| = max(xf — Xf-j).
i
If / is continuous on [я,ft] and {Гу} is a sequence of partitions of [a,b\ with
|Гу| -> 0, we shall see in (2.9) that V = liniy^^ 5/. Example 3 shows that
this equality may fail for functions which are discontinuous even at a single
point: if we take / and [я,ft] as in Example 3 and choose the Гу such that
x = 0 is never a partitioning point, then lim = 0, while if we choose the
Гу such that x = 0 alternately is and is not a partitioning point, then lim
does not exist.
Example 4. Let /be the Dirichlet function, defined by /(x) = 1 for rational x
and /(x) = 0 for irrational x. Then, clearly, L[n,ft] — -F oo for any interval
[a,ft].
Example 5. A function which is continuous on an interval is not necessarily
of bounded variation on the interval. To see this, let {ну} and {</}, j =
1,2,..., be two monotone decreasing sequences in (0,1] with ax — 1,
limy^ ny = hmy^ di = 0 and £ dj = -Foo. Construct a continuous/as
follows. On each subinterval [пу+1,Яу], the graph of / consists of the sides
of the isosceles triangle with base [ну+1,Пу] and height dj. Thus f(a^ = 0,
and if mj denotes the midpoint of [пу+1,Яу], then f(m^ = dj. If we further
define/(0) = 0, then/is continuous on [0,1]. Taking Гк to be the partition
defined by the points 0, {ny}yt{, and {my}y=1, we see that кУГк = 2 £y=i dj.
Hence, Г//;0,1] = -Foo.
We mention here that there exist functions which are continuous on an
interval but which are not of bounded variation on any subinterval. See
Exercise 26 of Chapter 3.
Example 6. A function/defined on \a,b\ is said to satisfy a Lipschitz condition
(2.2)
Functions ot bounaea voJibitoo:
on [a,b\, or to be a Lipschitz function on [a,6], if there is a constant C such
that
l/U) - /0)1 < c\x - y\ for all x,y e [a,b].
Such a function is clearly of bounded variation, with K[/; a,b\ < C(b — a).
For example, if f has a continuous derivative on [a,6], then (by the mean-
value theorem) f satisfies a Lipschitz condition on [a,b].
For more examples of functions of bounded variation, see the exercises
at the end of the chapter.
In the next two theorems, we summarize some of the simplest properties
of functions of bounded variation. The proof of the first theorem is left as an
exercise.
(2.1) Theorem
(i) Iff is of bounded variation on [a, b], then f is bounded on [a,6].
(ii) Let f and g be of bounded variation on [a,b\. Then cf {for any real
constant c),f + g andfg are of bounded variation on [a,&]. Moreover,
fig is of bounded variation on \a,b\ if there exists an e > 0 such that
|£(x)| > sfor xe[a,b].
Before stating the second result, we note that if Г is a refinement of Г,
that is, if Г contains all the partitioning points of Г plus some additional
points, then 5Г < This follows from the triangle inequality, and is most
easily seen in the case when Г consists of all the points of Г plus one addi-
tional point. The case of general Г can be reduced to this simple case by add-
ing one point at a time to Г.
(2.2) Theorem
(i) If [a',b'] is a subinterval of [a,b], then P[a',b'] < V[a,b]; that is,
variation increases with interval.
(ii) If a < c < b then P[a,b] = P[a,c] + У[с,Ь]; that is, variation is
additive on adjacent intervals.
Proof (i) Let I = [a,b\, Г = [af,bf], V = V[a,b\, and V' = V[a',b']. If
Г' is any partition of Г, let Г be Г' with the points a and b adjoined. Then Г is
a partition of I and 5rz[Z'] < 5r[Z]. Thus 5rz[Z'] < V, and therefore W < V.
(ii) Let I = [a,6], Ц = [a,c], I2 = [c,6], V — P[a,b], = P[a,c], and
V2 = F[c,Z>]. If П and Г2 are апУ partitions of It and Z2 respectively, then
Г - T1 u T2 is one of I, and Sr[Z] - 5Г1[Д] + Sr2[Z2]. Thus, £Г1[Д] +
Therefore, taking the supremum over Tt and T2 separately, we
obtain Vx 4- V2 < V.
Conversely, if Г is any partition of I, let Г be Г with c adjoined. Then
18
Functions of Bounded Variation
(2.3)
*Sr[Z] < 5г[Г|, and Г splits into partitions of f and Г2 of 72. Thus, we have
ад] ад] = вд] + ад/2] < + v2.
Therefore, V < Vt 4- K2, which completes the proof of (ii).
For any real number x, define
+ _ Jx if x > 0, _ __ f 0 if x > 0.
X [0 if x < 0, X ( —x if x < 0.
These are called the positive and negative parts of x, respectively, and satisfy
the relations
(2.3) x+,x" >0; |x| = x+ 4- x-; x = x+ — x~.
Given a function f and a partition Г = {xJ7=o °f \aJ>\, define
Pr = Prl/; = f (Ж) -Я^-1)]+,
i = 1
m
Nr = W; a,b] = £ [/(%,-) - Ж-,)]-.
i= 1
Thus, Pr is the sum of the positive terms of 5r, and — Nr is the sum of the
negative terms of 5r. In particular, by (2.3), Pr > 0, Nr > 0,
(2.4) Pr + Nr = Sr,
(2.5) Pr- Nr=f(b)-f(a).
The positive variation P and the negative variation N of/are defined by
P = P[f; a,b] = sup Pr,
г
N = N[f; a,b] = sup Nr.
г
Thus, 0 < P,N <4-oo.
(2.6) Theorem If any one of P, N, or V is finite, then all three are finite.
Moreover, we then have
P 4- N = V, P — N = f(b) — f(a),
or equivalently,
P = |[JZ + f{b) - f(a)], N = |[F - f(b) + /(«)].
Proof. By (2.4), Pv + Nr < V, and therefore, since Pv and Nr are non-
negative, P < V and N < V. In particular, P and N are finite if V is. By (2.4)
again; Sr < P + N and, therefore, К < P + N. However, if either P or N
is finite, so is the other by (2.5), and therefore, so is V. This gives the first
part of the theorem.
(2.9)
Functions of Bounded Variation
13
Now choose a sequence of partitions I/ so that РГк -> P. Then Nrk -> N
since Pr and Nr differ by a constant [see (2.5)]. Letting к oo in the relations
prk - ЛТГк = /(6) — f(a) and РГк + Nrk < V, we obtain P-N = f(b) -
f(a) and P + N < V. Since V < P T N was shown above, we have V =
P 4- N, and the theorem follows.
(2.7) Corollary (Jordan's Theorem) A function f is of bounded variation on
[a,b\ if and only if it can be written as the difference of two bounded in-
creasingfunctions on [a,b\.
Proof Suppose/ = Л — /2, where ff and f2 are bounded and increasing
on [я,ft]. Then ff and f2 are of bounded variation on \a,b], and therefore, by
(2.1)(ii), so is/
Conversely, suppose/is of bounded variation on \a,b\. By (2.2)(i),/is of
bounded variation on every interval [я,*], a < x < b. Let P(x) and N(x)
denote the positive and negative variations of/ on [#,x], respectively. By the
analogue of (2.2)(i) for P and #(see Exercise 3), it follows that P(x) and N(x)
are bounded and increasing on [a,ft]. Moreover, by (2.6) applied to [я,х],
/(x) = [P(x) + f(a)] — N(x) when a < x < ft. Since P(x) is bounded and
increasing, so is P(x) + f(a), and the corollary follows.
Note that since the negative of an increasing function is decreasing,
corollary (2.7) may be rephrased to say that / is of bounded variation if and
only if it is the sum of a bounded increasing function and a bounded decreas-
ing function.
We remark here that there exist continuous functions of bounded varia-
tion which are not monotone in any subinterval. See Exercise 27 of Chapter 3.
In the next theorem, we consider a continuity property of functions of
bounded variation. We recall from Chapter 1 that a discontinuity is said to
be of the first kind if it is either a jump or a removable discontinuity.
(2.8) Theorem Every function of bounded variation has at most a countable
number of discontinuities, and they are all of the first kind.
Proof Let/be of bounded variation on [я,ft]. By (2.7), we may assume/
is bounded and increasing on [я,ft]. Then the only discontinuities of/ are of
the first kind; in fact, they are all jump discontinuities. If D denotes the set
of all discontinuities of /, then D = {x :/(x + ) — f(x — ) > 1/k}.
Since / is bounded, each set on the right is finite (or empty); therefore D is
countable.
We now discuss a property of the variation of a continuous function.
(2.9) Theorem Iff is continuous on [a,b], then V ~ lim|rp0 £r; that is,
given M satisfying M < И, there exists ft > 0 such that Sr > M for
any partition Г of[a,b\ with |T| < ft.
20
Functions of Bounded Variat л
(2.10)
Proof. We recall the discussion following Example 3. Given M, M < И,
we must find <5 > 0 so that > M if |Г| < 6. Select ц > 0 such that M +
[i < V, and choose a fixed partition Г = {х7}у=0 such that > M 4- /i.
Using the uniform continuity off on \af\, find q > 0 so that
(i) l/W ~/(x')| < /W 4- 1) if |x - x'J < rj.
Now let Г be any partition which satisfies
(ii) 1П < ъ
(iii) |Г| < min;(x; - x^).
We claim that 5r > M, from which the theorem will follow by choosing <5
to be the smaller of rj and min7- (x; — Xj.J. Write Г = {xJ7=o and
sr = £ 1Ж) -Ж-.)1 = S' + S",
i=l
where E" is extended over all i such that (x^^Xf) contains some x;. By (iii),
any (xH1,xf) can contain at most one xJf and therefore the number of terms
of E" is at most к 4- 1. Let Г и Г denote the partition formed by the union
of the points of Г and Г. Then Г и Г is a refinement of both Г and Г. More-
over, 5ГиГ = E' 4- E'", where E"' is obtained from E" by replacing each term
by |/(xf) -/(x;)| 4- |/(x;) -/(Xi-JI, Xj being the point of Г in (xf_ 1?xf).
By (i) and (ii), each of these two terms is less than цЩк 4- 1), and therefore
Х"<2<‘+1>2(ХТ)"*‘'
Hence,
E' = Srur - E'" > SruF - д,
so that 5Г > 5Гк,г — Since Г и Г is a refinement of Г, 5ГиГ > This
gives Sr > Sr — [i > M and completes the proof.
(2.10) Corollary If f has a continuous derivative/' on [a,b\, then
V =f \f'\dx, P=f{f'}*dx, N = Г {/'}- dx.
J a J a J a
Proof. By the mean-value theorem,
sr = £ 1Ж) -Ж--1)! = £ - xf-i)
i = i i=i
J
for appropriate e (xz_ 1?xf), i = 1, . . ., m. Hence, by (2.9),
V = lim Sr = lim £ |/'(^)|(х; - л;_1)= [ dx,
|Г|-0 |Г|->0 i=l Ja
rwuu s Idui e uuivcs
by definition of the Riemann integral. Moreover, by (2.6),
p = |[Г + f(b) -/(a)] = |
•b
|/'(x)| dx +
ъ
f'(x) dx
a
f*b p
= 1 (l/'WI + A*)] dx = [/'(X)]+ dx.
J a J a
The formula for N follows similarly from the fact that
= /(&)+/(«)].
For an extension of (2.10), see (7.31) in Chapter 7.
In passing, we note that there are notions of bounded variation for open
or partly open intervals, as well as for infinite intervals. Suppose, for example,
that (a,b) is a bounded open interval. Let [д',bf] c (a,b), and define V(a,b) =
lim as «' -> л and bf -> b. If t(a,b) < +oo, we say f is of bounded
variation on (a,b). Similarly, if/is defined on (— oo,+ oo), let F( — oo, + oo) =
lim K[fl,fe] as a -> — oo and b -> 4-oo. Analogous definitions hold for [a,b),
(a, + oo), [fl,4-oo), etc. See Exercise 8.
We may also consider the notion of bounded variation for complex-
valued f defined on an interval. The definition is the same as for the real-
valued case, and we leave it to the reader to show that a complex-valued f is
of bounded variation if and only if both its real and imaginary parts are.
2. Rectifiable Curves
As an application of the notion of bounded variation, we shall discuss its
relation to rectifiable curves (initially, those in the plane). A curve C in the
plane is two parametric equations
(2.11) C: iX = , a < t < b.
(J = M)
The graph of C is {(x,y) : x = </>(/), у = a < t < b}. The graph may
have self-intersections and is not necessarily continuous or bounded. We
think of the curve itself as the mapping of [a,6] onto the graph.
Let Г = {a = t0 < tx < • • • < tm — b} be a partition of [a,6], and
consider the corresponding points Pt = i = 0, 1,. . . , m, on the
graph of C. Draw the polygonal (broken) line connecting Po to Pt, Pr to
P2, . . ., Pm_ t to Pm in order, and let
/(Г) = f ([</>(0 - ^(Гг-!)]2 + [^(O - ^-x)]2)1/2
i = i
denote its length. The length L of C is defined by the equation
(2.12) L = £(C) = sup /(Г).
22 Functions of Bounded Variation (2.13)
Thus, 0 < L < 4-oo. If the graph of C is discontinuous, then as we move
along the graph, the length of every missing segment will contribute to L.
Moreover, the possibility that the graph may be traversed more than once,
that is, that the mapping t -> (</>(*), a < t < b, may not be one-to-one,
will add to L.
We call C rectifiable if L < 4- oo.
(2.13) Theorem Let C be a curve defined by (2.11), Then C is rectifiable if
and only if both ф and ф are of bounded variation. Moreover,
У(ф), У(ф) <L< У(ф) + У(ф).
Proof We will use the simple inequalities
|a|, |Z>| < (a2 + b2)1/2 < |o| + |Z>|
for real a and b. Thus, if C is rectifiable and Г = {/J is any partition of
[a,2>], the inequality
/(D = E (Ж) - Ф^)]2 4- [ф&) - ф^)]2)1?2 < L
implies E |0(rf) - ф^1-г)\ < L and E \ф(^) - Hence, У(ф),
У(ф) < L. On the other hand, for any C,
/(Г) < E \Ф(^ - ф(^,)\ 4- E - ^i-0\ < У(Ф) +
Hence, L < У(ф) 4- У(ф).
It follows that if ф(г) is any bounded function which is not of bounded
variation on [я,6] (see Example 5 and Exercise 1), then the curve given by
x = у — ф(Т), a < t < b, is not rectifiable, even though its graph lies in a
finite segment of the line у = x. Thus, the length of the graph of a curve is
not necessarily the same as the length of the curve.
In the special case that C is given by a function у — f(x), (2.13) reduces
to the simple statement that C is rectifiable if and only if f is of bounded
variation.
Curves in R" can be treated similarly, and we shall be brief. By a curve C
in R" we mean a system xt — ф^),. . ., xn = фпФ)> for t *n some [#,&].
We consider a partition Г = {tJ7=o of laJA an<^ Ле length /(Г) of the cor-
responding polygonal line:
m m / n \ 1/2
/(Г) = X = z x [«Ш) - ^Vi-1)]2 •
i=l i=l\J=l /
J
The quantity L ~ sup /(Г) is called the length of C, and if L < 4- oo, C is
said to be rectifiable. As seen from the definition of /(Г), exactly as in the case
n = 2, C is rectifiable if and only if each ф} is of bounded variation.
(2.15)
The Riemann-Stieltjes Integral
23
3. The Riemann-Stieltjes Integral
Let f and ф be two functions which are defined and finite on a finite interval
[a,b\. If Г = {(7 = x0 < xi < ’ ’ * < xm ~ b} is a partition of [я,6], we
arbitrarily select intermediate points i satisfying xz_t < < xi9 and
write
m
(2.14) Rr = £ ЖИЖ) - ф{х,_ J],
i=l
Rr is called a Riemann-Stieltjes sum for Г, and of course depends on the points
f the functions f and ф and the interval [a,6], although we shall usually not
display this dependence in our notation.
If
(2.15)
I = lim Rr
|гно
exists and is finite, that is, if given a > 0 there is a <5 > 0 such that |Z — Rr| < a
for any Г satisfying |Г| < 5, then I is called the Riemann-Stieltjes integral of
f with respect to ф on [a,b], and denoted
7 =
fb rb
= fix) = /</</>.
J a J a
A necessary and sufficient condition for the existence of (1ф is that given
a > 0, there exist <5 > 0 such that |Rr — Rr| < aif |Г|, |Г'| < Ь. See Exercise
11.
We list four preliminary remarks about this integral.
1. If ф(х) = x, J*/с1ф is clearly just the Riemann integral J*/dx. In this case,
Theorem 5.54 in Chapter 5 gives a necessary and sufficient condition for
the existence of the integral.
2. If f is continuous on [я,&] and ф is continuously differentiable on [af\,
then $ьа/Лф = $ьа/ф' dx, [See also (7.32).] In fact, by the mean-value
theorem,
Rr = ^КШ(хд - Ж-i)] = - Хг_Д
with xf_! < rji < Xi. Using the uniform continuity of </>', we obtain
lim(rH0 Rr = $/ф' dx.
3. Let ф(х) be a step function; that is, suppose there are points a =
a0 < cq < • • • < = b such that ф is constant on each interval
(a-_ t,cq). Let
^>(az+) = lim ф(х), i = 0, 1, . . . , m — 1,
24
Functions of Bounded Variation
(2.15)
and
— lim ф(х), i — 1,., ., m,
X^Xi~
denote the limits from the right and left at af, and let dt = ф(аг + ) —
), i = 1, • • •, m - 1, d0 = <^(a0 + ) - </»(a0), and dm = ф(ат) -
ф(ат—) denote the jumps of ф. Then, for continuous f,
rb m
/с1ф = ^/(^а,
J a i~0
The existence of the integral can be verified directly or by appealing to
(2.24).
4. In definition (2.15), no condition other than finiteness is imposed on f or
</>, but we shall see later that the most important applications occur when
ф is monotone or, more generally, of bounded variation. We note now
that if \baf dcj) exists then f and ф have no common points of discontinuity.
To prove this, suppose that both f and ф are discontinuous at x, a < x < b.
Suppose first that the discontinuity of ф is not removable. Then there
is a fixed q > 0 such that, for every s > 0, there exist points xt and x2
with x — < xt < x < x2 < x 4- and |ф(х2) “ > Л- Let
Г = {xj be a partition of [a>b\ with |F| < a such that xfo_! = xt and
xio = x2 f°r some i0. Choose a point e [x^^xj for i z0 and two
different points £io and in [х1о_1?х/о]. Let Rr be the Riemann-Stieltjes
sum using in each [xf_1?xj, and let R'r be the sum using in [x^^xj
for i Ф z0 and £-0 in [x^.^xj. Then, clearly,
i*r - *ri = 1Ж) - ж,-di
Since /is discontinuous at x, we can choose and £'-0 subject to the re-
strictions above and such that |/(£io) — /(<^0)l > V for some /л > 0 in-
dependent of 8. It follows that |Rr - RJj exceeds a positive constant
independent of e, contradicting the assumption that Rr — R'r -> 0 as
1П, |Г| - о.
If the discontinuity of ф at x is removable, a similar argument can be
given. The main difference is that we consider Г with x as a partitioning
point xio and argue for either [x^-^x] or [x,xfo+1], depending on the
nature of the discontinuity of/at x. The arguments in the case where x is
either a or b are similar.
In the theorem which follows, we list some simple properties of the
Riemann-Stieltjes integral. The proofs are left as an exercise.
(2.20) v The Riemann-Stieitjes integral za
(2.16) Theorem
(i) If ^baf <1ф exists, then so do J* cf d(j) and jbf d[c(f)) for any constant
c, and
гь Cb p
cf d$ = f д(сф) = c fd($>.
J a J a J a
(ii) If\bafi d(f) and J&/2 dcfr both exist, so does (f\ + f2) d($), and
fb p p
(A + /2) # = А^ф + \ f2 dф.
J a J a J a
(iii) If\baf dф^ and j£/t/</>2 exist, so does ^baf d^t + ф2), and
f /^(01 + Ф2) = [ fttyl + f fd<l>2-
J a J a J a
The additivity of the integral with respect to intervals is given by the
following result. See also Exercise 14.
(2.17) Theorem If\b(J^ exists and a < c < b, then ^caf dф and \bf dф boti
exist and
f*b p f*b
fty = /</</> + f<ty.
J a J a J c
Proof In the proof, will denote a Riemann-Stieltjes sum cor-
responding to a partition Г of [я,й]. To show that dф exists, it is enough
to show that given e > 0, there exists b > 0 so that if rt and T2 are parti-
tions of [я,с] with ITJ, |Г2| < b then
(2.18) |ЯГ1[а,с] ~ ЯГ2[я,с]| < £.
Since \bf dф exists, there is a b > 0 so that for any partitions Ti and Г2 о
[a,b\ with |TiI, |Г'2| < b, we have
(2.19) ~ Яг/[л,6]| < e.
Let rt and Г2 be partitions of [я,с] with given sets of intermediate points
Complete Ti and Г2 to partitions Ti and T2 of [a,ft] by adjoining the same
points of [c,b]', that is, let Г' be a partition of [c,6] and let Ti — Гх о Г'
Г2 = Г2 u Г'. Select a set of intermediate points in [c,6] for Г', and let the
intermediate points of Ц and T2 consist of these together with the sets foj
Г\ and Г2, respectively. Then
p = «г,fee] + Дг'кЧ
^г/fe^] = ^r2fec] +
26
Functions of Bounded Variation
(2.21)
If we now assume that ITJ, |Г2| < <5 and choose F so that |F'| < then
|F'i|, |Г2| < <5, and (2.18) follows from (2.19) by subtracting the equations
in (2.20).
The proof of the existence of \bf с!ф is similar. The fact that
7^ = [Уйф + \bfd<t>
a J a J c
follows from (2.20). This completes the proof.
The next result is the very useful formula for integration by parts.
(2.21) Theorem If \baf (1ф exists, then so does J* ф df and
ь гь
/Лф = [ЛЬ)ф(Ь) -/(а)ф(а)] - ф df
a J a
Proof. Let Г = {a = x0 < xx < • • • < xm ~ b} and xI_1 < < xt.
Then
m mm
Rr = = Елем*.) - E леж-j
i=l i = l i=l
m m— 1
= Ележ^) - £ле1+1Ж^)
i=i i = o
m — 1
= - E ЯЯ)[Я£.+1) -ЛЭД +A^W) -Л^Ж«)
i= 1
since xm = b and x0 = a. If we subtract and add ф(а)[А£х) — /(#)] +
ф(Ь)У\Ь) — f(fnf\ on the right side of the last equality and cancel like terms,
we obtain // ~ — Tr + [/(Z>)</>(Z?) — ^а)ф(а)], where
m — 1
тг = e <x*.wi+1) -ле;)] + ши -лап + ф(ь)[ль) -ли
& i=l
Since the straddle the xf (successive £f’s may be equal), Tr is a Riemann-
Stieltjes sum for ф df. Observing that the roles of ф and f can be inter-
changed, and taking the limit as |Г| —> 0, we see that \bfdф exists if and
only if $ьаф df exists, and that $ь^ф = Щ))ф(Ь) - Да)ф(а)] - $^df This
completes the proof.
Now let /be bounded and ф be monotone increasing on \a,b\. If Г =
{*J7=o, let
= inf /(x), Mi = sup /(x),
Xi-i<,X<Xi J Xi-i<X<Xi
-t-г = E mit<Kxi) -
i=l
m
иг = E
(2.26)
the Rlemann-Stieitjes Integral
Since — oo < rrii < Mi < Too and ф(х^ — > 0, we see that
(2.22) Lr < Яг < Ur.
Lr and Ur are called the upper and lower Riemann-Stieltjes sums for Г,
respectively. The behavior of £r and Ur is somewhat more predictable than
that of Ar, as we now show.
(2.23) Lemma Let f be bounded and ф be increasing on [a,b\.
(i) IfT' is a refinement ofV, then Lr> > Lr and Ur> < Ur.
(ii) IfT\ and T2 are any two partitions, then ЬГ1 < Ur2.
Proof, To see (i) for upper sums, suppose that Г' has only one point x’
not in Г. If x' lies between xt_ t and of Г, then sup[x._ j >xj/(x), sup[x,x.-j/(x)
< Mb so that
sup /(х)[Ж) - </>(Xi-l)]
[Xi-lX]
+ sup /(x)[</>(xf) - ф(х')] < М[ф(х>) -
[x'.Xf]
Hence, Ur < Ur. Since Г' can be obtained by adding one point at a time to
Г, an extension of this reasoning proves (i) for upper sums. The argument
for lower sums is similar.
To show (ii), note that Tj и Г2 is a refinement of both and T2. Hence,
by part (i) and (2.22), we obtain £Г1 < £Г1иг2 ^г^г2 - ^r2> which com-
pletes the proof.
We now come to an important result which gives sufficient conditions for
the existence of f/t/ф.
(2.24) Theorem If f is continuous on \a,b} and ф is of bounded variation on
[a,b], then ftfс1ф exists. Moreover,
\[^ф < (sup|/iW;^].
I J a [«,bJ
Proof To prove the existence, we may suppose by (2.7) and (2.16)(iii)
that ф is monotone increasing. Then, by (2.22), Lr < Rr < Ur, and it is
enough to show that 1ш1|ф0 Lr and lim^o Ur exist and are equal. This is
clear if ф is constant on [a,b]. If ф is not constant, let Г = {xj and note that
given s > 0, the uniform continuity off implies there exists <5 > 0 such that
if |T| < b, then Mi — mt < е/[ф(6) — ф(а)]. Hence, if |Г| < S,
(2.25) 0 < Ur — Lr — £ (Mi — mflf^Xi) — ф(х1_1)] < s.
Therefore,
(2.26) lim (Ur - Lr) - 0,
|Г|->0
28 Functions of Bounded Variation (2.27)
and it is enough to show that limJrp0 Ur exists. This is immediate since
otherwise there would exist an e > 0 and two sequences of partitions, {T£}
and {T£}, with norms tending to zero such that UVk, — иГк„ > e. In view of
(2.26), we would then have, for к large enough, £Гк, — иГк. > > 0, con-
tradicting the fact that Lr, < for any Г' and Г" [lemma (2.23)].
To complete the proof, note that the inequality |J*/с!ф\ < (sup[a>b] |/|)
У[ф; a,b] follows from a similar one for 7?r by taking the limit.
Combining (2.21) and (2.24), we see that \baf с1ф exists if either f or ф is
continuous and the other is of bounded variation.
(2.27) Theorem (Mean-value Theorem) If f is continuous on [a9b\ and ф is
bounded and increasing on \a,b\9 there exists £ e [a,b\ such that
rb
f<W = /©[</>(*) -
J O
Proof. Since ф is increasing, we have
(тт/)[ф(6) - ф(а)] < Rr < (max/)[</>(&) - ф(а)]
[а.Ы
for any Rr. Since $bf dф exists [see (2.24)], it must satisfy
rb
(тт/)[ф(Ь) - ф(а)] < f <1ф < (max/)[</>(7>) - ф(а)].
[a,b] J a [a,b]
The result now follows immediately from the intermediate value theorem for
continuous functions.
In passing, we note that Riemann-Stieltjes integrals can be defined on
open or partly open bounded intervals and on infinite intervals. If f and ф
are defined on (a,b) for example, let a < a’ < br < b and define
гь rb'
\ f d^ ~ lim f dф,
J a a'~+a J a'
b'-*b
provided the limit exists. Similarly, let
Г + оо rb *
f dф = lim f dф
J — oo a-* — oo J a
b-> + oo
if the limit exists. Analogous definitions can be given for [«,6), (я,4-оо),
[a, + oo), etc.
4. Further Results About Riemann-Stieltje^ Integrals
We will discuss a variant of the definition of ^/Лф in the case where /is
bounded and ф is increasing. Note that it then follows from part (ii) of (2.23)
(2.2В) Fu эг Kesuits About Kiemann-btieltjes integrals 23
that — oo < supr£r < infr UT < + co. It is natural to ask if the existence
of JJ/d(/> in this case is equivalent to the statement that
(2.28) sup Lr = inf Ur,
г г
which we know to be an equivalent definition in the case of Riemann integrals
[see (1.20)]. Unfortunately, the answer in general is no, as the following
example shows. Let [a,b\ = [—1,1] and
. [0 if — 1 < x < 0,
- V if Osxi 1,
,z 4 [0 if — 1 < x < 0.
- tl if 0 < x < 1.
Since/ and ф have a common discontinuity, does not exist. In fact,
if Г straddles 0, that is, if < 0 < xfo for some z0, then Rr = /(£ro) for
< £fo < xio. Hence, Rr may be 0 or 1, and thus cannot have a limit.
On the other hand, it is easy to check that Ur = 1 for any Г, and that Lr = 0
if Г straddles 0 and Lr = 1 otherwise. Hence, neither lim|rp0 JRr nor
limjrpo Lr exists, but supr Lr = infr Ur = 1.
In the following two theorems, we explore relations between (2.15) and
(2.28).
(2.29) Theorem Let f be bounded and ф be monotone increasing on [a,b\. If
lbaf ^Ф exists, then lim^r^0 Lr and Hm^r^Q Ur exist, and
lim Lr = lim Ur = sup Lr = inf Ur — \ f dф.
|Г| — 0 |Г|->0 Г Г Ja
Proof Let I = \baf dф. By hypothesis, given £ > 0, there is a <5 > 0 such
that \I - Rr| < £ for any Rr with |Г| < <5. Given Г = {xj?=o with |Г] < b,
choose and in [x^^xj, i — 1,. .. , m, such that
£ £
° - Mi ~ < ф(Ь) - ф(а) and ° - ~m'< ф(Ь) - ф(аУ
Let R'r = ^/(^)[ф(х1) - ф(Х[_ !>] and R’p = “ <Kxi- i)l- Then
И - < e, |7 - Я?| < a,
g
0 < Up-Rr< ^Ф(Ь) _ ф(а)^ - ^.-1)] =
and
g
» S «Г - i.- s _ ф{а)1ФЫ - - «.
О
30
Functions of Bounded Variation
(2.30)
Combining inequalities, we obtain
|(7r - Z| < |£7r - AH + |Ap - Z| < e + e = 2e
and
|£r - Z| < |£r - Ar) + |Ap - Z| < e + £ = 2fi.
Hence, lim(rp0 Ur — limir^oAr = Z. Since, by (2.23), Lr < supr Lr <
infr Ur < Ur, the theorem follows.
(2.30) Theorem Let f be bounded, and let ф be monotone increasing and
continuous on [a,6]. Then lim^^Q Lr and liih^^Q Ur exist, and
lim Lr — sup Lr, lim Ur = inf Ur.
|Г|->0 Г |Г|->0 r
In particular, if in addition supr Lr = infr Ur, then \baf dф exists, and
sup Lr — inf Ur = \ f dф.
Г r . Ja
Proof The proof is similar to that of (2.9). It is enough to show that
lim|rp0 Lr = supr Lr and 1т1|Г|^0 Ur = infr Ur since the last assertion of
the theorem will then follow by (2.22). We will give the argument for the
upper sums; the one for the lower sums is similar. Let infr Ur = U. Given
E > 0, we must find b > 0 such that Ur < U -I- e if |F| < b. Choose Г =
{xj}j=o such that Uf < U + |£, and let M = sup[flZj] \f\. By the uniform
continuity of ф, there exists t] > 0 such that
g
\ф(х) - ф(х')\ < 4(A, +
if |x — x'| < t].
Now let Г = {xJ7=o be any partition for which |Г| < t] and |Г| <
miny (Xj — X/-1). It is enough to show that Ur < U + £. Write
m
= X МШ - </>(*,_,)] = X' + E",
i = l
where S" is as in the proof of (2.9). Then (7ГиГ = S' + S'", where S'" is
obtained from S" by replacing each of the terms М^ф^х^ — ф^х^^] by
(2.31) sup /(х)[ф(х^ - <£(•*.-i)l + sup/(x)[0(x,) - </>(x;)],
Xj being the point of Г in (x^^Xf). Hence Ur — Ur^r — ~ At least
one of sup^. ^ *.]f and sup^. ^/equals If it is the first, the difference
between Л/^ф^х,.) — ^(x^J] and (2.31) is easily seen to be
(Ml - sup/)[</>(*,) - ф(х^)]-
(2.31) Fur r Results About Riemann-Stieltjes integrals 41
If it is the second, the difference is
(Mi - sup /)[</>(x7) - <^(Xi-i)].
In either case, the difference is at most 2Me/4(A: 4- V)M = e/2(^ + 1) in
absolute value. Hence, Ur — Ur^T < (k + l)e/2(k 4- 1) = Moreover,
(7ri>r < Uf < U + Therefore, Ur < U 4- 4- = U + s, and the
theorem follows.
Exercises
1. Let f(x) = x sin (1/x) for 0 < x < 1 and/(0) = 0. Show that /is bounded and
continuous on [0,1], but that V[f; 0,1] =4-00.
2. Prove theorem (2.1).
3. If [a',6'] is a subinterval of [a,£>], show that P[a',b'] < Р[а,Ь] and N[a'tb'] <
N[a,b].
4. Let {fk} be a sequence of functions of bounded variation on [a,b]. If L[/k; a,b]
< M < 4-°o for all к and if fk —*/pointwise on [a,6], show that/is of bounded
variation and that Lf/; a,b] < M. Give an example of a convergent sequence
of functions of bounded variation whose limit is not of bounded variation.
5. Suppose/is finite on [л,£>] and of bounded variation on every interval [a 4- £, b],
s > 0, with K[/; a 4- £, £] < M < 4-°°. Show that V[f\a,b} < 4-00. Is
V[f\ a,6] < Ml If not, what additional assumption will make it so?
6, Letf(x) = x2 sin (1/x) for 0 < x < 1 and/(0) = 0. Show that V[f; 0,1] <4-00.
[Examine the graph of /, or use Exercise 5 and (2.10).]
7. Suppose / is of bounded variation on [a,b]. If / is continuous on [a,6], show
that V(x), P(x), and N(x) are also continuous. (If Г = {xj, note that И*/ - i,*d
- |/to-i) - Sr.)
8. The main results about functions of bounded variation on a closed interval
remain true for open or partly open intervals and for infinite intervals. Prove,
for example, that if / is of bounded variation on (—00,4-00), then / is the
difference of two increasing bounded functions.
9. Let C be a curve with parametric equations x = and у — ^(t), a < t < b.
(a) If ф and у/ axQ of bounded variation and continuous, show that L —
lim|ri-0 /(Г).
(b) If ф and у/ are continuously differentiable, show that L ~ fa([<^z(t)]2 4-
[y, W)1/2 dt,
10. If < A2 < • • • < is a finite sequence and — 00 < j < 4-00, write
Ел as a Riemann-Stieltjes integral. [Take/(x) = e~sx, ф to be an appro-
priate step function, and [a,b] to contain all the 2k in its interior.]
11. Show that JJ/exists if and only if given 8 > 0, there exists 3 > 0 such that
|Rr-RrJ < e if |F|, |Г'| <3.
л
1
32
Functions of Bounded Variation
12. Prove that the conclusion of (2.30) is valid if the assumption that ф is continuous
is replaced by the assumption that f and ф have no common discontinuities.
(Instead of the uniform continuity of ф, use the fact that either f or ф is con-
tinuous at each point Xj of Г.)
13. Prove theorem (2.16).
14. Give an example which shows that for a < c < b, \caf с!ф and \bf с!ф may both
exist but \baf (1ф may not. Compare (2.17). [Take [#,£>] = [—1,1], c = 0, and/
and ф as in the example following (2.28).]
15. Suppose / is continuous and ф is of bounded variation on [a,b]. Show that the
function i/djx) = \af ^Ф is of bounded variation on [a,6]. If g is continuous on
[a,Z>], show that $bg dy = fbagf ^ф.
16. Suppose that ф is of bounded variation on [a,b\ and that / is bounded and
continuous except for a finite number of jump discontinuities in [a,6]. If ф is
continuous at each discontinuity of/ show that Jbf dф exists.
17. If ф is of bounded variation on ( — 00,4-00), f is continuous on ( — 00,4-00),
and lim|X|_> + oo/(x) = 0, show that ft^/dф exists.
18. Let/(z) = Sr=o akzk be a power series. Show that if £ |afc| < +00, then f(z)
is of bounded variation on every radius of the circle |z| = 1. [If for example
the radius is 0 < x < 1 and the ak are real, then f(x) = 2 ak xk — £ akxk.]
Chapter 3
Lebesgue Measure and Outer Measure
In this chapter, we will define and study the Lebesgue measure of sets in R".
This will be the foundation for the theory of integration to be developed later.
We shall base the presentation on the notion of the outer measure of a set.
1. Lebesgue Outer Measure; the Cantor Set
We consider closed «-dimensional intervals I = {x : aj < Xj < bj, j =
1,. . ., n} and their volumes v(F) = f]j=i ~~ aj)- ($ee P- ?•) To define the
outer measure of an arbitrary subset E of Rn, cover E by a countable collec-
tion S of intervals Ik, and let
= X v(Ik).
IkeS
The Lebesgue outer measure (or exterior measure) of Д denoted |2?|e, is defined
by
(3.1) |£|e = inf <7(S),
where the infimum is taken over all such covers S of E. Thus, 0 < |£|е <
+ oo.
(3.2) Theorem For an interval Z, |Z|e = v(I).
Proof. Since I covers itself, we have |Z|e < v(Z). To show the opposite
inequality, suppose that 5 = {41?= i is a cover of Z. Given e > 0, let Z* be
an interval containing Ik in its interior such that v(/*) < (1 4- s)vf4). Then
I cz IJk and since Zis compact, the Heine-Borel theorem implies there
is an integer N such that I cz (J*=1 /*. Clearly, v(Z) < t>(Z*). Hence,
v(F) < (1 + v(Ik) < (1 + e)o-(S). Since s can be chosen arbitrarily
small, it follows that v(I) < a(5), and therefore, that v(I) < |/|e. This com-
pletes the proof.
Note that the boundary of any interval has outer measure zero.
о
33
34
Lebesgue Measure and Outer Measu _
(3.3)
The following two theorems state simple but basic properties of outer
measure.
(3.3) Theorem If Et c E2, then |£Je < \E2\e.
The proof follows immediately from the fact that any cover of E2 is also
a cover of Er.
(3.4) Theorem If E — |J Ek is a countable union of sets, then |2?|e <
E
Proof We may assume that |Ek|e < +oo for each к — 1, 2, , since
otherwise the conclusion is obvious. Fix e > 0. Given k, choose intervals
such that Ek cz (Jy Zjfc) and Ej vUjk)) < |Fk|e + e2~*. Since E c (Jk Z^*\
we have |E|e < Therefore,
|£|e + e2-t> = E + 8,
к к
and the result follows by letting e -» 0.
We see in particular that any subset of a set with outer measure zero has
outer measure zero, and that the countable union of sets with outer measure
zero has outer measure zero. Since any set consisting of a single point clearly
has outer measure zero, it follows that any countable subset of R" has outer
measure zero. For example, the set consisting of all points each of whose
coordinates is rational has outer measure zero, even though it is dense in R".
There are sets with outer measure zero which are not countable. As an
illustration, we will construct a subset of the real line with outer measure
zero which is perfect, and therefore uncountable, by (1.9). Variants of the
construction and analogues for R", n > 1, are given in the exercises.
Consider the closed interval [0,1]. The first stage of the construction is to
subdivide [0,1] into thirds and remove the interior of the middle third; that is,
remove the open interval (|,|). Each successive step of the construction is
essentially the same. Thus, at the second stage, we subdivide each of the re-
maining two intervals [0,|] and [|, 1] into thirds and remove the interiors,
and of their middle thirds.We continue the construction for each
of the remaining intervals. The sets removed in the first three successive
stages are indicated below by darkened intervals:
-------------|-------------------------(-------------------------}--------------------------L
0 i f 1
-------------f----------------).-------(-------------------------)---------(
A _L .2 JL 2. _7_ _8 i
U 9 9 3 3 9 9 1
(3.5)
Leuesgue uuwr ivieubuic, шс
The subset of [0,1] which remains after infinitely many such operations is
called the Cantor set C: thus, if Ck denotes the union of the intervals left at
the £th stage, then
(3.5) C = Q Ck.
k=l
Since each Ck is closed, it follows from theorem (1.7) that C is closed. Note
that Ck consists of 2k closed intervals, each of length 3“k, and that C contains
the endpoints of all these intervals. Any point of C belongs to an interval in
Ck for every к and is therefore a limit point of the endpoints of the intervals.
This proves that C is perfect. Finally, since C is covered by the intervals in
any Ck9 we have |C|e < 2k3“k for each к. Therefore, |C|e = 0.
We now introduce a function associated with the Cantor set which will
be useful later. If Dk = [0,1] — Ck9 then Dk consists of the 2k — 1 intervals
Ik (ordered from left to right) removed in the first к stages of construction of
the Cantor set. Let fk be the continuous function on [0,1] which satisfies
/k(0) = o, A(l) = 1, fk(x) = j2~k on Ik9 j = 1, . . ., 2k - 1, and which is
linear on each interval of Ck. The graphs ofand f2 are shown in the foliow-
ing illustration:
By construction, each fk is monotone increasing, Д+1 = fk on Ik, j = 1
2,... , 2* — 1, and IA - A+il < 2"fc. Hence, g (A - A+i) converge
uniformly on [0,1], and therefore, {fk} converges uniformly on [0,1]. Le
f — lim^^ /fc. Then /(0) ~ 0, /(1) = 1, f is monotone increasing and con
tinuous on [0,1], and/is constant on every interval removed in constructin
C, This / is called the Cantor-Lebesgue function.
36
Lebesgue Measure and Outer MeastU
(3.6)
The next two theorems give useful relations between the outer measure
of a set and the outer measures of open sets and Gd sets (see p. 6) which
contain it.
(3.6) Theorem Let E cz R". Then given £ > 0, there exists an open set G
such that E cz G and |G|e < |£|e + £. Hence,
(3.7) |£|e = inf |G|e,
where the infimum is taken over all open sets G containing E.
Proof. Given £ > 0, choose intervals Ik such that E cz (Jk°=i Ik and
^°=1 г(/к) < |£|e + Let I* be an interval containing Ik in its interior (I*) ° .
such that v(Ik) < v(Ik) + e2~k~1. If G — (Z*)°, then G is open and con-
tains E. Furthermore,
OO 00 00
|G|e < £ v(4*) < £ v(O + S £ < |E|e + ie + = |E|e + e.
k= 1 k=l k=l
This completes the proof.
(3.8) Theorem If E c R", there exists a set H of type G6 such that E cz H
and |£% = |Я|е.
Proof By (3.6), there is for every positive integer к an open set Gk^ E
such that \Gk\e < |£|e + 1/k. If H == Q£°=1 Gk, then H is of type Gd and
/7 id E. Moreover, for every k, |E|e < \H\e < |Gfc|e < |£|e + 1/k. Thus,
\E\e = \H\e.
The significance of (3.8) is that the most general set in R” can be included
in a set of relatively simple type, namely Gd, with the same outer measure.
In defining the notion of outer measure, we used intervals I with edges
parallel to the coordinate axes. The question arises whether the outer measure
of a set depends on the position of the (orthogonal) coordinate axes. The
answer is no, and to prove this we will simultaneously consider the usual
coordinate system in R” and a fixed rotation of this system. Notions per-
taining to the rotated system will be denoted by primes. Thus, Г denotes an
interval with edges parallel to the rotated coordinate axes, and |Я|' denotes
the outer measure of a subset E relative to these rotated intervals:
(3.9) |E|; = inf£r(za
where the infimum is taken over all coverings of E by rotated intervals Ik.
The volume of an interval is clearly unchanged by rotation. (See p. 7.)
(3.10) Theorem — \E\efor every E cz R".
Proof NJq first claim that given Г and £ > 0, there exist {It} such that
Г cz (J Ц and £ u(/z) < v(Z') + £. To see this, let be an interval containing
Г in its interior, such that v(/0 < v(E) + e. By theorem (1.11), the interior of
(3.12) ' Lebesgue Measurable Sets 3/
I[ can be written as a union of nonoverlapping intervals It, Hence, Г cz (J It.
Moreover, since the It are nonoverlapping and (J f=1 Ц c /j for every positive
integer JV, we have v(^i) - Therefore, a(/z) < v(I[) <
v(L) + s, which proves the claim. A parallel result is that given I and e > 0,
there exist {//} such that / с (J f and £ r(/j) < v(I) + 8.
Let E be any subset of R”. Given e, choose {4}^=i such that E cz |J Ik
and £ u(/fc) < |£|e + For each k, choose such that Ik cz 7^,/
and Yj v(Jk,i) К4) + s2-k“1. Thus,
+ l^le + 8.
k,l к
Since E cz we obtain \E\'e < |^|в 4- s. Hence, \E\’e < |E|e, and by
symmetry, |£|; = |£V
For a related result, see Exercise 22.
2. Lebesgue Measurable Sets
A subset E of R” is said to be Lebesgue measurable, or simply measurable,
if given 8 > 0, there exists an open set G such that
E cz G and |G — E\e < 8,
If E is measurable, its outer measure is called its Lebesgue measure, or simply
its measure, and denoted |£|:
> (3.11) |E| = |£|e, for measurable E.
The condition that E be measurable should not be confused with the
conclusion of (3.6), which states that there exists an open G containing E
such that |(7|e < |^|е + e- Fi general, since G = E u (G — E) when E cz Gf
we only have |G|e < |E|e 4- \G — E\e, and we cannot conclude from |G|e <
|E|e + 8 that |G - E\e < 8.
We now list two simple examples of measurable sets. A nonmeasurable
set will be constructed in (3.38).
Example 1. Every open set is measurable.
This is immediate from the definition.
Example 2. Every set of outer measure zero is measurable.
For suppose that |E|e = 0. Then given e > 0, there is by (3.6) an open G
containing E with |G| < 4- 8 — 8. Hence,
\G — E\e < M <e.
(3.12) Theorem The union E = (J Ek of a countable number of measurable
sets is measurable, and
< X 1^1.
38 Lebesgue Measure and Outer Measure . (3.13)
Proof. Let 8 > 0. For each к = 1,2,..., choose an open set Gk such
that Ek с Gk and \Gk — Ek\e < e2~k. Then G = (J Gk is open and E c G.
Moreover, since G — E c (J (Gk — Ek), we have
|G - E\e < |(J (Gk - £t)|e < £ |Gk - Ek\e < e.
This proves that E is measurable. The fact that |E| < £ \Ek\ follows from
(3.4).
(3.13) Corollary An interval I is measurable, and \I\ — v(F).
Proof I is the union of its interior and its boundary. Since its boundary
has measure zero, the fact that I is measurable follows from (3.12) and the
results of Examples 1 and 2. Theorem (3.2) shows that 11| = v(I).
(3.14) Theorem Every closed set is measurable.
In order to prove this, we will use (1.11) and the next two lemmas.
(3.15) Lemma If {Ik}k=i ™ a finite collection of nonoverlapping intervals,
then (J Ik is measurable and ||J Ik\ = £ |/k|.
Proof The fact that (J Ik is measurable follows from (3.13). The rest of
the lemma is a minor extension of (3.2), and its proof is left as an exercise.
We recall from Chapter 1 that the distance between two sets Ex and E2 is
defined as d(Ei,Ef) — inf {|xj — x2| : xt e Et, x2 e E2}.
(3.16) Lemma If d(Er,E2) > 0, then |Ei u E2\e = |Et |e + \E2\e.
Proof By (3.4), lEi и E2\e < |EJe + |E2|e. To prove the opposite in-
equality, suppose 8 > 0, and choose intervals {Ik} such that Ег и E2 c |J Ik
and £ |7J < |Et u E2|e + 8. We may assume that the diameter of each Ik is
less than J(E1?E2). (Otherwise, we divide each Ik into a finite number of sub-
intervals with this property.) Hence, {Ik} splits into two subsequences {Ik}
and {/£}, the first of which covers Et and the second, E2. Clearly,
|£lle + l^le X 141 + X 14'1 = X 141 < 1^1
Therefore, |EX|e + |E2|e < |Ej и E2|e, which completes the proof.
A special case of this result will be used in the proof of (3.14). If Et and
E2 are compact and disjoint, then d(EXrEf) > 0 by Exercise 1(1), p. 13;
therefore, |Et u E2\e = |EJe H- |E2|e if Et and E2 are compact and disjoint.
Proof of theorem (3.14). Suppose first that F is compact. Given 8 > 0,
choose an open set G such that F c G and'|G| < |F|e + &. Since G — F is
open, theorem (1.11) implies there are nonoverlapping closed intervals Ik,
к = 1,2,..., such that G — F = |J Ik. Therefore, |G — F|e < У |/J, and
(3.19)
Lebesgue Measurable Sets
39
it suffices to show that £ \Ik\ < e. We have G = Fu ((Jfc°=
(IJfcL t Ik) for every positive integer N. Therefore,
|G| >
= |F|e + (J Ik
k= 1
by (3.16), F and (Jk=1 Ik being disjoint and compact. Since ||J*=i 4L =
141 by (3.15), we obtain ££=1 \Ik\ < |G| — |F|e < s for every N, so that
Efc°=i 141 e> as dfesired. This proves the result in the case where F is
compact.
To complete the proof, let F be any closed subset of R” and write F =
|J Fk, where Fk = F n {x : |x| < к], к = 1,2,.... Each Fk is compact and,
therefore, measurable. Hence, Fis measurable by (3.12).
(3.17) Theorem The complement of a measurable set is measurable.
Proof. Let E be measurable. For each positive integer /<, choose an open
set Gk such that E c Gk and |Gk — E\e < 1/k. Since CGk is closed, it is
measurable by (3.14). Let H = (Jfc CGk. Then His measurable and H c CH.
Write CH = H и Z, where Z = CH - H. Then Z c CH - CGk = Gk - H,
and therefore, \Z\e < 1/k for every k. Hence, |Z|e = 0 and, in particular,
Z is measurable. Thus, CH is measurable since it is the union of two mea-
surable sets.
The following two theorems are corollaries of (3.12) and (3.17).
(3.18) Theorem The intersection E — (~}kEk of a countable number of
measurable sets is measurable.
Proof. Since Ek is measurable, CEk is measurable by (3.17). However,
CH = C(QkHk) = CEk, and hence, CH is measurable. Therefore, by
another application of (3.17), E is measurable.
(3.19) Theorem If Er and E2 are measurable, then Ex — E2 is measurable.
Proof. Since Ex — E2 — Ex n CH2, the result follows from (3.17) and
(3.18).
As a consequence of (3.12) and (3.17) and their corollaries (3.18) and
(3.19), it follows that the class of measurable subsets of R” is closed under
the set-theoretic operations of taking complements, countable unions, and
countable intersections. Such a class of sets is called a a-algebra; that is,
a nonempty collection E of sets E is called ъв-algebra of sets if it satisfies the
following two conditions:
(i) CHeSif HeS.
(ii) U*-Ek g S if Ek e E, к - 1,2,....
40 Lebesgue Measure and Outer Measur. (3.20)
Note that it follows from the relation C(Qk Ek) = (J* CEk that Ek e Z
if Z is a ст-algebra and Ek e Z, к == 1,2,.... Moreover, it is easy to see that
the entire space and the empty set belong to any ст-algebra.
The following result is just a reformulation of (3.12) and (3.17).
(3.20) Theorem The collection of measurable subsets of R" is a о-algebra.
Note, for example, that if Ek, к = 1, 2, . . . , are measurable, then so are
limsup Ek and liminf Ek since
limsup Ek = P| |J Ek and liminf TT* = |J P Ek.
J=1 k = j j=l k~j
If and 4>2 are two collections of sets, we say that is contained in ^2
if every set in is also in ^2. If is a family of ст-algebras Z, we define
Z to be the collection of all sets E which belong to every Z in J5'. It is
easy to check that Z is itself a ст-algebra, and is contained in every Z
in
Given a collection of subsets of R", consider the family EF of all ст-
algebras which contain and let F — Pse^ Z. Then F is the smallest a-
algebra containing ; that is, S is a сг-algebra containing and if Z is any
other сг-algebra containing then Z contains <?.
The smallest ст-algebra of subsets of R" containing all the open subsets of
R" is called the Borel a-algebra of R”, and the sets in & are called Borel
subsets ofW. Sets of type Fa, Gd, FffS, Gda (see p. 6), etc., are Borel sets.
(3.21) Theorem Every Borel set is measurable.
Proof Let Л be the collection of measurable subsets of R". By (3.20),
Л is a ст-algebra. Since every open set belongs to Л, and $ is the smallest
ст-algebra containing the open sets, is contained in Л.
3. Two Properties of Lebesgue Measure
The next two theorems give properties of Lebesgue measure which are of
fundamental importance. To prove them, we first need the following char-
acterization of measurability in terms of closed sets.
(3.22) Lemma A set E in R” is measurable if and only if given e > 0, there
exists a closed set F с E such that \E — F\e < г.
Proof E is measurable if and only if CE is measurable, that is, if and
only if given e > 0, there exists an open G such that СЕ ci G and \G — CE\e
< a. Such G exists if and only if the set F = CG is closed, F с E, and
I# — F|e < e (since G — СЕ = E — F).
(3.26) Two Properties of Lebesgue IVteasure 41
(3.23) Theorem If {Ek} is a countable collection of disjoint measurable sets,
then
IIM1 =S|E*|.
к k
Proof First, suppose each Ek is bounded. Given e > 0 and к = 1,2,...,
use (3.22) to choose a closed Fk <= Ek with )Ек — Fkl < s2~k. Then <
IFjJ + s2~k by (3.12). Since the Ek are bounded and disjoint, the Fk are com-
pact and disjoint. Therefore, by (3.16), |(J*=1 Fk\ = |Ffc| for every m.
The fact that (J?=1 Fk <= (J”= i then implies X*=i |Ft| < £fc|.
Hence,
0 £fc > £ |Ft| > £ (|£k| - £2-k) = £ |£J - s,
k=l k=l k=l k = l
so that |IJr=i Ek\ St°=i Since the opposite inequality is always true,
the theorem follows in this case.
For the general case, let Ij, j = 1, 2,. . , be a sequence of intervals in-
creasing to R”, and define S\ — Ц and Sj ~ Ij — for j > 2. Then the
sets Ekj — Ekr\ Sj, k,j = 1,2, ..., are bounded, disjoint, and measurable;
Ek = (J; Ektj and (Jfc Ek = |Jk,j EkJ. Therefore, by the case already estab-
lished, we have
|U Ek\ = |(J EkJ\ = X = X (X |£4I) = 11^1-
к k,j k,j к f к
(3.24) Corollary If {Ik} is a sequence of nonoverlapping intervals, then
lU 41 = Z141-
Proof. It is clear that |(J 41 X 141- On the other hand, the Ik being
disjoint, we have ||Ji4l lU Al = X I Al = X 141- Thus, |(J /J = £ 141-
(3.25) Corollary Suppose Ex and Ег are measurable, F2 cz £1? and |F2| <
+ oo. Then\Er - £2| - |FJ - |F2|.
Proof. Since Ex = F2 u (Ft - F2), - |F2| + - F2| by (3.23).
Since |F2| < +oo, the corollary follows.
The second basic property of Lebesgue measure concerns its behavior for
monotone sequences of sets.
(3.26) Theorem Let {Ek}fL r be a sequence of measurable sets.
(i) If Ек/ E, then lim^^ |£fc| = |£|.
(ii) If Ek\ E and |EJ < +oo for some k, then lim^^ |Et| = |£|.
Proof (i) We may assume that \Ek\ < +oo for all k; otherwise both
lim^^oo \Ek\ and |F| are infinite. Write
E = E, и (E2 - £,) и • • и (Ek - Ek^) и • • •.
•о
42 Lebesgue Measure and Outer Measu (3.27)
Since the terms in this union are measurable and disjoint, we have by (3.23)
that
|E| = |E,| + |E2 - EJ + ••• + + ....
By (3.25),
|E| = |EJ + (|E2| - |EJ) + • • • + (|Et| — lE^I) + • • • = lim |Efc|.
00
(ii) We may clearly assume that [| < 4- oo. Write
E2 = Eu (E2 - E2) и • • • и (Et - Et+i) и • • •.
Since the terms on the right are disjoint measurable sets, and since each Ek
has finite measure, we have
, |£il = |E| + (|Ei| - |E2|) + • • • + (|EJ - |Ek+ J) + • • •
= |E| + |Ej| - lim |EJ.
fc-> 00
Therefore, l^l = lim^^ |2?k|, which completes the proof.
The restriction in (ii) that |Ek\ < + oo for some к is necessary, as the fol-
lowing example shows. Let Ek be the complement of the ball with center 0
and radius k. Then \Ek\ = + oo for all к and Ek \ 0, the empty set. There-
fore, lim^^ \Ek\ = Too, while |0| = 0.
Although we are interested almost exclusively in measurable sets, proving the
measurability of a given set is occasionally difficult in practice, and it may be
desirable to apply theorems about outer measure. A particularly useful result is
the following modification of part (i) of (3.26). (See also Exercise 21.)
(3.27) Theorem If Ek / E, then lim^^ = |E|e.
Proof. For each k, let Hk be a measurable set (e.g., a set of type G6} such that
Ek c Hk and |/ffc| = lEfcle. For m = 1,2, . . ., let Vm = #*• Since the Vm
are measurable and increase to V = (J Vmt it follows from (3.26) that limm_<30 | Vm\
= | F|. Since Em c Vm c Hm, we have < \ Vm\ < \Hm\ = |£m|e. Hence,
\Vm\ = lim^oo |Em|e exists, and lim^^ |Ем|е = |F|. However, V = (J Vm
э (J Em = E, and therefore, lim^^ iFje > |£je. The opposite inequality is
obvious since Em с E, and the theorem follows.
4. Characterizations of Measurability
Lemma (3.22) was a characterization of measurability in terms of closed
subsets of a set. The next three theorems give some other characterizations.
The first one states that the most general/measurable set differs from a Borel
set by a set of measure zero.
(3.28) Theorem
(i) E is measurable if and only if E — H — Z, where H is of type Gd
and [Z| = 0.
(3.30)
Characterizations от меаьшашту
(ii) Е is measurable if and only if E = H Z, where H is of type Fa
and \Z\ = 0.
Proof If E has the representation expressed in either (i) or (ii), it is
measurable since H and Z are.
Conversely, to prove the necessity in (i), suppose that E is measurable.
For each к = 1,2,..., choose an open set Gk such that E cz Gk and |Gfc — E\
< l/к. Set H - Gk. Then His of type Gd, E cz Я, and H - E cz Gk - E
for every k. Hence, \H — E\ = 0, and (i) is proved.
The necessity of (ii) follows from that of (i) by taking complements:
if E is measurable, so is CZ, and therefore CE = Q Gk — Z, where the Gk
are open and |Z| = 0. Hence, E = ((J CGJ u Z, which completes the proof.
(3.29) Theorem Suppose that |Z|e < + oo. Then E is measurable if and only
If given e > 0, E = (S и — N2, where S is a finite union of non-
overlapping intervals and |^|e, |Я2|е < £-
The proof is left as an exercise.
Our final characterization of measurability states that the measurable
sets are those which split every set (measurable or not) into pieces that are
additive with respect to outer measure. This characterization will be used in
Chapter 11 to construct measures in abstract spaces.
(3.30) Theorem (Caratheodory) E is measurable if and only if for every set A
\A\e = \A n E\e + \A - E\e.
Proof Suppose that E is measurable. Given J, let Я be a set of type G5
such that A cz H and |Л|е = |Я|. 'Since Я = (Hn £)u (Я - Z), and
since Я n Z and H—E are measurable and disjoint, \H\ — |Яп Z| +
\H - Z|. Therefore, \A\e ~ \H r\ E\-I- \H — E\ > \A n E\e + \A - E\e,
Since the opposite inequality is clearly true, we obtain |Л|в = \A n E\e +
\Л - E\e.
Conversely, suppose that Z satisfies the stated condition for every A.
In case |Z|e < Too, choose a Gd set Я such that Z cz Я and |Я| = |Z|e.
Then H = Ev (H - Б), and by hypothesis, \H\ = |Z|e + |Я — Z|e. There-
fore, \H ~ E\e = 0; so the set Z = Я — Z is measurable, and consequently,
Z is measurable.
In case |Z|e = Too, let Bk be the ball with center 0 and radius k,k —
1,2,..., and let Ek = Z n Bk. Then each Ek has finite outer measure and
E = (J Ek. Let Hk be a set of type Gd containing Ek with \Hk\ = \Ek\e. By
hypothesis, \Hk\ = \Hk n E\e + \Hk - E\e > \Ek\e + \Hk - E\e. Therefore,
\Hk — E\ — 0. Let Я ~ |J Hk. Then H is measurable, Z cz Я, and H — E =
|J (Hk — Z). In particular, H—E has measure zero, and since Z = Я —
(Я Z), Z is measurable. This completes the proof.
44 Leoesgue Measure and Outer Measui (3.31)
As a special case of Caratheodory’s theorem, we obtain the following
result.
(3.31) Corollary If E is a measurable subset of a set A, then |Л|е = |Я| 4-
И - E\e. Hence, if\E\ < + oo, \A - E\e = \A\e - |F|.
We can now prove a stronger version of (3.8).
(3.32) Theorem Let E be a subset of R". There exists a set H of type G6 such that
E с H and for any measurable set M, \E л M\e = \H n M\.
If M = R", this reduces to (3.8).
Proof, Consider first the case when |2Г|е < 4-oo. Let IT be a set of type G6 such
that E с Я and |£|e = |Я|. If M is measurable, then by Carathdodory’s theorem
|F|e = |Ел M|e + |F - M\e and |Я| = \Hc\M\ + \H - M\. Therefore,
|ЯлМ|е + \E — M\e = \HnM\ 4- \H-M\.
Since all these terms are finite, and since the inclusion E — M с H — M implies
that |E — M\e < \H — M\, we have |FnM|e > |Я л A7|. The opposite inequality
is also true since E л M с H n M, and the theorem follows in this case.
In case |E|e = 4-oo, we write E = (J Ek with |Д|е < +oo and Ek f E. For
example, Ek could be the intersection of E with the ball of radius к centered at
the origin. By the case already considered, there is a set Uk of type G6 such that
Ek c Uk and n M\e = \Uk n M\ for any measurable M. Let Hk = Um.
Then Hk is measurable, Hk / H = (J Hk, and Ek c Hk a Uk. Hence, Ekc\M
с Hkc\ M a Ukri M, and therefore, \Ek n M\e = \Hk л M\ for measurable M.
Since Ek / E and Hk / H, we have ЕкслМ / E c\ M and Hkc\ M / H n M.
By (3.27), IF n M\e = |Яп M\ for measurable M.
Note that our set Я is of type G6a. To obtain a set of type G6, use (3.28) to write
Я = Hi — Z, where Hr is of type G6 and \Z\ ~ 0. Then E c Hi. Moreover,
Hi r\ M = (H n M) u(Z n M), and since |Z| = 0, we have |Я1 л M\ =
[Ял M\, so that \E л M\e — |Я1 л M\. This completes the proof.
5. Lipschitz Transformations of R"
Theorem (3.10) showed that the notion of outer measure is independent of
the orientation of the coordinate axes. Since measurability and measure are
defined in terms of outer measure; it follows that these too are independent
of rotation of the axes. We wish to study the effect of other transformations
of R" on the class of measurable sets; that is, we seek a condition on a trans-
formation T: R" -» R" such that the image ТЕ = {у : у = Гх, х е Е} of
every measurable set E is measurable. We note that a continuous transforma-
tion may not preserve measurability: see Exercise 17.
A transformation у = Tx of Rn into itself is called a Lipschitz transforma-
tion if there is a constant c such that
|Tx — 7x'| < c|x — x'|.
(3.36)
Lipschitz Transformations of FL
45
If у. = fj(x), j = 1, ... ,n, are the coordinate functions representing T
(see p. 11), it follows that T is Lipschitz if and only if each fj satisfies a
Lipschitz condition |/j(x) — //x')| < cjx — x'|. For example, a linear
transformation of R" is clearly a Lipschitz transformation. More generally,
a mapping j = 1,.. ., n, for which each has bounded first
partial derivatives in Rn is a Lipschitz mapping.
(3.33) Theorem If у = Tx is a Lipschitz transformation ofW, then T maps
measurable sets into measurable sets.
Proof We will first show that a continuous transformation sends sets of
type Fff into sets of type Fa. A continuous T maps compact sets into compact
sets by (1.17); therefore, since any closed set can be written as a countable
union of compact sets, T maps closed sets into sets of type Fa. [Here, we have
used the relation T(Q £k) = (J TEk, which holds for any T and {£к}.] It
follows that T preserves the class of Fa sets.
We will next show that a Lipschitz transformation T sends sets of measure
zero into sets of measure zero. Since |Tx — Tx'| < c|x — x'|, the image of a
set with diameter of d has diameter at most cd. It follows by covering with
cubes that there is a constant c'sqch that |T/| < c'|/| for any interval/. (TI
is measurable since I is closed.) If |Z| = 0 and s > 0, choose intervals {Ik}
^covering Z such that£ \Ik | <£. £ince TZcz \JTIk, we have |TZ|e |Т/Л|
<с'%\1к\<с'1. Hence, \TZ\ = 0.
If E is a measurable set, we use (3.28) to write E = H u Z, where H is of
type Fa and |Z| = 0. Since ТЕ = TH и TZ, ТЕ is measurable as the union
of measurable sets. This completes the proof. ,
For an extension of (3.33) when n = 1, see Exercise 10 of Chapter 7.
In the special case that T is linear, we will derive a formula for the measure
of ТЕ. It may be helpful to think of T as an n x n matrix and of Tx as the
vector resulting from the matrix action of T on x. If P is a parallelepiped
(see p. 7), arguments like those used in proving (3.2) and (3.10) show that
(3.34) |P | = u(P)
(see Exercise 16). Hence, by p. 7, |P| is the absolute value of the n x n
determinant whose rows are the edges of P.
(3.35) Theorem Let T be a linear transformation of R", and let E be mea-
surable. Then \TE\ = <5|E|, where b is the absolute value of the deter-
minant ofT.
Proof Let b — |det T|. Therj for any interval 7, \TI\ = <5|Z| by p. 7.
If E is any subset of 1Г and c > 0, choose intervals {Ik} covering E with
E |/J < |£’|e + s. Then |ТЕje < £ = <5 g 141 < <5(K + e)-Therefore,
(3.36) |T£|e < <5|£|e.
46
Lebesgue Measure and Outer Measure
(3.37)
To see that \TE\ = 5|E| for measurable E, choose an open set G containing
E with — E| < s. Using (1.11), write G as a union of nonoverlapping
intervals {Ik}. By (3.36), we may assume <5 > 0. Then since the TIk are non-
overlapping parallelepipeds, |TG| = E 1^41 = <5 E 141 ^1^1- Now, since
E cz G, we have <5|E| < <5|G| = |ZG|. Also, '
\TG\ < \TE\ + \T(G - E)\ < \TE\ + <5e,
by (3.36). Combining inequalities, we get <5|E| < |ZE| + <5e, so that <5|£j <
\TE\. Hence, by (3.36), (5|E| = \TE\.
6. A Nonmeasurable Set *
We will now construct a nonmeasurable subset of R1; the construction in R",
n > 1, is similar. We will need the axiom of choice in the following form.
Zermelo's Axiom: Consider a family of arbitrary nonempty disjoint sets
indexed by a set A, {Ex : a e A}. Then there exists a set consisting of exactly
one element from each Ea, a g A.
We also need the following lemma.
(3.37) Lemma Let E be a measurable subset ofR1 with |E| > 0. Then the
set of differences {d : d ~ x — y, x g E, у e E} contains an interval
centered at the origin.
Proof Given e > 0, there exists an open set G such that E c G and
|G| < (1 + 8)|£j. By (1.11), G can be written as a union of nonoverlapping
intervals, G ~ JJ Ik. Letting Ek = E n Ik, it follows that E — \J Ek, that the
Ek are measurable, and that two different Eks have at most_one point in
common. Therefore, |G| = E 141 and |E| = E Since |G| < (1 + 8)|ZT|,
we must have |40l < (1 + e)|Efco| for some k0. Choosing 8 = | and letting I
and £ denote the sets 40 and Efco, respectively, we have c / and |<f | > ||Z|.
We claim that if $ is translated by any number d satisfying \d\ < ||Z|, the
translated set has points in common with S. Otherwise, since $ u $d is
contained in an interval of length |Z| 4- \d\, we would have |<f| + |^d| <
|Z| + |rf|, or 2|(^| < |Z| + |J|. [Here, we have used the fact that |EJ = |E|
(see Exercise 18).] However, the last inequality is false if \d\ < ||Z| since
|^| > ||Z|. This proves the claim, and thus the lemma.
(3.38) Theorem ( Vitali) There exist nonmeasurable sets.
Proof We define an equivalence relation on the real line by saying x and
у are equivalent if x — у is rational. The equivalence classes then have the
form Ex = {x + r : r is rational}. Two classes Ex and Ey are either identical
or disjoint; therefore, one equivalence class consists of all the rational
numbers, and the other distinct classes are disjoint sets of irrational numbers.
(3.39)
A Nonmeasurable Set
The number of distinct equivalence classes is uncountable since each class is
countable but the union of all the classes is uncountable (this union being
the entire line).
Using Zermelo’s axiom, let E be a set consisting of exactly one element
from each distinct equivalence class. Since any two points of E must differ
by an irrational number, the set {d : d — x — у, x e E, у e E} cannot con-
tain an interval. By (3.37), it follows that either E is not measurable or |£| =
0. Since the union of the translates of E by every rational number is all of
R1, R1 would have measure zero if E did. We conclude that E is not mea-
surable.
(3.39) Corollary Any set in R1 with positive outer measure contains a non-
measurable set.
Proof. Let A satisfy |Л|е > 0, and let £ be the nonmeasurable set of (3.38).
For rational r, let Er denote the translate of E by r. Then the Er are disjoint
and (J Er = (-oo,+ 00). Thus, A = (J (A n Er) and |Л|е < £ \A n Er\e.
If A n Er is measurable, then ]A n £,| = 0 by lemma (3.37). Since
|Л|е > 0, it follows that there is some r such that A n Er is not measurable.
This completes the proof.
Exercises
1. There is an analogue for bases different from 10 of the usual decimal expansion
of a number. If b is an integer larger than 1 and 0 < x < 1, show that there
exist integral coefficients ck, 0 < ck < b, such that x = ckb~k. Show
that this expansion is unique unless x = cb~k, in which case there are two
expansions.
2. When b — 3 in Exercise 1, the expansion is called the triadic or ternary expansion
of x. Show that the Cantor set C consists of all x such that % has some triadic
expansion for which every ck is either 0 or 2.
Let f(x) be the Cantor-Lebesgue function: see p. 35. Show that if x E C
and x ~ S ck3~k, where each ck is either 0 or 2, then fix') ~ S {Y2Ck)2~k.
3. Construct a two-dimensional Cantor set in the unit square {(x,y) : 0 < x,y < 1}
as follows. Subdivide the square into nine equal parts and keep only the four
closed corner squares, removing the remaining region (which forms a cross).
Then repeat this process in a suitably scaled version for the remaining squares,
ad infinitum. Show that the resulting set is perfect, has plane measure zero,
and equals С x C.
4. Construct a subset of [0,1] in the same manner as the Cantor set by removing
from each remaining interval a subinterval of relative length 0, 0 < & < 1
48
Lebesgue Measure and Outer Measin^
Show that the resulting set is perfect and has measure zero.
5. Construct a subset of [0,1] in the same manner as the Cantor set, except that
at the £th stage, each interval removed has length <53 “k, 0 < 3 < 1. Show that
the resulting set is perfect, has measure 1 — d, and contains no intervals.
6. Construct a Cantor-type subset of [0,1] by removing from each interval re-
maining at the /cth stage a subinterval of relative length 0k, 0 < вк < 1. Show
that the remainder has measure zero if and only if S = + °<х (Use the
fact that for ak > 0, П*°=1 ak converges, in the sense that lim^oo U?si ak exists
and is not zero, if and only if logak converges.)
7. Prove (3.15).
8. Show that the Borel сг-algebra in R" is the smallest сг-algebra containing
the closed sets in Rn.
9. If {Ek}k^r is a sequence of sets with £ |2Г*|е < +oo, show that limsupE^
(and so also liminf Ek) has measure zero.
10. If E^ and E2 are measurable, show that \Et uE2| + |£i n E*2| = l^il + lE^I.
11. Prove (3.29).
12. If Et and E2 are measurable subsets of R1, show that Ex x E2 is a measurable
subset of R2 and [Ei x E2| = \Et||2T21. (Interpret 0-oo as 0.) (Hint: Use a
characterization of measurability.)
13. Motivated by (3.7), define the inner measure of E by = sup |F|, where the
supremum is taken over all closed subsets F of E. Show that (i) < |£je,
and (ii) if |E|e < -(-oo, then E is measurable if and only if |£jf = |£je. [Use
(3.22).]
14. Show that the conclusion of part (ii) of Exercise 13 is false if |E|e = -(-oo.
15. If E is measurable and A is any subset of Ey show that [F| = + \E — J|e.
(See Exercise 13 for the definition of |Л||.)
16. Prove (3.34).
17. Give an example which shows that the image of a measurable set under a
continuous transformation may not be measurable. (Consider the Cantor-
Lebesgue function and the pre-image of an appropriate nonmeasurable subset
of its range.)
18. Prove that outer measure is translation invariant} that is, if Eh = {x -J- h: x g E}
is the translate of E by h, h g R", show that |Eh|e = |£je. If E is measurable,
show that Еъ is also measurable. [This fact was used in proving (3.37).]
19. Carry out the details of the construction of a nonmeasurable subset of R",
n > 1.
20. Show that there exist disjoint Ely E2y. .., Eky... such that |(J Ek\e < 2 |Ejt|e
with strict inequality. (Let E be a nonmeasurable subset of [0,1] whose rational
translates are disjoint. Consider the translates of E by all rational numbers
r, 0 < r < 1, and use Exercise 18.)
21. Show that there exist sets Ely E2l . . ., Ek, . . . such that Ek \ E, < -(-oo,
and Hindoo |Efc|e > |E|e with strict inequality.
Exercises
4Э
22. Show that the outer measure of a set is unchanged if in the definition of outer
measure we use coverings of the set by parallelepipeds with a fixed orientation
(i.e., with edges parallel to a fixed set of n linearly independent vectors), rather
than coverings by intervals.
23. Let Z be a subset of R1 with measure zero. Show that the set {x2: x g Z} also
has measure zero.
24. Let O.aia2 • • • be the dyadic development of any x in [0,1], that is, x = ai2“x
+ a22“2 + • • •, at — 0,1. Let k19 k2,... be a fixed permutation of the positive
integers 1, 2,. .. , and consider the transformation T which sends x ==
O.«i«2 * ’ ’ to Tx = О.аЛ1ак2 • • •. If E is a measurable subset of [0,1], show that
its image ТЕ is also measurable and that \TE\ = |Z?|. [Consider first the special
cases of E a dyadic interval [s2~fc, (s + 1)2“*], s = 0, 1,. . ., 2* — 1, and then
of E an open set (which is a countable union of nonoverlapping dyadic inter-
vals). Also show that if E has small measure, then so has ТЕ.]
25. Construct a measurable subset E of [0,1] such that for every subinterval 7,
both Ec\l and I — E have positive measure. [Take a Cantor-type subset of
[0,1] with positive measure (see Exercise 5), and on each subinterval of the
complement of this set, construct another such set, and so on. The measures
can be arranged so that the union of all the sets has the desired property.]
26. Construct a continuous function on [0,1] which is not of bounded variation on
any subinterval. [The construction follows the pattern of the Cantor-Lebesgue
function with some modifications. At the first stage, for example, make the
corresponding function increase to 2/3 (rather than 1/2) in (0,1/3), then make
it decrease by 1/3 in (1/3,2/3), and then increase again 2/3 in (2/3,1). The
construction at other stages, is similar, depending on whether the preceding
function was increasing or decreasing in the subinterval under consideration.]
27. Construct a continuous function of bounded variation on [0,1] which is not
monotone in any subinterval. (The construction is like that in the preceding
exercise, except that the approximating functions are less steep. For example,
at the first st age,: let the function increase to 1/2 + e, then decrease by 2e,
and then increase again by 1/2 + e. Choose the c’s at each stage so that their
sum converges.)
Chapter 4
Lebesgue Measurable Functions
We will use the concept of Lebesgue measure to introduce a rich class of
functions and a method of integrating these functions. In this chapter,
we describe the class of functions.
Let f be a real-valued function defined on a set E in R", that is, — oo <
/(x) < + oo, x e E. Then f is called a Lebesgue measurable function on E,
or simply a measurable function, if for every finite a, the set
{X e E :/(x) > a}
is a measurable subset of R". In what follows, we shall often use the abbrevia-
tion {/ > a} for {x e E :/(x) > a}. Since
(00 \
U {/> -к} ,
к=1 )
the measurability of E is equivalent to that of {/ = — 00} if we assume that
/is measurable. For simplicity, we shall always assume that {/ = —00} is
measurable, and so consider only measurable functions defined on measurable
sets.
As a varies, the behavior of the sets {/ > a} describes how the values of
/are distributed. Intuitively, it is clear that the smoother/is, the smaller the
variety of such sets will be. For example, if E = R" and/is continuous in R",
then {/> a} is always open. A function/defined on a Borel set E is said to
be Borel measurable if {/ > a} is a Borel set for every a. Thus, every Borel
measurable function is measurable. —'
One further comment will be helpful later. Let Ж denote the class of
measurable subsets of R". Much of the development of measurable functions
given below depends only on the и-algebra structure of JI and the properties
of Lebesgue measure. Thus, a measurable function inherits its elementary
properties from those of measurable sets. It is reasonable to expect, therefore,
that many of the methods of this chapter can be used to develop results in
50
(4Л)
th Hilary rropeirxies Ol runvuvn»
more general settings, for spaces other than R” and сг-algebras other than Л.
This will be done in Chapter 10. To save too much repetition there, it will be
helpful if the reader notices which properties of Л and Lebesgue measure are
used in the proofs here. These will be discussed at the end of the chapter.
1. Elementary Properties of Measurable Functions
(4.1) Theorem Let f be a real-valued function defined on a (measurable) set
E. Then f is measurable if and only if any of the following statements
holds for every finite a:
(i) {f > a} is measurable,
(ii) {/< a} is measurable,
(iii) {f < a} is measurable.
Proof Since {/>«} = ПГ=1 {f > a ~~ V&}> the measurability of f
implies (i). Since {/< a} is the complement of {f > a}, it follows that (i)
implies (ii). Since {/ < a} = Q£°=1 {f < a + \jk}, we see that (ii) implies
(iii). Finally, since {f > a} is the complement of {f < a}, it follows that/is
measurable if (iii) holds.
The proof of the following is left as an exercise.
(4.2) Corollary If f is measurable, then {f >~ °°},{/< L°°}, {f =
{<2 </</?},{ / — a}, etc., are all measurable. Moreover, f is measur-
able if and only if {a < f < is measurable for every finite a.
Also, observe that {a < f < b} = {/ > a} - {/ > b}.
Let /be defined in E. If S is a subset of R1, the inverse image of S' under
/ is defined by
Г1(5) = {хб£:/(х)е5}.
(4.3) Theorem / is measurable if and only if for every open set G in R1, the
inverse imagef~l(G) is a measurable subset ofRn,
Proof If G = (a,+°°)> then f'\(G) ={a <f < 4-oo}. Hence, if /’’ (G) is
measurable for every such G,f must be measurable. To prove the converse,
suppose that /is measurable and let G be any open subset of R1. By (1.10),
G can be written G ~ (J* (a*A)- equals {ak </ < bk}, and is
therefore measurable. Since f~i(G) = (Jit/ ^AA))» it follows that f~x(G)
is measurable too.
We observe that the proof above also shows that /is Borel measurable if
and only iff~1(G) is Borel measurable for every open G c: R1.
(4.4) Theorem Let A be a dense subset of R1. Then f is measurable if
{f > a} is measurable for all a 6 A,
Proof. Given any real a, choose a sequence {ak} in A which converges to
Б2 Lebesgue Measurable Functiouo (4.5)
a from above: ak e A, ak> a, lim^^ ak = a. Then {/> a} = IJ {/ > ak},
and the theorem follows.
A property is said to hold almost everywhere in E or, in abbreviated form,
a.e., if it holds in E except in some subset of E with measure zero. For
example, the statement “/ = 0 a.e. in E” means that/(x) = 0 in E, with the
possible exception of those x in some subset Z of E with |Z| =0.
' The next few theorems give some simple properties of the class of mea-
surable functions.
(4.5) Theorem If f is measurable and if g = f a.e., then g is measurable and
l{<7 >«}l =
Proof. If Z — {f / g}, then {g > a} и Z = {/ > a} и Z. Therefore,
f being measurable, {g > a} и Z is measurable, and since this differs from
{g > a} by a set of measure zero, g is measurable. Finally,
\{9 > *}l = \{g >a}uZ\ = \{f> a}uZ\ = |{/> a}|.
In view of the last theorem, it is natural to extend the definition of mea-
surability to include functions-which are defined only a.e. in E, by saying
that such an/is measurable on E if it is measurable on the subset of E where
it is defined. Note also that if/is measurable on E, then it is measurable on
any measurable Et c= E since {x e Er:/(x) > a} = {x e E:/(x) > a} n Ex.
If ф and / are real-valued measurable functions defined on R1 and Rn,
respectively,1 their composition <£(/(x)) may not be measurable (see Exercise
5). If ф is continuous, however, we have the following result.
(4.6) Theorem Let ф be continuous on R1 and let fbe finite a.e. in E, so that,
in particular, </>(/) is defined a.e. in E. Then ф(/) is measurable if f is.
Proof. We may assume that / is finite everywhere in E. We will use the
fact that since ф is continuous, the inverse image ф~1(б) of an open set G is
open. [The reader who is not already familiar with this fact can find related
material in both (4.15) and Exercise 10.] By (4.3), it is enough to show that
for every open G in R1, {x : /(/(x)) g G} is measurable. However, {x: ф(/(х))
=/“1(c/>"1(G)), and since ф~1(Сг) is open and / is measurable,
is measurable by (4.3).
Remark: The cases which arise most frequently are </>(/) = |/|, |/P(p > 0),
ect, etc. Thus, V
IЛ \ЛР(Р > 0), ecf,
are measurable if/is (even if we do not assume that/is finite a.e., as is easily
seen). Another special case worth mentioning is that of
f+ = max {/,0}, f~ = -min {/,0}.
It is enough to observe that the functions x+ and x~ are continuous.
(4Л2) Е юпшу ---
(4,7) Theorem Iff and g are measurable, then {f > g} is measurable.
Proof. Let {rk} be the rational numbers. Then
{f> g} = U {/> rt > g} = U «/> rk} ^{g < nJ),
к к
and the theorem follows.
(4.8) Theorem Iffis measurable and Л is any real number, then f 4- 2 and
ff are measurable.
The proof is left as an exercise.
We note that the sum f + g of two functions f and g is well-defined
wherever it is not of the form 4-00 4- (—co) or —oo 4- oo. In the
next theorem, we assume for simplicity that f 4- g is well-defined everywhere.
(For extensions, see Exercise 6.)
(4.9) Theorem Iff and g are measurable, so is f + g.
Proof Since g is measurable, so is a — g for any real a, by (4.8). Since
(fig > a} = {f > a ~ g}> the result follows from (4.7).
A corollary of (4.8) and (4.9) is that a finite linear combination Xxfx 4-
• • • 4- INfx of measurable functions fx, . . . ,f^ is measurable. Thus, the class
of measurable functions on a set E forms a vector space.
In the theorem which follows we consider products of functions. In
addition to the familiar conventions about products of infinities, we adopt
the convention 0- ± oo = ± oo -0 = 0.
(4.10) Theorem If f and g are measurable, so is fg. If g Ф 0 a.e., fig is
measurable.
Proof By (4.6) and the remark following it, f2(= |/|2) is measurable if
/is. Hence, if/and g are measurable and finite, the formula fg = [(/ 4- g)2 —
(f — g)2]/^ implies that fg is measurable. The proof when / and g can be
infinite and the proof of the second statement of the theorem are left as
exercises.
(4.11) Theorem If {A(x)}fc°= i IS a sequence of measurable functions, thei
supfc/Xx) and infk/k(x) are measurable.
Proof. Since inffcA = — supfc (—A), it is enough to prove the result fo:
sup*/*. But this follows easily from the fact that {supft A > a} = IJ* (Л >
As a special case of the preceding theorem, we see that if A,. . . ,fN an
measurable, then so are maxfcA and minkA- In particular, if/is measurable
then so are /+ = max {/,0} and /4 ~ —min {/0}, a fact we have alread
observed [see the remark following (4.6).]
(4,12) Theorem If {A} *s K±a sequence of measurable functions, the
limsup^^fk and liminfk.^xfk are measurable. In particular, if
lim^nfi exists a.e., it is measurable.
Proof. Since
limsup/k = inf {sup fk}, liminffk — sup {inf/^J,
k->co j k>j fc->co J k>j
the first statement follows from (4.11). The second statement is then a corol-
lary of (4.5) since wherever lim^^/j. exists, it equals limsup^^/^
The characteristic function, or indicator function, /£(x), of a set E is defined
by
z 4 f 1 if x e E
Xe(x) - |o if X££-
Clearly, %E is measurable if and only if E is measurable. %E is an example of
what is called a simple function: a simple function is one which assumes only
a finite number of values, all of which are finite. If f is a simple function
taking (distinct) values al9. . . , aN on (disjoint) sets Er, . . . , EN, then
N
/(x) = £ ak%Ek(x).
k=l
We leave it as an exercise to show that such an f is measurable if and only
if El9 . . ., En are measurable.
(4.13) Theorem
(i) Every function f can be written as the limit of a sequence {f} of
simple functions.
(ii) If f > 0, the sequence can be chosen to increase to f, that is, chosen
such that fk < fk+1 for every k.
(iii) If the function f in either (i) or (ii) is measurable, then the fk can be
chosen to be measurable.
Proof. We will prove (ii) first. Thus, suppose that f > 0. For each k,
к = 1,2,..., subdivide the values of f which fall in [0,A:] by partitioning
[0Л] into subintervals [(y - l)2"k,y2"k],y = 1,..., k2k. Let
^2 if < ftx) <1, j = 1,.. ., k2k.
^(X)=J 2k 2k ’ 2k J
- к if /(x) > k.
Each/^ is a simple function defined everywhere in the domain off Clearly,
fk Л+i in passing from fk to fk+1, each subinterval [(y - l)2~k,j'2~k]
is divided in half. Moreover, fk~> f since 0 < f — fk < 2"k for sufficiently
large к wherever f is finite, and fk^k-+ Too wherever / = + co. This
proves (ii).
To prove (i), apply the result of (ii) to each of the nonnegative functions
f+ and /", obtaining increasing sequences {f'k} and {f^} of simple functions
such that/j ->/+ and/',; -*f~. Then Л - f"k is simple and/* - f’L ->/+
-Г =/
Finally, it is enough to prove (iii) for / > 0 since otherwise we may con-
sider f+ and /“.In this case, however,
к2к j - 1
fk = 2k +
This is measurable if/is since all the sets involved are measurable.
Note that if / is bounded, the simple functions above will converge
uniformly to /
2. Semicontinuous Functions
We now study classes of functions / whose continuity properties on a set E
can be characterized by the nature of {/ > a} or {/ < a}. Let/be defined on
E, and let x0 be a limit point of E which lies in E. Then / is said to be upper
semicontinuous at x0 if
limsup/(x) < /(x0).
x—>xo;xc£
We will usually abbreviate this by saying that f is use at x0. Note that if
/(x0) = + oo, then /is automatically use at x0; otherwise, the statement that
/ is use at x0 means that given M > /(x0), there exists <5 > 0 such that
/(x) < M for all x e E which lie in the ball |x — x0| < <5. Intuitively, this
means that near x0 the values of/do not exceed/(x0) by a fixed amount.
Similarly, /is said to be lower semicontinuous at x0, or Isc at x0, if
liminf /(x) > /(x0).
x-*xo;xe£
Thus, if /(x0) = —oo, / is Isc at x0, while if /(x0) > — oo, the definition
amounts to saying that given m < /(x0), there exists д > 0 such that/(x) > m
if x e E and |x — x0| < d. Equivalently,/is Isc at x0 if and only if —/is use
at x0.
It follows that/is continuous at x0 if and.only if |/(x0)| < + oo and/is
both use and Isc at x0. As simple examples of functions which are use every-
where in R1 but not continuous at some x0, we have
z 4 (0 if x < x0, ' 4 (0 if x / x0.
H|W= I ifx>X = ifx_x
Hence, — zq and — u2 are Isc everywhere in Rh The Dirichlet function of
p. 16, Example 4, is use at the rational numbers and Isc at the irrationals.
A function defined on E is called use (Isc, continuous) relative to E if it is
teoesgue Measurable function
(4.14)
□o
use (Isc, continuous) at every limit point of E which is in E. The next theorem
characterizes functions which are semicontinuous relative to a set.
(4.14) Theorem
(i) A function f is use relative to E if and only if {xg E : /(x) > a} is
relatively closed [equivalently, {x e E : /(x) < a} is relatively open]
for all finite a.
(ii) A function f is Isc relative to E if and only if {x e E : /(x) < a} is
relatively closed [equivalently, {x e E : /(x) > a} is relatively open]
for all finite a.
Proof Statements (i) and (ii) are equivalent since f is use if and only if
—/is Isc. It is therefore enough to prove (i). Suppose first that /is use relative
to E. Given a, let x0 be a limit point of {x e E : /(x) > a} which is in E.
Then there exist xk e E such that xk -> x0 and/(xk) > a. Since/is use at x0,
we have /(x0) > limsup^ ^/(xj. Therefore,/(x0) > a, so that x0 e {x e E :
f(x) > a}. This shows that {x e E :/(x) > a} is relatively closed.
Conversely, let x0 be a limit point of E which is in E. If/is not use at x0,
then /(x0) < + oo, and there exist M and {xk} such that /(x0) < M, xk g E,
xk -> x0, and /(xk) > M. Hence, {x e E :/(x) > M} is not relatively closed
since it does not contain all its limit points which are in E.
(4.15) Corollary A finite function f is continuous relative to E if and only
if all sets of the form {x e E: /(x) > a} and {x 6 E :/(x) < a} are
relatively closed [or, equivalently, all {xgE :/(x) > a} and {x e E :
/(x) < a} are relatively open] for finite a.
(4.16) Corollary Let E be measurable, and let f be defined on E. If f is use
(Isc, continuous) relative to E, then f is measurable.
Proof Let / be use relative to E. Since {x e E : /(x) > a} is relatively
closed, it is the intersection of E with a closed set. Hence, it is measurable,
and the result follows from (4.1).
The results in (4.14)-(4.16) deserve special attention in certain cases.
Suppose, for example, that E = Rn and / is use everywhere in R". Since
{/> a} = (J/=i {/> a + 1/k}, it follows from (4.14) that {/> a} is of type
Fa. Since an Fa set is a Borel set, we see that a function which is use (similarly,
Isc or continuous) at every point of R" is Borel measurable.
3. Properties of Measurable Functions:
Egorov’s Theorem and Lusin’s Theorerii
Our next theorem states in effect that if a sequence of measurable functions
converges at each point of a set E, then, with the exception of a subset of E
with arbitrarily small measure, the sequence actually converges uniformly.
This remarkable result cannot hold, at least in the form just stated, without
some further restrictions. For example, if E = R” and fk = x{x:jx|<k}, then fk
converges to 1 everywhere but does not converge uniformly outside any
bounded set. Again, if the fk are finite but the limit f is infinite in a set of
positive measure, then |/fc — f\ is also infinite in this set. The difficulties in
these examples can be easily overcome: the missing ingredient in the first
case is that |F| < + oo, and in the second, that |/| < + oo a.e. Adding these
restrictions, we obtain the following basic result.
(4.17) Theorem (Egorov's Theorem) Suppose that {f} is a sequence of
measurable functions which converges almost everywhere in a set E of
finite measure to a finite limit f Then given e > 0, there is a closed
subset F of E such that \E — F| < g and {fk} converges uniformly to
f on F,
In order to prove this, we need a preliminary result which is interesting in
its own right.
(4.18) Lemma Under the same hypothesis as in Egorov's theorem, given
8,q > 0, there is a closed subset F of E and an integer К such that
\E — F\ < q and |/(x) — /fc(x)| < cfior x e F and к > К,
Proof Fix z,q > 0. For each m, let Em = {(/- fk\ < & for all к > m}.
Thus, Em = {|/- fk\ < e}, so that Em is measurable. Clearly, Em cz
Fm+1. Moreover, since fk-+ f a.e. in E and/is finite, Em / E — Z, |Z| = 0.
Hence, by (3.26), |Fm| -► \E — Z| = |F|. Since |F| < Too, it follows that
\E — Em\ -> 0. Choose m0 so that |F — Emo| < %q, and let F be a closed
subset of Emo with |FWo - F| < %q. Then \E - F\ < q and \f~ fk\ < g in
F if к > m0.
Proof of Egorov's theorem. Given g > 0, use (4.18) to select closed
Fm cz E, m > 1, and integers Km£ such that \E — FJ < g2~w and \f — fk\ <
1/m in Fm if к > Km>£. The set F = Fm is closed, and since F cz Fm for all
m, fk converges uniformly to f on F. Finally, E — F = E — Q Fm —
(J (F - Fm) and, therefore, \E — F| < £ \E — Fm\ < g. This completes the
proof.
See Exercises 13 and 14 for an analogue of Egorov’s theorem in the
continuous parameter case; i.e., in the case when/y(x) -►/(x) as у y0.
We have observed that a continuous function is measurable. Our next
result, Lusin’s theorem, gives a continuity property which characterizes
measurable functions. In order to state the result, we first make the following
definition. A function / defined on a measurable set E has property $ on F
if given 8 > 0, there is a closed set F с E such that
(i) IF - F| < g,
(ii) / is continuous relative to F.
We recall that condition (ii) means that if x0 and {хл} belong to F and
x* -> x0, then/(xfc) ->/(x0). In case F is bounded (and, therefore, compact),
(ii) implies that the restriction of f to F is uniformly continuous [theorem
(1.15)].
(4.19) Lemma A simple measurable function has property
Proof. Suppose that/is a simple measurable function on E, taking distinct
values al9 . . . , aN on measurable subsets E{,... , EN. Given 8 > 0, choose
closed Fj c Ej with \Ej — Fj\ < e/N. Then the set F = Fj is closed,
and since E — F = |J Ej — (J Fj cz (J (Ej ~ Fj), we have \E — F| <
£ \Ej — Fj\ < &. It remains only to show that / is continuous on F. Note
that each Fj relatively open in F, so the only points of Fin a small neighbor-
hood of any point of Fj are points of Fj itself. The continuity on F of /follows
from this since /is constant on each Fj.
Property is actually equivalent to measurability, as we now show.
(4.20) Theorem (Fusin's Theorem) Let f be defined andfinite on a measurable
set E. Then f is measurable if and only if it has property on E.
Proof. If /is measurable, then by (4.13) there exist simple measurable
fk, к = 1,2,..., which converge to /. By (4.19), each fk has property so
given e > 0, there exist closed Fk cz E such that \E — Fk\ < and fk is
continuous relative to Fk. Assuming for the moment that |£| < 4- oo, we see
by Egorov’s theorem that there is a closed Fo cz E with \E — Fo| < such
that {/J converges uniformly to / on Fo. If F = Fo n (Q/= i Fk), then F is
closed, each fk is continuous relative to F, and {/} converges uniformly to /
on F. Hence,/is continuous relative to F by (1.16). Since
00
\E - F| < |E - Fo| + £ \E - Fk\ < le + is = e,
fc = l
it follows that/has property # on E. This proves the necessity of property #
for measurability if |F| < +oo.
If |F| =+oo, write E = j Ek, where Ek is the part of E in the ring
{x : к — 1 < |x| < k}. Since |FJ < 4-oo, we may select closed Fk cz Ek
such that \Ek — Fk\ < e2'k and/is continuous relative to Fk. If F = (J/= j Fk,
it follows that |F — F| < £ №k ~~ Fk\ < 8> and that/is continuous relative
to F. A simple argument shows that Fis closed, and therefore/has property
# on E.
Conversely, suppose that/has property^ on E. For each k,k = 1,2,...,
choose a closed Fk cz E such that |F — Fk\ < 1/k and the restriction of /to
ii |
Fk is continuous. If H = (j£L t Fki then H cz E and the set Z = E — H has
measure zero. We have
{x e E : /(x) > a} == {x e H : /(x) > a} u {x e Z : /(x) > a}
= U {x e Fk :/(x) > a} и {x e Z :f(x) > a}.
k^l
Since {x e Z :/(x) > a} has measure zero, the measurability of f will follow
from that of each {x e Fk :/(x) > a}. However, since /is continuous relative
to Fk and Fk is measurable, {x e Fk :/(x) > a} is measurable by (4.16). This
completes the proof of Lusin’s theorem.
4. Convergence in Measure
Let/and {/J be measurable functions which are defined and finite a.e. in a
set E. Then {fk} is said to converge in measure on £ to / if for every s > 0,
lim |{x e E : |/(x) - A(x)| > e}| = 0.
k-*oo
We will indicate convergence in measure by writing
This concept has many useful applications in analysis. Here we will discuss
its relation to ordinary pointwise convergence; the first result is basically a
reformulation of (4.18).
(4.21) Theorem Let f and fk9 к = 1, 2, ..., be measurable and finite a.e.
in E. If fk -+ f a.e. on E and |£| < + oo, then fk f on E.:
Proof. Given > 0, let F and К be as defined in lemma (4.18). Then if
к > К, {хе £: |/(x) — /*(х)| > e} cz £ — £, and since |£ — £| < ц, the
result follows.
We recall that the conclusion above may not hold if |£| = -f-oo, as
shown by the example £ = Rw,/fc = Z{X:|x| <*p an^/“ 1-
Convergence in measure does not imply pointwise convergence a.e., even
for sets of finite measure. To see this, take n = 1 and let {Д} be a sequence of
subintervals of [0,1] satisfying the following conditions:
(i) Each point of [0,1] belongs to infinitely many Ik.
(ii) lim*-^ |Zfc| = 0.
For example, let the first interval be [0,1], the next two be the two halves
of [0,1], the next four be the four quarters, and so on. Then iffk — we have
fk 0, while fk diverges at every point of [0,1].
There is, however, the following partial converse to (4.21).
(4.22) Theorem Iff. f on E> there is a subsequence fkj such that fk. -> f
a.e. in E.
Proof. Since fk f given j = 1,2,..., there exists kj such that
[/-Л1
< 2J
for к > kj. We may assume that kj /. Let E} = {|/ — fkj\ > 1//}, and Hm =
\Jf=mEj. Then < 2"< |ЯШ| < - 2"" + 1, and \f-fkj\ < 1/J
in E - Ej. Thus, if j > m, \f-fk.\ < 1/j in E - IIso that fk. ->/ in
E — Hm. Since \Hm\ -> 0, it follows that fk ->/a.e. in E. This completes the
proof.
The following theorem gives a Cauchy criterion for convergence in
measure.
(4.23) Theorem A necessary and sufficient condition that {fk} converge in
measure on E is that for each e > 0,
lim |{x e £ : |/k(x) ~/(x)| > e}| — 0.
k,l-+ oo
Proof. The necessity follows from the formula
{1Л -fi\ > e} <= {1Л - /I > и {\ft -Л > 1П
and the fact that the measures of the sets on the right tend to zero as kJ oo
if f^f-
To prove the converse, choose Nj9 j = 1, 2,. . . , so that if kJ > Nj9 then
\{\fk ~/zl > < 24 We may assume that Nj /. Then |/Nj + 1 - /N.| <
2"i except for a set £}, \Ej\ < 2~J. Let If — (J JL £ Дг Then
I /nj+/*) - fNj(*)\ < 2~j for j > i and x ф If.
It follows that £ (/^ t — fN^) converges uniformly outside If, and therefore,
that {fN.} converges uniformly outside If. Since |Z/| < ^7->/2^7' = 2~l + 1,
we obtain that {fNj} converges a.e. in E and, letting/ — lim fNj, that fy. f
on E. In order to show that / / on E, note that
{\fk -/| > e} c {|/ -Д| > |8} U {\fNj -f\> M
for any Nj. To show that the measure of the set on the left is less than a
prescribed q > 0 for all sufficiently large k, select Nj so that the first term on
the right has measure less than ^q for all large к (here, we use the Cauchy
condition), and so that the measure of the second term on the right is also
less than -^q. This completes the proof.
As pointed out at the beginning of the chapter, many of the results above
depend on only a few basic properties of Lebesgue measurable sets and
Lebesgue measure. This is especially true for the elementary properties of
measurable functions [theorems (4.1)-(4.12)] and the section about con-
vergence in measure, which use only the fact that the class of measurable
subsets of R" is a a-algebra, and in (4.5), that subsets of a set of measure
zero are measurable. Egorov’s theorem uses two additional facts: the funda-
mental result (3.26) concerning monotone sequences of sets, and (3.22)
about the approximability of measurable sets by closed sets. Actually, even
(3.22), which is a topological property of Lebesgue measure, is not needed
in the proof of Egorov’s theorem if instead of requiring that F be closed,
we merely require that it be measurable.
The rest of the chapter, namely, the material on semicontinuous functions
and Lusin’s theorem, uses somewhat more restrictive topological properties
of R” and Lebesgue measure. For example, about R", we have used metric
properties, and about Lebesgue measure, we have used (3.22) [e.g., in
(4.19)] and the fact that Borel sets are measurable [e.g., in (4.16)].
Exercises
1. Prove corollary (4.2) and theorem (4.8).
2. Let/be a simple function, taking its distinct values on disjoint sets EL ..., EN.
Show that /is measurable if and only if Eit. . ., EN are measurable.
3. Theorem (4.3) can be used to define measurability for vector-valued (e.g.,
complex-valued) functions. Suppose, for example, that / and g are real-valued
and defined in R”, and let F(x) = (/(x),^(x)). Then Fis said to be measurable
if F~\G) is measurable for every open G c R2. Prove that Fis measurable if
and only if both / and g are measurable in R".
4. Let /be defined and measurable in Rn. If T is a nonsingular linear transforma-
tion of R", show that /(Tx) is measurable. [If Er — {x :/(x) > a} and E2
= {x :/(Tx) > a}, show that E2 = T~1Ei.]
5. Give an example to show that ^(/(x)) may not be measurable if ф and / are
measurable. (Let F be the Cantor-Lebesgue function and let / be its inverse,
suitably defined. Let ф be the characteristic function of a set of measure zero
whose image under F is not measurable.)
6. Let / and g be measurable functions on E.
(a) If / and g are finite a.e. in E, show that / 4- g is measurable no matter how
we define it at the points when it has the form + oo 4- (—oo) or — oo 4- 00.
(b) Show that fg is measurable without restriction on the finiteness of / and g.
Show that / 4- g is measurable if it is defined to have the same value at
every point where it has the form 4-00 + (—00) or —00 4- 00. (Note that
a function h defined on E is measurable if and only if both {h = 4-°°}
and {h = —00} are measurable and the restriction of h to the subset of E
where h is finite is measurable.)
7. Let / be use and less than 4-00 on a compact set E. Show that / is bounded
62
Lebesgue Measurable Functions
above on E. Show also that f assumes its maximum on E, i.e., that there exists
x0 g E such that /(x0) > /(x) for all x g E.
8. (a) Let f and g be two functions which are use at x0. Show that f + g is use
at x0. Is/ — g use at x0 ? When is fg use at x0 ?
(b) If {fk} is a sequence of functions which are use at x0, show that infA/A(x)
is use at x0.
(c) If {fk} is a sequence of functions which are use at x0 and which converge
uniformly near x0, show that lim fk is use at x0.
9. (a) Show that the limit of a decreasing (increasing) sequence of functions use
(Isc) at x0 is use (Isc) at x0. In particular, the limit of a decreasing (increas-
ing) sequence of functions continuous at x0 is use (Isc) at x0.
(b) Let /be use and less than +oo on [a,b}. Show that there exist continuous
fk on [a,b\ such that/fc \/.
10. (a) If /is defined and continuous on E, show that {a <f < b} is relatively
open, and that {a < / < b} and {/ = a} are relatively closed.
(b) Let /be a finite function on R". Show that/is continuous on R" if and only
if f~r(G) is open for every open G in Rl, or if and only if f~\F) is closed
for every closed F in R1.
11. Let / be defined on Rn and let B(x) denote the open ball {y : |x — y| < r}
with center x and fixed radius r. Show that the function g(x) = sup {/(y) :
у g E(x)} is Isc and that the function A(x) = inf {/(у) : у g B(x)} is use on R".
Is the same true for the closed ball {y : |x — y| < r} ?
12. If/U/xGR1, is continuous at almost every point of an interval [tz/], show that
/ is measurable on [a,b}. Generalize this to functions defined in R". [For a
constructive proof, use the subintervals of a sequence of partitions to
define a sequence of simple measurable functions converging to /a.e. in [g,b].
Use (4.12). See also the proof of (5.54) in Chapter 5.]
13. One difficulty encountered in trying to extend the proof of Egorov’s Theorem
to the continuous parameter case fy{x) ~^f(x) as у y0 is showing that the
analogues of the sets Em in (4.18) are measurable. This difficulty can often be
overcome in individual cases. Suppose, for example, that f{x,y) is defined and
continuous in the square 0 < x < 1, 0 < у < 1, and that/(x) = lim^o/fcy)
exists and is finite for x in a measurable subset E of [0,1]. Show that if z and 3
satisfy 0 < e, 3 < 1, the set Ee6 = {x g E : |/(x,y) — f(x)\ < г for all у < <5}
is measurable. [If yk, к = 1, 2, ..., is a dense subset of (0,<5), show that E&
= {xeE: |/(x,yfc)-/«I < ^.]
14. Let f(x,y) be as in Exercise 13. Show that given e > 0, there exists a closed
F с E with \E — F\ < 8 such that /(x,y) converges uniformly for x g Fto /(x)
as у —> 0. [Follow the proof of Egorov’s Theorem, using the sets Е^цт defined
in Exercise 13 for the sets Em of (4.18).]
is. Let {fk} be a sequence of measurable functions defined on a measurable E with
|E| < 4-ос. If |/(x)| < Л/ < fioo for all к for each x g E, show that given
e > 0, there is a closed F a E and a finite M. such that |E — F\ < e and
|/i(x) | < M for all к and all x g F.
Exercises
63
16. Prove that fk f on E if and only if given 0, there exists К such that
|{|/“ Al > £}| < £ if & > К Give an analogous Cauchy criterion.
17. Suppose that fk~~^f and gk g on К Snow that fk 4- gk~^ f 4- g on E and,
if 1^1 <4-oo, that fkgk fg on EZIf, in addition, gk -+ g on E, g 0 a.e.,
and JjE7| <4-00, show that fklffk* fig on [For the product fkfgk, write
fkdk - fg= (fk ~f\9k - g) +f(gk ~ g) 4- g(fk -ff Consider each term
separately, using the fact that a function which is finite on E, |E| < 4-сю, is
bounded outside a subset of E with small measure.]
18. If/is measurable on E, define o)f(g) = |{f > a} | for — oo < a < 4- oo. If/k / f,
show that / ojf. If/fc / show that a>fk -+ a>f at each point of continuity
of a)f. [For the second part, show that if fk f then limsup^ да u>fk(g) <
a>f(a — e) and liminfk_>oo Mrk(a) > of(a + e) for every e > 0.]
19. Let f(x,y) be a function defined on the unit square 0<x<l,0<y<l
which is continuous in each variable separately. Show that / is a measurable
function of (x,y).
20. If / is measurable on [a,b}9 show that given г > 0, there is a continuous g on
[a,b] such that |{x :/(x) Ф ^(x)}| < -e. (See Exercise 18 of Chapter 1.)
Chapter 5
The Lebesgue Integral
1. Definition of the Integral of a Nonnegative Function
There are several equivalent ways to define the Lebesgue integral and develop
its main properties. The approach we have chosen is based on the notion that
the integral of a nonnegative f should represent the volume of the region under
the graph of f.
We start then with a nonnegative function/, 0 </< +oo, defined on a
measurable subset E of R". Let
Г(/Е) {(x»)eR” + 1 :xe£,/(x) < +oo},
R(/,£) = {(x,j) g R" + 1 : x g £, 0 < у < /(x) if/(x) < + oo
and 0 < у < -Foo if /(x) = +oo}.
T\fE) is called the graph of f over E and R(f>E) the region under f over E,
If 7?(/£) is measurable (as a subset of Rn+1), its measure |R(/,£’)|(„+ n is
called the Lebesgue integral of f over E, and we write
1Ж^)1(и+1)= f/(x)t/x.
Je
Usually, one of the abbreviations
f fdx or [ f
Je Je
is used, and at times the lengthy notation
• • • /(x15. . ., xn) dxr • • • dxn
J E J
is convenient/We stress that the definition applies only to nonnegative/;
a definition for functions which are not nonnegative will be given later.
We also note that the existence of the integral is equivalent to the measur-
64
ability of R(fE) and does not require the finiteness of |R(fE)|(n +1}. The next
theorem is of basic importance.
(5.1) Theorem Let f be a nonnegative function defined on a measurable set
E. Then $Ef exists if and only if f is measurable.
We will show here only that the integral exists if/is measurable, postpon-
ing a proof of the converse until Chapter 6, (6.11). We need several lemmas,
the first of which proves the theorem for functions which are constant on E.
In this case, R(f,E) is a cylinder set; that is, it has the form {(x,y) : x g E,
о < У < a}.
(5.2) Lemma Let E be a subset of R", 0 < a < + oo, and define Ea =
{(x,y) : x g E, 0 < у < a} for finite a and = {(x,y) : x g E,
0 < у < + co}. If E is measurable (as a subset of Rn\ then Ea is mea-
surable (as a subset ofW+l} and [£'а|(„+= я|£|(п).*
Proof. The result follows from a series of simple observations. First,
assume that a is finite. If |£| = 0 or if E is an interval which is either closed,
partly open, or open, the result is clear. Next, if E is an open set, then by
(1.11), it can be written as a disjoint union of partly open intervals, E = |J Ik.
Therefore, Ea = (J ZM, and since the Ik,a are measurable and disjoint, Ea is
measurable and |Ea\ = £ |ZM| = £ a\Ik\ = a\E\.
Let E be of type Gd, E ~ t Gk, with | | < + oo. We may assume that
Gk\E by writing E = Gt n ((/ n G2) n (G^ n G2 n G3) n • • •. There-
fore, by (3.26), |GJ -> |£| as co. Moreover, Gk a is measurable, |Gk a\ =
a|GJ, and Gk a \ Ea. Therefore, Ea is measurable and |£J = lim^^ |Gk a| =
fllimboo \Gk\ = a\E\.
If E is any measurable set with |E| < + oo, then by (3.28), E = H — Z,
where |Z| = 0, H is a set of type G5, H = Gk, and |GJ < Too. Since
Ea — Ha — Za, we see that Ea is measurable and \Ea\ = \Ha\ = a\H\ = a|£j.
Finally, if |£j = + oo, the result follows by writing E as the countable union
of disjoint measurable sets with finite measure. This completes the proof in
case a is finite.
If a — + oo, choose a sequence {ak} of finite numbers increasing to + oo.
The conclusion then follows easily from the fact that Eak / E^.
(5.3) Lemma If f is a nonnegative measurable function on E, 0 < |£j <
+ oo, then T(fE} has measure zero.
Proof Given e > 0 and к = 0, 1, ... , let Ek = {k& < f < (к + l)e}.
The Ek are disjoint and measurable, and their union is the subset of E where
/is finite. Hence, Г(/£) = (J Г(/Л). Since |Г(/Л)|е < by (5.2), we
obtain
|Г(/Е)1е < Z 1П/Ш < SE 1^1 S^l-
*Here and below, 0 • oo should be interpreted as 0.
If |E| < -boo, thi^ implies that T(fE) has measure zero. If |E| = +oo,
write E as the countable union of disjoint measurable sets with finite measure.
Then T(/E) is the countable union of sets of measure zero, and the lemma
follows.
Proof of the sufficiency in (5.1). Let /be nonnegative and measurable on
E. By (4.13), there exist simple measurable fk / f. Therefore, R(fk,E)v
T(f,E) / R(f,E), and since T(/E) is measurable (with measure zero), it is
enough to show that each R(fk,E) is measurable. Fix к and suppose that the
distinct values of / are ax, . . ., aN, taken on measurable sets . . . , EN,
respectively. Then R(fk,E) = Ej,aj- Therefore, R(fk,E) is measurable by
(5.2), and the proof is complete.
(5.4) Corollary If f is a nonnegative measurable function, taking values
ar, . . . , aN on disjoint sets Et, ... , EN, respectively, and if E = (J Ej,
then
f f= i «ж-i-
JE J=i
Proof Clearly R(fE) = (JJ=i Ej,aj- Since the Ej are measurable and
disjoint, so are the Ejta. Therefore, J£/ = an<^ ^le corollary
follows from the fact that
2. Properties of the Integral
(5.5) Theorem
'Z (i) Iff and д are measurable and if 0 < д < f on E, then \Eg < J£/.
In particular, (inf/) < j£/
(ii) Iff is nonnegative and measurable on E and if \Ef is finite, then
f < + oo a.e. in E.
(iii) Let Ex and E2 be measurable and Er c E2. Iffis nonnegative and
measurable on E2, then J£1/ < JE2/.
Proof. Parts (i) and (iii) follow from the relations R(g,E) cz R(f,E) and
R(f,Ej) cz R(fEj), respectively. To prove (ii), we may assume that |E| > 0.
If f ~ + oo in a subset Er of E with positive measure, then by (i) and (iii),
we have f£/ > ^f > f£1 a = djEJ, no matter how large a is. This con-
tradicts the finiteness of J£/.
(5.6) Theorem (Monotone Convergence Theorem for Nonnegative Functions)
7/' {/} is a sequence of nonnegative^ measurable functions such that
fk 7 f on E, then
Proof. By (4.12),/is measurable. Since R(fk,E) и Г(/Е) / R(fE) and
T(f,E) has measure zero, the result follows from (3.26).
(5.7) Theorem Suppose that f is nonnegative and measurable on E, and
that E is the countable union of disjoint measurable sets Ej, E = (J Ej.
Then
f r=lf f-
Je Jej
Proof The sets R(fEj) are disjoint and measurable. Since R(f,E) =
(J R(fEjf the result follows from (3.23).
The next four theorems are corollaries of the results just proved. The
first one provides an alternate definition of the integral which will be useful
in Chapter 10 as a motivation for defining integration with respect to abstract
measures.
(5.8) Theorem Let f be nonnegative and measurable on E. Then
f f = sup£[inf/(x)]|£'J|,
Je j *eEj
where the supremum is taken over all decompositions E = (J j Ej of E
into the union of a finite number of disjoint measurable sets Ej.
The reader will observe that the formula resembles the definition of the
Riemann integral if the Ej are taken to be subintervals.
Proof. If E = (Ji Ej is such a decomposition, consider the measurable
function д taking values а} = infye£j/(y) on Ej, j — 1, . . . , N. Since 0 <
д < f we have by (5.4) and (5.5) that - Je/- Therefore,
sup X < f f.
j Ej Je
To prove the opposite inequality, consider for к = 1,2,... , the sets
{Ejk)},j = 0, 1, ... , k2k, defined by
= {44 -f< й’7'=ь ’k2k; E'k} = {f-k}’
and the corresponding measurable functions
ft = E (inf /)Xe/x> •
i EjW
[Compare the simple functions in (4.13)(ii).] Then 0 < fk / f, and by the
monotone convergence theorem, -> JE/. Since = Ej (inf^.w/)^j0!,
it follows that
supE(inf/'M'l [ f
J Ej JE
which completes the proof.
(5.9) Theorem Let fbe nonnegative on E. If\E\ == 0, then iEf = 0.
This can be proved in many ways; for example, it follows immediately
from the last result.
We can now slightly strengthen the statement of part (i) of (5.5).
(5.10) Theorem If f and д are nonnegative and measurable on E and if
V g<fa.e. in E, then fEg < fEf. z
In particular, if f and g are nonnegative and measurable on E and if
f=g a.e. in E, then \Ef = g.
Proof. Write E = A о Z, where A and Z are disjoint and Z — {g > f}.
Then |Z| = 0. Therefore, by (5.7) and (5.9), J£/ = j\/ + Jz/ = $Af Since
the same is true for g, and since f > g everywhere on A, the result follows.
In defining f£/ we assumed that/was defined everywhere in E. In view
of (5.10), J£/is unchanged if we modify /in a set of measure zero. Hence,
we may consider integrals J£ / where/is defined only a.e. in E, by completing
the definition of / arbitrarily in the set Z of measure zero where it is un-
defined. As a result of (5.9), this amounts to defining f^/to be fE_zf
Similarly, we may extend the definition of the integral and the results a-
bove to measurable functions which are nonnegative only a.e. in E.
(5.11) Theorem Let f be nonnegative and measurable on E. Then \Ef = 0
if and only if f = 0 a.e. in E.
Proof. If / = 0 a.e. in E, then J£/ = 0 by (5.10). Conversely, suppose
that / is nonnegative and measurable on E and that J£/ = 0. For a > 0,
we have by (5.4) and (5.5) that
a|{x e E :/(x) > a}| < Г / < Г / = 0.
J{xeE:/(x) >a} • JE
Therefore, {x e E : /(x) > a} has measure zero for every a > 0. Since the set
where/ > 0 is the union of those where/ > l/Л, it follows that/ = 0 a.e. in
E.
The proof above also establishes the following useful result.
(5.12) Corollary (Tchebyshev's Inequality) Let f be nonnegative and mea-
surable on E. If a > 0, then
\{xe E :f(x) > a}\ < 1 f f.
a JE
The significance of this inequality is that it estimates the “size” of / in
terms of the integral of /.
The next two theorems establish the linear properties of the integral for
nonnegative functions.
i к UpGS UiCO V 6 J.HG к I I ucy i QI
J (5.13) Theorem If f is nonnegative and measurable, and if c is any non-
negative constant, then
[с/ = c f f.
Je Je
Proof, If f is simple, then so is cf, and the theorem follows in this case
from the formula for integrating simple functions [see (5.4)]. For arbitrary
measurable f > 0, choose simple measurable fk with 0 < fk / f Then
0 < cfk / cf and
cf = lim cfk = lim cl fk = cl f
Je k-+oo JE oo JE JE
V (5.14) Theorem If f and д are nonnegative and measurable, then
[ (/ + 9) = [ f + f 9-
Je Je Je
' Proof First suppose that f and д are simple: / = EEi ядХл£ and д —
Yj=i bjXBf Then f + д is also simple, taking values at + bj on At n If:
f + 9 = EuOi + Thus,
f (/ + 9) = E (hi + bj^Ai n Bj\ ==ZaiY\Ai^ Bj\ +'ZbjY \Ai П Bj\
Je i,j i j j i
= 2>1Л1 + E W = [/ + L.
Je Je
For general nonnegative measurable /and g, choose simple measurable
fk and gk such that 0 < fk Z f and 0 < gk Z 9- Then fk + gk is simple and
0 <fk + 9kSf + 9- Therefore,
I (/+ 9) = Ит (fk + gk) = lim ( Д + \ gk) = /+ g,
Je fc-*oo Je k-+oo \ Je Je / Je Je
which completes the proof.
(5.15) Corollary Suppose that f and ф are measurable on E, 0 < f < ф, and
Je/bsfinite. Then
f (Ф-Л- f Ф - f /.
Je Je Je
Proof By (5.14), we have j£/ + (Ф — f) ~ $еФ- Since f£/ is finite,
the result follows by subtraction.
(5.16) Theorem If fk, к = 1,2, , are nonnegative and measurable, then
i’ / co \ 00 (*
ЕЛ) = Е fk-
e \fc=i / k=i Je
Proof, The functions defined by FN = Uk=\fk are nonnegative and
measurable, and increase to j fk. Hence,
CO \ I* ZV p CO p
Ё fk) = lim Fn = lim £ \ fk = £ fk.
k=l / N~*oo JE N->ook=lJE Jt=ljE
Note that the preceding theorem is essentially a corollary of the monotone
convergence theorem (5.6). Conversely, the monotone convergence theorem
can be deduced from this result. Verification is left to the reader.
The monotone convergence theorem gives a sufficient condition for inter-
changing the operations of integration and passage to the limit: lim fk =
lim J£/fc. It is an important problem to find other conditions under which
this is true. First, we show that some restriction other than the mere con-
vergence of fk to f is necessary. Let E be the interval [0,1], and for
к — 1, 2, . . ., let fk be defined as follows: when 0 < x < \/k, the graph
of fk consists of the sides of the isosceles triangle with altitude к and base
[0,1/A:]; when 1/k < x < l,/k(x) = 0.
Clearly, fk 0 on [0,1], but JJ/* = 1(1 /fc)(Ar) = | for all k. Hence, lim fk *
Jo lim/v
(5.17) Theorem (Fatou's Lemma) If {fk} is a sequence of nonnegative
measurable functions on E, then
I (liminffk) < liminf \
Je k-*oo л-+оо Je 5
Proof First note that the integral on the left exists since its integrand is
nonnegative and measurable. Next, let gk = inf {A>A+i>- • •} ^or each
Then gk / liminf/^ and 0 < gk < fk. Therefore, by (5.6) and (5.10),
gk -> (liminfA), \ gk fk,
JE JE ’ JE JE
so that
i (liminffk) = lim gk < liminf fk.
e Je Je
(5.18) Corollary Let fky к = 1, 2,. . . , be nonnegative and measurable on E,
and let fk fa.e. in E. If^Efk < Mfor all k, then jEf < M.
Proof By Fatou’s lemma, JE (liminf/к) < M. Since liminffk = lim fk =
f a.e. in Ey the conclusion follows.
We now prove a basic result about term-by-term integration of convergent
sequences.
(5.19) Theorem (Lebesgue's Dominated Convergence Theorem for Nonnegative
Functions) Let { fk} be a sequence of nonnegative measurable functions
on E such that fk -> f a.e. in E. If there exists a measurable function ф
such that Л<ф a.e. for all к and if fE ф is finite, then
Je Je
Proof. By Fatou’s lemma,
liminfliminf
Je
and the theorem will follow if we show that
Jf > limsup fk.
e Je
To prove this inequality, apply Fatou’s lemma to the nonnegative functions
Ф — obtaining
f liminf (ф - fk) < liminf f (ф - fk).
JE JE
Since fk-+ f a.e., the integrand on the left equals ф — / a.e., so that the
integral on the left is JE ф — JE/by (5.15). The right-hand integral equals
liminf I \ ф — I fk1 = ф — limsup I fk.
\ Je Je J Je Je
Combining formulas and cancelling jE ф, we obtain the inequality jEf >
limsup JE/k, as desired.
3. The Integral of an Arbitrary Measurable f
Let/'be any measurable function defined on a set E. Then f ~ f± — f~ and,
by the remark following (4.11), f+ and f~~ are measurable. Therefore, the
integrals Je/+(x) dx and J£/ (x) dx exist and are nonnegative, possibly hav-
ing value 4- oo. Provided at least one of these integrals is finite, we define
[ /(x) dx = f /+(x) dx - f /”(x) dx,
Je Je Je
and say that the integral Je/(x) dx exists. If f > 0, then f — f+, and this
definition agrees with the previous one. As in the case when f > 0, we will
use the abbreviations ^Efdx and J£/
The definition clearly applies if f is defined only a.e. in E, asjn the case
when f > 0 (see p. 68). For the sake of simplicity, we shall usually assume
that f is defined everywhere in E. /
If Je/exists then, of course, — oo < W< + oo. If Ly exists and is finite,
we say that f is Lebesgue integrable, or simple integrable, on E and write
fe L(E). Thus,
L(E) = < f: I /is finite >.
L Je J
If J£/exists, then
[ f < f f+ 4- f Г = f (Л +Г),
Je Je Je Je
by (5.14). Since/+ + /” = |/|, we obtain the inequality
J (5.20)
I* f dx < f |/| dx.
Ie Je
\f (5.21) Theorem Let fb^ measurable on E. Then f is integrable over E if and
onlyif\f\is.
Proof It follows immediately from (5.20) that /e L(E) if | f | e £(£). If
feL(E), then the difference Je/+ — Je/” is finite, and therefore, since at
least one of JE/+ or Je/” is finite, both must be finite. Hence, their sum is
finite. Since this sum is J£ (/+ + /“) = J£ |/|, it follows that |/| e L(E).
The simple properties of J£/for general/follow from the results already
established for / > 0. As a first example, we have the following theorem.
(5.22) Theorem Iffz L(E), then f is finite a.e, in E.
Proof. If f e L(E), then |/|eZ(E), and the result follows from (5.5)(ii).
(5.23) Theorem
(i) If both \Ef and J£ д exist and if f < д a.e. in E, then ^Ef < J£ g.
In particular, if f — g a.e. in E, then J£/ = jE g.
(ii) If J£2/exists and is a measurable subset of E2, then J£1/exists.
H i I. > U i > M U r-i i U i U 4 U ! у IViGaiUjUMiG/
Proof.
(i) The fact that / < g a.e. implies that 0 < /+ < g+ and 0 < g~ < f~ a.e.
in E. By (5.10), we then have J£/ + < J£ g* and J£/“ > J£ g~, and (i)
follows by subtracting these inequalities.
(ii) If J£2/exists, at least one of J£2/+ or J£2/“ is finite. If Ex cz E2, then by
(5.5)(iii), at least one of f£1/+ or J£1/~ is finite. Therefore, J£1/exists.
(5.24) Theorem If \Ef exists and E — (Jk Ek is the countable union of dis-
joint measurable sets Ek, then
f/=x[ /
Je к jEk
Proof Each J£k/ exists by (5.23)(ii). We have
f f= fr - [/" r r
Je Je Je jEk JEk
by (5.7). Since at least one of these sums is finite, we obtain
f/=z([ /+ - f /) = zf /•
Je \ . Ek J^k / J Ek
(5.25) Theorem If\E\ = 0 or iff ~ 0 a.e. in E, then §Ef = 0.
Proof. The theorem follows by applying (5.9) or (5.11) to/+ and f~.
The next few results deal with the linear properties of the integral.
(5.26) Lemma If f£/is defined, then so is JE (—/), and J£ (—/) = — \Ef
<3
Proof. Since (—/)+ = f~ axvifi-fY = f +, and at least one of J£/~ or
Je/+ is finite, we have J£(-/) = ]Ef~ ~ Se/+ = -Je/-
7 (5.27) Theorem If\Ef exists and c is any real constant, then J£ (cf) exists and
[ (cf)=J /
Je Je
Proof. If c > 0, (c/)+ — cf+ and (cf)~ — cf~. Therefore, by (5.13),
JeG/) + ~ c\e/+ and JeG/)~ = Je/~- It follows that JE(c/) exists and
Je (<f) = c(Je/+ “ Je/~) = c J£/ If c = — 1, the theorem reduces to (5.26).
For any c < 0, we have c — (— l)(|c|), and the result follows from the cases
c = — 1 and c > 0.
(5.28) Theorem If fig g L(E), then f + g e L(E) and
f (/+^)- f/+ f
Je Je Je
Proof. Since |/4- g\ < |/[ 4- we have from (5.23)(i) and (5.14) that
I I/ + g\ < [ (1/1 + Ы) = f l/l + f \g\ < +co.
Je Je Je Je
Hence f + g e L(E).
To prove the rest of the theorem, first suppose that each of /, g9 and
f 4- g has constant sign on E. There are then six possibilities: (1) f > 0,
g > 0 (so that/4- g > 0); (2)/> 0, g < 0,/4- g > 0; (3)/> 0, g < 0,
/+£<0; (4) / < 0, £>0, /+0>O; (5)/<0, £ > 0, /4-£<0;
(6)/< 0, g < 0 (so that/ 4- g < 0). Note that these possibilities are mutually
exclusive. [Note also that in some cases the inequalities can be refined: for
example, in case 2,/ can never be zero.] The result in case 1 is just (5.14).
Cases 2-6 are all similar, and we shall consider only case 2. Then / > 0,
—g> 0,/ 4- g > 0, and since by (5.22) each function is finite a.e., (/+#) +
(-0) =/a.e. Hence, by (5.23)(i) and (5.14), we have f£ (f + g) + =
The result in case 2 now follows from (5.26) and the fact that all the
integrals involved are finite.
For arbitrary f and g in £(£), we subdivide E into at most six measurable
sets, Ei9 . . ., E6, where possibilities (1), . . ., (6) hold, respectively. Since
Ei and Ej are disjoint for i J, we have
6 p 6 / p p \ p p
(/ + g) = E (/ + g) = E ( f + g I = / + \ g-
e j—ijEj j=i\jEj Jej j Je Je
This completes the proof.
It follows that if fk e L(E), к = 1, . .., N, and if ak are real constants,
then Ek=i akfk e L(E) and
p / N \ N p
( E akfk) = E fk-
Je \*==i / л=i Je
(5.29) Corollary Let f and ф be measurable on E, f> ф a.e. and ф g L(E).
Then
Proof. First, note that fEf exists since f~ < ф~ a.e. implies that fEf~ is
finite. Next, fE(f~ Ф) exists since f - ф > 0 a.e. If /g L(E). the corollary
follows from (5.28). If f ф L(E), then since f~ e L(E), we must have $ Ef =
4-00. The fact that ф e L(E) implies that f - ф ф L(E), so that fE(f - ф) =
+°°, since f- ф > 0 a.e. This completes the proof.
In Chapter 8, we will study conditions on / and g which imply that
fg g L(Ef For now, we have the following simple result.
(5.30) Theorem Iffs L(E)f g is measurable on E, and there exists a finite
constant M such that < M a.e. in E, then fg e L(E).
Proof. Since \fg\ < M\f\ a.e. in E, we have by (5.10) and (5.27) that
fcl/31 fr M\f\ = M JE|/|..Hence,/5eZ(£).
yj (5.31) Corollary If f e L(Ef / > 0 a.e., and there exist finite constants
a and p such that a < g < (3 a.e. in E, then
f<{ fg < p[f
JE JE JE
Proof By (5.30), fg gL(E). Since0 a.e., we have af <fg<(3f a.e. in
E, and the conclusion follows by integrating.
We now study conditions which imply that \Efk -> J£/ if fk -+f in E.
Most of the results are extensions of those we derived for nonnegative
functions.
* (5.32) Theorem (Monotone Convergence Theorem) Let {/J be a sequence
* of measurable functions on E.
(i) If fk / f a.e. on E and there exists ф e L(E) such that f > ф a.e.
on E for all к, then fEfk-> fEf
(ii) If fk\f a.e. on E and there exists ф e L(E) such that fk ^Ф a.e.
on E for all k, then fEfk -*fEf
Proof To prove (i), we may assume by (5.25) that fk7 f and /'к>ф
everywhere on E. Then 0 <fk -ф/f-ф on E, so that by (5.6), fE (f - ф)
-> fE (f - Ф). Therefore, by (5.29), fEfk - /Еф-> fEf - /Еф, and since
ф e L(E), the result follows.
We can deduce (ii) from (i) by considering the functions — fk. Details are
left to the reader.
(5.33) Theorem (Uniform Convergence Theorem) Let fk e £(£), к = 1,2,...,
and let {/k} converge uniformly to f on E, |E| < +oo. Then f e L(E)
and $Efk -> fE/-
Proof. Since |/| < IAI + I/- Al and {/i} converges uniformly to f on E,
we have |/| < |AI + 1 on E if к is sufficiently large. Since |£j < +oo, it
follows that f e L(E). From (5.28) and (5.20), we obtain
The last integral is bounded by (supx6E |/(x) - A(x)l)l£|, which by hypothesis
tends to 0 as к -» co. This proves the theorem.
(5.34) Theorem (Fatou’s Lemma) Let {fk} be a sequence of measurable
functions on E. If there exists ф e L(E) such that Гк>ф a.e. on E
for all k, then
1»
(liminf/k) < liminf
E k~»oo k-»oo
KU.ои/
Proof. The result follows by first applying (5.17) to the sequence {fk — ф}
of nonnegative functions, and then using (5.29). Details are left to the reader.
(5.35) Corollary Let {fk} be a sequence of measurable functions on E. If
there exists ф e L(E) such that fk ф a.e. on E for all k, then
(limsup/*) > limsup fk.
JE Л-+00 k-+oo JE
Proof This follows from Fatou’s lemma since -fk > - ф a.e. and
liminif (-fk) = -limsup4.
(5.36) Theorem (Lebesgue's Dominated Convergence Theorem) Let {fk} be a
sequence of measurable functions on E such that fk -> f a.e. in E. If there
exists ф g L(E) such that |/*| < ф a.e. in E for all k, then $Efk -> $Ef.
Proof. By hypothesis, — ф < fk < ф a.e. in E. Therefore, 0 < fk 4- ф <
2ф a.e. in E. Since 2ф g £(£*), we conclude from (5.19) that J£ (fk 4- ф) -+
J£(/ + Ф)- Since ф,/ and all the fk are integrable on E, the result follows from
(5.28).
The following special case of the dominated convergence theorem is
often useful if E has finite measure.
(5.37) Corollary (Bounded Convergence Theorem) Let {fk} be a sequence of
measurable functions on E such that fk f a.e. in E. If \E\ < 4-00 and
there is a finite constant M such that < M a.e. in E, then $Efk -> J£/.
In later chapters, we will consider the integrals of complex-valued func-
tions. Here we mention only the definition. (See p. 125 for some further
remarks.) If/ = fi 4- if2 with and /2 real-valued, we define
(For the measurability of such f see p. 61, Exercise 3.) Many basic properties
of the ordinary integral are valid in this case.
4. A Relation Between Riemann-Stieltjes and
Lebesgue Integrals; the If Spaces, 0 < p < 00
It turns out that there is a remarkably simple and useful representation of
Lebesgue integrals Over subsets of R" in terms of Riemann-Stieltjes integrals
(over subsets of R1, of course). In order to establish this relation, we must
first study the function
co(a) = ro/jE(a) = |{xe£ :/(x) > oe}|,
(S.39) ГЫ a: i Between Riemann-Stieltjes and Lebesgue Integrals 77
where f is a measurable function on E and — co < a < 4- co. We call
the distribution function of f on E.
Some properties of co were given in Exercise 18, Chapter 4. Clearly, it is
not affected by changing/in a set of measure zero, and it is decreasing. As
a -> 4- 00,
{x e E -.f(x) > a} \ {x e E :/(x) =4-00};
hence, if we assume that/is finite a.e. in E, then by (3.26)(ii)
lim m(a) = 0,
a—» + co
unless co(a) =4-00. Similarly,
lim co(a) = |7T|.
a-+ — 00
We will assume from now on that |£| <4-00. This insures that co is bounded,
that lima_ + oo m(a) = 0, and that co is of bounded variation on (—00,4-00)
with variation equal to |Ej. The assumption is made only to simplify the
properties of co, and is not entirely necessary (see, e.g., Exercise 16); in fact,
the case |£| = + is often important.
In the results below, / is a measurable function which is finite a.e. in E,
IE\ < +00, and we write
m(a) = co/>£(a), {/ > a} = {x e E :/(x) > a}, etc.
(5.38) Lemma If a. < fi, then |{a </< ft}\ — co(d) — co(/?).
Proof We have {/>/?}<={/> «} and {a < f < fi} = {f > a} —
{ f > /?}. Since |{/ > P}\ <4-00, the lemma follows from (3.25).
Given a, let
co(a4-) = lim co(a 4- fi), m(a —) = limco(a — e)
e\0 e\0
denote the limits of co from the right and left at a.
(5.39) Lemma
(a) m(a4-) = co(a); that is, co is continuous from the right,
(b) m(a-) = |{/> a}|.
Proof If ek \ 0, then {/ > a 4- ek} Z {/ > a} and {/ > a — sk} \
{f > a}. Since these sets have finite measures, it follows from (3.26) that
co(a 4- ek) co(a) and co(a — £fe) -> |{/ > a}|. This completes the proof.
We now know that co is a decreasing function which is continuous from the
right. It may have jump discontinuities, with jumps co(a —) — ca(a), and
intervals of constancy. These possibilities are characterized by the behavior
of/ stated in the following result.
/8
the Lebesgue integral
(5.40) Corollary
(a) a)(a —) — co(a) = |{/ = a}|; in particular, co is continuous at a if
and only if\{f — a}| = 0.
(b) w is constant in an open interval (a,ft) if and only if \{a < f < fi}\ =
0, that is, if and only if f takes almost no values between a and ft.
Proof Since \{f > a}| = |{/> a}| 4- |{/ = a}|, part (a) follows im-
mediately from (5.39)(b). To prove part (b), note that |{a < f < fi}\ =
|{/> a}| — |{/> p}\ = m(a) — со(р-). This is zero if and only if co is
constant in the half-open interval [a,/?). However, since co is continuous from
the right, it is constant in (a,/?) if and only if it is constant in [a,p).
The rest of the theorems in this section give relations between Lebesgue
and Riemann-Stieltjes integrals. As always, f is measurable and finite a.e.
in £, |E| < + oo and co = co/>£.
(5.41) Theorem If a < /(x) < b (a and b finite) for x e E, then
• p
f = — a cZco(a).
JE J a
Proof The Lebesgue integral on the left exists and is finite since f is
bounded and |E| < +oo. The Riemann-Stieltjes integral on the right exists
by (2.24). To show that they are equal, partition the interval [a,b\ by cz =
ao < ai < ’ “ <<** = £, and let Ej = {ccj-i < f < aj}. The Ej are disjoint
and E = (Jj=i Ej. Hence, jEf = hjf anct therefore,
к (* к
Z a.-JEJ < /< E afEfi
j=l JE j=l
By (5.38), |Ej| — — co(ay) — — [co(ay) — co(ay-r)]. Hence, both
sums above are Riemann-Stieltjes sums for — J^acZco(a). Since these sums
must converge to — J* a dco(a) as the norm of the partition tends to zero, the
conclusion follows.
Theorem (5.41) can be extended to the case when/is not bounded on E
as follows.
(5.42) Theorem Let fbe any measurable function on E, and let Eab = {x e E :
a < /(x) < b} (a and b finite). Then
Г P
f = — a dafix).
J^ab v
Proof Let coab(a) = |{xe£ab :/(x) > a}|. Then ojab is the distribution
function of/on Eab. By (5.41), we have \Eabf ~ ~~ a Wc claim that
the last integral equals a dco(a). By taking limits of Riemann-Stieltjes
sums which approximate these integrals, we only need to show that
. i-> M U vv > > .I.UIIIUIHI WUI^SLJOU UliU ь_ ч, МО О M О HIlGyiUiO / M
coab(d) — ^abtP) = co(oc) — co(fi) for a < a < /? < b. By (5.38), this is equiv-
alent to showing that |{x e Eab : a < /(x) < fi}\ = |{xeE:a < /(x) < /?}|
for such a and /?. However, by the definition of Eab and the restrictions on
a and /?, {x e Eab : a < /(x) < /?} = {x g E : a < /(x) < fl}. This proves the
claim and the theorem too.
In both (5.41) and (5.42), the integrals of f are extended over sets where
/is bounded. This restriction is removed in the next theorem, where we define
(see p. 28)
r + oo Cb
a do (a) = lim a dafia),
J — oo a-> — oo J a
b~> 4- oo
if the limit exists.
(5.43) Theorem If either \Ef or jt” a <7co(a) is finite, then the other exists
and is finite, and
/ = — I a <7co(a).
Je J — oo.
Proof. By (5.42), \EaJ = -\baa.dco(f). If fe L(E), then \Ef as
я— oo, Z?-> +oo, since this holds for both /+ and /“. Therefore,
lim^-^ + я [ — j* a <s/co(c0] exists and equals J£/ which proves half of
the theorem.
Now suppose that jt” a dco(a) exists and is finite. Then j/ a dco(f) is
finite, and we claim that J£/ + = — fo a da)(a). By (5.42), for b > 0, J£ob/ =
— jo a rfco(a). Therefore, as b -> Too, J£ob/-> — Jo % daj(a). On the other
hand, as b increases to + oo, EQb Z {0 < / < + oo}. Therefore,
f / = f f+ -* [ /+ = [ /+,
jEob JEqj, J{0</<4-oo} Je
and the claim follows. A similar argument, using the sets Ea0 with a -* — oo,
shows that J£/“ = да a <7co(a). Since all the integrals are finite, it follows
that \Ef = \Ef+ - $Ef~ = a. doi(a).
Two measurable functions/and g defined on E are said to be equimeasur-
able, or equidistributed, if
^/.eC06) ~ WgfE(a) for all a.
We may intuitively think of two equimeasurable functions as being “rear-
rangements” of each other. For such functions, we have
|{a < f < b}\ = \{a < g < b}|, |{/= a}| = |{g = «}|, etc.
We also have the following corollary.
80
The Lebesgue Integral
(5.44)
(5.44) Corollary If f and g are equimeasurable on E and f e LfEf then
g e L{E) and
The method used to derive (5.41)—(5.43) illustrates a basic difference
between Riemann and Lebesgue integrals. The Riemann integral is defined
by a limiting process whose initial step involves pardoning the domain of f.
On the other hand, we saw in (5.41) that the Lebesgue integral can be ob-
tained from a process which partitions,the range off In order to define this
process more clearly, let f be a nonnegative measurable function which is
finite a.e. in E, |£| < +oo. Let Г = {0 = a0 < ax < • • •} be a partition of
the positive ordinate axis by a countable number of points -> + oo, and let
|Г| = supt(afc+1 - a*). Let Ek = {ak <f< at+1} and Z = {/ = +oo}.
Then the Ek are measurable and disjoint, ]Z| = 0 and E = (0 Ek) и Z, so
that |£j = £ |£t|. Let
sr = X, ‘S'r = X ak+ 1
(5.45) Theorem Let f be a nonnegative measurable function which is finite
a.e. in E, |jE7| < 4-00. Then
f — lim xr = lim 5r.
Je |г|->о |Г|->о
Proof. We may assume that f is finite everywhere, since changing it in a
set of measure zero does not affect the expressions above. Given Г, define
functions фг and фг by setting фг = ak in Ek and фг = ak+1 in Ek, к =
0, 1,. . . . Then 0 < фг < f < фг, J£ фг = sr and J£ фг = Sr. Hence,
Je
If vr < + 00, then
0 < Sj- - ,vr = £(at+1 - |Г||£|,
so that Sr < +00 and Sr — 5r -> 0 as |Г| -> 0. The conclusion of the
theorem now follows easily in case $Ef < +00. On the other hand, if J£/ =
+ 00, then 5Г = +00, and therefore also 5Г = +00, which completes the
proof.
Theorem (5.45) is the origin of an anecdote which compares the methods that
Lebesgue and Riemann might have used to count coins. - The story goes
that Lebesgue would have been a better banktcller. To see why, imagine coins
placed at various points along the x-axis (there may be coins of equal value at
different points), and think of f(x) as the value of the coin at x. Suppose that we
(5.47) Relat Between Riemann-Stieltjes and Lebesgue Integrals 81
want to determine the total value of all the coins. In Lebesgue’s method, partition-
ing the ordinate axis and forpiing the sets Ek corresponds to sorting the coins
according to value; computing |Ek| corresponds to counting the number with a
given value. Thus, S ak|£k| = ]*/ represents the total value. Riemann’s method
is less efficient; it approximates the total by arbitrarily grouping the coins (parti-
tioning the x-axis), and then summing the products of the number of coins in a
given group by the value of any chosen coin in the group.
The relation between Lebesgue and Riemann-Stieltjes integrals can be
extended in a useful way to give Riemann-Stieltjes representations for inte-
grals of the form Je <£(/), where f and E are subject to the usual restrictions,
and ф is assumed to be continuous. This last assumption assures the mea-
surability of ф(/) by (4.6).
(5.46) Theorem If a < f < b (a and b finite) in E and ф is continuous on
[a,b], then
f <A(T) = - f
Je J a
Proof. Since ф is bounded and E (as always) has finite measure, we see
that <£(/) e L(E). Since ф is continuous, the Riemann-Stieltjes integral exists
by (2.24). Write f as the limit of simple measurable fk with a < fk < b as
follows: for к = 1,2,..., let a = <4k) < a(k) < • • • < = b be parti-
tions of [afi\ with norms tending to zero, and let /k(x) = a(k) when a^2 2 <
/(x) < a^k). Then ф(^) -> </>(/) in E. Since the ф^к) are uniformly bounded
and IjE/J < 4-oo, it follows from the bounded convergence theorem that
Je <Kfk) Je <£(/)• However, ф(/к) is simple, taking values ^(a^) on {otfl t <
f < a^k)}. Therefore, by (5.38),
f ФШ = -ХйлШ’) - «(aj-i)]>
Je j
so that as к -> oo, J£ ф(/к) -> — J„ ф(а) dafia). This completes the proof.
In the next theorem, let
C + oo pb
ф(а) dco(a) = lim ф(а) dofiafi
J — oo a-* — oo J a
b-> + oo
if the limit exists (cf. p. 28).
(5.47) Theorem Let ф be continuous on (—00,4-00). If ф^)еЕ(Е), then
Jt ф(а) dofif) exists and
p p + 00
</>(/) = - ф(а) dco(a).
Je J -00
Proof Since the proof is similar to that of part of (5.43), we shall be
82
The Lebesgue Integral
(5.48)
brief. For finite a and b, a < b, let Eab and wab be as in (5.42). By (5.46),
Ьаъ Ф(Л = “К Ф(а) MO- Therefore, as in the proof of (5.42), j£a!, </>(/) =
~ Ja ^(a) da)(a). The theorem now follows by letting a -> — oo and b 4- oo.
We remark that if ф is continuous and nonnegative, then the equality
p p + OO
Ф(Л = - Ф(а) M«)
Je J — oo
holds without restriction on the finiteness of either side. To see this, let
a -> — oo, b -> Too in the equation fEab </>(/) = — ф(а) dafx).
Thus, for any continuous ф, we have
p p + 00
\Ф(Л\ = - !</>(«)! Ma).
Je J — oo
Taking ф(а) = |a|p, 0 < p < <x>, it follows that
f |/|₽= _ Г°°ИМа).
Je J — oo
If f is nonnegative, we obtain
(5.48) I fp = - |°°ap Jco(a).
Je Jo
Hence, for any measurable f,
[ \f\P = ~ p&ya).
Je Jo
Given ф > 0, let denote the class of measurable f such that </>(/) e
L(E). If ф(а) = |а|р, 0 < p < oo, the standard notation is
L\E) = J/: f \f\p < +ool,0 <p < a).
I Je J
Note that LfE) = L(E). We will systematically study the Lp classes in
Chapter 8. For now, we only want to complete (5.48).
First, note that for measurable f there is an Lp version of Tchebyshev"s
inequality:
(5.49) M) < -t f f”, a > 0.
06 Jtf>«}
The proof is left as an exercise. Hence, if f is in Lp, apa>(a) remains bounded
asa-т +oo. A stronger result is actually true.
(5.50) Lemma If f e Lp{Ef then
lim арсо(а) = 0.
a-> + oo
(5.52)
Riemann ana teoesgue
Proof. This will be a corollary of (5.49) if we show that i{f>3.}fp 0 as
a -> 4-oo. We may suppose that a runs through a sequence uk 4-co. Let
fk = f wherever f > uk and fk = 0 elsewhere. Then \{f>ak}fp = f£/?. Since
f is finite a.e., fk -> 0 a.e. Moreover, 0 < /£ < |/|pe£(£), and the result
follows from the dominated convergence theorem.
In the following theorem, we use (5.50) to integrate the Riemann-Stieltjes
integral in (5.48) by parts.
(5.51) Theorem If f > 0 and f e LP(E\ then
f fp == — Г cP doj(u) = p | od^c^a) du.
Je Jo Jo
Proof. The first equality is just (5.48). For the second, if 0 < a < b <
4- co, we have
rb гь
— up d<o(u) = —bpco(b) 4- apco(a) 4- p up~lco(u) du,
J a J a
by (2.21) and the fact that up is continuously differentiable on [#,&]. Now let
a 0 and b -> 4- co. Then bpa)(b) -> 0 by (5.50), and apafd) -> 0 since |R| <
4-co (see also Exercise 14). The theorem follows immediately.
For an extension of (5.51), see Exercise 16.
5. Riemann and Lebesgue Integrals
We now study a relation between Lebesgue and Riemann integrals over finite
intervals [я,й] of R1, and give a characterization of those bounded functions
which are Riemann integrable. The Lebesgue integral will be denoted
by \baf and the Riemann integral by (R) \baf.
(5.52) Theorem Let fbe a bounded function which is Riemann integrable on
[<z,Z>]. Then f e L[a,b] and
rb rb
f = (*) /.
J a J a
Proof. Let {Гк} be a sequence of partitions of [я,й] with norms tending to
zero. For each k, define two simple functions as follows: if x^f < < • • •
are the partitioning points of Гк, let lk(x) and uk(x) be defined in each semi-
open interval [x^x^Vi) as the lower and upper bounds of f on [х^),х(Д) J,
respectively. Then lk and uk are uniformly bounded and measurable in [#,£>),
and if Lk and Uk denote the lower and upper Riemann sums off correspond-
ing to Fk, we have
4 —
wk = Uk.
his цеикьуие Hiie^rcti
Note also that lk < f < uk and, if we assume that Г&+ j is a refinement of Tk,
that 4 Z and uk \. Let / = lim^^ lk and и — 11тк^да uk. Then / and и are
measurable, I < f < и and, by the bounded convergence theorem, Lk ->
I and Uk -> But since f is Riemann integrable, Lk and Uk both con-
verge to (R) fa/by (2.29). Therefore,
p p
(fi) /= / =
J a J a
U.
Since и — I > 0, (5.11) implies that I = f — и a.e. in [a,Z>]. Therefore, / is
measurable and (A) \baf = f^/ which completes the proof.
Theorem (5.52) says that any function which is Riemann integrable is
also Lebesgue integrable, and that the two integrals are equal. There are, of
course, bounded functions which are Lebesgue integrable but not Riemann
integrable. One such is the Dirichlet function defined for 0 < x < 1 by
letting f(x) = 1 if x is rational and f(x) = 0 if x is irrational. Since f = 0
except for a subset of [0,1] of measure zero, its Lebesgue integral is 0. On the
other hand, its Riemann integral does not exist since every upper Riemann
sum is 1 and every lower Riemann sum is zero.
The practical value of (5.52) is that it allows us to compute the Lebesgue
integral of Riemann integrable (e.g., continuous) functions. Using the
monotone convergence theorem, we can easily extend (5.52) to include
improper Riemann integrals of nonnegative functions. Special as it is, the
following result is useful in applications.
(5.53) Theorem Let f be nonnegative on a finite interval [«/] and Riemann
integrable {so, in particular, bounded) over every subinterval [a + s, b],
£ > 0. Suppose that the improper integral *
(*b
I = lim (A) /
8-»0 Jfl + e
exists and is finite. Then f e L[a,b] and
rb
'/=/.
J a
Proof. Observe that by (5.52), \ba+ef = (A) fa+e/, a > 0. The result then
follows from the monotone convergence theorem by letting e -> 0.
We note in passing that the finiteness of the improper Riemann integral
of an/ which is not nonnegative does not in general imply that/is integrable
(see Exercise 7).
Our final result is a characterization of those bounded functions that are
Riemann integrable.
(Ьф4) LAVI _
\
(5,54) Theorem A bounded function is Riemann integrable on [a,b] if and
only if it is continuous a.e. in [a,&].
Proof Suppose that f is bounded and Riemann integrable. Let Гк, 7fc,
wfc, etc. be as in the proof of (5.52). Let Z be the set of measure zero outside
which I = f = u. We claim that if x is not a partitioning point of any and
if x ф Z, then/is continuous at x. In fact, if/is not continuous at x and x is
never a partitioning point, there exists e > 0, depending on x but not on k,
such that uk(x) — lk(x) > s. This implies that u(x) — l(x) > e, which is im-
possible if x ф Z. Therefore, /is continuous a.e. in [a,Z>].
To prove the converse, let / be a bounded function which is continuous
a.e. in [a,b\. Let {Ц} be any sequence of partitions with norms tending to
zero, and define the corresponding Гк, u'k, L'k, and Uk as in (5.52). Note that
{lk} and {uk} may not be monotone since Г*+ i may not be a refinement of Ц.
However, by the continuity of /, both lk and uk converge a.e. to /. Hence,
by the bounded convergence theorem, J* lk and j* uk both converge to \baf.
Since Lk = lk and Uk = J* uk, it follows that the upper and lower Riemann
sums Converge to the same limit. Therefore, / is Riemann integrable.
Exercises
1. If/is a simple measurable function (not necessarily positive) taking values aj
on Eh j = 1, 2,. .., N, show that \Ef = [Use (5.24).]
2. Show that the conclusions of (5.32) are not true without the assumption that
ф e L(Ef [In part (ii), for example, take/fc = z(fc, + oo).)
3. Let {fk} be a sequence of nonnegative measurable functions defined on E.
If fk -*/and/fc / a.e. on E, show that f£/fc /£/.
4. If/gL(0,1), show that xkf(x) e £(0,1) for к = 1,2,..., and JJxk/(x) dx —* 0.
5. Use Egorov’s theorem to prove the bounded convergence theorem.
6* Let /(x,y), 0 < x,y < 1, satisfy the following conditions: for each x, f(x,y)
v is an integrable function of y, and (df(x,y)ldx) is a bounded function of (x,y).
Show that (df(x,y)/bx) is a measurable function of у for each x and
7. Give an example of an /which is not integrable, but whose improper Riemann
integral exists and is finite.
8. Prove (5.49).
9. If p > 0 and f E\f — /fc|p -* 0 as к —> oo, show that fk-^ f on E(and thus that
there is a subsequence fkj /a.e. in E).
10. Ifp > 0, f£|/ — fk\p 0, and JeIAP’ < Л/for all A:,show that /£|/|р < M.
11. For whichp > 0 does 1/x g£p(0,1)? LP{\^)1 £p(0,oo)?
12. Give an example of a bounded continuous f on (0,oo) such that
— 0 but f f Lp(0,oo) for any p > 0.
13. (a) Let {fk} be a sequence of measurable functions on E. Show that fk
converges absolutely a.e. in E if £ Je|A| < 4-oo. [Use theorems (5.16)
and (5.22).]
(b) If {rk} denotes the rational numbers in [0,1] and {ak} satisfies 2 1^1 < 4-°°,
show that ak\x — rk\~1/2 converges absolutely a.e. in [0,1].
14. Prove the following result (which is obvious if |£| < Ч-oo), describing the
behavior of apa)(a) as я-+0~Ь. If feLp(E), then lima^0+ apa>(a) = 0. (If
/> 0, e > 0, choose d > 0 so that J{/<<$} fp < £. Thus, ap[co(a) — w(<5)] <
i{a<f<6}fp < « for 0 < a < <5. Now let a —* 0.)
15. Suppose that/is nonnegative and measurable on E and that co is finite on (0,oo).
If Jo ocpda is finite, show that lima^0+ apa)(a) = limd^ + 30 />pco(Z?) = 0.
(Consider f“l2 and Jg/2.)
16. Suppose that/is nonnegative and measurable on E and that co is finite on (0,oo).
Show that (5.51) holds without any further restrictions [that is,/need not be in
LP(E) and |£| need not be finite] if we interpret Jo ap da>(a) — lima^0+ J2-
ft-* + OO
[Use Еаь to obtain the relation J£/p = — Jo ap c/<*>(a). If either Jo ap dco(a)
or Jo ap~1co(a) da is finite, use (5.50) and the results of Exercises 14 or 15 to
integrate by parts.]
17. If/> 0 and co(a) < c(l 4- a)~p for all a > 0, show that/gLr, 0 < r <p.
18. If/> 0, show that f eLp if and only if Zk=°°-.oo 2kpco(2k) < +oo. (Use Exercise
16.)
19. Derive analogues of (5.52) and (5.54) for integrals over intervals in R", n > 1.
20. Let у = Tx be a nonsingular linear transformation of R”. If J£/(y) dy exists,
show that
f /(У) dy = |det f i f(Tx)dx.
JE JT-^E
[The case when / = /£1, Er с £, follows from integrating the formula /£1(Tx)
= Zt-ie/x) °ver T~lE, and then applying (3.35).]
21. If J^/= 0 for every measurable subset A of a measurable set E, show that
/ = 0 a.e. in E.
Chapter 6
Repeated Integration
Let f(x,y) be defined in a rectangle
I = {(x,y) : a < x < b, с < у < d}.
If f is continuous, we have the classical formula
f(x,y) dx dy
f(x,y) dy
dx,
and there is an analogous formula for functions of n variables.
The first two sections of this chapter extend this and related results on
repeated integration to the case of Lebesgue integrable functions. The last
section contains some applications.
1. Fubini’s Theorem
We shall use the following notation. Let x = (.q, . . . , xn) be a point of an
«-dimensional interval
Д = {x = (%i, . . ., x„) : at < Xi < bi9 i = 1,. . . ,«},
and let у be a point of an т-dimensional interval I2,
Л = {y = (Ji, • • • > y«d Cj yj dj, 1,. .., m}.
Here Ц and I2 may be all of Rn and Rm, respectively. The Cartesian product
I = x I2 is an (n + m)-dimensional interval consisting of points (jq, . . .,
xn, y1} . . . , ym). We shall denote such points by (x,y). A function f(xx, . . . ,
xn, yl9 . . ., ym) defined in / will be written /(x,y), and its integral j*,/ will
be denoted by JJr/(x,y) dx dy.
(6.1) Theorem (Fubini’s Theorem) Let f(x,y) e L(I), I = f x I2. Then
(i) for almost every xe f, /(x,y) is measurable and integrable on I2
as a function of y;
87
(ii) as a function of x, f/2/(x,y) dy is measurable and integrable on 7\,
and
/(x,y) dx dy =
JI Л1
/(x,y)rfy
U2
dx.
Setting/ = 0 outside /, we see that it is enough to prove the theorem when
f = R", I2 — Rw, and I = Rn+m. For simplicity, we then drop /,I2, and I
from the notation and write J/(x,y)t/x for jZ1/(x,y) rfx, L(dx) for Iff),
L(dx dy) for L(I), etc.
We will prove the theorem by considering a series of special cases. The
first two lemmas below will help in passing from one case to the next. In these
lemmas, we say that a function / in L(dx dy) for which Fubini’s theorem is
true has property JF.
(6.2) Lemma A finite linear combination of functions with property JF has
property JF.
This follows immediately from (4.9) and (5.28).
(6.3) Lemma Letfi,f2,. . . ,fk,. . . , have property JF, Iffk / f or fk \f
and felfdx dy), then f has property JF.
Proof. Changing signs if necessary, we may assume that fk / f. For each
k, there exists by hypothesis a set Zk in R" with measure zero such that
A(x,y) e L(dy) if x ф Zk. Let Z = (Jk Zk, so that Z has Rn-measure zero.
If x ф Z, then/fc(x,y) e L(dy) for all k, and therefore, by the monotone conver-
gence theorem applied to {fk(x,y)} as functions of y,
^(x) = j/fc(x,y) dy / h(x) = j/(x,y) dy (x ф Z).
By assumption, hk(x) e L(dx), fk e L(dx dy) and j JЛ(х,у) dx dy = J hk(x) dx.
Therefore, using the monotone convergence theorem again, we obtain that
f J/(x,y) dx dy = J h(x) dx. Since fe L(dx dy), it follows that helfdx),
which implies that h is finite a.e. This completes the proof.
The next three lemmas prove special cases of Fubini’s theorem.
(6.4) Lemma IfE is a set of type G5, viz., E = Q/= i Gk, and if Gt has finite
measure, then %E has property JF.
Proof Case 1. Suppose that E is a bounded open interval in R"+w:
E = x J2, where and J2 are bounded open intervals in R" and Rm,
respectively. Then (jE’I = |Ji11J2|, where |/Ц and |J2| denote the measures of
Ji and J2 in R" and RZM. For every x, %£(x,y) is measurable as a function of y.
If h(x) = f /£(x,y) dy, then /z(x) = |/2| for x e J{, and /?(x) = 0 otherwise.
Therefore, j h(x) dx = [JJI/J- but also, J J /£(x,y) dx dy = |E| = 1ЛНЛ1>
and the lemma is proved in this case.
Case 2. Suppose that E is any set (of type G3 or not) on the boundary of
an interval in Rn+w. Then for almost every x, the set {у : (x,y) g E} has Rm-
measure zero. Therefore, if й(х) = j X£(x,y) dy, it follows that /z(x) = 0 a.e.
Hence, j Л(х) dx = 0. But also, f J /£(x,y) dx dy = IE) = 0.
—-
Case 3. Suppose next that E is a partly open interval in R”+m. Then E is
the union of its interior and a subset of its boundary. It follows from cases 1
and 2 and (6.2) that %E has property
Case 4. Let E be an open set in R*+w with finite measure. Write E =
|J I7, where the are disjoint, partly open intervals. If Ek = (J*=1 Ij, then
= Yj= i Xip so that XEk has property 3? by case 3 and (6.2). Since %Ek /
Xe> Xe has property & by (6.3).
Case 5. Let E satisfy the hypothesis of (6.4). We may assume that Gk\ E
by considering the open sets Gt, Gr л G2, Gr л G2 л G3, etc. Then /Gk \ x£,
and the lemma follows from case 4 and (6.3).
(6.5) Lemma If Z is a subset of R”+w with measure zero, then %z has
property 3F. Hence, for almost every x g R”, the set {y : (x,y) g Z} has
W1-measure zero.
Proof Using (3.8), select a set Hof type G3 such that Z a Я and |Я| = 0.
If H = Q Gk, we may assume that Gr has finite measure, so that by (6.4),
j jzn(x,y) dy dx = jjxH(x,y) dx dy = 0.
Therefore, by (5.11), |{y : (x,y) e H}\ = f xH(x,y) dy = 0 for almost every
x. If {y : (x,y) g H} has Remeasure zero, so does {y : (x,y) g Z} since
Z a H. It follows that for almost every x, %z(x,y) is measurable in у and
J Xz(x,y) dy = 0. Hence, J[J xz(x,y) dy] dx = 0, which proves the lemma
since j f Xz(x,y) dx dy = \Z\ = 0.
(6.6) Lemma Let E cz Rn+m. If E is measurable with finite measure, then
yE has property 3F.
Proof Using (3.28), write E = H — Z, where H is of type G3 and Z has
measure zero. If H = Q Gk, choose with finite measure [see the proof of
(3.28)]. Since Хе ~ Xh ~ Zz> the lemma follows from (6.4), (6.5), and (6.2).
Proof of FubinVs theorem. We must show that every feL(dxdy) has
property 3F. Since f = f+ — f~, we may assume by (6.2) that f > 0. Then,
by (4.13), there are simple measurable fk / ffk > 0. Each fk g L(dx dy), and
by (6.3), it is enough to show that these have property 3F Hence, we may
assume that /^is simple and integrable, say f = vjXej- Since each Ej
must have finite measure, the result follows from (6.6) and (6.2).
If/g L(Rn+m), then by Fubini’s theorem,/(x,y) is a measurable function
of у for almost every x e R". We now show that the same conclusion holds if
/is merely measurable.
(6.7) Theorem Let /(x,y) be a measurable function on R”+w. Then for
almost every x e R",/(x,y) is a measurable function of у e Rm.
In particular, if E is a measurable subset ofR?+m. then the set
Ex = {У (x,y) e E}
is measurable in Rm for almost every xeR".
Proof Note that if / is the characteristic function of a measurable
E c Rn+m, then the two statements above are equivalent. To prove the result
in this case, write E — H u Z, where His of type Fa in R” + m and |Z|„+ni = 0.
Then £x = Hxu Zx. Hx is of type Fa in Rm, and for almost every x e Rrt,
\Zx\m = 0 by (6.5). Therefore, Ex is measurable for almost every x.
If/is any measurable function on Rn+z”, consider the set E(a) — {(x,y) :
/(x,y) > a}. Since E(a) is measurable in Rn +m, the set £(a)x = {y ; (x,y) e
E(a)} is measurable in R7” for almost every x e R". The exceptional set of Re-
measure zero depends on a. The union Z of these exceptional sets for all
rational a still has Rn-measure zero. If x ф Z, then {y : /(x,y) > a} is mea-
surable for all rational a, and so for all a by (4.4). This completes the proof.
We will now extend Fubini’s theorem to functions defined on subsets of
+ m
(6.8) Theorem Let /(x,y) be a measurable function defined on a measurable
subset E ofRn+m. and let Ex = {у : (x,y) e E}.
(i) For almost every x e R",/(x,y) is a measurable function of у on Ex.
(ii) 7//(x,y) e L(E), then for almost every x e R",/(x,y) is integrable on
Ex with respect to у; moreover. J£x^(x,y) dy is an integrable function
of x and
f(x.y)dxdy
/(x,y) dy
dx.
Proof Let / be the function equal to / in E and to zero elsewhere in
R" + w. Since /is measurable on E. f is measurable on R”+w. Therefore, by
(6.7), J(x,y) is a measurable function of у for almost every x e R”. Since Ex
is measurable for almost every x e R", it follows that/(x,y) is measurable on
almost every Ex. This proves (i).
If/e L(E), then/e L(Rn+m) and.
I J(x,y) dxd'y =
7(x,y) dy dx.
|R”1 (ii)
Since Ex is measurable for almost every x, we obtain by theorem (5.24) that
(Ь.У)
JRm/(x?y) с/у = Уех/(х,у)^у for almost every xeR". Part (ii) follows by
combining equalities.
2. Tonelli’s Theorem
By Fubini’s theorem, the finiteness of a multiple integral implies that of the
corresponding iterated integrals. The converse is not true, even if all the
iterated integrals are equal, as shown by the following example.
(6.9) Example Let я = m = 1, let 7 be the unit square and {7J be the infinite
sequence of subsquares shown below:
Subdivide each Ik into four equal subsquares by lines parallel to the x and у
axes.
7<3>
4<2>
Ik
For each /с, let f = 1/|7J on the interiors of and 7^3), and let/ = - 1/|/|
on the interiors of 7£2) and I^\ Let f = 0 on the rest of 7, that is, outside
(J 4 and on the boundaries of all the subsquares. Clearly, JJ/(x,y)<Tp = 0
for all x9 and JJ/(x,y) dx = 0 for all y. Therefore,
u
о
f{x9y) dy dx =
о
f(x,y) dx dy = 0.
о
However,
f f 1/(АУ)1 dx dy = £ j* [ |/(x,j')| dx dy = £ 1 = + oo.
J Jr к J Jlk к
Hence, finiteness of the iterated integrals does not in general imply that
of the multiple integral. However, for nonnegative /, we have the following
basic result.
(6.10) Theorem (TonellTs Theorem) Let /(x,y) be nonnegative and measur-
able on an interval I — Ц x /2 c>/Rn+m. Then, for almost every x e 4,
/(x,y) is a measurable function of у on /2. Moreover, as a function of x,
fr2/(x,y) dy is measurable on Il9 and
/(x,y) dx dy
f(*j)dy
h2
dx.
Proof This is actually a corollary of Fubini’s theorem. For к — 1,2,.,.,
let A(x,y) - 0 if |(x,y)| > к and/k(x,y) = min {k,f(x,y)} if |(x,y)| < k. Then
fk > 0, fk / f on I and 4 g L(I) (/л is bounded and vanishes outside a com-
pact set). Hence, Fubini’s theorem applies to each4- The statement concern-
ing the measurability of J\2/(x,y) dy then follows from its analogue for fk;
in fact, by the monotone convergence theorem, $i2fk(x,y) dy / J,2/(x,y) dy.
[The measurability of/(x,y) as a function of у was proved in (6.8).] By the
monotone convergence theorem again,
(a)
(b)
4(x,y) dx dy -> /(x,y) dx dy,
U J Jr
dx.
Since4eL(I),the left-hand sides of(a)and(b)are equal. Therefore, so are the
right-hand sides, and the theorem follows.
An extension of Tonelli’s theorem to functions defined over arbitrary
measurable sets E is straightforward.
Since the roles of x and у can be interchanged above, it follows that if f
is nonnegative and measurable, then
dx -
In particular, we obtain the important fact that for / > 0, the finiteness of
(6.12)
Аррисанопь Qi ruuiliia inGUiUiii
any one of FubinPs three integrals implies that of the other two. Hence, for any
measurable f the finiteness of one of these integrals for |/| implies that /is
integrable and that all three Fubini integrals of / are equal.
An easy consequence of Tonelli’s theorem is that the conclusion
/(x,y)Jy dx
hi L J12
of Fubini’s theorem holds for measurable / even if j*fz/~ ±00 (that is, it
holds if JJ/jf merely exists). In fact, if JJZ/ = H-oo, we have £fz/+ = +00
and/- e L(f). By Tonelli’s theorem,
f/+ = f ( f f+dy\dx,
Ji Jii \ J12 /
[ f = f ( f f dy\ dx.
11 Jii \ J12 /
Since ffz/ is finite, the desired formula follows by subtraction.
3. Applications of Fubini’s Theorem
We shall derive several important results as corollaries of Fubini’s and
Tonelli’s theorems. The first one is the necessity of the condition in theorem
(5.1). Using the notation of Chapter 5, we will prove the following result.
(6.11) Theorem Let f be a nonnegative function defined on a measurable set
E c R". If R(fE), the region under f over E, is a measurable subset of
R" + 1, then f is measurable.
Proof. For 0 < у < -h oo, we have
{x e E :/(x) > y} = {x : (x,.y) e R{f,E)}.
Since R(fE) is measurable, it follows from (6.7) that {x e E: /(x) > j>} is
measurable (in R") for almost all (linear measure) such y. In particular,
{x e E :/(x) > y] is measurable for all у in a dense subset of (0,oo). If у is
negative, then {x 6 E : /(x) > у} = E, which is measurable. We conclude
that/is measurable [cf. (4.4)].
As a second application of Fubini’s theorem, we will prove a result about
the convolution of two functions. More general results of this kind will be
proved in Chapter 9. If / and д are measurable in R”, their convolution
Cf * #)(x) defined by
(/* ff)(x) = [ /(X - t)^(t) dt,
JR"
provided the integral exists.
We first claim that/* д = д */ that is, that
(6.12) f /(x - t)gf(t) dt = [ f(t)g(x - t) dt.
* JR" JR"
This is actually a special case of results dealing with changes of variable.
In this simple case, however, it amounts to the statement that if x g R”, then
(6.13) r(t) dt = r(x —- t) dt
jRn JRm
when r(t) = /(x - t)#(t). For fixed x, x - t ranges over R” as t does. There-
fore, for any measurable r > 0, (6.13) follows from the geometric interpreta-
tion of the integral. [See (5.1) and the definition of the integral of a non-
negative function.] For any measurable r, it follows by writing r = r+ — r~.
(For the effect of a linear change of variables, see Exercise 20 of Chapter 5.)
The result we wish to prove for convolutions is the following.
(6.14) Theorem If fe L(Rn) and g g £(R”), then (f * g)(x) exists for almost
every x e R”. Moreover, f * g e L(Rn) and
f I/* g\ dx < f f 1/l^xVf \g\dx
jRn \ JRn / \ JR"
In order to prove this, we need a lemma.
(6.15) Lemma If/(x) is measurable in R”, then the function F(x,t) = f(x — t)
is measurable in R" x R" = R2”.
Proof. Let Fi(x,t) = /(x). Since/is measurable, it follows from (5.2) that
Г/хД) is measurable in R2n: in fact, the set {(x,t) : Ffx,t) > a}, which equals
{(x,t) :/(x) > a, t g R"}, is a cylinder set with-measurable base {x : /(x) > a}
in R”. For (£,iy) g R2m, consider the transformation x = £ — ly, t = £ + iy.
This is a nonsingular linear transformation of R2”, and therefore, by; (3.33)
(see Exercise 4 of Chapter 4), the function F2 defined by F2(£,»y) = Ffg — ly,
£ -h jy) is measurable in R2". Since F2(£,/y) = /(£ — //), the lemma follows.
Proof of Theorem (6.14). Suppose first that both / and g are nonnegative
on R”. By (6.15),/(x — t)#(t) is measurable on R” x R” since it is the product
of two such functions. Hence, the integral
J = /(x - t)g(t)dtdx
is well-defined. By Tonelli’s theorem and (6.13),
I
/(x - t)^(t) dt dx
f(x — t) dx dt =
The first of these equations can be written I = j (/ * g)(x) dx, so that
j(/* 0)(x) dx =
(6.17)
Applications of Fubini's Theorem
This proves the theorem for f > 0 and g > 0. The general case can be
reduced to the nonnegative case by observing that |/* g\ < \f\ * |g|. Hence,
\f*g\ dx < j(|/| * |0|) dx = Q|/| dxj^ j|0| dx
In the proof above, we showed the following useful fact.
(6.16) Corollary Iff and g are nonnegative and measurable on R", then
(/* 0) dx = I I f dx j I g dx
JRn \ JRn J \ JRH
Our last application of Fubini’s theorem is an important theorem of
Marcinkiewicz concerning the structure of closed sets. For simplicity, we
will restrict our attention to the one-dimensional case. Given a closed subset
F of R1 and a point x, let
<5(x) = 8(x,F) = inf {|x — у I : у e F}
denote the distance of x from F. Thus, <5(x) = 0 if and only if x e F. By (1.10),
the complement of Fis a union (Jft (ak,b^ of disjoint open intervals. At most
two of these intervals can be infinite. The graph of <5(x) is thus an irregular
sawtooth curve: over any finite interval (ak,b^, the graph is the sides of the
isosceles triangle with base (ak,b^) and altitude %(bk — як); outside the terminal
points of F the graph is linear. If we move from a point x to a point y, the
distance from F cannot increase by more than |x — y|. Hence,
|<5(x) - <5(y)l < I* - tI;
that is, <5 satisfies a Lipschitz condition.
We shall prove the following theorem. [See also Exercises 7-9, and
theorem (9.19).]
(6.17) Theorem (Marcinkiewicz) Let F be a closed subset of a bounded
open interval (a,b), and let d(x) = d(x;F) be the corresponding distance
function. Then, given Л > 0, the integral
Cb 5A(y)
Mx(x) = MK(x',F) = j +
is finite a.e, in F, Moreover, Mx e L(F) and
мх <
where G = (a,b) — F,
Before the proof, we note that what makes the finiteness of Mfx) re-
markable is the singular behavior of <5A(^)/|x — j/p as у -> x. Since <5(.r)
So
Repeated Integration
(8.17)
<5(x) as у -» x, it follows that Мл(х) = + oo if x ф F (see Exercise 9). If x e F,
then <5(x) = 0, but the mere Lipschitz character of <5 in the estimate
m = ш - w ix - У\л = i
|x->i1+z |л-^|1+л ~|x-j>|1+A |x-j|
is not enough to imply that Мл(х) is finite since f* dyj\x — y| = Too. The
key to the convergence of Мл at a point x is the fact that <5(l) vanishes not
only at x but also at every у g F. Thus, roughly speaking, the finiteness of
Мл(х) means that F is “very dense” near x.
Proof. Since <5 — 0 in F, integration in the integral defining Мл can be
restricted to the set G = (a,b) — F without changing Мл. Thus,
Г (*/ Г dx \
Mfx)dx^ dfy)( --------------7гп)^
Jf Jg \ JF Iх “ Л /
the change in the order of integration being justified by Tonelli’s theorem.
To estimate the inner integral, fix у e G and note that for any x e F, we have
|x — y| > <5(y) > 0. Thus,
Г dx f dx Г00 dt г х-л
I Iy — vl1+A “ lr Л+Л 2/1 &(У)
Jf Iх Л J|x-y|><5(y) Iх “ Л J d(y) 1
In particular,
| Mfx) dx
b\y)[2rld(yyx]dy = 2/T1|G| < Too.
Jg
Exercises
1. (a) Let E be a measurable subset of R2 such that for almost every xgR1,
{у : (x,y) e E} has Remeasure zero. Show that E has measure zero, and
that for almost every у e R1, {x : (x,y) e E} has measure zero.
(b) Let /(x,y) be nonnegative and measurable in R2. Suppose that for almost
every x g Rx,/(x,y) is finite for almost every y. Show that for almost every
у e RL/fx,^) is finite for almost every x.
2. If f and g are measurable in R", show that the function A(x,y) — /(x)^(y) is
measurable in R" x R". Deduce that if Et and E2 are measurable subsets of Rn,
then their Cartesian product Er x E2 = {(x,y) :xe£bye E2} is measurable
in R” x Rn, and X £2| = |Ei||E2|.
3. Let f be measurable on (0,1). If /(x) — f(y) is integrable over the square
0<x<l,0<y<l, show that f e L(0,l).
4. Let /be measurable and periodic with period 1: /(/ -f 1) /(0- Suppose that
there is a finite c such that
nf(a + t)~f(b + t)\dt<c
J о
Exercises
97
for all a and b. Show that f g £(0,1). (Set a = x, b = —x, integrate with respect
to x, and make the change of variables £ = x + t, j? = — x 4- t.)
5. (a) If f is nonnegative and measurable on E and w(y) = |{x g E:/(x) > y}[,
у > 0, use Tonelli’s theorem to prove that $Ef = fo dy. [By defini-
tion of the integral, we have Je/ = |R(/,F)| = JJk^e) dx dy. Use the
observation in the proof of (6.11) that {xgE :/(x) > y} — {x : (x,y) g
F(/,E)}, and recall that co(y) = |{x g E :f(x) > y}| unless у is a point of
discontinuity of co.]
(b) Deduce from this special case the general formula
f fp = P f yp~'1o>(y)dy (/> 0, 0 <p < oo).
J E JO
6. For f g ZXR1), define the Fourier transform f of /by
/(x) = dt (x e R1).
(For a complex-valued function F= Fo 4- iFt whose real and imaginary
parts Fo and Fi are integrable, we define j* F = j* FQ + i J Fb) Show that if
f and g belong to £(IU), then
= ?(x)g(x).
7. Let F be a closed subset of R1 and let <5(x) = b(x,F) be the corresponding
distance function. If A > 0 and/is nonnegative and integrable over the comple-
ment of F, prove that the function
f Wfr)
Jr* |x -Я1+а У
is integrable over F, and so is finite a.e. in F. [In case / = Xta.b), this reduces
to (6.17.)]
8. Under the hypotheses of (6.17) and assuming that b ~ a<l, prove that the
function
M0(x) = f [log l/<5(y)] 4* - у| 1 dy
J a
is finite a.e. in F.
9. Show that MA(x; F) = 4- oo if x ф F, A > 0.
10. Let vn be the volume of the unit ball in R". Show by using Fubini’s theorem thai
J о
(We also observe that the integral can be expressed in terms of the Г-function:
Г0) - J? e-'F'1 dt, s > 0.]
11. Use Fubini’s theorem to prove that
f e~lx*2 dx = 7in/2.
J R'1
[For л=1, write (j* tS e~x2 dx)2 = f t S fiS e~x2~y2 dx dy and use pola
^coordinates. For n > 1, use the formula <?~|xl2 = • * -e“x”2 and Fubini’:
theorem to reduce to the case n = 1.]
Chapter 7
Differentiation
The main results in this chapter deal with problems of differentiability. A
variety of topics is considered, but for the most part, the results are related
to the analogue for Lebesgue integrals of the fundamental theorem of
calculus.
1. The Indefinite Integral
If/is a Riemann integrable function on an interval [a,b\ in R1, then the
familiar definition of its indefinite integral is
F(x) = f f(y) dy, (a < x < b).
J a
The fundamental theorem of calculus asserts that Ff = f if f is continuous.
We will study an analogue of this result for Lebesgue integrable / and higher
dimensions.
We must first find an appropriate definition of the indefinite integral.
In two dimensions, for example, we might choose
Я*1,*2) = /Ош) dyt dy2.
J ai J a2
It turns out, however, to be better to abandon the notion that the indefinite
integral be a function of point and adopt the idea that it be a function of set.
Thus, given /e £(Л), where A is a measurable subset of R”, we define the
indefinite integral of f to be the function
= [ /
JE
where E is any measurable subset of A.
Fis an example of a set function, by which we mean any real-valued func-
tion F defined on a n-algebra L of measurable sets such that
98
(7.1)
I he maerintte integral
(i) F(E) is finite for every Ее S,
(ii) F is countably additive; i.e., if E = (Jfc Ek is a union of disjoint Ek g E,
then
F(F) = £F(Ffc).
к
By (5.5) and (5.24), the indefinite integral of an /g L(A) satisfies (i) and (ii)
for the ci-algebra of measurable subsets of A. We shall systematically study
set functions in Chapter 10.
We now discuss a continuity property of the indefinite integral. Recall
(from p. 5) that the diameter of a set E is the number
sup {|x - y| : x,y g E}.
A set function F(E) is called continuous if F(E) tends to zero as the diameter
of E tends to zero; that is, F(E) is continuous if, given 8 > 0, there exists
d > 0 such that |F(F)| < 8 whenever the diameter of E is less than d. An
example of a set function which is not continuous can be obtained by setting
F(E) = 1 for any measurable E which contains the origin, and F(E) — 0
otherwise.
A set function F is called absolutely continuous with respect to Lebesgue
measure, or simply absolutely continuous, if F(E) tends to zero as the measure
of E tends to zero. Thus, Fis absolutely continuous if given 8 > 0, there exists
<5 > 0 such that |F(F)| < 8 whenever the measure of E is less than <5.
A set function which is absolutely continuous is clearly continuous. The
converse, however, is false, as shown by the following example. Let A be the
unit square in R2, let D be a diagonal of A, and consider the сг-algebra of
measurable subsets E of A for which E n D is linearly measurable. For such
E, let F(E) be the linear measure of E n D. Then F is a continuous set func-
tion. However, it is not absolutely continuous since there are sets E con-
taining a fixed segment of D whose R2-measures are arbitrarily small.
(7.1) Theorem Iffz L(A), its indefinite integral is absolutely continuous.
Proof. We may assume that f > 0 by considering/4* and f~. Fix к and
write f = д 4- h, where д = f whenever f < к and д = к otherwise. Given
8 > 0, we may choose к so large that 0 < h < and, a fortiori, 0 <
J£ h < |8 for every measurable E cz A. On the other hand, since 0 < д < к,
we have 0 < j*E д < if |F| is small enough. Thus
0 < I / = g+\h<&+&=E
JE JE JE
if |E| is small enough. This completes the proof.
We remark here that (7.1) has the following converse: If F(£) is a set
function which is absolutely continuous with respect to Lebesgue measure,
100
Differentiation
(7.2)
then there exists an integrable f such that F(E) = $Ef for measurable E.
A proof of this fact, known as the Radon-Nikodym theorem, is given in
Chapter 10.
In the case of the real line, there is an alternate notion, also termed ab-
solute continuity, which pertains to ordinary functions. This notion and its
relation to the integral dy are discussed in Section 5 of this chapter.
2. Lebesgue’s Differentiation Theorem
We now come to a fundamental theorem of Lebesgue concerning differentia-
tion of the indefinite integral. For f e L(Rn), let F be the indefinite integral of
/, and let Q denote an «-dimensional cube with edges parallel to the coordi-
nate axes. Given x, we consider those Q centered at x, and ask whether the
average
F(Q) 1 f f( ' ,
\q\ iei]/y )dy
converges to /(x) as Q contracts to x. If this is the case, we write
r Г< '
lim = Л*),
Q\x 1^1
and say that the indefinite integral of f is differentiable at x with derivative /(x).
In case n ~ 1, the question is whether
1
llm Th dy =
h-*0 £n J x-h
which we shall later see is essentially equivalent to
lim I [ dy =
h->0 П J x
that is, to djdx \xaf(y) dy = f(x).
Since f can be changed arbitrarily in a set of measure zero without affect-
ing its indefinite integral, the best we can hope for is that Fis differentiable to
f almost everywhere. This is actually the case.
(7.2) Theorem (Lebesgue's Differentiation Theorem) If f e £(R”), its in-
definite integral is differentiable with derivative /(x) at almost every
xeR".
The proof of this basic result is difficult and requires several new ideas
with wide applications. One of them is to consider the function
/*(x) = sup Tj | |/(y)| dy,
(7.3)
Lebesgue's Differentiation Theorem
iUi
where the sup is taken over all Q with center x. This function plays an
important role in analysis.
Let us first observe that the theorem is easy to prove for continuous func-
tions. In fact, if /is continuous at x and Q is a cube with center x, then
tL f /(У) dy - f(x)
1У1 Jq
7777 [ [/(у) -r(x)]rfy
ILjI jq
< 77m [ l/(y) - /(x)| dy < sup |/(y) - /(x)|,
IVI JQ yeQ
which tends to zero as Q shrinks to x.
The strategy of the proof is to approximate a given/e £(R") by continuous
functions Ck. This approximation is stated in lemma (7.3), and is global in
nature. Hence, it will be necessary to find a way to control the local behavior
(i.e., the averages) of f — Ck by this global estimate. This step is carried out
in lemma (7.9), and consists of estimating the size of /* in terms of J |/|.
Lemma (7.4) is a crucial covering lemma used to prove (7.9).
(7.3) Lemma Iffs L(R"), there exists a sequence {Ck} of continuous func-
tions with compact support such that
I/ — Cfc| t7x 0 as £ oo.
JR”
Proof If / is an integrable function for which the conclusion holds, we
will say that / has property j/. We will prove the lemma by considering a
series of special cases. To help in passing from one case to the next, we first
show that
(1) A finite linear combination of functions with property stf has property <2/.
(2) If {fk} is a sequence of functions with property j/, and if JRn|/ — /J -> 0,
then /has property <2/.
To prove (1), it is enough to show that any constant multiple, af of a
function with property has property j/, and that the sum,^ + /2, of two
functions with property j/ has property «2/. These facts follow easily from
the relations
J|#/—aC| = |a| JlZ- Cb
p(/i + /2) - (Ci + c2)| < p/i - CJ + j|/2 - C2|.
To prove (2), let {fk} and/ satisfy the hypotheses of (2). First note that
since/. is integrable and J ]f \ < J |/-/| + f |/k|, it follows that/is integrable.
102
Differentiation
(7.4)
Next, given 8 > 0, choose kQ so that f \f — /fto| < 8/2. Then choose a continu-
ous C with compact support such that j |/ko - C| < s/2. Since
JlZ - Cl < p/-Aol + JlAo - C| < 8/2 + 8/2 = 8,
we see that /has property j/.
To prove the lemma, let/e L(R"). Writing/ = f+ —f~, we may assume
by (1) that/ > 0. Then, by (4.13), there exist nonnegative simple fk /* f. Thus,
y^e£(Rn) and j \f — fk\ -> 0, so that by (2), we may suppose that / is an
integrable simple function. Hence, by (1), we may assume that/ = |£"| <
-f-oo. Given 8 > 0, choose an open G such that E c G and [G — £| < s.
Then
I/g - Xe\ = \G - E\ < e,
so we may assume that/ = %G for an open G with |G| < 4- oo. Using (1.11),
write G = |J Ik, where the Ik are disjoint, partly open intervals. If we let fN
be the characteristic function of Zk, we obtain
f* OO
I/-AI = E 141-о
J k — N+l
since 1/1 = |Gj < + oo. By (2), it is thus enough to show that each
fN has property j/. But fN is the sum of %1к, к = 1, . . ., N, so it suffices by
(1) to show that the characteristic function of any interval I has property j/.
This is practically self-evident: if /* denotes an interval which contains the
closure of I in its interior and which satisfies |Z* — Z| < 8, then for any con-
tinuous C, 0 < C < 1, which is 1 in Z and 0 outside Z*, we have
IZr - C| < |Z* - < 8.
This completes the proof of (7.3).
The lemma which follows is a preliminary version of a covering lemma
due to Vitali [theorem (7.17)] which has many applications.
(7.4) Lemma (Simple Vitali Lemma) Let E be a subset of\ln with |E|e <
4- oo, and let К be a collection of cubes Q covering E. Then there exist a
positive constant fl, depending only on n, and a finite number of disjoint
cubes Qr, . . . , Qn in К such that *
E \Qj\ > P\E\e.
Leoesgue s Mineremiduun inuuieni
(7.4)
Proof We will index the size of a cube Q e К by writing Q = (?(/), where
t is the edge length of Q. Let Kr ~ К and
r* = sup{/:2 = 2(0^}.
If /* = 4-oo, then Kt contains a sequence of cubes Q with |g| -> +oo. In
this case, given fi > 0, we simply choose one Q g Kr with \Q\ > p\E\e. If
t* < Loo, the idea is still to pick a relatively large cube: choose Q{ =
e Kt such that tr > Now split Kr ~ K2 и K2, where K2 consists
of those cubes in Kt which are disjoint from Q19 and K2 of those which
intersect Qt. Let Q* denote the cube concentric with 2i whose edge length
is 5f. Thus, | 2* | = 5"| Qi |, and since 2f > Z*, every cube in K2 is contained
ь er
Starting with j = 2, continue this selection process for j = 2, 3, . . . , by
letting
ft = sup {/ :e = 2(0 e^.},
choosing a cube 2/ = 2/0) e with 0 > and splitting Kj = KJ+1 u
K'j+1, where Kj+l consists of all those cubes of Kj which are disjoint from
Qj. If A?j + 1 is empty, the process ends. We have r* > Z*+1; moreover, for
each j, the 2i> • • • > 2j are disjoint from one another and from every cube in
Kj+ ls and every cube in K'j+ x is contained in the cube 2* concentric with Qj
whose edge length is 5/y. Note that 12*1 “ 5"| 2jl •
Consider the sequence t* > t* > • • •. If some KN+l is empty (that is,
if /* = 0 for j > N + 1), then since
ATj = K2 cj K2 — * ’ — । kj । kj * * kj K2,
and E is covered by the cubes in Kl9 it follows that E is covered by the cubes
in K'N+i и • • • и K2. Hence, E c (jy=1 2*> so ^at
\E\e < £ ie*i = 5- £ ie,-i.
J=1 J = 1
This proves the lemma with ft = 5~n.
On the other hand, if no t* is zero, then either there exists a 3 > 0 such
that t J > 3 for allj, or /* -> 0. In the first case, tj > %3 for all j and, therefore,
Xj=i 12,1 +00 as °0- Given any fl > 0, the lemma follows in this
case by choosing N sufficiently large.
Finally, if t* -> 0, we claim that every cube in Kr is contained in (Jy Q*.
Otherwise, there would be a cube Q = Q(t) not intersecting any Qj. Since
this cube would belong to every Kj, t would satisfy t < t* for every J and,
therefore, t = 0. This contradiction establishes the claim. Since E is covered
by the cubes in Къ it follows that
104
Differentiation
(7.5)
\E\e < цел = 5" цел
j j
Hence, given /? with 0 < P < 5"”, there exists an N such that £j=1 |2y| >
P\E\e. This completes the proof.
We stress that the lemma above does not presuppose the measurability of
E, and that the.proof can be shortened if E is measurable. In fact, if E is
measurable, we can suppose it is closed and bounded [see, e.g., (3.22)]. Hence,
assuming as we may that the cubes in К are open, it follows from the Heine-
Borel theorem (1.12) that E can be covered by a finite number of cubes.
For 21» we then choose the largest cube; similarly, in subsequent steps, we
take Qj to be the largest cube disjoint from Ql9 . . . , 2/-i- Thus, E c: |J gj,
and the lemma follows.
Before stating the last lemma, we make a definition and a few remarks.
If f is defined on R” and integrable over every cube Q, let
(7.5) /*(x) = sup + j |/(y)| dy,
where the supremum is taken over all Q with edges parallel to the coordinate
axes and center x. The function /*, called the Hardy-Littlewood maximal
function of f is a gauge of the size of the averages of |/| around x. It clearly
satisfies
(i) 0 </*(x) < +oo,
(7.6) (ii) (/ + 0)*(x) < /*(x) + 0*(x),
(iii) (c/)*(x) = |c|/*(x).
If /*(x0) > a for some x0 e R" and a > 0, it follows from the absolute
continuity of the indefinite integral that/*(x) > a for all x near x0. Hence,
according to (4.14),/* is lower semicontinuous in R". In particular, it is
measurable.
We now investigate the size of/*. For any measurable Д
J|£n Q\ 1
Xe(x) = sup I : Q has center x >.
If E is bounded and denotes the smallest cube with center x containing E,
then
i£ne*i iei
m , rn*
It follows that there are positive constants ct and c2 such that
(7.7) ci jfji Ze(x) < c2 for large |x|.
(7.9)
Lebesgue's Differentiation theorem
In particular, if |£| > 0, Xe is not integrable over R". We leave it as an exercise
to show that for any measurable f which is different from zero on a set of
positive measure, there is a positive constant c such that
/*(x)>j|p; for |x[ > 1.
Therefore,/* is never integrable unless f = 0 a.e.
To find a way to estimate the size of /*, recall that by Tchebyshev’s
inequality
|{x 6R" : |/(x)| > a}| < i f |/(x)| dx.
a jRn
Hence, if/e L(Rn), there is a constant c independent of a such that
(7.8) |{x g R” : |/(x)| > a}| < - (a > 0).
Any measurable/ integrable or not, for which (7.8) is valid is said to belong
to weak L(Rn). Thus, any function in L(R”) is also in weak L(Rn). The func-
tion |x| ~n is an example of a function in weak L(R”) which is not in £(R”).
(7.9) Lemma {Hardy-Littlewood) If f g L(R”), then f* belongs to weak
Lift"). Moreover, there is a constant c independent of f and a such that
|{x g R” :/*(x) > a}| < - f |/|, a > 0.
a JRn
Proof. First suppose that in addition to being integrable, / has compact
support. Then, as in (7.7), there is a constant q depending on f such that
/*(х) < С1|хГл for sufficiently large |x|. In particular, {/* > a} has finite
measure for every a > 0. Now fix a > 0 and let
£={/*> a}.
If x g E, then by the definitions of E and/*, there is a cube gx with center x
such that I6J-1 fQJ/| > a. Equivalently,
ISxl < ~ f 1/1.
ajQx
The collection of such Qx covers E, so by (7.4), there exist $ > 0 and xb ..., xy
in E such that Qxt,. . ., QXN are disjoint and |£j < [Г1 i I6xj- Therefore,
i^i<4zИ Q
= PaJUj = iQxj Pa JR”
This proves the result for such f with c =
L./ i 1 i C! & U L i Ct 11 U H
t/.-IO)
Given any f e L(R"), we may assume without loss of generality that/ > 0
since replacing/by | f | does not alter/*. Let {fk} be a sequence of integrable
functions with compact support such that 0 < fk / f. Then there is a constant
c independent of к and a > 0 such that
|{x e R" :/*(x) > a}| < f /*<£[ /
JRn QC JRH
Since/* /f*> it follows from (3.26) that
|{x e R":/*(x) > a}| < f /
a Jr'1
which completes the proof.
Proof of Lebesgue's theorem. Given f e L(R”), there exists by (7.3) a
sequence of continuous integrable Ck such that |/ — Ck\ -> 0. Let F(Q) =
ff and Fk(Q) = Ck. Then for any k,
limsup
Q\x
F(Q) .. .
I Q\ /(X)
< limsup
Q\x
+ limsup
Q\x
FjQ) Fk(Q)
iei 121
^_C(x)
iei t(}
+ |Ck(x) ~/(x)|,
where the limsup is taken for cubes with center x which shrink to x. Since
Ck is continuous, the second term on the right is zero. Moreover,
F(Q) Fk(Q)
121 121
4£i/- - (f~СкШ>
and therefore, the first term on the right is majorized by (/ — Cfc)*(x). Hence,
for every k9
(7.10)
limsup
<2\x
F(Q) „ x
121 /(x)
< (/- Q)*(x) + |/(x) - Ct(x)|.
Given e > 0, let E. be the set on which the left side of (7.10) exceeds e.
By (7.10),
Fe c |x : (/- CJ*(x) > || и jx : |/(x) - C/x)| >
Applying (7.9) to the first set on the right and Tchebyshev’s inequality to the
second, we obtain
\Ee\e<c(~] 4 + 4 I/-CJ.
\z/ Jr'1 \z/ Jr”
Since c is independent of k, it follows by letting к -> oo that [jE’de = 0.
. I *+; t ucucbyue ь ии iciciiiiduuii ineuiein iu/
Let E be the set where the left side of (7.10) is positive. Then E = {JkE£k for
any sequence ek -> 0, and therefore |E| = 0. This means that limQ4x F(Q)/\Q\
exists and equals /(x) for almost every x, which completes the proof.
We now list several extensions and corollaries of Lebesgue’s theorem.
(I) A measurable function f defined on R” is said to be locally integrable on R"
if it is integrable over every bounded measurable subset of R".
(7.11) Theorem The conclusion of Lebesgue's theorem is valid if instead of
being integrable, f is locally integrable on R".
Proof It is enough to show that the conclusion holds a.e. in every open
ball. Fix a ball and replace/by zero outside it. This new function is integrable
over RM, its integral is differentiable a.e., and since differentiability is a local
property, the initial function/is differentiable a.e. in the ball. This completes
the proof.
(II) For any measurable E, note that
JLf _ n 61
isi ’
By (7.11), the left-hand side tends to Xje(x) a-e- as Q \ x; that is,
|£n0|
(7.12) lim = x£(x) a.e.
Q\x 1^1
A point x for which this limit is 1 is called a point of density of E, and a
point for which it is zero is called a point of dispersion of E. Since
\QryE\ \Q n CE\ = [SI
iei iei iei ’
every point of density of E is a point of dispersion of CL, and vice versa.
Formula (7.12) can be restated as follows.
(7.13) Theorem Let Ebe a measurable set. Then almost every point of E is a
point of density of E.
Thus, roughly speaking, a set “clusters” around almost all of its points.
(Ill) The formula limQ4x (1/121) fo/(y) ^У = /(x) can be written
lim iTn f [Г(у) “ dy = °’
Q\x IVl JQ
and is valid for almost every x if / is locally integrable. A point x at
which the stronger statement
(7.14) lim L [ |/(y) - /(x)| dy = 0
q\x Ikfl J(2
is valid is called a Lebesgue point of f and the collection of all such
points is called the Lebesgue set of f,
(7.15) Theorem Let f be locally integrable in Rn. Then almost every point of
R” is a Lebesgue point of f\ that is, there exists a set Z (depending on f)
of measure zero such that (7.14) holds for хф Z.
Proof Let {rfe} be the rational numbers, and let Zk be the set where the
formula
limi7ji ( 1/(У) ~ = l/(x) ~ 'kl
is not valid. Since |/(y) — rk\ is locally integrable, we have |Zfc| = 0. Let
Z = IJ Zk; then |Z| = 0. For any Q, x, and rk,
£ l/(y) -/(x)l dy < £ |/(y) - rj dy + £ |/(x) - rj dy
= i^j £ l/(y) - nJ dy + |/(x) - rj.
Therefore, if x ф Z,
limsup -T f |/(y) - /(x)| dy < 2|/(x) - rj
Q\X ISZI
for every rk, For an x at which/(x) is finite (in particular, almost everywhere),
we can choose rk such that |/(x) — rk\ is arbitrarily small. This shows that the
left side of the last formula is zero a.e., and completes the proof.
(IV) So far, the sets contracting to x have been cubes centered at x with
edges parallel td the coordinate axes. Many other sets can be used. A
family {5} of ineasurable sets is said to shrink regularly to x provided
(i) The diameters cjf the sets S tend to zero.
(ii) If Q is the smallest cube with center x containing S, there is a con-
stant к independent of 5 such that
\q\ < ад
The sets 5 need not contain x.
(7.16) Theorem Let f be locally integrable in R\ Then at every point x of the
Lebesgue set of f (in particular, almost everywhere),
£ l/(y) - /(x)l dy 0
U-»//
i I i«3 V i LCii S
uvcnny i-cisiiiia
for any family {S} which shrinks regularly to x. Thus also
Tf/(y)^y-/(x) a.e.
1^1 JS
Proof If 5 cz g, we have
f l/(y) -/(x)l dy f l/(y) - /(x)l dy.
Js Jq
Hence, if {5} shrinks regularly to x and Q is the least cube with center x
containing S, then
[ l/(y) - /(x)l dy < Ш! f l/(y) - /(x)| dy
|o| Js |o I 1У1 Jq
< 4 Иdy'
If x is a Lebesgue point off the last expression tends to zero, and the theorem
follows.
In particular, for functions of a single variable, we obtain
1
hm т f(y) dy = f(x') a.e.
h->0 n J x
3. The Vitali Covering Lemma
The theorem which follows is a refinement of the simple Vitali lemma (7.4).
We shall assume that each point of a set is covered not just by a single cube,
but by a sequence of cubes with diameters tending to zero. In this case, it
turns out that we can cover almost all points of the set by a sequence of
disjoint cubes. The result will be essentially a corollary of (7.4).
A family К of cubes is said to cover a set E in the Vitali sense if for every
x g E and q > 0, there is a cube in К containing x whose diameter is less
than rj.
(7.17) Theorem (Vitali Covering Lemma) Suppose that E is covered in the
Vitali sense by a family К of cubes and that 0 < IjEJe < +°o« Then,
given s > 0, there is a sequence {Qj} of disjoint cubes in К such that
\E - u Qj\ = °> Z ICjl < 0 + s)lE\e-
j J
Proof The second relation is automatically satisfied if we choose an open
set G containing E with |G| < (1 + s)|£je and cpnsider only those Q in К
which lie in G.
1 1U
Differentiation
(7.18)
By (7.4), there exist a constant /?, 0 < ft < 1, depending only on the
dimension, and disjoint Qx, . .., QNi in К such that "£!i=i \Qj\ > P\E\e-
Therefore,
e - U Qj
J=1
= |G| - E \Qj\ < |£|e(I + 6 - /?)•
j=i
Hence, by considering from the start only those s with 0 < s < Д/2, we have
E- {JQj
J=1
Thus, the part of E not covered by the cubes obtained from the simple Vitali
lemma has outer measure less than |£|e(l — /7/2). We now repeat the process
for the set Ex = E — which is still covered in the Vitali sense by
those cubes in TsT which are disjoint from Q19 ... , QNi. We obtain QNl + 1, . . . ,
QN2, disjoint from each other and from Q19 . . ., QNl, such that
N2 N2 ( B\ ( B\2
E - U Qi = Et- (J Qj < 1 < |E|e 1 -P- ].
J=1 e j = ^1 + i e \ A J \ LJ
Continuing in this way, we obtain at the mth stage disjoint Qt, . . . , QNm in
К such that
E- (jQi < We 1 "Я .
j = l e \
Since (1 — /?/2)m 0 as //w oo, there is a sequence of disjoint cubes in К
with the desired properties.
(7.18) Corollary Suppose that E is covered in the Vitali sense by a family К
of cubes and 0 < |^|е < 4-oo. Then, given £ > 0, there is a finite col-
lection Q19. . . , QN of disjoint cubes in К such that
E-UQj <e,
j= 1 e
E \Qj\ < (1 + e)|£|e.
J = 1
This is part of the proof of Vitali’s lemma.
Note by Caratheodory’s theorem that
N
|E|e = E - J
Есл (J Qj
J=i
so that for E and {(2j}‘J=i as in (7.18), we have
(7.19)
In particular,
(7.20)
|£|e-e<
Erfj Qj
- e < E 1СЛ
J = 1
4. Differentiation of Monotone Functions
As an application of Vitali’s covering lemma, we will prove a basic result
concerning the differentiability of monotone functions on R1. If f(x), x e R1,
is a real-valued function defined and finite in a neighborhood of x0, consider
the four Dini numbers (or derivates)
n f( \ г /(xo + h) - f(x0)
Dx f(Xo) = limsup-------7-------,
h->0+ n
D2f(x0) = hminf---------7------
h->0 + n
n f( . r Ж> + Л)-/(х0)
Dif{x0) = limsup--------г-------,
h->0- “
n Г( \ r f^X° + - Л*о)
= hminf---------7-------.
h->0- n
Clearly, D2f < D\f and D$f < Dzf. If all four Dini numbers are equal, we
say that f has a derivative at xo, and call the common value the derivative
/'(x0) at xo-
(7.21) Theorem Let f(x), xeR1, be monotone increasing and finite on an
interval (a,b). Then/has a measurable, nonnegative derivative f' almost
everywhere in (a,b). Moreover, I
(7.22) 0 < Г/' </(/>-)-/(«+).
J a
Proof. We may assume that (a,b) is finite; the general case follows from
this by passage to the limit. We will show that the set {x e (a,b}: Dlf(x) >
D4f(x)} has measure zero. A similar argument will apply to any two Dini
numbers of f. It is enough to show that each set
Ar,s = {xe(a,b) : DJ(x) > r > s > D4f(x)}
has measure zero since the original set is the union of these over rational r
and 5. We may assume r, s > 0 since f is increasing.
Fix r and s,r > s > Q, write A = Ar s, and suppose that | A |e > 0. If x e A,
Д the fact that D^fix) < s implies the existence of arbitrarily small h > 0 such
that
/(x-A)-/(x)
——h—<5-
By (7.18) and (7.19), given s > 0, there exist disjoint intervals [x7 — Zzy, xj,
j ~ 1, . . . , N, such that
&f(xj)-f(xj-hj)<shj,J^ 1,...,A.
(ii) 14 n (J/=l IXj - hP > Mie - e-
112
Differentiation
(7.22)
(iii) hj < (1 + £)H|e.
Combining (i) and (iii), we obtain
(iv) Z?=1 [f(Xj) -ftXj - h>j\ < 5(1 + £)И|е.
Let В = A n (JJ=1 [xy — hp xj. For every у e В which is not an end-
point of some [x7- — hj9 xj, the fact that Dxf(y) > r implies that there exist
arbitrarily small к > 0 such that [у9 у + k] lies in some [xy — xj and
f(y + k)-f(y)
к
Hence, by (7.18) and (7.20), there exist disjoint [yh yt 4- &J, i = 1, . . . , M,
such that
. (v) Each [yf, yt 4- кД lies in some [x7- — hj9 xj.
(vi) f(y{ + k^ - f(yt) > rk„i = 1,. . . , M.
(vii) > |2?L - e > \A\e - 2e [by (ii)].
Therefore,
(viii) £Г=1[/(Л + kt) -f(yt)] > r(\A\e - 2k).
Since/is increasing, it follows from (v) that
E [ZO.- + kt) - f(yt)] < Z [/(*,) - Kb - A;)].
i=l j=l
Combining this inequality with (iv) and (viii), we obtain r(M|e — 2e) <
5(1 4- я)И|е- Since e > 0 is arbitrary, this gives r < s, which is a contradic-
tion. Hence, |Л|е = 0.
Since an analogous argument applies to any two Dini numbers, it follows
that f’(x) exists for almost every x in (a,b). Extend the definition of/by setting
/(x) = f(b—) for x > b, and let
/tM _№+*)-№> Л 1\
J h \ kJ
for x e (a,b) and к = 1,2,.... Each/. is nonnegative and measurable, and
fk f' a-e- *n (я,6)- Hence,/' is measurable on (a,b), and by Fatou’s lemma,
[ /' < lim inf [ fk.
J a fc->oo J a
Iff(b~) is finite (otherwise, the result is obvious), we have
1 fHft ra-h/j i р + й
/-sj. /
(7.25)
Differentiation of Monotone i-uncuons
Since /(a+) < (W Ja+V f(.a + ^), we obtain
pb pb pb
/' < liminf fk = lim fk = f(b-) - f(a+),
J a fc-*oo J a Hooja
which proves the theorem.
The inequality in formula (7.22) cannot in general be replaced by equality,
even if f is continuous on [a,6]. To see this, let f be the Cantor-Lebesgue
function on [0,1] (p. 35). Then f is continuous and monotone increasing on
[0,1],/(0) = 0,/(l) = 1, and since/is constant on every interval removed in
constructing the Cantor set,/' = 0 a.e. Thus, JJ/' = 0, while/(1) — /(0) = 1.
We shall return in theorem (7.29) to the question of equality in (7.22).
By (2.7), any function of bounded Variation can be written as the difference
of two bounded monotone increasing functions. Hence, we obtain the
following result. (See also Exercise 9.)
(7.23) Corollary If f is of bounded variation on then /' exists a.e. in
[a,Z>], and f' e L[a,b].
If/is of bounded variation on [a9b\ and V(x) denotes its (total) variation
on [a,x], a < x < b, then V is monotone increasing by (2.2). The following
result gives an important relation between V' and/'.
(7.24) Theorem Iff is of bounded variation on [a,6] and V(x) is the variation
6ff on [a,x], a < x < b9 then
F'(x) = |/'(at)| for a.e. x e [«/].
We will prove this with the help of the next lemma, which is of independent
interest.
(7.25) Lemma (Fubini) Let {/J be a sequence of monotone increasing func-
tions on [a9b\. If the series s(x) = £/k(x) converges on \a9b\9 then
s'(x) = £/*(*) a.e. in \a,b\.
Proof. Let Sm = ^k=lfk, rm = ХГ=т+1Л- Then and rm are monotone
increasing functions and s — sm + rm. With the exception of a set Zm, |ZJ =
0, these three terms together with /ь . . . 9fm are differentiable and s' =
s'n 4- r'm. In particular, s' > s'm = ^k=1/k except in Zm. It follows that
oo
E fk
except in Z - J^Z,, |Z| =0.
To prove that in the last inequality we actually have equality a.e., it is
enough to show that r'm 0 a.e. for m running through a sequence of values
< ’' * • Select {mf increasing so rapidly that £ rm.(x) converges at
114
Differentiation
(7.25)
both x = a and x = b. This implies the convergence of £ {rm.(b) — rm.(a)}
and also, in view of the monotonicity of rmj9 of .(/? — ) — rm.(« + )}.
By (7.22), we have
’b рь
° < JX = L 4, < E - rmj(a+y}-
J a J a
Thus, £ r'm. is integrable over (a,b), and therefore, it is finite a.e. in (a,b).
Thus, r'mj -> 0 a.e., and the proof is complete.
Proof of Theorem (7.24). Let /be of bounded variation on [a9b\9 and let
T(x) = P[a,x] be its variation on [a,x], a < x < b. Then V(a) = 0 and V(b)
is the variation of f on [a9b\. Select a sequence {Ffc : Tfc = {xk}} of partitions
of \a9b\ such that 0 < V(b) — SVk < 2~k, where
j
For each k9 define a function fk on [a,ft] as follows: if x e let
f(x] = f /(*) + CJ if Ж)
I -/(*) + ckj if Ж) </(^-0,
where the ck are constants chosen so that fk(a) = 0 and fk is well-defined
(i.e., single-valued) at Xj for every j. Then, for all к and /
Л(*>) -A(^-i) = \f(^)
so that
^ = Е[Ж)-Л(^-1)] =AW-
j
Hence, for any k, we obtain 0 < V(b) — fk(b) < 2~k.
We will show that each У(х) — fk(x) is an increasing function of x. This
amounts to showing that if x < y, then fk(y) - fk(x) < V(y) - V(x). If x
and у both belong to the same partitioning interval of Tfc, then fk(y) — fk(x) <
|/(y) — /(x)|, and therefore,
АОО - Ш < V[x,y] = P(y) - T(x).
In the general case, if xk, xk+19 . . ., xk are the points of Tfc between x and y9
the result follows by adding the inequalities for the intervals [x,x?L
[4.Л Since Г(а)=Л(я) = 0 and V(b) - fk(b) <
it follows that 0 < V(x) — fk(x) <2 k for all x e [a9b\. Hence,
E[^) -AW] < E2’*1 < +ю
for xe[a,b], so that by (7.25) the series ^[F'(x) — fk(x)] converges a.e. in
[a9b\. Hence, fk -* V' a.e. However, |/J — |/'| a.e., so that |/'| = |K'| a.e.
(7.27)
>olutely Continuous and Singular Functions
rib
The theorem now follows from the fact that V is nonnegative wherever it
exists.
5. Absolutely Continuous and Singular Functions
We now turn to the question of equality in formula (7.22). As we know,
the Cantor-Lebesgue function is an example of an increasing continuous f
whose derivative is integrable on [0,1], but for which JJ/' / /(1) — /(0).
A finite function f on a finite interval [a,b] is said to be absolutely continuous
on [a,b\ if given s > 0, there exists <5 > 0 such that for any collection {[^Д]}
(finite or not) of nonoverlapping subintervals of [a9b\9
Е1Ж)-Ж)1 <*ifE(^-«,)<<5.
For example, if g is integrable on [a,6] and G(x) = J* g denotes its in-
definite integral, then
G(^)| < [ \g\
for any nonoverlapping By (7.1), |#| is an absolutely continuous set
function, and therefore, G is an absolutely continuous function. Thus, an
indefinite integral is an absolutely continuous function. One of the main
results proved below is the converse: Every absolutely continuous function is
an indefinite integral.
Another example of an absolutely continuous function is any f which
satisfies a Lipschitz condition:
(7.26) mx)_ f(yy < C|x-y|
for x,y e [a,6].
On the other hand, the Cantor-Lebesgue function f is an example of a
continuous function which is not absolutely continuous, since if Ck =
(J; [aj,bj] denotes the intervals remaining at the Ath stage of construction of
the Cantor set, then |Ck| -> 0, while
Е(Ж)-Ж)] = i
J
for every к.
A function which is absolutely continuous is clearly continuous. We will
also show that if/is absolutely continuous on [a,b]9 then/' exists a.e. in [я,/>].
This is an immediate corollary of (7.23) and the following theorem.
(7.27) Theorem If f is absolutely continuous on [a,b], then it is of bounded
variation on [a,b].
Proof Choose b so that £ |/(6f) — f(al)\ < 1 for any collection of non-
116
Differentiation
(7.28)
overlapping intervals with (b, — a^ < <5. Then the variation off over any
subinterval of [af] with length less than d is at most 1. Hence, if we split
[a,b] into N intervals each with length less than then V[a,b] < N.
The assumption above that \aj>\ is finite is necessary as shown by the
example /(x) = x.
A function f for which/' is zero a.e. in [#,Z>] is said to be singular on [a,b\.
The Cantor-Lebesgue function is an example of a nonconstant, singular
function on [0,1].
(7.28) Theorem If f is both absolutely continuous and singular on [a,b],
then it is constant on [a,b].
Proof. It is enough to show that f(a) = f(b) since this result applied to any
subinterval proves that/is constant. Let £ be the subset of (rz/), where f — 0,
so that |£| = b — a. Given e > 0 and x e £, we Jiave [x, x + h] cz («/) and
\f(x + Л) -- f(x)\ < eh for all sufficiently small h > 0. Let d be the number
corresponding to e in the definition of the absolute continuity of / By (7.18)
and (7.20), there exist disjoint Qj = [x;, Xj + /?J, j = 1, . . . , N, in (a,b)
such that
(i) |/(x7 + hj) -/(xy)| <
(ii) \Qj\ > (b - a) - <5.
By (i),
E \f(xJ + Ijj) < 6 £ I2J e(b - a)-
j=l
Moreover, since by (ii) the total length of the complementary intervals is less
than b, the sum of the absolute values of the increments of/over them is less
than e. Thus, the sum of the absolute values of the increments of f over the
Qj and the complementary intervals is less than e(b — a) + e. Hence, |/(Z>) —
f(a)\ < e(b — a) + e, so that f(b) ~ f(a). This completes the proof.
(7.29) Theorem A function f is absolutely continuous on \a,b\ if and only if
f' exists a.e. in [a,Z>],/' is integrable on [a,b] and
f(x) ~~ f(a) ~ Г/' (a < x < b).
J л
Proof. We have already observed (see p. 115) that any function of the
form (7(x) — j* g, g e LfaJPy is absolutely continuous on [a,b]. Hence, the
sufficiency of the condition follows.
Conversely, if/is absolutely continuous, let F(x) = f*f'. Fis well-defined
by virtue of (7.27) and (7.23). Moreover, by (7.16), F! = /' a.e. in [«/].
It follows that F(x) — /(x) is both absolutely continuous and singular on [a,b\.
Hence, by (7.28), we obtain F(x) — /(x) = F(a) — f(g) for x e [a,b\. Since
F(a) = 0, the proof is complete.
(7.31)
□solutely Continuous and Singular functions
I a /
(7.30) Theorem If f is of bounded variation on [«,&], then f can be written
f = g 4- Л, where g is absolutely continuous on [a,b] and h is singular
on [a,b]. Moreover, g and h are unique up to additive constants.
Proof. Let g(x) = \*f' and h =f- g. Then h' = f - д' = f - f =
0 a.e. in [a,b], so that h is singular, and the formula f — g + h gives the
desired decomposition. If / = gr + ht is another such decomposition, then
g — gY = — h. Since g — gr is absolutely continuous and hx — h is
singular, it follows from (7.28) that g — gY — hr — h~ constant, which
completes the proof.
The next theorem, which is an extension of (2.10), gives formulas for the
variations of an absolutely continuous function.
(7.31) Theorem Let f be absolutely continuous on [a,Z>], and let V(x), P(x),
and N(x) denote its total, positive and negative variations on [a,x],
a < x < b. Then V, P, and N are absolutely continuous on [a,b], and
V(x) = Г |/1 Л*) = Г {/'}! and N(x) = Г {/'}-.
J a J a J a
Proof. We will first show that V is absolutely continuous. If [ct,p] is a
subinterval of [a,6] and Г = {xk} is a partition of [a,/?], then
L(j8) - 7(a) = = sup £ 1Ж) -
Г
= sup£
Xk
Xk~l
Hence, if is a collection of nonoverlapping subintervals of [a,b\, then
нть n^)] f \n
ЛДаь/П]
From this inequality and (7Л), it follows that V is absolutely continuous on
[a,Z?]. Therefore, by (7.29) and the fact that V(a) = 0, we have V(x) = J* V'.
Since V' = |/'| a.e. by (7.24), we obtain V(x) = J* |/'|.
The fact that P and N are absolutely continuous and the formulas for P
and N now follow from the relations V(x) = J* |/'|, f(x) = f(a) 4- faf',
P(x) = 1[F(X) + ftx) - f(d)]> and N(x) = l[K(x) - f(x) + f(a)]. [See (2.6).]
This completes the proof. ‘
In Chapter 2, p. 23, we proved that if g is continuous on [я,6] and/is con-
tinuously differentiable on [я/], then
b Cb
9 df = gf' dx.
a j a
This is a special case of the first part of the following theorem.
118
Differentiation
(7.32)
(7»32) Theorem {Integration by Parts)
(i) If д is continuous on [a,b] and f is absolutely continuous on [a,b],
then
rb rb
\ gdf=\ gf'dx.
J a J a
(ii) If both f and g are absolutely continuous on [я,6], then
Cb rb
gf'dx = g(b)f(b) - g(a)f(a) - g'f dx.
J a J a
Proof To prove (i), first note that the integrals in the conclusion exist
and are finite by (2.24) and (5.30). Let Г = {xj be a partition of [a,b] with
norm |Г|. Then
gf' dx = £ gf' dx = e(xi-i) /'(*) dx
Ja JXi-i Jxi-i
+ X [ [#(*) - g(.xi-i)]f'(x)dx-
J Xi-1
The first term on the right equals £g(xf_ ~ f(xt-i)L which converges
to J* g df as |Г| -> 0. The second term on the right is majorized in absolute
value by
Pxi rb
[ sup |gf(x) - g(y)|] £ |/'| dx = [ sup |gr(x) - gr(j)|] dx.
|х-у|^|Г| Jxi-i |Х-У|^|Г| J«
Since g is uniformly continuous on \a,b\, the last expression tends to 0 as
|T| -> 0. This proves (i).
We can easily deduce (ii) from (i) by using (2.21). In fact, if f and g are
absolutely continuous, then
rb rb rb
gf'dx = I g df = g(b)f(b) - g(a)f(a) - fdg
J a J a J a
ГЬ
= g(b)f(t>) - g(a)f(a) - fg' dx.
J a
This proves (ii).
For a generalization of part (i) above, see (9.14) in Chapter 9.
6. Convex Functions
Let ф be defined and finite on an interval (a,b). Then ф is said to be convex
in (a,b) if for every [xnx2] (я>£), the graph of ф on [x15x2] lies on or below
the line segment connecting the points (х1,ф(х1)) and (x2,0(x2)) of the graph
of ф.
(7.36)
Convex t-unctions
I i a
Let у = L(x) be the equation of this line segment. Then since every x e
[xbx2] can be written x = 0Xi 4- (1 — 0)x2 for appropriate 0, 0 < 0 < 1,
the condition for convexity is
ф(0х1 4- (1 - 0)x2) < L(9x1 4- (1 - 0)x2)
for [x j ,x2] 02 (a,b) and 0 < 0 < 1. Since
L(0xr + (1 - 0)x2) = 0L(x^ 4- (1 - 0)L(x2) = 0ф(х^ 4- (1 - 0)ф(х2),
it follows that ф is convex in (a,b) if and only if
(7.33) ф(0х, + (1 - 0)x2) < Оф^) 4- (1 - 0)</>(x2)
for a < Xj < x2 < b, 0 < в < 1. Equivalently, ф is convex in (a,b) if and
only if
„ „ - JPlXl + P2X1\ „ + P2<K*2)
(7.34) ф\-----------:----- <-----------;--------
\ P1+P2 J Pl + P2
for a < xt < x2 < b, > 0, p2 > 0, pt + p2 >
(7.35) Theorem {Jensen"s Inequality) Let ф be convex in (a,b). Let {x7}j=1
be points of (a,b) and {Pj}j=i satisfy pj > 0 and ^Pj > 0. Then
'HLPjXj] YPj4(Xj)
^^PjJ' YPj *
The proof follows by repeated application of (7.34).
(7.36) Theorem
(i) If Ф1 and ф2 are convex in (af), then фх 4- ф2 is convex in (a,b).
(ii) If ф is convex in (a,b) and c is a positive constant, then сф is convex
in (a,b).
(iii) If фк, к ~ 1,2,..., are convex in (a,b) and фк ф in (a,bf then
ф is convex in (a,b).
The proof is left as an exercise.
(7.38)
(7.37) Theorem If ф' exists and is monotone increasing in (a,b), then ф is
convex in (a9b). In particular, if ф" exists and is nonnegative in (a9b)9
then ф is convex in (a,b).
Proof We shall use the following inequality: If bi9b2 > 0, then
. fat a2l at 4- a2 (a< afl
mm V’/Tf < г—, ~~ < max<~->^k
Pi 62J bt + b2 pi Z>2J
To prove the first inequality in (7.38), let m — min {«ipi,u2p2}. Then
mZ>i < «1 and mb2 < a2, so that т(Ьг + b2) < ax + a2. This is the desired
result. The second inequality is proved similarly.
The slope of the line connecting two points (а,ф(а)) and (р,ф(Р)) is
[ф(Р) — $(«)]/(/? ~ °0- Hence, to show that ф is convex, we must show that
if a < Xi < x < x2 < b9 then
#(*2) - ^(xj > ф(х) - ф(хх)
X2 — “ X — %1
Write
<Х*2) - <ft(*i) = 1Ф(х2) - ф(х)] + [ф(х) - </>(%!)]
X2 - *1 [x2 - x] + [x - xr]
Since ф' exists in (a,b), the mean-value theorem implies that there are £ t e
(xt,x) and £2 e (x,x2) such that
ф(х) - ф(х,) ф(х2) - ф(х)
—Т— -----------= Ф (Ci)> —;-----------— = Ф (£г)-
Л А1 А2 — А
Since ф' is increasing, ф'^г) < ф'^2\ and it follows from the first
inequality in (7.38) that
<X*2) - </>(x,) > ф(х) - ф(х{)
X2 — Xt ~ X — Xi
This completes the proof.
As a corollary of (7.37), we see that
(i) xp is convex in (0,oo) if p > 1.
(7.39) (ii) eax is convex in ( — oo,+ oo).
(iii) log 1/x ~ — logx is convex in (0,oo).
(7.40) Theorem ^ф is convex in (a,b), then ф is continuous in (a,b). More-
over, ф' exists except at most in a countable set and is monotone in-
creasing.
Proof, Since ф is convex, the slope [ф(х 4- h) — ф(х)]/Л, h > 0, decreases
with h. Hence, the derivative on the right,
n+ , , л г + h) - ф(х)]
D ф(х) = lim ------7---------,
h-*0 + "
exists and is distinct from 4- oo in (a,bf Similarly, the derivative on the left,
D ф(х) = lim ------г---------,
h-*0+ "
exists and is distinct from — oo in (a,b). Since [ф(х) — ф(х — А)]/Л <
[ф(х + h) — ф(х)]//г, h > 0, we obtain
(7.41) — oo < В~ф(х) < В+ф(х) <4-оо.
This shows in particular that ф is continuous in (a,b).
We next claim that
(7.42) В+ф(у) < В~ф(х) if a < у < x < b.
In fact, if у < x, then as seen from the discussion above, we have
£> Ф(у) < —rzy- D
This proves the claim. We therefore obtain В+ф(у) < В~ф(х) < В+ф(х)
if у < x, which shows that В+ф is monotone increasing. Similarly, В~ф is
monotone increasing.
To complete the proof of the theorem, note that В+ф can have at most a
countable number of discontinuities since it is monotone and finite on (a,b).
If x is a point of continuity of Z>+$, then letting у -> %— in the last inequalities,
we obtain В+ф(х) = В~ф(х). Therefore, ф' exists at every point of continuity
of В+ф, and the theorem follows.
^(7.43) Theorem If ф is convex in (a,b), then it satisfies a Lipschitz condition
on every closed subinterval of (a,b). In particular, if a < xt < x2 < b,
we have
ф(х2) - ф(х1) = I Ф'.
J XI
Proof Let [x^] be a closed subinterval of (a,b), and let xr < у < x <
x2. Then as before
n+ ,, x Ф(х) ~ Ф(У) n- , / ч
D+Ф(У) < ——- < D ф(х),
л У
иптегегшаиоп
(/.44)
so that, since й+ф and D ф are monotone increasing,
D - —7TT— - D ^2)-
л у
Hence, |ф(х) — ф(у)\ < C|x — j|, where C is the larger of |7)+ф(х1)| and
|Z>“ф(х2)|. This shows that ф satisfies a Lipschitz condition on [x1?x2], and
the rest of the theorem follows.
The next result is a useful version of Jensen’s inequality for integrals.
We shall need the notion of a supporting line: If ф is convex on (a,b) and
x0 e (a,b), a supporting line at x0 is a line through (x0,<^(x0)) which lies on
or below the graph of ф on (a,b). It follows from the discussion preceding
(7.41) that any line through (х0,ф(х0)) whose slope m satisfies Z>"^(x0) <
m < 1)+ф(х0) is a supporting line at xQ.
(7.44) Theorem (Jensen's Integral Inequality) Let f and p be measurable
functions finite a.e. on a set A c R\ Suppose that fp and p are inte-
grable on A,p > 0, and fAp >0. If ф is convex in an interval contain-
ing the range off then
Jhfp\ L <Kf)p
X$ap) ~ Lp
Proof. By hypothesis,/is finite a.e. in A. Choose (a,Z>), — 00 < a < b <
+ °o, so that ф is convex in (a,b), and so that a </(x) < b for every x at
which fix) is finite. The number 7 defined by
is finite and satisfies a < у < b. If m is the slope of a supporting line at у and
a < t < b, then ф(у) + m(t — y) < </>(!)• Hence, for almost every x e A,
Ф(у) + w[/(x) - y] < ^(/(x)).
Multiplying both sides of this inequality by p(x) and integrating the result
with respect to x, we obtain
</>(?) \ P + m ( \ fp - у p ) < (f>(f)p.
JA \JA JA / JA
Here the existence of f^if)p follows from the integrability of p and fp.
[The continuity of ф implies that 0(/) is measurable.] Since fAfp ~~ 7/AZ;
- 0, the last inequality reduces to
i»AG!
£ ей. а
Ф(у) \ р Ф(Лр>
JA JA
which is the desired result.
Exercises
1. Let /be measurable in R” and different from zero in some set of positive meas-
ure. Show that there is a positive constant c such that /*(x) > c|x|“" for
|x| > 1.
2. Let <^(x), x e R”, be a bounded measurable function such that ф(х) — 0 for
|x| > 1, and f ф = 1. For £ > 0, let ^E(x) = £~пф(х1£). (фе is called an approxi-
mation to the identity If/e L(R"), show that
lim (/*Ox) = /(x)
E-*0
in the Lebesgue set of /. [Note that f фс = 1, e > 0, so that
(M)(x) -/(x) = j[/(x - y) -/(x)]«y)rfy.
Use (7.16).]
3. Show that the conclusion of (7.4) remains true for the case of two dimensions
if instead of being squares, the sets Q covering E are rectangles with x-dimension
equal to h and ^-dimension equal to h2. (Of course, h varies with Q.)
Show that the same conclusion is valid if the ^-dimension is any increasing
function of h. Generalize this to higher dimensions.
4. If, Ei and E2 are measurable subsets of R1 with JEJ >0 and |E2| > 0, prove
that the set {x : x = Xi — x2, eEb x2 gE2} contains an interval, [cf.
(3.37).]
5. Let / be of bounded variation on [u,Z>]. If / = g + A, where g is absolutely
continuous and h is singular, show that
\\f'dx + \^dh,
J a J a J a
for any continuous ф.
6. Show that if а > 0, xa is absolutely continuous on every bounded subinterval
of [0,oo).
7. Prove that f is absolutely continuous on [a,b] if and only if given e > 0, there
exists b > 0 such that |£ [/(/>,) ~-/W]| < £ for any finite collection
of nonoverlapping subintervals of [л,6] with £ (bt — at) < d.
8. Prove the following converse of (7.31): If f is of bounded variation on [л,/>],
and if the function F(x) = И[л,х] is absolutely continuous on [л,6], then f is
absolutely continuous on [zz,Z>].
9. If/is of bounded variation on [g,b], show that
£ I f'\ < И[а,6].
Show that if equality holds in this inequality, then/is absolutely continuous
124
Differentiation
on [a,Z>]. (For the second part, use (2.2)(ii) and (7.24) to show that Их) is
absolutely continuous, and then use the result of Exercise 8.)
10. Show that if f is absolutely continuous on [a,6] and Z is a subset of [a,6] of
measure zero, then the image set defined by /(Z) = {и>: w = /(z), z eZ) also
has measure zero. Deduce that the image under f of any measurable subset of
[a,/>] is measurable. [Compare (3.33).] [Hint: use the fact that the image of an
interval ДД] is an interval of length at most — Е(я,).]
11. Prove the following result concerning changes of variable. Let g(t) be monotone
increasing and absolutely continuous on [a,/?], and let /be bounded and meas-
urable on [a,b], a = #(a), b = g(fi). Then f(g(t)) is measurable on [a,/?] and
(bf(x)dx= [*f(g(t))g'(t)dt.
J a J a
(Consider the cases when/is the characteristic function of an interval, an open
set, etc.)
12. Use Jensen’s inequality to prove that for a,b > 0, p.q > 1, (1/p) + (l/<7) = 1,
we have
ab < — 4-----.
P Q
More generally, show that
v» aJ Pj
’ * aN < 2 -d-d
j=i Pj
where aj > 0, pj > 1, £j=i (J/Pj) = L (Write aj = exdpjand use the convexity
of ex.)
13. Prove theorem (7.36).
14. Prove that ф is convex on (a,b) if and only if it is continuous and
JX1 ф(хд 4- ф(х2)
" 2
for Xi,*2 e (a,b).
15. Theorem (7.43) shows that a convex function is the indefinite integral of a
monotone increasing function. Prove the converse: If ф(х) — jpf(f)dt 4- ф(а)
in (a,b) and/is monotone increasing, then ф is convex in (atb). (Use Exercise 14.)
16. Show that the formula
for integration by parts may not hold if/is of bounded variation on (— oo, 4- °°)
and g is infinitely differentiable with compact support. (Let / be the Cantor-
Lebesgue function on [0,1], and let / = 0 elsewhere.)
17. A sequence {фк} of set functions is said to be uniformly absolutely continuous
if given e > 0, there exists b > 0 such that if E satisfies |E | < <5, then | фк(Е) | < £
for all k. If {fk} is a sequence of integrable'Tunctions on (0,1) which converges
pointwise a.e. to an integrable / show that fol/ — Al * 0 if and only if the
indefinite integrals of the fk are uniformly absolutely continuous.
Chapter 8
Lp Classes
1. Definition of £p
If E is a measurable subset of R" and p satisfies 0 < p < oo, then LP(E)
denotes the collection of measurable/for which J£ |/|p is finite, that is,
LP(E) = I \f\p < Tool, 0 <p < oo.
I Je J
Here, f may be complex-valued. (See Exercise 3 of Chapter 4 for the defini-
tion of measurability of vector-valued functions.) In this case, if f = /x 4- if2
for measurable real-valued and/2, we have \f\2 = f2 + /2, so that
I/1M/2I < l/l I/il + I/2I.
It follows that /e LP(E} if and only if both /1?/2 e LP(E). (See Exercise 1.)
We shall write
У IIГ !l P IП (0 </> < co);
thus, LP(E) is the class of measurable/ for which ||/||Pj£ is finite. Whenever
it is clear from context what E is, we will write Lp for LP(E) and ||/||p for
ll/llp.E- Note that = L-
In order to define L^^E), let /be real-valued and measurable on a set E
of positive measure. Define the essential supremum of / on E as follows:
If |{x e E : /(x) > a}| > 0 for all real a, let ess£sup/ = + 00; otherwise, let
ess sup/ = inf {a : |{x e E :/(x) > a}| = 0}.
E
Since the distribution function m(a) = |{x e E : /(x) > a} | is continuous from
the right [see (5.39)], it follows that co(ess£sup/) = 0 if ess£sup/ is finite.
Therefore, ess£sup/is the smallest number M, — 00 < M < +co, such that
f(x) < M except for a subset of E of measure zero.
\ 125
I <co
uj Classes
(8.1)
A real- or complex-valued measurable f is said to be essentially bounded,
or simply bounded, on E if ess£sup |/| is finite. The class of all functions that
are essentially bounded on E is denoted by LX(E). Clearly,/belongs to LK(E)
if and only if its real and imaginary parts do. We shall write
ll/lloo = ll/lloo.E = esssup|/|.
E
Thus, II/IIqo is the smallest M such that |/| < M a.e. in E> and
= +oo}.
The following theorem gives some motivation for this notation.
(8.1) Theorem If\E\ < Too, then Ц/L = lim^ ||/||p.
Proof, LetM= ll/lloo. If M' < M9 then the set A = {xeE: |/(x)| > M'}
has positive measure. Moreover, ||/||p > (JA |/|p)1/p > 7И'|Л|1/р. Since
|Л|1/Р -> 1 as p -> oo, it follows that liminf^||/||p > so that
liminfp^oo 11/1/ ЛТ- However, we also have ||/|/ < (jE Mp)1/p = M\E\1/p.
Therefore, lim sup^^ 11/1/ which completes the proof.
Remark: This result may fail if lEI = +oo. Consider, for example, the con-
stant function/(x) = c, c / 0, in (0,oo). Clearly,/eL00 but/^Lp for 0 <p<co.
We will how study some basic properties of the Lp classes.
(8.2) Theorem TfO < pt < p2 < oo and \E\ < -boo, then LP2 c LP1.
Proof Write E = Er u E29 Er being the set where |/| < 1, and E2 the
set where |/| > 1. Then
[ \f\p = [ \f\p + f l/Г, 0<p< oo.
JE JEi JE2
The first term on the right is majorized by 1^1; the second increases with p
since its integrand exceeds 1. It follows that if/e LP2, p2 < oo, then/e LPl,
pt < p2. If p2 = oo, then/is a bounded function on a set of finite measure,
and so belongs to LP1.
Remarks:
(i) The hypothesis above that E have finite measure cannot be omitted:
for example, x~1/pi belongs to LP2(l,oo) ifp2 > p19 but does not belong
to LP1(l,oo). Again, any nonzero constant is in £°°, but is not in LPl(E)
if H - + oo and p{ < oo.
(ii) A function may belong to all LPl withp/< p2 and yet not belong to LP2.
In fact, if p2 < oo, x“1/P2 belongs to LP1(O,1), p{ < p2, but does not
belong to Z/2(O,1); log l/xisin^/OJjfor^j < oo, but is not in L00^,!).
(iii) We leave it to the reader to show that any function which is bounded on
(0.0/
turner ь HiuquaiiLy, iyuii\uwaKi ь snequditiy
4 £-i
E (|£| < 4-oo or not) and which belongs to LP1 also belongs to LP2,
Pi > Pi-
The next theorem states that the Lp classes are vector (i.e., linear) spaces.
Its proof is left as an exercise.
(8.3) Theorem Iffg e LP(E), p > 0, then f 4- g e LP(E) and cfe LP(E) for
any constant c.
2. Holder’s Inequality; Minkowski’s Inequality
In order to discuss the integrability of the product of two functions, we will
use the following basic result.
(8.4) Theorem (Young’s Inequality) Let у = ф(х) be continuous, real-
valued, and strictly increasing for x > 0, and let ф(0) = 0. Ifx =
ф(у) is the inverse of ф, then for a,b > 0,
Га ГЬ
ab < ф(х) dx 4- ф(у) dy.
Jo Jo
Equality holds if and only if b = ф(а).
Proof A geometric proof is immediate if we interpret each term as an
area and remember that the graph of ф also serves as that of ф if we inter-
change the x and у axes. Equality holds if and only if the point (a,b) lies on
the graph of ф.
If ф(х) = xa, а > 0, then ф(у) — y1/a, and Young’s inequality becomes
ab < я1+7(1 4- a) 4- bi + 1/a/(l 4- 1/a). Setting/? = a 4- 1 and/?' = 1 4- 1/a,
we obtain
ap bp' .11
(8.5) ab <--------1—т if a,b > 0, 1 < p < oo, and - 4—7=1.
P P P P
Two numbers p and p' which satisfy 1//? 4- 1//?' = 1, /?,/?' > 1, are called
conjugate exponents. Note that p' = p/(p - 1), and that 2 is self-conjugate.
We will adopt the conventions that /?' = co if p = 1, and /?' = 1 if /? = co.
'! Zb
is Glasses
(b.b)
(8.6) Theorem (Holder's Inequality) If 1 < p < oo and 1/p 4- 1/p' = 1,
then < ||/||p||fir||p.; that is,
- Г /г \ vp / r \
I/0I m l^r (1 <P < oo);
JE \JE J \JE /
f Ш < (ess sup |/|) [ |^|.
Je e Je
Proof. The last inequality, which corresponds to the case p = oo, is
obvious. Let us suppose then that 1 < p < oo. In case ||/||p = ||д||р, = 1,
(8.5) implies that
Г in f
\fg\ < I — + —г I = 4- —7-
Je Je \ P P / P P
=7+7=1=
For the general case, we may assume that neither ||/||p nor ||#||pz is zero;
otherwise fg is zero a.e. in E, and the result is immediate. We may also
assume that neither ||/||p nor ||д||р> is infinite. If we set fY = //||/||p and gr =
glWgWp'i then ||/i||p = = 1. Therefore, by the case above, we have
b 1/1011 1; i.e., JE|/3l < 11/М0lip., as desired.
The case p = p' = 2 is a classical inequality.
(8.7) Corollary (Schwarz’s Inequality)
p / p \ 1/2 / p \ 1/2
\fg\ < l/l2 Ы2 •
Je \ Je / \ Je J
The theorem which follows is usually referred to as the “converse of
Holder’s inequality.” (See also Exercise 15 in Chapter 10.)
(8.8) Theorem Let f be real-valued and measurable on E, let 1 < p < 00
and 1/p 4- l/pf = 1. Then
(8.9) ll/llp = sup f fg,
JE
where the supremum is taken over all real-valued g such that ||g||p/ < 1
and Ie/9 exists.
Proof. That the left-hand side of (8.9/ majorizes the right-hand side
follows from Holder’s inequality. To show the opposite inequality, let us
consider first the case of / > 0, 1 < p < co.
If ll/llp = 0, then f = 0 a.e. in E, and the result is obvious. If 0 < \\f\\p <
+ oo, we may further assume that ||/||p = 1 by dividing both sides of (8.9) by
H/llp. Now let g = fp/p'. It is easy to verify that ||g||p, = 1 and \Efg = 1,
which completes the proof in this case.
If || f || p = +oo, define functionson E by setting
fk(x) = 0 if |x| > k, fk(x) = min [f(x),k] if |x| < k.
Then each fk belongs to Lp and ||/J|P -> ||/||p = + oo. By the case already con-
sidered, we have ||/J|P = $Efkgk for some gk > 0 with ||рл||р, = 1. Since
f > fk, it follows that
\ f9k £ fkdk -’+<».
Je Je
This shows that
sup \ fg = +00 = \\f\\p.
Шр'=1 Je
To dispose of the restriction f > 0, apply the result above to |/|. Thus,
there exist gk with ||gfc||p, = 1 such that
\\f\\p = lim f \f\gk = lim f fgk,
Je Je
where gk = gk (sign/). (By sign x, we mean the function equal to +1 for
x > 0 and to —1 for x < 0.) Since = 1, the result follows.
The cases p = 1 and oo are left as exercises.
We have already observed that the sum of two Lp functions is again in Lp.
The next theorem gives a more specific result when 1 < p < oo.
(8.10) Theorem (Minkowski's Inequality) If 1 < p < oo , then \\f + g\\p <
ll/llP + ll^llp; that is,
a\ 1/p / r \ 1/p / f \ 1/p
E\f+9\P) +(JE|0|J
ess sup I/ + g\ < ess sup |/| + ess sup |p|.
E EE
Proof. If p = 1, the result is obvious. Ifp = oo, we have |/| < Ц/Цда a.e.
in E and |p| < HpL a.e. in£. Therefore, |/+ g\ < ||/IL + UpL a.e. in E,
so that ||/+ 0||да < H/IL + HpL.
For 1 < p < co,
ll/+<= C I/+0|P= f I/+0Г11/+d
Je Je
< [ + f 1/+0ГхЫ.
Je Je
130
Lp Classes
(8.11)
In the last integral, apply Holder’s inequality to \f + g\p 1 and |g| with
exponents p' — p/(p — 1) and p, respectively. This gives
f \f+ дГ'\д\ < И+<7ИГ1Ш,-
JE
Since a similar result holds for J£|/ + ^lp-1|/l? we obtain ||/T g\\p <
II/ + ^llp”1 (ll/llp + 1Ы1Д and the theorem follows by dividing both sides by
ll/T^llp”1- [Note that if ||/+ g\\p = 0, there is nothing to prove; if
II/ + g\\P = +°o, then either ||/||p = + oo or ||/IP = Too by (8.3), and the
result is obvious again.]
Remark: Minkowski’s inequality fails for 0 < p < 1. To see this, take
E = (0,1),/- x(0 x) and g = %(x Then ||/ + g\\p = 1, while \\f\\p + \\g\\p
= + 2~l/p = 21~1/p < 1. [See also (8.17).]
3. Classes lp
Let a = {ak} be a sequence of real or complex numbers, and let
llallp = (E 0 < P < co; Hah = sup |<zt|.
к к
Then a is said to belong to lp, 0 < p < oo, if ||/|p < + oo, and to belong to
/°° if hL < +«).
Let us show that if 0 < pY < p2 < oo, then lPt <= lP2. [The opposite
inclusion holds for LP(E), |£| < + oo, by (8.2).] For p2 = oo, this is clear,
and for p2 < oo, it follows from the fact that if |6zfc| < 1, then |#JP2 < |як|Р1.
An example of a sequence which is in lP2 for a given p2 < oo but which is not
in lpi for pY < p2 is {(А/к log2 к)1/Р2 : к > 2}. Any constant sequence {ak},
ak — c =£ 0, belongs to Z00 but not to lp for p < oo. The same is true for
{1/log к : к > 2}, whose terms even tend to zero.
(8.11) Theorem If a = {ak} belongs to lp for some p < oo, then Птр^да ||a||p
= Halloo.
Proof If a g lPo, then aelp for p0 < p < oo. Since |ak| -> 0, there is a
largest |ak|, say |aio|. Thus, Цац» = |aj. Write £ |at|p = |afto|₽ £ \ak/ako\p.
Since |tffc/zko| < 1, we see that £ |як/яко? decreases (and so is bounded) as
p -► oo. Hence, there is a constant c such that |яко|р < ||fl||£ < c|#ko|p. Since
c1/p -> 1 as p -> oo, the theorem follows.
The next two results are analogues for series of Holder’s and Minkowski’s
inequalities. Their proofs are left as exercises. If a — {a^ and b — {bk}, we
use the notation
ab — {й/Д}, a + b = {ak + bk}, etc.
(8.12) Theorem (Holder's Inequality) Suppose that 1 < p < oo, 1/p T 1/p'
(8.13) 1 Banach and Metric Space Properties 131
= 1, a = {aj, b = {bk}, and ab = {«Д}. Then ||aZ>||j < ||«llpll^!lp'
that is,
Z l«AI S (Z 1«J₽)1/P (Z l^lp')1/p' (1 < p < oo);
Z \akbk\ (sup |at|)Z Д1-
(8.13) Theorem (Minkowski's Inequality) Suppose that 1 < p < oo, a =
{ak},b = {bk},anda + b = (ak + bk}. Then ||a + 6||p < ||a||p + ||Z>||p;
that is,
(Z + ^lp)1/p < (Z Ыр)1/р + (Z IMP)1/P (1 P < <*>);
sup \ak + bk\ < sup |aj + sup Д|.
Even though Minkowski’s inequality fails when p < 1 (see Exercise 3),
lp is still a vector space for 0 < p < 1; that is, a + b g lp and aa = {&ak} g lp
if a,b g lp and a is any constant.
4. Banach and Metric Space Properties
We now define a notion which incorporates the main properties of Lp and lp
when p > 1. A set У is called a Banach space over the complex numbers if it
satisfies the following conditions:
(BJ X is a linear space over the complex numbers C; that is, if x,y g X and
a g C, then x -4- у g X and ax g X.
(B2 ) X is a normed space; that is, for every x g X there is a non-negative
number ||x|| such that
(a) ||x|| = 0 if and only if x is the zero element of X.
(b) ||ax|| = |a| ||x|| for a g C; x g X.
(C) h + я < INI + IK
If these conditions are fulfilled, ||x|| is called the norm of x.
(B3 ) X is complete with respect to its norm; that is, every Cauchy sequence
in Xconverges in X, or if ||xfc — xm|| -> 0 as k,m oo, then there is an
x g X such that ||xk — x|| -> 0.
A set X which satisfies (BJ and (B2), but not necessarily (B3), is called a
normed linear space over the complex numbers. A sequence {vfc} such that
\\xk — x|| -> 0 as к -> oo is said to converge in norm to x.
Restricting the scalars a in (BJ and (B2) to be real numbers, we obtain
definitions for a Banach space over the real numbers and for a normed linear
space over the real numbers. Unless specifically stated to the contrary, we will
take the scalar field to be the complex numbers.
If JU is a Banach space, define d(x,y) = ||x — j|| to be the distance between
x and y. Then,
132
Lp Classes
(MJ d(x,y) > 0; d(x,y) = 0 if and only if x = y.
(M2) d(x,y) = d(y,x).
(M3) d(x,y) < d(x,z) -4- d(z,y) (triangle inequality).
Any set which has a distance function d(x,y) satisfying (MJ, (MJ,
and (M3) is called a metric space with metric d. Therefore, a Banach space
is a metric space whose metric is the norm. Moreover, by (B3), a Banach
space Xis a complete metric space; that is, if d{xk,xf) -> 0 as kpn -> oo, then
there is an x e X such that d(xk,x) -> 0.
(8.14) Theorem For 1 < p < oo, LP(E) is a Banach space with norm
ll/ll = ll/llp.E-
Proof. Parts (B J and (B2) in the definition of a Banach space are clearly
fulfilled by LP(E\ parts (a) and (c) of (B2) being (5.11) and Minkowski’s
inequality, respectively. [Regarding part (a), we do not distinguish between
two Lp functions which are equal a.e.; thus, the zero element of LP(E) means
any function equal to zero a.e. in E.]
To verify (B3), suppose that {/J is a Cauchy sequence in LP(E). Ifp = oo,
then |A -fm\ < ||А - ЛИ® except for a set Zt>m of measure zero. If Z =
Ut,m zk,™ then z has measure zero, and \fk - fm\ < || fk -/X outside Z
for all к and m. Hence, {fk} converges uniformly outside Z to a bounded limit
f and it follows that \\fk — fW^ -> 0. (Note that convergence in £°° is equiva-
lent to uniform convergence outside a set of measure zero.)
In case 1 < p < oo, Tschebyshev’s inequality (5.49) implies that
|{xe^:|A(X)-4(x)|>£}| <в~р
E
Hence, {fk} is a Cauchy sequence in measure. By (4.22) and (4.23), there is a
subsequence {fk} and a function f such that fk -> f a.e. in E. Given e > 0,
there is a К such that
/ r \ i/p
1Д-А1₽ = IIA;-Allp<eifM>^
\JE /
Letting kj -> oo, we obtain by Fatou’s lemma that ||/ — AHP < & if к > К.
Hence, ||/-Allp->0 as £-> oo. Finally, since ||/||p < \\f - fk\\p + II Allp <
+ oo, it follows that/e LP(E\ which completes the proof.
A metric space X is said to be separable if it has a countable dense subset;
that is, X is separable if there exists a countable set {xj in X with the property
that for every x e X and every в > 0, there is an xk with d(x,xk) < в. In the
next theorem, we will show that Lp is separable if 1 < p < oo. Note that L00
is not separable: take Z/°(0,l), for example, and consider the functions
/(%) = X(o,o(x)> 0 < / < 1. There are an uncountable number of these, and
ll/t — /tdloo = 1 if / A C. See also Exercise 10.
(8.17)
Banach and Metric Space Properties
133
(8.15) Theorem If\ < p < oo, LP(E) is separable.
Proof. Suppose first that E = R", and consider a class of dyadic cubes
in R”. Let D be the set of all (finite) linear combinations of characteristic
functions of these cubes, the coefficients being complex numbers with rational
real and imaginary parts. Then D is a countable subset of ZZ(R"). To see that
D is dense, use the method of successively approximating more and more
general functions: First, consider characteristic functions of open sets [every
open set is the countable union of nonoverlapping dyadic cubes by (1.11)],
of G5 sets, and of measurable sets with finite measure; then consider simple
functions whose supports have finite measure, nonnegative functions in
Z/(R”), and, finally, arbitrary functions in Lp(Rn). The details are left to the
reader [cf. lemma (7.3)]. This proves the case E = R".
For an arbitrary measurable E, let Dr denote the restrictions to E of
the functions in D. Then D' is dense in LP{E), 1 < p < oo. In fact, given
p an&feLp(E), let/ = / on E and = 0 off E. Then fr eF(Rn), so
that given s > 0, there exists g e D with (JR„ \fx — g\p dxfilp < s. Therefore,
(Je I/ “ g\pdxf/p < £. This shows that D' is dense in LP(E) and completes
the proof.
As we have already noted, Minkowski’s inequality fails when 0 < p < 1.
Therefore, || • is not a norm for such p. However, we still have the follow-
ing facts.
(8.16) Theorem If 0 < p < 1, LP(JE) is a complete, separable metric space,
with distance defined by
d(fig) = II/- g\\p,E.
Proof With d(fig) so defined, properties (MJ and (M2) of a metric space
are clear. To verify (M3), which is the triangle inequality, we first claim that
(a + b)p < ap + bp if a,b > 0, 0 < p < 1.
If both a and b are zero, this is obvious. If, say, а ф 0, then dividing by ap,
4ve reduce the inequality to (1 + t)p < 1 + tp, t > 0 (t = b/d). This is clear
since both sides are equal when t = 0 and the derivative of the right side
majorizes that of the left for t > 0.
It follows that |/(x) — #(x)|p < |/(x) — 7z(x)|p -4- |Л(х) — g(fi)\p if 0 <
p < 1. Integrating, we obtain \\f- g\\p < ||/- h\\p + ||/z - g\\p, which is
just the triangle inequality. The proofs that Lp is complete and separable with
respect to || • ||p are the same as in (8.14) and (8.15).
It is worth noting that the triangle inequality is equivalent to the basic
estimate
(8.17) ||/+ g\\* < ИЛ' + IK (0 </> < 1).
The analogous results for series are listed in the following theorem.
134
Lp Classes
(8.18)
(8.18) Theorem
(j) If 1 < p < oo, lp is a Banach space with ||a|| = ЦлЦ^. For 1 <
p < oo, Zp is separable’, F° is not separable.
(ii) T/'O < p < 1, lp is a complete, separable metric space, with distance
d(a,b) = ||a -
Proof. We will show that lp is complete and separable when 1 < p < oo
and that Z00 is not separable. The rest of the proof of (i) and the proof of (ii)
are left to the reader.
Suppose that 1 < p < oo, я(1) = {^°} g lp for z = 1,2,..., and
||a(0 _ a(7)||p -> 0 as i,j -» oo. Since ||я(1) — a(7)||p > — <4J)| for every k,
it follows that |a[l} — а^\ 0 for every к as i,j -> oo. Let ak = Нт^да a(ki}
and a ~ {rzA}. We will show that aelp and ||#(/) — a\\p -> 0. Given e > 0,
there exists N such that,
(L 14° - «mi/₽ = ll«(i) - < £ if i,j > N.
к
Restricting the summation to к < M and letting / -> oo, we obtain
м . \ 1 ip
£ \a[l) — ak\p I < e for any M, if i > N.
k= i J
Letting M -+ oo, we get ||я(0 ~ a||p < £ if i > N; that is, ||я(1) — a||p -+ 0.
The fact that ||a||p < ||a — a(l)||p T ||tz(l)||p shows that a e lp.
To prove that lp is separable when p < oo, let D be the set of all sequences
{dk} such that (a) the real and imaginary parts of dk are rational, and (b) dk =
0 for к > N (N may vary from sequence to sequence). Then D is a countable
subset of lp. If a = {ak} g lp and £ > 0, choose N so that 1 < e/2.
Choose dt, ... , dN with rational real and imaginary parts such that
^k=i \ak “ dk\p < e/2. Then d — {dr, . . . , dN,0, . . .} belongs to D and
||a — d\\p < £. It follows that D is dense in lp, and therefore that lpis separable.
To see that Z00 is not separable, consider the subclass of sequences a =
{ak} for which each ak is 0 or 1. The number of such sequences is uncountable,
and \\a — «/Ида = 1 for any two different such sequences. Hence, Z00 cannot
be separable.
We know from Lusin’s theorem that measurable functions have continuity
properties. The next theorem gives a useful continuity property of functions
in Lp.
(8.19) Theorem {Continuity in If) Iff e U(W), 1 < p < oo, then
lim [|/(x + Ii) - /(x)||p = 0.
|h|-0
Proof. Let Cp denote the class of/e Lp such that ||/(x + h) — f(x) ||p -> 0
as ]h| -> 0. We claim that (a) a finite linear combination of functions in Cp is
(8.20) The Space L2; Orthogonality 135
in Cp, and (b) fkeCp and \\fk — /||p -> 0, then /g Cp. Both of these facts
follow easily from Minkowski’s inequality; for (b), for example, note that
||/(x + h) - /(x)||p
< ||/(x + h) -A(x + h)||„ + ||A(X + h) -/(x)||p + ||/fc(x) -/(x)||p
= ||A(x + h) -A(X)||P + 2||Л(х) ~/(x)||p.
Since fkeCp, we have limsup|hH0 ||/(x + h) -/(x)||p < 2||/t(x) - /(x)||p,
and (b) follows by letting к -> oo.
Clearly, the characteristic function of a cube belongs to Cp. Hence, in
view of the fact that linear combinations of characteristic functions of cubes
are dense in LP(R”) [see (8.15)], it follows from (a) and (b) that Lp is contained
in Cp.
We remark without proof that this result is also true for 0 < p < 1.
(Use the same ideas for || • ||p.) It fails, however, for p = oo, as shown by the
function % = X(o,oo)(x) on (-oo, + oo). In fact, % g Lc0(—oo, +oo) but
||x(x 4- h) - x(4lloo = 1 for all h =£ 0.
5. The Space L2; Orthogonality
For complex-valued measurable /, f = 4- if2 with fx and /2 real-valued
and measurable, we have J£/ = J^/j 4- i J£/2 (see P- 76). We will use the
fact that |J£/| < J£ |/|. (See Exercise 1.)
Among the Lp spaces, L2 has the special property that the product of any
two of its elements is integrable (Schwarz’s inequality). This simple fact leads
to some important extra structure in L2 which we will now discuss.
Consider L2 = L2(E), where E is a fixed subset of Rn of positive measure,
and write ||/|| = ||/||2>£, Je/ = J/ etc. For /# G define the inner product
of f and g by
(8.20) </#> = j/£,
where g denotes the complex conjugate of g. Note that by Schwarz’s in-
equality,
K/<7>l < \\f\w\g\\-
Moreover, the inner product has the following properties:
(a) <g,f> = <//>•
(b) </1 +f2, g> = + </2,0>, </,01+ 02> = </,0t> + </,02>-
(c) <a/,0> = = a<f,g>, a. e C.
(d) <//>1/2 = ||/||.
fo </#> = then f and g are said to be orthogonal. A set {фа}аеЛ *s
orthogonal if any two of its elements are orthogonal; {фа} is orthonormal if
it is orthogonal and \\фа\\ = 1 for all a. Note that if {фа} is orthogonal and
i^o is biasses /
\\фа\\ Ф 0 for every a, then {</>a/|| </>a ||} is orthonormal. Henceforth, we will
assume that \\фа\\ / 0 for all a for an orthogonal system {фа}. This implies
that no element is zero and that no two elements are equal.
(8.21) Theorem Any orthogonal system {фа} in L2 is countable.
Proof. We may assume that {фа} is orthonormal. Then for a / /?, we have
- <M2 = = Ш2 + ll-Ы2 = 2,
so that ||фа — фр\\ = y/2. Since L2 is separable, it follows that {фл} must be
countable.
A collection фг, . . . , фы is said to be linearly independent if акФк(х)
= 0 (a.e.) implies that every ak is zero. An infinite collection of functions is
called linearly independent if each finite subcollection is. No function in a
linearly independent set can be zero a.e.
(8.22) Theorem If {фк} is orthogonal, it is linearly independent.
Proof. Suppose that ap//kl T • * • + aNil/kN = 0. Multiplying both sides
by if/kl and integrating, we obtain by orthogonality that at = 0. Similarly,
a2 ~ “ aN ~ 0-
The converse of (8.22) is not true. However, the next result shows that if
{фк} is linearly independent, then the system formed from suitable linear
combinations of its elements is orthogonal.
(8.23) Theorem (Gram-Schmidt Process) If {фк} is linearly independent,
then the system {фк} defined by
Ф1 = Ф1
Ф2 = а21Ф1 + ф2
Фк = ак1Ф\ + + + фк
is orthogonal for proper selection of the a^.
Proof. Having фг = proceed by induction, assuming that фг,... ,
фк-г have been chosen as required. We will determine constants bki,. . . ,
bk,k-i so that the function фк defined by
Фк = Ьк1Ф1 T ’ ‘ * T Ьк>к_1фк^1 -4- фк
is orthogonal to фг, . . ., фк-1. If j < к,
by orthogonality. Since Ф 0, bkj can be chosen so that (фк9ф^ — 0,
rourier benes; rarsevai s rormuia
id/
j < k. Since each with j < к is a linear combination of ф19 . . ; , фр the
theorem follows.
When the фк are selected by the Gram-Schmidt process, we shall say that
they are generated from the фк. Note that the triangular character of the
matrix in (8.23) means that each фк can also be written as a linear combina-
tion of the фр j < k.
We call an orthogonal system {фк} complete if the only function which
is orthogonal to every фк is zero; that is, {фк} is complete if </,</>k> = 0 for
all к implies that f = 0 a.e. Thus, a complete orthogonal system is one
which is maximal in the sense that it is not properly contained in any larger
orthogonal system.
The span of a set of functions {фк} is the collection of all finite linear
combinations of the фк. In speaking of the span of {фк}9 we may always
assume that {фк} is orthogonal by discarding any dependent functions and
applying the Gram-Schmidt process to the resulting linearly independent set.
A set {фк} is called a basis for L2 if its span is dense in L2; that is, {фк}
is a basis if given f g L2 and s > 0, there exist N and {ak} such that Ц/ —
^k= i акФк\\ < e- The ak can always be chosen with rational real and imaginary
parts. Any countable dense set in L2 is of course a basis. It follows that L2 has
an orthogonal basis.
(8.24) Theorem Any orthogonal basis in L2 is complete. In particular, there
exists a complete orthonormal basis for L2.
Proof Let {фк} be an orthogonal basis for L2. We may assume that {фк}
is orthonormal. To show that it is complete, let = 0 for all k. Then
</,/> = </>/“ ak^k) for any N and ak. By Schwarz’s inequality,
|</,/>| < И1Ц1/- E^=i аЛН> and so, since the term on the right can be
chosen arbitrarily small, </,/) = 0. Therefore, f = 0 a.e., which completes
the proof.
6. Fourier Series; Parseval’s Formula
Let {фк} be any orthonormal system for L2. If/G L2, the numbers defined by
= ck(f) = = f f$k
JE
are called the Fourier coefficients off with respect to {фк}. The series Ед скфк
is called the Fourier series of f with respect to {фк}, and denoted 5[/J =
Efc скФк- We also write
/ ~ E скФк-
к
The first question we ask is how well S[/], or more precisely, the sequence
offiits partial sums, approximates f Fix Nand let L = укфк be a linear
combination of ф1? . . . , We wish to know what choice of }q, . . . , yN
makes ||/~ L|| a minimum. Note that since {фк} is orthonormal, ||L||2 =
<L,L) = E"=i Ш2- Hence,
и- ah2 = [(7- ЬаУ/- e укФк)
J \ k=l / \ k=l /
= ll/ll2 - E (УЛ + УЛ) + E 1Ул1%
k=l k=l
where the ck are the Fourier coefficients of f Since
|Q - yk\2 = (c* - 7k)(Q - У*) = IqI2 - (ykck + У/Д) + lyj2>
we obtain
II/-ill2 = ll/ll2 + E lQ-yJ2 - E Ы2-
k=l k=l
Therefore,
(8.25) min ||/-Z||2 = ll/ll2 - E kfcl2;
yi,-'-,7N k=l
that is, the minimum is achieved when yk = ck for к = 1, . . . , N, or equiva-
lently, when L is the Mh partial sum of 5[/]. Writing sN = sN(f) = JJL t скфк,
we have from (8.25) that
(8.26) II/--UP = ll/ll2 - E iQl2-
k=l
(8.27) Theorem {фк} be an orthonormal system in L2, and let feL2.
(i) Of all linear combinations укфк with N fixed, the best L2-
approximation to f is given by the partial sum sN = скфк of the
Fourier series of f
(ii) (Bessel's inequality) The sequence {q} of Fourier coefficients of f
belongs to I2 and
(oo \ 1/2
E № ИИ-
k — 1 /
Proof Part (i) has been proved. Note that since ||/ — ^л||2 > 0, Bessel’s
inequality follows from (8.26) by letting N -+ oo.
If jFis a function for which equality holds in Bessel’s inequality, that is, if
/co \ 1/2
(8-28) E Ы2 =0/11.
\k = 1 /
then/is said to satisfy Parseval'sformula. From (8,26), we immediately obtain
the next result.
i UUHCi о G 8 i G Й» f raibGVQl й Г V s t I I U I d
i OiJ
(8.29) Theorem ParsevaTs formula holds for f if and only if |kN — /Ц —► 0,
i.e., if and only if S[/] converges to f in L2 norm.
The following theorem is of great importance.
(8.30) Theorem {Riesz-Pischer Theorem) Let {фк} be any orthonormal
system, and let {ck} be any sequence in I2. Then there is an f g L2 such
that S[/] = скФк> l-e^ such thctt tck} Is sequence of Fourier coeffi-
cients of f with respect to {фк}. Moreover, f can be chosen to satisfy
ParsevaTs formula.
Proof Let tN = ££= j скфк. Then if M < N,
W^N ~~ II2
N 2 N
E скФк = Ё №•
M + l M+l
The fact that {q} e/2 implies that {tN} is a Cauchy sequence in L2. Since
L2 is complete, there is an f e L2 such that || f — -> 0. If N > k,
/Фк = j (/- tMk + 1хФк = j (/“ tM>k + ck.
Since the integral on the right is bounded in absolute value by ||/ — M \\фк\\ =
||/- tN\\, we obtain by letting N -+ oo that $f$k = ck. Thus, S[/] = ^k скфк,
so that tN = sN(f), and it follows from (8.29) that Parseval’s formula holds
for /. This completes the proof.
There is no guarantee that the Fourier coefficients of a function uniquely
determine the function. However, if {фк} is complete, we can show that the
correspondence between a function and its Fourier coefficients is unique;
that is, iff and д have the same Fourier coefficients with respect to a complete
system, then f = д a.e. This is simple, since the vanishing of all the Fourier
coefficients off — д implies that f — д — 0 a.e. An important related fact is
the following.
(8.31) Theorem Let {фк} be an orthonormal system. Then {фк} is complete
if and only if ParsevaTs formula holds for every f g L2.
Proof Suppose that {фк} is complete. If/e L2, Bessel’s inequality implies
that its Fourier coefficients {ck} belong to I2. Hence, by the Riesz-Fischer
theorem, there exists а д in L2 with £[#] = скФк and ||#|| = (X Iq|2)1/2«
Since/and д have the same Fourier coefficients and {фк} is complete, we see
that/ = д a.e. Hence, ||/|| = ||#|| = (£ |q|2)1/2, which is Parseval’s formula.
Conversely, suppose that Parseval’s formula holds with respect to {фк}
for every fe L2. If (/,фк) = 0 for all k, then ||/|| = (£ |</,^>|2)t/2 = 0.
Therefore, / = 0 a.e., which proves that {фк} is complete.
Suppose that {фк} is orthonormal and complete and that f,g g L2. Let
Ск = <f,<K>’ dk = <0,фл>, с = {ej, d = {dk}, ||c|| = (£ |ck|2)1/2 and (c,d) =
£ ckdk. We claim that
(8.32) Ш> = (c,d).
To prove this, observe that by Parseval’s formula, </ + g,f + дУ =
(c + d, c + d), or
ШУ + <g,g> + 2 Re </,#> = (c,c) 4- (d,d) + 2 Re (c,d),
where Re z denotes the real part of z. Cancelling equal terms, we see that
Re {fdf) ~ Re (c.d). Applying this to the function if gives Re (ifgy =
Re(zc,d). But Re {ifgy = Re i{fgy = —Im </,#>. Similarly, Re (ic,d)
= — Im (c,d). Therefore, Im {fgy — Im (c,d), and (8.32) is proved.
Another corollary of (8.31) is given in the next result. First, we make
several definitions. Let Xt and X2 be metric spaces with metrics dt and d2,
respectively. Then Xt and X2 are said to be isometric if there is a mapping T
of Xi onto X2 such that
d^f.g) = d2{Tf, Tg)
for all fg g Xr. Such a Fis called an isometry. Thus, an isometry is a mapping
which preserves distances. An isometry is automatically one-to-one, and two
isometric metric spaces may be regarded as the same space with a relabeling
of the points. For example, two L2 spaces, L2(E) and Z2(E"), are isometric if
there is a mapping T of L2(E) onto L2(£') such that ||/ — g||2>£ = IITf —
Tg\\ 2> Ef for all fg g L2(E). The isometries we shall encounter will be linear,
i.e., will satisfy
Г(а/ -4- pg) = uTf + pTg for all scalars a, fi.
If Г is a linear map of L2(E) onto L\E'f then since Tf — Tg = T(f — g),
it follows that T is an isometry if and only if
ll/U = Г/кг
for all f g L\E). Similarly, a linear map of L\E) onto /2 is an isometry if
and only if ||/||2>£ = i|Tf\\l2 for all/GL2(E).
(8.33) Theorem All spaces L\E) are linearly isometric with I2, and so with
one another.
Proof For a given E, define a linear correspondence between L2(E) and
I2 by choosing a complete orthonormal system {фк} in L2(E) and mapping
an f e L2(E) onto the sequence {</,</>*>}., of its Fourier coefficients. This
mapping is onto all of I2 by the Riesz-Fischer theorem, and is an isometry
by (8.31).
7. Hilbert Spaces
A set H is called a Hilbert space over the complex numbers C if it satisfies the
following three conditions:
(Ht) Я is a vector space over C; that is, iffig e H and a e C, then f + g € H
and а/ e H.
(H2) For every pair f\g e Я, there is a complex number fifig), called the inner
product of f and g, which satisfies
(a) (g,f) = №).
(b) (Л +Л,0) = (А,0) + (f2,g).
(c) (af,g) = <x(f,g) for a e C.
(d) (f,f) > 0, and (f,f) = 0 if and only if/ = 0.
Notice that (a), (b) and (c) imply that (/ + g2) = (/gj + (f,g2) and
(/,«£) = «(/#)• Define
II/II = (//)1/2-
Before stating the third condition, we claim that
(8.34) |(//)| < H/IIhll (Schwarz's inequality).
If H0II = 0, this is obvious. Otherwise, letting 2 = — (/,g)/||g||2, we obtain
0 < ( f 4- Xa f + la) - II f II2 21№)|2 + l№)|2 II f II2 l(/’0)|2
о<(/+л9,/+ад-и11 -2 b|i + ||J|2 - ll/ll - b||2 .
and Schwarz’s inequality follows at once.
We will show that || • || is a norm on H by proving the triangle inequality.
In fact,
< II/ + fill2 = (/+0,/+ g) = ll/ll2 + 2Re(/,0) + Ы12-
Since |Re (fi,g)\ < |(/g)| ll/ll ||#||, it follows that the right side is at most
(ll/ll + hll)2- Taking square roots, we obtain ||/ + 0|| < ||/|| + Ш, as
desired. Hence, Я is a normed linear space.
We also require
(H3) Я is complete with respect to || • ||.
In particular, a Hilbert space is a Banach space.
As for L2 spaces, a linear map T of a Hilbert space Я onto a Hilbert
space Hr is an isometry if and only if \\fi\\H = ||7/||H, for all f e Я.
A Hilbert space is called infinite dimensional if it cannot be spanned by a
finite number of elements; hence, an infinite dimensional Hilbert space has
an infinite linearly independent subset. The space L2 with inner product
142
Lp Classes
(8.35)
(/#) “ lf9 and Ле space I2 with (c,d) = ^k ckdk are examples of separable
infinite dimensional Hilbert spaces. In fact, there are essentially no other
examples, as the following theorem shows.
(8.35) Theorem All separable infinite dimensional Hilbert spaces are linearly
isometric with I2, and so with one another.
Proof. The proof is a repetition of the ideas leading to (8.33), so we shall
be brief. Let Я be a separable infinite dimensional Hilbert space, and let
{ek} be a countable dense subset. Discarding those ek which are spanned by
other e'h we obtain a linearly independent set {e^ with the same dense span
as (X). Since H is infinite dimensional, {efc} is infinite. Using the Gram-
Schmidt process, we may assume that {ел} is orthonormal: (ef,cj = 0 for
i Ф к and ||efc|| = 1 for all k. It follows that {ek} is complete; in fact, if
ffiefi = 0 for all k, then
f - E akek
= ll/ll2 + E Ы2 > II/II2.
k~ 1
If f were not zero, the span of the ek could not be dense. Hence, / = 0, which
shows that Я has a complete orthonormal system {ek}.
Next, we will show that Bessel’s inequality and the Riesz-Fischer theorem
hold for {efc}. If/e Я, let ck = (/,efc). Then
о < f - E cket
k=l
2 N
= ил2 - E IqI2-
k=l
Letting N co, we obtain Bessel’s inequality (£ fik[2)1/2 < ||/||. In par-
ticular, {q} belongs to I2.
To derive the Riesz-Fischer theorem, let {yj be a sequence in I2, and set
Ук^к- Then
1км " M2
X w2 о as M,N co, M < N.
k = M + 1
Since H is complete, there is a g g H such that ||# — tN\\ -> 0. We have
(g,ek) = (g - h, + (W*) = (g - tN, ek) + yk (к < AT).
Letting N oo, it follows from Schwarz’s inequality that (#,efc) = yk. Hence,
tN = sN(g) and ||g - M2 = 1Ы2 ~ E*=i l^l2- Letting N -> co in the last
equation, we see that g satisfies Parseval’s formula ||gr|| = (E 2)1/2. This
gives the analogue of the Riesz-Fischer theorem.
Now let f g H, and let ck = (/ej. Talcing yk = ck above, we see by the
completeness of {ek} that g = f so that Parseval’s formula holds: ||/|| =
(У kJ2)1/2- The fact that H is linearly isometric with I2 now follows as in the
proof of (8.33).
Exercises
143
Exercises
1. For complex-valued, measurable f, f ~ fi 4- if2 with fr and f2 real-valued and
measurable, we have f £ / = J E fi 4- i $Ef?> Prove that f £/is finite if and only if
f£ |/| is finite, and If£/| < f£ |/|. (Note that |f£/l = К/1)2 + (b/2)2]1/2,
and use the fact that (a2 -h b2)112 = a cos a 4- b sin a for an appropriate a,
while (a2 4- b2)1/2 > ja cos a 4- b sin a| for all a.)
2. Prove the converse of Holder’s inequality for p = 1 and 00. Show also that
for real-valued f$Lp(E\ there exists a function g^Lp,(E\ lip 4- lip' = 1,
such that fg ф If (E). (Construct g of the form 2 akgk for appropriate ak and
gk, with gk satisfying $Efgk — 4-00.)
3. Prove Theorems (8.12) and (8.13). Show that Minkowski’s inequality for series
fails when p < 1.
4. Let f and g be real-valued, and let 1 < p < 00. Prove that equality holds in
the inequality |f fg\ < ||/||р|Ы1р' if and only if fg has constant sign a.e. and
|/|p is a multiple of \g\p' a.e.
If II/4- g||p = II/||p 4- ll^llp and g Ф 0 in Minkowski’s inequality, show
that /is a multiple of g.
5. For 1 < p < 00 and 0 < |E| < 4-00, define
/lr \ltp
L Prove that if pr < p2, then 7VP1[/] < WP2[/].
Prove also that Np[f + g} < Np\f] 4- NP[g], (1/|E|) f£ \fg\ < NP[f]Np> Ы,
1/p 4- 1/p' = 1, and that lim^^ 7Vp[/] = ||/||oo. Thus, Np behaves like || • ||p,
but has the advantage of being monotone in p.
6. Prove the following generalization of Holder’s inequality. If £*=i 1/pt = 1/r,
Pi,r > 1, then
||/i---AI|r <11Лк -Uli,,
(See also Exercise 12 of Chapter 7.)
7. Show that when 0 < p < 1, the neighborhoods {/: ||/||p < e} of zero in
£p(0,l) are not convex. (Let/= z(0,^) and g = z(ep,2eP). Show that \\f\\p = ||^||p
= £, but that || J/4- lg\\P > £.)
8. Prove the following integral version of Minkowski's inequality for 1 < p < 00:
[ j I fdx jP dy\ lP ~ / [j I Ip /у] P dx,
[For 1 < p < сю, note that the pth power of the left-hand side is majorized by
Jf [f |/(2,z)|<7z]₽_1 |/(x,y)| dxdy. Integrate first with respect to у and apply
Holder’s inequality.]
9. If / is real-valued and measurable on E, define its essential infinmm on E by
css inf/ = sup {« : |{xe£:/(x) < a}| = 0}.
E
If/> 0, show that ess£ inf/ = (ess£ sup I//)-1.
10. Prove that L™(E) is not separable for any E with |£| > 0. (Construct a sequence
of decreasing subsets of E whose measures strictly decrease. Consider the
characteristic functions of the class of sets obtained by taking all possible
unions of the differences of these subsets.)
11. If fk in Lp, 1 < p < co, gk —> g pointwise, and ||#л||< M for all k,
prove that fkgk -> fg in Lp.
12. Let f{R} eLp. Show that if \\f- fk\\p — 0, then \\fk\\p — ||/||p. Conversely,
if A — /a.e. and ||/л||р -> ||/||p, 1 < p < oo, show that ||/-/k||P — 0.
13. Suppose that /k —► /a.e. and that fk,f&Lp, 1 < p < oo. If ||/k||p < M < +oo,
show that j* fkg —> j* fg for all g eLp> , 1/p -j- 1/p' = 1.
14. Verify that the following systems are orthogonal:
(a) {>, cos x, sin x,..., cos kx, sin kx,. . .} on any interval of length 2л;
(b) k = 0> ±lj ±2> on
15. If /g Z2(0,2ti), show that
lim f(x) cos kx dx = lim f(x) sin kx dx = 0.
k-* oo J О к-» oo J О
Prove that the same is true if/e Е^О^л) (This last statement is the Riemann-
Lebesgue lemma. To prove it, approximate/in L1 norm by L2 functions. See
(12.21).)
16. A sequence {fk} in Lp is said to converge weakly to a function / in Lp if Jfkg
$ fg for all g elf. Prove that if fk — / in Lp norm, 1 < p < oo, then {fk}
converges weakly to /in Lp. Note by Exercise 15 that the converse is not true.
17. Suppose that /fc,/g L2 and that J fkg j fg for all g e L2 (that is, {/} converges
weakly to/in Z2). If ЦАЦг — II/II2, show that A — /in L2 norm.
18. Prove the parallelogram law for Z2:
II/+0II2 + II/-#ll2 = 2||/||2 + 2||^||2.
Is this true for Lp when p 2 ? The geometric interpretation is that the sum
of the squares of the diagonals of a parallelogram equals the sum of the squares
of the sides.
19. Prove that a finite dimensional Hilbert space is isometric with R” for some n.
20. Construct a function in Z1^--oo, + °o) which is not in Z2(a,£) for any a < b.
(Let g(x) = x"1/2 on (0,1) and g(x) = 0 elsewhere, so that f = 2. Consider
the function f(x) = 2 akg(x — rk), where {rk} is the rational numbers and
{ak} satisfies ak > 0, £ ak < -Foo.)
21. If /g Zp(R"), 0 < p < 00, show that
JQ>Z(y) -z(x)l₽f/y -Oae-
Note by Exercise 5 that this condition for a given p implies it for all smaller p.
22. Let be a complete orthonormal system in Z2, and let m = {mk} be a
given sequence of numbers. If/gZ2,/ ~ скфк, define Tfby Tf ~ Z ткскфк.
Show that T is bounded on Z2, i.e., that there is a constant c independent of /
such that [|7/||2 < c||/||2 for all /gZ2, if and only if m g Z00. Show that the
smallest possible choice for c is ||/n||i*>.
v Chapter 9
Approximations of the Identity;
Maximal Functions
1. Convolutions
The convolution of two functions f and g which are measurable in R" is
defined by
(/* 0)(x) = [ f(f)g(* - t) dt (x e R").
jRn
lA-Chapter 6, (6.14), we saw that \\f * || t < Ц/^ ||g|| P Moreover, according
to (6.16), ||/* g||i = ll/ll JIg||i if / and g are nonnegative. In this section,
we will study some additional properties of convolutions, beginning with the
following theorem.
(9.1) Theorem Let 1 < p < oo, /e ZP(R") and g eZ?(R"). Then f * g e
LP(R") and
\\f*g\\P < ll/llpllg||p
Proof. Since \f * g\ < If I * |g|, we may assume that / and g are non-
negative. We may also suppose that 1 < p < oo, since when p = 1 the result
is just (6.14). If p = oo, then
(/* g)(x) < Г ll/lloo0(x - 0Л = ll/lla, f 0(x - t)dt = H/ILH0II1.
JRn JRn
Therefore, ||/* ^IL ll/ILIIвII1, as claimed.
If 1 < p < 00, we write
(/* 0)(x) = I" [/(t)^(x - t)1/p]g(x - t)1/p' dt, -^ + ^7=1.
JR" P P
By Holder’s inequality with exponents p and p',
145
- •.• .su'iio м» inc WttHiiiy, iVidAihidl J GUOGS
/ r \ a/p / r \
(/*g)(x)<l f(t)pg(x - t) dt) ( gf(x-t)Jt)
= (fp * <7)(х)1/%111/<
Now raise the first and last terms in this inequality to the pth power and
integrate the result. Since JR„ (fp * g) dx = ||/||£||#|| t by (6.16), we obtain
ll/*< < ll/ll>lir°’/₽') = ll/ll>llf-
The theorem follows by taking pth roots.
Theorem (9.1) is an important special case of the next result, whose proof
is left to the reader. (See Exercise 2.)
(9.2) Theorem (Young’s Convolution Theorem) Let p and q satisfy
1 < p,q < oo and 1/p + 1/q > 1, and let r be defined by 1/r =
1/p + 1/q - 1. If ft Lp(Rn) and g g Lq(Rn), thenf* g g Lr(Rn) and
II/* < < 11/llplK
Note that when q = 1, Young’s theorem reduces to (9.1). See also Exercise
3.
A convolution f * К with К fixed defines a transformation T \f -> f * К
which is called the convolution operator with kernel K. Theorem (9.1) states
that a convolution operator with an integrable kernel maps functions in Lp
into the same Lp. The next result shows an effect that convolution operators
with smooth kernels have on Lp,
For a positive integer m, we denote by Cm the class of functions /(x),
x g R”, whose partial derivatives up to and including those of order m exist
and are continuous. The subset of Cm of functions with compact support is
denoted C™. Similarly, C00 is the class of infinitely differentiable functions,
and Cq is the corresponding subset of functions with compact support.
(For the existence of such functions, see Exercise 4.) Finally, if a = (a t,..., a„),
where the ak are nonnegative integers, then the ath partial derivative of f is
denoted by
(daf\ / 6ai da" \
(9.3) Theorem If 1 < p < аз, fe Z/(Rn) and Ke then f* KeCm and
D“(f* K)(x) = (/* D“K)(x),
a = (aj,. . ., a„), aj + • • • + a„ < m.
J
Proof. We first claim that if К is any continuous kernel with compact
support, then f * К is continuous. In fact,
GOHVOaUIiOnS
14/
к/н)(х+ь)-(/*ад
/(t)X(x + h-t)rft- /(t)X(x-t)A
Jr" Jr*
= f /(x - tw + h) - ад dt
jRn
( r \ Vp / f \ 1/p'
< |/(x - t)\pdt |x(t + h) - ад*' л
\ JR'1 / \ JR" j
= \\f\\p\\K(t + h) - K(t)\\p..
The last expression tends to zero as |h| -> 0 since К is uniformly continuous
and has compact support. [Note that in case pf < co, this also follows from
(8.19) since К e Lp'. See also Exercise 3.]
Next, let Кe C%, m > 1. Fix i, i — 1, . . . , n, and let h = (0, . . . , 0, h,
0, . . . , 0), where h is in the zth coordinate position. Note that
(/* K)(x + h) - (/* 7Q(x) = Г f/C(x - t + h) - *(x - t)l
h Jr" I h J
Г dK
= /(0 T- (x - t + h') dt,
Jrh
by the mean-value theorem, where h' = (0, . . . , 0, h', 0,. . . , 0) for some h'
depending on x and t which is between 0 and h. Hence, as h -> 0, (dKIdx?)
(x — t + h') converges to (Э7^/сх-)(х — t) uniformly in t. Since SKjSxi has
compact support, it follows from the uniform convergence theorem that the
last integral converges to [f * (dK/dxf](x). Therefore, d(f * JCfxy/dxt exists
and equals [f * (dK/dxf](x), which is continuous by our earlier remarks.
The proof of the theorem for m — 1 is now complete. The proof for m —
2, 3, . . . follows by repeated application of the case m = 1.
It follows from this result that f^KeC^ if /eLp, 1 < p < oo, and
KeCq. If, in addition, / has compact support, then so has f * K. In fact,
if is the support of К and S2 is the support of/, then the formula (/* A)(x)
= fs2/(O^(x “ 0 implies that (/* K)(x) — 0 unless there are points
t e S2 for which x — t e St. Hence, the support of f * К is contained in
{x : x = Sj, + s2, Sj e 5i, s2 e S2}, and so is bounded. An application of this
fact is given in Exercise 5.
(9.4) Theorem Iff& £(R") and К is bounded and uniformly continuous on
R”, then f * К is bounded and uniformly continuous on Rn.
The proof is similar to the first part of the proof of (9.3) and is left as an
exercise. See also Exercise 3.
Mppiuxinidiiuiib oi ui(jiufcfiuiy; iviaximai ry .sons
2. Approximations of the Identity
Given К and e > 0, let
ВД - e"W-) = £~пк(^, . . . Л"1
For example, if A(x) = Z{|x|<i}(x), then A£(x) = e"nZ{|x|<£}(x)- In this case,
taking successively smaller values of 8 produces kernels with successively
higher peaks and smaller supports. The effect on any positive К with com-
pact support is roughly the same.
In general, K£ has the following basic properties.
(9.5) Lemma If Ke L^R”) and s > 0, then
(i) JRn K£ = JRn K.
(ii) J|xj ><5 \KS\ -> 0 as s -> 0, for any fixed <5 > 0.
Proof. Part (i) follows immediately from the change of variables у = x/a.
(See Exercise 20, Chapter 5.) For part (ii), fix <5 > 0, and let у — x/s. Then
f 1ВД1 dx = dx = f |Ду)| dy.
Since Ke L and <5/e —> + oo as a -> 0, it follows that the last integral tends to
zero as £ -> 0. This completes the proof.
Note that for К > 0, property (i) above means that the areas under the
graphs of К and K£ are the same, while (ii) means that for small £ the bulk
of the area under the graph of K£ is concentrated in the region above a small
neighborhood of the origin.
For any KeL, we can expect from (ii) that the effect of letting e -> 0 in
the formula (/ * jQ(x) = j f(x — t)A?e(t) dt will be to emphasize the values
off(x — t) corresponding to small t. In fact, the next four theorems show that
(f * K£)(x) -> /(x) in various senses (e.g., in norm or pointwise) as 8 -> 0, if
К is suitably restricted. A family {K£ : 8 > 0} of kernels for which f * K£-+ f
in some sense is called an approximation of the identity.
In what follows, we shall use the notation /£(x) for the convolution
(/* K£)(x).
(9.6) Theorem Let f£ = f * K£, where К e L^R”) and JRM К = 1. Iffe
£P(R"), 1 < p < co, then
||/c-/||p^0as£->0.
Proof. By (9.5)(i),
/(x) =/(x) [ O) dt = f /(x)^(t) dt.
JRn JR"
Therefore,
|Дх) -/(X)!
= [/(x-t) -/(xM(t)A
jRn
< f |/(x - t) - /(x)| IW'W)!1''' dt,
Jr"
where 1/p + 1/p' = 1 (1/p' = 0 if p = 1). Applying Holder’s inequality with
exponents p and p', and then raising both sides to the /?th power and integrat-
ing with respect to x, we obtain
t) -/(x)|q^£(t)l dt
~ p/p'
t dx
= ИИ?7'' I/(x - t) - /(x)|p|XE(t)| dt
jRn L jRn
dx.
Changing the order of integration in the last expression (which is justified
since the integrand is nonnegative), we obtain
11Л-/11? ИН?/₽' jRn K(W(t) dt,
where <£(t) = JR„ |/(x - t) - /(x)|₽ Jx = ||/(x - t) - /(x< For d > 0,
write
4 = [ |^(W(t) dt = f + [ = Ae>i + Be>i.
Given g > 0, we can choose <5 so small that ф(€) < g if |t| < <5 [note that
0(t) -> 0 as |t| -> 0 by (8.19)]. Then
A.U l*e(0l Л<^И111
for all e. Moreover, ф is a bounded function by Minkowski’s inequality
[note that ||</>|| «> < (2||/||p)p], so that Be d is less than a constant multiple of
|A?£(t)| dt, which tends to zero with s. This proves that Ц -» 0 as s -> 0,
and the theorem follows.
(9.7) Corollary For 1 < p < oo, Cq is dense in ZP(R”).
Proof. Let f e Lp, 1 < p < oo. Given g > 0, write f = g 4- h where g
has compact support and \\h\\p < g. Choose a kernel Ke Cq with = 1,
and let ge = g * K£. Then gE e C$ and, by (9.6), \\g - g£\\p -> 0. By
Minkowski’s inequality, \\f~ g£\\p < \\g - g£\\p + \\h\\p < \\g - g£\\p + g.
Choosing e so that \\g — ge\\p < g, we obtain ||/— g£\\p < 2g, and the corol-
lary follows.
150
Approximations of the Identity; Maximal Ft tions (9.8)
The next result is a substitute for (9.6) in case f 6 L00.
(9.8) Theorem Let ft=f* where fe Lx(Rn), К e Ll (Rn) and fR„ К = 1.
Then fe-> f as e -> 0 at every point of continuity of f.
Proof As before,
|/£(x) - /(x)| < f |/(x - t) - /(x)||X£(t)| dt.
Jrh
If f is continuous at x, then given q > 0, there exists <5 > 0 such that
|/(x - t) ~/(x)| < rj if |t| < 3. Hence,
f |/(x - t) -/(х)1Ю)1 dt < J K(t)| dt + 2II/L f Ю)1 dt.
JR" J|t|<5 J|t|;><3
Since |^£(t)| dt < ||A^|| t and |^£(t)| dt 0 as s -> 0 for any fixed
<5, it follows that |/Xх) — /(x)| -> 0.
Before stating the next result, we introduce some useful notation. If
i/<(x) and ф(х) are defined in a neighborhood of x0 and if ф > 0 there, we say
that
ф(х) = О(ф(х)) as x -> x0
if there is a constant c such that \ф(х)/ф(х)\ < c near x0. If in addition,
lim^ ф(х)/ф(х) = 0, we say that
ф(х) = о(ф(х)) as x -> x0.
In particular, the expressions ф(х) = 0(1) and ф(х) = <?(1) as x -> x0 mean
respectively that ф is bounded and that ф -> 0 as x -> x0. The notation most
commonly occurs when x0 = 0 or ±oo. A similar notation is used when x
is a discontinuous variable, say a sequence of integers tending to 4-oo. For
example, ak — 0(1) and ak = o(l) as к -> +oo mean respectively that {ak}
is a bounded sequence and that ak -> 0 as к -> 4- oo.
The following theorem concerns the pointwise convergence of fz when
f elf See Exercise 12 for the case f e Lp.
(9.9) Theorem Let f=f*Kz, where feL^W1), К e Z?(Rn) n L°°(R"),
JRn К = 1, and K(x) = €>(|x|”") as |x| + oo. Then fE-> f as г -> 0
at each point of continuity of f
Proof If f is continuous at x, then given > 0, choose 5 > 0 such that
|/(x — t) — /(x)| < r] if |t| < d. As usual, (
\Ш - /(x)| < // f Ю)1 dt + f | f(x - t) - f(x)\\KE(t)\ dt
<»/И111 + ( l/(x - t)|K(t)| dt + |/(x)| [ K(t)|A.
(9.10)
Approximations of the Identity
The last term on the right tends to zero with £ by (9.5). It is enough, therefore,
to show that the second term tends to zero with s. Write |A7(x)| = /г(х)|х|-п,
where /z(x) 0 as |x| -> + co. Then
f |/(X - t)|K(t)| dt = [
< sup ~ ) г I l/(x “ 0I dt.
Note that sup|t|S^(t/£)->0 as e->0. Hence, since f|t|Sa |/(x - t)| dt < ||/|| 1;
the last expression above tends to zero with e, and the theorem follows.
There are many classical kernels which satisfy the restrictions we have
imposed. Three important examples are listed below for the case n = 1.
(9.10) The Poisson kernel. Let
K(x) = P(x) = j [x g (- co, + oo)].
Then 7>eZ,1(—oo, + oo) r> £°°(-оо, + оо), Jt“ P = 1, P is positive
and P(x) = nd*!-1) as |x| -+ +co. In fact, P(x) = O(|x|~2) as |x| ->
4- oo. We have
PE is called the Poisson kernel, and the convolution
1 Г + °° £
ЛИ - (/. Л)(л) - - ] m d<
is called the Poisson integral of f.
Setting e = у and letting f(x,y) = fy(x), we obtain a function f(x,y)
defined in the upper half-plane {(x,y) : — oo < x < Too, у > 0]. Notice
that ylfy1 4- л2) is the imaginary part of — 1/z, z = x 4- iy, and so is har-
monic in the upper half-plane; that is, Pfx) satisfies Laplace’s equation
f d2 d2\
We leave it as an exercise to show that if fe Lp, 1 < p < oo, then
/ d2 d2\ f + co / d2 d2\
(aP + = J + Py(x ~ r) dt’
so that (d2/dx2 + d2!dy2)f(x,y) = 0 for у > 0. Hence,/(х,у) is also harmonic
in the upper half-plane.
If/is integrable on (-oo, + oo), it follows from (9.9) that/(x,y) -+/(*) as
i
мрргохкпанонБ оч ine laeniny; ivtaximai r
VLiOHS
у -+ 0 wherever/is continuous. Thus,/(x,j) solves the Dirichlet problem for
the upper half-plane; that is, if /(x) is continuous and integrable on
(—oo, + oo), then f(x,y) defines a function which is harmonic in the upper
half-plane and which tends to /(x) as у -> 0.
(9.11) The Fejer kernel Let
z . 1 /sin x\ 2
K(x) = n\~x~J [* e (- 00,4- 00)].
Then К satisfies the same conditions as in (9.10), and KE(x) =
(l/я) [s sin2 (x/&)/x2]. Setting w = 1/s, we obtain the Fejer kernel
4 1 sin2 wx
F(x,w) —--------г-.
71 WX
If /e £(— 00,4- 00) and if/is continuous at x, then by (9.9)
г 1 Г + °° xsin2W? /у x
hm - f(x- t) ,2 dt =f(x).
w-* 4-OO J — OO Wl
(9.12) The Gauss-Weierstrass kernel The function
ТЦх) — —r= e~x2 [x e (— 00,4- 00)]
V71
also satisfies all the required conditions (see Exercise 11, Chapter 6).
Here, X/x) = (l/^/Tieje-^2, and letting 8 = ^Jy,y > 0, we obtain the
Gauss- Weierstrass kernel
W(x,y) = ~e~x2/y.
If /is integrable on ( — 00,4- 00) and continuous at x, then
lim ~7^, f ~ 0е~‘2,у dt = /(*)•
y-*o + у J -00
Notice that W(x,y) satisfies the heat equation
д2 d
dx dy
If we strengthen the condition K(x) = о(|х|“л), |x| -> 4-00, used in (9.9),
we can obtain the convergence of/e to / almost everywhere. The following
result is fairly typical of theorems of this kind. Its hypotheses are met by any
of the three examples just listed.
(9.13) Theorem Suppose that feLljSC), К is bounded, K(x) = O(|x|)~'I“A)
as jxf —> Too for some A > 0, and fR„ К = 1. If fE =. f * Ke, then
f-^fat each point of the Lebesgue set of f.
Proof Let x0 be a point of the Lebesgue set of /[see (7.14)], so that
P~n J|X|<P l/(xo + x) - /(xo)l <7x -> 0 as p -» 0. By considering the function
/(x0 T x), we may assume that x0 = 0. Since the hypothesis on К implies
that K(x) = u(|x|-”), the conclusion follows from (9.9) if/is continuous at 0.
Hence, subtracting from/a continuous function with compact support which
equals /(0) at 0, we may suppose that /(0) = 0.
The hypotheses |Дх)| < M and K(x) = O(|x|-n-A) can be combined
into a single estimate:
1Дх)| < (1 + )х))д+л.
Hence,
|a;(x)I < лл(е + |х|)п+л-
Therefore,
f
1Л(0)| < jRn |/(x)| j^rfx,
and it remains to show that the integral tends to zero. We will use the follow-
ing lemma, which is of some independent interest.
(9.14) Lemma Suppose that /(x) is integrable over a spherical shell a <
|x| < b and that ф(р) is continuous for a < p < b, 0 < a < b < T oo.
Let F(p) = (X|/(x) dxfor a < p < b. Then
f 7(х)ф(|х|) Jx = f ф(р) dF(p),
Ja^|x|^b J a
the integral on the right being a Riemann-Stieltjes integral.
Proof Note that this reduces to the formula in (7.32)(i) in case n = 1.
In any case, writing / = /+ — /“, we see that F is the difference of two
bounded increasing functions. Hence, F is of bounded variation on [a,b\ and
ф dF is well-defined. We may assume that / > 0. Let I = Ja^|x|^b/(X)
</>(|x|) <7x, and let {a = p0 < Pi < * ” < Pk = b} be a partition of [a,b\.
Then
fc r
I = E /(х)ф(|х|) dx,
JPi- 1^|х|^р£
and since / > 0,
E mt I* /(x) dx < I < £ [ f(x) dx,
i=l Jpi-i<|x]^pi i=l Jpf-i^|x]^pi
154
Approximations of the Identity; Maximal Fu ions (9.14)
where and Mt are respectively the minimum and maximum of ф in
[pf_ This can be rewritten
£ - Яр..-J] < I < f 7W;[F(Pi) -
i=l i=l
By (2.24), the extreme terms in this inequality both converge to J* ф(р) dF(p)
as the norm of the partition tends to zero, and the lemma follows.
Returning to the proof of (9.13), let F(p) = f|x|<p |/(x)| dx. The hypotheses
that x0 = 0 is a Lebesgue point of f and that /(0) — 0 imply that given
t] > 0, there is a <5 > 0 such that F(p) < црп if p < <5. Write
f f f
l/(x)| (P 4- lvlT"+* dx = + = A + B-
Jrh Vs -r |X|) J|x|<<5 J|x|><5
Taking ф(р) = ел/(е + р)”+л and [a,b\ — [0,<5] in (9.14), we have
A = Jo(e + p)n + kdF{p}'
Integrating by parts and observing that F(0) = 0, we obtain
ел Г5 ел
A=(£ + sy^F® + (и + Л) ]ог(р)(е + рГЛ1ф-
The first term on the right tends to zero as e -» 0. The definition of d and the
change of variables p = Et show that the second term is at most
(« + j0^n(g + + dp = (« + j Q + ^п + Л-И^-
Hence,
limsup A < (n + Л)А , - дт+л+Т =
£-+o Jo U + u
where c(— сл>п) is finite since A > 0.
Finally, to estimate B, note that if |x| > d then e + |x| > <5, so that
ЕЛ Г ЕЛ
в <^+л| dx 3^+А 11/111-
Непсе, lim^o В = 0. Combining these estimates, we obtain limsup^o (A + B)
< сц, and the theorem follows by letting'^ -> 0. See Exercise 12 for the
case f e Lp, p > 1.
The kernels {Ke} for К satisfying the kinds of conditions above are ex-
amples of approximations of the identity.
(9.16)
e Hardy-LiTXiewooa тахнпш гипииии
3. The Hardy-Littlewood Maximal Function
Let/* denote the Hardy-Littlewood maximal function of /:
/*(x) = suPj^j j 1/(у)1
where the supremum is taken over all cubes Q with center x and edges parallel
to the coordinate axes [see (7.5)].
We observed in Chapter 7, p. 105, that/* is not integrable over R" (unless
f = 0 a.e.), but does satisfy the weak-type condition
(9.15)
|{xeR” :/*(x) > a}| <^IL/b
(a > 0),
where c depends only on n [lemma (7.9)]. The behavior of /* on the other Lp
spaces, 1 < p < oo, turns out to be better. For example, it is clear from the
definition of/* that/*(x) < ||/|| „> for all x. Thus,/* is bounded if/is, and
II/*IIoo ll/lloo- The following theorem describes the behavior of/* when
feLp.
(9.16) Theorem Let 1 < p < oo and f e Lp(Rn). Then f* e LP(R") and
ll/*llp < dl/llp,
where c depends only on n and p.
Proof. We may assume that 1 < p < oo. The idea is to obtain informa-
tion for Lp by interpolating between the known results for L1 and L00. Thus,
for a > 0, let
m(a) = |{xgR" :/*(x) > a}|
denote the distribution function of/*. Fix a > 0 and define a function g by
g(x) = /(x) when |/(x)| > a/2 and g(x) = 0 otherwise. Then since |/(x)| <
|^(x)| + a/2,
/*(x) < sup £ I^(y)| dy + ~ = £*(x) +
In particular,
( a
{xeR" :/*(х) > к} c jxeR” : #*(x) > 5
so that, by (9.15),
co (a) <
x e R” : 0*(x) >
4^11.
[ l/(x)l с/х.
a J{xcRn: |/(x)| >a/2}
158 Approximations of the Identity; Maximal Func ms (9.17)
We have the formula fRn/*p dx = p fo ap 1co(a) da, which was stated on
two occasions: Exercise 16, Chapter 5, and Exercise 5, Chapter 6. Hence
Г2с Г "I
1ЖН dx da.
La J{xcRn: |/(x)| >a/2} J
f*p dx < p \ ap 1
/Rn J 0
Interchanging the order of integration in the expression on the right (which
is justified since the integrand is nonnegative), we obtain
p p / Г21Л*)1 \
f*p dx < 2cp\ |/(x)|l ap 2da\dx.
JR” JRn \ J 0 J
Sincep — 2 > — 1 (i.e., p > 1), the inner integral equals (2|/(x)|)p~ lf{p — 1)
a.e. (wherever /(x) is finite), so that
f 2ppc f 2ppc
f*pdx<—4 [f(x)\”dx = —4 и л?-
Jr« P ~ 1 Jr« P “ 1
Taking /?th roots, we see that ||/*||p < -4p||/||p, where Ap = 2ppcj{p — 1).
This completes the proof. Note that the constant Ap tends to 4- oo as p -> 1,
and is bounded as/? -> oo.
The Hardy-Littlewood maximal function plays an important role in many
parts of analysis concerned with operator theory and differentiation. It arose
naturally in Chapter 7 in connection with Lebesgue’s differentiation theorem.
As another illustration of its usefulness, we have the following result.
(9.17) Theorem Let K(x) be nonnegative and integrable on R”, and suppose
that K(x) depends only on |x| and decreases as |x| increases [that is,
K(x) = ф(|х|), where ф(/), t > 0, is monotone decreasing]. Then
sup К/* л;)(х)| < c/*(x),
fi>0
with c independent of f.
Proof. We first remark that there is a constant c depending only on n
such that
sup <5 f |/(x - y)| dy < cf*(x).
<5>0 J|y|<<5
This follows by enclosing the ball |y| < b in the cube with center 0 and edge
2b and observing that the ratio of the two volumes is bounded independent
of b.
To prove the result, we will use a method based on Fubini’s theorem.
Fix e, and let E = {(у/) : у g Rn, t > 0, Kfy) > t}. Then E is a measurable
subset of R"+1 by (5.1), and
PXe(y) poo
Ae(y) = dt = ZE(y,f) dt.
0
(9.18)
I he iviarcmKifevv^*- uuuy.u.
Hence,
К/* ОД| =
f f(x - y)A;(y) dy
JRn
f l/(x - y)l
Jr”
Хе(У,0 dt
Changing the order of integration in the last expression, we obtain
l(/* 7Q(x)| <
Jo
0
f l/(x - y)lxE(y,0 dy dt
JRn
[ l/(x - y)| dy dt.
J(y:Ke(y)>*}
Let Et = {y : Kc(y) > t}, t > 0. Since K(y) depends only on |y| and de-
creases as |y| increases, E, is a ball with center 0. Hence, by our earlier remark,
1CZ* JQ(x)|
0
0
f IXх “ У)1 d? dt
|Д|с/*(х) dt = c/*(x) | |£J dt.
J о
Finally, note that |£f| is the distribution function of KE9 so that
\Et\dt= ikil = mil-
J о
Therefore, |(/* £e)(x)| < c||£||i/*(x), and the theorem follows by taking the
sup over 8 > 0.
We leave it to the reader to show that (9.17) can also be derived from
(9.14); see, e.g., the proof of (12.61) in Chapter 12.
In particular, for the kernel K(x) = 1/(1 4- |x|n+A), A > 0, Theorem (9.17)
gives
(9.18)
sup
€>O
6
L/(x _
< c/*(x),
a fact which will be used in the next section.
Note that the conclusion of (9.17) is valid for any К which is majorized
in absolute value by a kernel satisfying the hypothesis of (9.17). This includes
any К satisfying the hypothesis of (9.13).
4. The Marcinkiewicz Integral
We recall from Chapter 6, (6.17), that if F is a closed subset of a bounded
open interval (a,b) in R1, and if <5(x) denotes the distance from x to F, then
the Marcinkiewicz integral
f6 <Зл(у)
Мл(х) = jj-Y -~y|TOf/>' > °)
aic kucatiity, iVtckXliHdt Г? lions
is integrable over F. More generally, in Exercise 7 of Chapter 6, we considered
the expression
f ь\уЖу) , m
L к - г-4, (A>0’-
where f is nonnegative and integrable over the complement of F. If f =
Х(а}д), this reduces to Mk(x).
Now consider the integral
f <5A(y)/(y) , , B„4
Л(/)(х) = —гпт+л^У (xeR),
Jrh |A — y|
and the modified form
„ < - f <^л(у)/(у) ,
' Jr» Iх ~ УГ + Л + <5"+A(x) У'
Here again, 2 > 0, <5(x) denotes the distance from x to a closed set F cz R",
and f is nonnegative and measurable on R". Notice that Hk(f) and Jk(J} are
equal in F since 5 is zero there. For the same reason, J^f) and are in-
dependent of the values of f on F. Therefore, we may assume for simplicity
that f = 0 on F.
We will prove below that if f e Lp(Rn — F), 1 < p < oo, then e
LP(R"). This implies the basic fact that JA(/) eLp(F). (In general, Jx(f) diverges
outside F: see, for example, Exercise 9 of Chapter 6.) For the proof, it will be
convenient to consider one more modification of \ namely,
Г\( \ _ f ^A(y)/(y) ,
aU)W “ L |X - y|"+A + <5П+Л(У) y-
As we move from a point x to another point y, the distance from F does
not increase by more than |x — y|. Hence,
|<5(x) - 5(y)| < |x - y|.
It follows that <5(y) < |x — y| + <5(x), so that we have
<5"+A(y) < 2"+a[|x - y|n+A + <5n+A(x)],
as well as a similar inequality with x and у interchanged. We immediately
obtain that
2_„_Л_1Я,(/)(Х) < W)(x) 2и+л+1я;(/)(х).
Thus, inequalities for Н'л lead to ones for but H[ is easier to deal with.
(9.19)
The Marcinkiewicz integral
(9.19) Theorem If fe L^n\ 1 < p < oo, and A > 0, then H^f) e Lp(Rn)
and
IIW)HP < C||/HP,
where c is independent of f In particular, || JA(/)||P>F < c||/||p.
Proof Fix p, 1 < p < oo, and let g be any nonnegative function with
ll^llp' < 1, where 1/p + 1/p' = 1. By interchanging the order of integration,
we obtain
£ d' - £№).s\y)[ £ Гх- у-|-.Й£ 5.„(y) л] dy.
The outer integration on the right can be restricted to Rn — F without chang-
ing the value of the integral. However, if у g R" — F, then <5(y) > 0, and it
follows from (9.18) that the inner integral on the right is bounded by
с<5-л(у)д*(у). Combining this estimate with Holder’s inequality and (9.16),
we obtain
f ЛгК/)(х)й'(х) dx < c f /(y)0*(y) d'j
JRn JRn
< Cill/llpll^ll,' < Clll/llp-
By (8.9), the supremum of the left side for all such g is ||Л](/)||р, so that
1|Я1(/)||Р < cJI/Hp, and the theorem follows.
Exercises
1. Use Minkowski’s integral inequality (see Exercise 8, Chapter 8) to prove (9.1).
2. Prove Young’s theorem (9.2). [For f,g > 0 and p,q,r < oo, write
(/*^)(x) = - t)’/r-- t)’(1/’-1/r» dt,
and apply Holder’s inequality for three functions (Exercise 6, Chapter 8) with
exponents r, Pi, and p2, where 1/pi = 1/p ~ 1/r, l/p2 = 1/q — 1/r.]
3. Show that if/gZ/(R") and KeLp’(Rn), 1 < p < oo, 1/p + 1/p' = 1, then
f *K is bounded and continuous in R".
4. (a) Show that the function h defined by h(x) = e~1/x2 for x > 0 and h(x) = 0
for x < 0 is in C00.
(b) Show that the function g(x) = h(x — d)h(b — x),a < b, is C00 with support
(c) Construct a function in C^(R") whose support is a ball or an interval.
160
Approximations of the identity; Maximal Fu Sons
5. Let G and be bounded open subsets of R" such that Gr c G. Construct
a function heCb such that h = 1 in Gr and h = 0 outside G. (Choose an open
G2 such that Gt c G2, G2 c G. Let h = %g2*K for a К g C°° with suitably
small support and f К = 1.)
6. Prove theorem (9.4).
7. Let/g£p(—00,4-00), 1 < p < oo. Show that the Poisson integral of/,/(*,y),
is harmonic in the upper half-plane^ > 0. [Show that ((d2/dx2) 4- (d2/dj>2))/(x,.y)
= JiSWM + (а2й2)Ж - t) dt.]
8. {Schur's lemma) For s,t > 0, let K(s,t) satisfy К > 0 and K(fs,kt) = Л-1^(5,Г)
for all 2 > 0, and suppose that t~VpK{\,t) dt — у < 4-oo for somep,l <p
< oo. For example, K(s,t) = l/(s + t) has these properties. Show that if
(W) = J"/(z)K(s,t) dt (f > o),
then ||Г/||Р < y||/||p. [Note that K(s,t) = s~lK(\,tls\ so that = ft f(ts)
K(l,t) dt. Now apply Minkowski’s integral inequality (see Exercise 8, Chapter
8).]
9. The maximal function is defined as /*(x) = sup ICl"1^ |/|, where the su-
premum is taken over cubes Q with center x. Let /**(x) be defined similarly,
but with the supremum taken over all Q containing x. Thus, /*(х) < /**(x).
Show that there is a positive constant c depending only on the dimension such
that /**(x) < c/*(x).
10. Let T :f-+ Tf be a function transformation which is sublinear', that is, T has
the property that if T/j and Tf2 are defined, then so is T(/i + f2), and
|T(/i 4-/2)(x)| < |(ГЛ)(х)| + \(Tf2)(x)\.
Suppose also that there are constants Ci and c2 such that T satisfies || Tf | |oo <
cJI/lloo and |{x: |(7/)(x)| > a}| < c2a"1||/||i, a > 0. Show that for 1 <p < oo,
there is a constant c3 such that ||7/||p < c3||/||P. This is a special case of an
interpolation result due to Marcinkiewicz. (An example of such a T is the
maximal function operator Tf = /*, and the proof in the general case is like
that for/*.)'
11. Generalize theorem (9.6) as follows: Let f = f*K„KeL\Rn) and К = у.
If /e£p(R"), 1 < P < 00, show that Ц/ — yf\\p — 0. Derive analogous results
for (9.8), (9.9), and (9.13). [The case у # 0 follows from the case у = 1 by
considering K(x)/y.]
12. Show that the conclusions of (9.9) and (9.13) remain true if the assumption that
f gL1 is replaced by f gLp, p > 1.
13. Let/e£p(0,l), 1 < p < oo, and for each к = 1,2,..., define a function fk on
(0,1) by letting IkJ = {x'. (j — l)2~fc < x < j2~k}, j = 1,. . ., 2*, and setting
fk(x) equal to |/M|“1 JIk f for x g IkJ. Prove that fk — /in Z/(0,l) norm. [Ex-
ercise 18 of Chapter 7 may be helpful for the case/? = !.]
Chapter 10
Abstract Integration
In the preceding chapters, we developed a theory of integration based on a
theory of measurable sets. The notion of the measure of a set was in turn
based on the primitive and classical notion of the measure (or volume) of an
interval in R”; this led almost automatically by the process of covering to the
notion of measure for more general sets.
In this chapter, we follow an alternate approach. We will consider a
family of sets in Rn and assume that they all have “measures”, that is, assume
that with each member of the family, we can associate a nonnegative number
satisfying elementary and natural requirements which justify calling it a mea-
sure. Starting with this assumption, we will develop a theory of integration
which follows the pattern of Lebesgue integration. The advantage of this
method is that it can be applied not only to R", but also to general abstract
spaces with much less geometric structure than R”. Thus, it is important for
applications. There are new questions which arise in the abstract setting,
but many of the theorems and proofs are practically the same as those for
Lebesgue measure in Rn. In such cases, we will usually refer to earlier chapters
for proofs.
It is natural to ask how we can construct such measures. One possible
approach is to start with the more elementary notion of an “outer measure”
in an abstract space, and, as in the case of R” discussed in Chapter 3, select a
subclass of sets on which the outer measure has additional properties, qualify-
ing it as a measure. This idea will be developed in Chapter 11.
1. Additive Set Functions; Measures
Let У7 be a fixed set, and let S be a o-algebra of subsets of ; that is, let S
satisfy
(а) У7 g Z.
161
KUbuaui iiiusyidiiuii
(b) If £gE, then its complement CE ( — — £)gE (i.e., E is closed
under complements). \
(c) If Ek e S for к = 1,2,..., then (J Ek e E (i.e., E is closed under count- j
able unions). {
It is easy to see that the definition is unchanged if condition (a) is replaced !
by the assumption that E be nonempty; see also p. 39. Another widely used
term for a a-algebra is a countably additive family of sets.
Immediate consequences of the definition are that the following sets
belong to E:
(1) The empty set 0 (= СУ7);
(2) n^if£fc6E,fc= 1,2,... ;
(3) limsup Ek (= Q“=l |J^=m Ek) and liminf Ek (= |J”=1 Ek) if each
(4) - £2 (= r> CE2) if Et,E2 e S.
We recall the basic fact that the collection of Lebesgue measurable sub-
sets of R" is a a-algebra: see (3.20). In general, the elements £ of a cr-algebra
E are called H-measurable sets, or simply measurable sets if it is clear from
context what E is.
If E is a a-algebra, then a real-valued function ф(£), £ g E, is called an
additive set function on £ if
(i) ф(Е) is finite for every £ g E, and
(ii) <£((J £0 — ^Ф(Ек) for every countable family {£k} of disjoint sets in E.
SinceIJ Ek is independent of the order of the Ek \ the series in (ii) converges
absolutely.
We obtain a simple example of a set function by letting £ be the a-algebra
of all subsets of У’ and defining ф(Е) = xe(xq) for a fixed x0 g У7. As another
example, let £ be the collection of all Lebesgue measurable subsets of R”,
and define ф(Е) = JE /, where f g £(R").
A function /z(£) defined for £ in £ is called a measure on E if |
I
(i) 0 < ju(£) <4-oo, and
(ii) ju(|J £*) = E/z(£k) for every countable family {Ek} of disjoint sets in E.
The choices д = 0 or д == 4- oo are always possible, but of little interest.
If д is a measure on E, then the triplet (У\Е,р) is called a measure space, f
For example, Lebesgue measure together with the class of Lebesgue mea- \
surable subsets of R” is a measure space. As another example, let У* be any
countable set, — {xk}, and let {ak} be a sequence of nonnegative numbers.
Let E be the family of all subsets of and define 8(E) — £ akj if £ = {xkj}.
Then (ffff8) is a measure space. Such a space is called a discrete measure
space.
The distinction between a measure and an additive set function is that a
rtMUILIVC I UiibHUliO, IViCdSUItib
1 QJ
measure is nonnegative, but may be infinite, while an additive set function
may take both positive and negative values, but is finite. There are similarities
between many of the properties of set functions and measures. If Ex cz E2
and д is a measure, then /z(E2 — EJ + ц(Ех) — ^(Ef), so that ц(Е2 — EJ =
/z(EJ — /z(EJ if /z(EJ is finite. If ф is an additive set function, then the
formula ф(Е2 — EJ = ф(Е2) — 0(EJ, Ex cz E2, always holds. Choosing
Et = E2, we see that ф(0) — 0 for an additive set function, and also
ц(0) = 0 for a measure, unless ^u(E) = + co for all E. Moreover, if Ex cz E2,
then /z(EJ < /r(EJ even if ju(EJ = -boo, and </>(EJ < ф(Е2) if ф > 0.
The next few results concern limit properties and a basic decomposition
for additive set functions. Both LP and S are fixed.
(10.1) Theorem 7/*{Ек} is a monotone sequence of sets in S (that is, if Ek / E
or Ек \ E) and ф is an additive set function, then ф(Е) = lim^^^ Ф(Ек).
Proof If Ek / E, then E = |J Ek = Ex u (E2 - EJ u (E3 - EJ u • • •.
Hence,
oo N
ф(Е) = </>(£,) + E «Ж - ^-i) = <Ж) + Hm E [ф(Ек) - ф(Ек_ф]
к —2 N~><X)k = 2
= lim ф(Е„).
N-+ oo
On the other hand, if Ek \ E, then LP — Ek / LP — E. Therefore, by
the case already considered, we have ф(£Р — Ek) -> ф(<Р — E). Since
ф(У - Ek) = ф(У) - ф(Ек) and ф(У - E) = ф(У’) - ф(Е), the result
follows.
The next theorem is similar to Fatou’s lemma (5.17).
(10.2) Theorem Let ф be a nonnegative additive set function, and let {Ek} be
any sequence of sets in 2. Then
^(liminf EJ < liminf ф(Ек) < limsup ф(Ек) < «^(limsup EJ.
Л-* oo k~> oo
Proof The sets Hm = f\f=mEk increase to liminfEk. Therefore, by the
preceding theorem, ^(liminf EJ = lim ф(Нт). Since Hm cz Em, we have
ф(Нт) < ф(Ет) and lim ф(Нт) < liminf ф(Ет). Therefore, 0(liminf EJ <
liminf ф(Ет), which proves the first inequality. The proof of the third one is
similar, and the second is obvious.
If E e S, the collection of sets E n A as A ranges over S forms a ст-algebra
S' of subsets of E. In fact, S' is just the collection of all S-measurable subsets
of E. If ф is an additive set function on S, then its restriction to S' is additive
on S'. On the other hand, if ф is an additive set function on S', then the
function defined by ф(А) = ф(А n E) is additive on S.
Now, let ф be an additive set function on the measurable subsets of a set
EgL Define
164
Abstract Integration
(10.3)
V(E) = У(Е-ф) = sup ф(А), У(Е) = У(Е;ф) - -inf ф(А),
(10.3) ЛА^
V(E) = У(Е;ф) = V(E) + f (£)
to be the upper, lower and total variation of ф on E, respectively. Note that all
three are nonnegative since ф(0) = 0. Moreover, as is easy to see from the
definitions, each variation is monotone increasing with E; that is, if Ex c E2,
then F(Ej) < F(E2), etc. We will show that if ф is an additive set function
on S, then so are V, V, and V. The first step in proving this is the following
lemma.
(10.4) Lemma If ф is an additive set function on S, then each of its three
variations is countably subadditive; that is, if Ekel, к — 1,2,...,
then
F((J Ek) < Y V(Ek),
with similar formulas for V and V.
Proof Let Hx — Ex, H2 — E2 — Ex, H3 = E3 — E2 — E1? .... Then
the Hk are disjoint and (J Ek = (J Hk. If A g E and A cz (J Ek, then A —
|J (A n ETk) and ф(А) — Хф(А n Hk). Therefore, since A n Hk cz Ek, we
have ф(А) < £ F(Ek). Hence,
7((J Ek) = sup ф(А) < £ V(Ek),
A^UEk,Ae£
which proves the result for V. The proof for V is similar, and the result for V
follows by adding.
(10.5) Lemma If ф is an additive set function on E, then its variations L(E),
F(E), and V(E) are finite for every Eel.
Proof It is enough to show the result for V. Suppose that K(E) = 4- oo
for some E. We claim that there would then exist sets Ek g E, к ~ 1,2,...,
such that Ek \ and |</>(Efc)| > к — 1. To see this, we argue by induction.
Let Et = E, and suppose that Ex zo E2 zo • • • =z> En have been constructed
with \ф(Ек)\ > к - 1 and И(ЕЛ) = + oo for к = 1, . . . , N. Since V(EN) =
Too, there exists A gE such that A cz EN and |ф(Т)| > \ф(Е^\ -F N. If
У (A) = +oo, let En+x — A, noting that |ф(Т)| > N. If У(А) < Too, let
EN+l = En — A. Then F(E^+1) = -Foo since by (10.4) we have V(EN) <
L(En+1) -F У(А). Furthermore,
|</>(£w+1)| = \ф(Е„) - ф(А)\ > \ф(А)\ - \ф(Е„)\ > N.
This establishes the existence of sets Ek with the desired properties. Thus,
by (10.1), we obtain |^(Q Efc)| = lim |ф(Ек)| = -Foo, contradicting the
finiteness of ф.
(10.9)
Additive Set Functions; Measures
1 bb
The final step in proving that the variations are additive set functions is
given in the next lemma.
(10.6) Lemina If ф is an additive set function on E and {Ek} is a sequence of
disjoint sets in E, then 7(|J Ek) — £ У(Ек). Similar formulas hold
for V and V.
Proof By (10.4), we have 7(|J E^ < £ V(E^. To show the opposite
inequality, given s > 0, choose Ak с Ek with V(Ef) < ф(Ак) 4- c2~k. This is
possible since 7(EJ is finite by the previous lemma. Since the Ek are disjoint
so are the Ak, and we obtain
X Ж) < £ <KAk) + e = <K(J Ak) + 8 < F(U Ek) + 8.
Since 8 is an arbitrary positive number, the result for У follows. The analogous
formula for V is proved similarly, and the one for V follows by adding.
Combining (10.5) and (10.6), we immediately obtain the next theorem.
(10.7) Theorem If ф is an additive set function on E, then so are its variations
7, V, and 7.
The result which follows is basic and gives a decomposition of an additive
set function into the difference of two nonnegative additive set functions.
It may be compared to (2.6).
(10.8) Theorem {Jordan Decomposition) If ф is an additive set function on E,
then
ф(Е) = 7(E) - 7(E) (Ее E).
Proof If A с E and A e E, then ф(Е) = ф(А) + ф(Е — A) = ф(А) —
(— ф(Е — Aj), Since ф(Е) is fixed, we have ф(Ак) -> V(E) for a sequence of
sets Ak e E if and only if - ф(Е - Ak) -> 7(E). Hence, ф(Е) = 7(E) - 7(E),
as claimed.
A sequence {Ek} of sets is said to converge if limsup Ek = liminf Ek, Thus,
{Ek} converges if each point which belongs to infinitely many Ek belongs to
all Ek from some k on. If {Ek} converges, it is said to converge to the set
E = limsup Ek = liminf Ek,
As a simple corollary of the Jordan decomposition, we obtain the follow-
ing result.
(10.9) Corollary If a sequence of sets {Ek} of Ъ converges to E, and if ф is
an additive set function on E, then lim^^ ф(Ек) = ф(Е).
Proof. If ф > 0, we may apply (10.2). Since the extreme terms there both
equal ф(Е), it follows that all four equal ф(Е). Hence, lim ф(Ек) exists and
equals ф(Е). For arbitrary ф, the result therefore holds for 7 and 7, and so,
by the Jordan decomposition, for ф itself.
166
Abstract Integration
(10.10)
Let be a measure space. We have already observed that p satisfies
ptE^ < p(E2) tf Et E2, ЕъЕ2 e S. Another basic property of /z is given
in the next theorem.
(10.10) Theorem Let (ff,lL,p) be a measure space, and let {Ek} be any sequence
of measurable sets. Then
xlM) < Zx^.
Proof. Write (J Ek as a disjoint union as follows:
U Ek = Ei V (E2 - -Cl) u (E3 - E2 - £\) и • • •.
Then
XU Ek) = //(£1) + X^2 - Et) + M(E3 -E2-E1)+
< n(EJ + ц(Е2) + (J.(E3) + • • • = £ ц(Ек),
which completes the proof.
By definition, a measure is countably additive on disjoint measurable
sets [cf. (3.23)]. The next result shows that it shares another basic property
of Lebesgue measure [see (3.26)].
(10.11) Theorem Let (^,S,//) be a measure space, and let {Ek} be a sequence
of measurable sets.
(i) If Ek / E, then lim^» /z(Et) = ц(Е).
(ii) If Ek\ E and //(Ako) < + oo- for some kQ, then lim^^ p(Ek) =
KE).
Proof Suppose that Ek / E. If //(Efc) < + 00 for all k, we may use the
same argument used to prove the first part of (10.1). If p(Ek) = +00 for
some k, then lim p(Ek) — KE) = +00. To prove the second part, we may
assume that k0 = 1 and use the argument for (10.1).
(10.12) Corollary Let (fP^p) be a measure space and let {Ek} be a sequence
of measurable sets. Then
(i) Xliminf Efc) < liminf^ /z(Ek).
(ii) If XUS Ct) < + 00 for some k0, then ju(Iimsup Ek) >
limsup^» p(E*).
Proof Part (ii) is an immediate corollary of (10.2). For part (i), let
Am = Ek, m=l,2, .... Then Am Z liminf Ek, and by (10.11),
//(liminf Ek) = lim„,_>00 p(Atn). Since Am cz Em, we have д(Лш) < p(Em) and
lim^oo p(Am) < liminf^^ p(Em). The result follows by combining in-
equalities.
( ш. I4j
iVifcct&ui auic t uir
2. Measurable Functions; Integration
We will now develop the notions of measurable functions and integration
in a measure space. These will be used later in the chapter to prove several
important results for set functions.
Let S be a fixed ст-algebra of subsets of У7, and let f(x) be a real-valued
function defined for x in a measurable set E. (As usual, f may take the
values ±oo.) Then f is said to be Immeasurable, or simply measurable, if
{x g E : /(x) > a} is measurable for — oo < a < + oo. We will state some
familiar results whose proofs depend only on the fact that the class of mea-
surable sets forms a ст-algebra. The proofs are therefore similar to those in
Chapter 4 for Lebesgue measurable functions, and details are left to the
reader.
(10.13) Theorem
(i) If f and g are measurable on a set Ее L, then so are f + g, cf
for real с, ф(/) if ф is continuous on R1, /+, f~, \f\p for p > 0,
fg, and 1/f if f / 0 in E.
(ii) If {fk} are measurable on Ее E, then so are supfc fk, inffc fk,
limsup^ fk, liminf^ fk, and, if it exists, lim^ fk.
(iii) If f is a simple function taking values vlf . . . , vN on disjoint sets
Et,... , En, respectively, then f is measurable if and only if each
Ek is measurable. In particular, %E ™ measurable if and only if E is.
(iv) If f is nonnegative and measurable on Eel, then there exist non-
negative, simple measurable fk / f on E.
If (If,Y,p) is a measure space, a measurable set E is said to have p-
measure zero, or measure zero, if p(E) = 0. A property is said to hold almost
everywhere in E with respect to p, or a.e. (p), if it holds in E except at most
for a subset of measure zero.
We have the following analogue of Egorov’s theorem.
(10.14) Theorem (Egorov’s Theorem) Let (jf£,p) be a measure space, and
let E be a measurable set with p(E) < + oo. Let {fk} be a sequence of
measurable functions on E such that each fk is finite a.e. in E and {fk}
converges a.e. in E to a finite limit. Then, given s > 0, there is a measur-
able set A cl E with p(E — A) < & such that {fk} converges uniformly
on A.
In general, we cannot choose A to be closed; in fact, If has very little
structure, and the notion of a closed set may not even be defined. The proof
is similar to that for Lebesgue measure, and is left as an exercise.
Let / be nonnegative on a measurable set E. Define the integral of f over
E with respect to p by
Aosiracx integration
(10.15) fdg = sup£ [infftx)]/^) (/> 0),
JE j xeEj
where the supremum is taken over all decompositions E = (J £, of E into the
union of a finite number of disjoint measurable sets Ej. We adopt the con-
vention 0- oo^oo-O = 0 for the terms of the, sum in (10.15). By (5.8), the
definition reduces to the usual Lebesgue integral in case У’ = R", E is the
class of Lebesgue measurable sets, p is Lebesgue measure and f is Lebesgue
measurable. Although definition (10.15) does not require the measurability
of /, many of the familiar properties of the integral are valid only for mea-
surable functions. All functions considered in the rest of this section are
assumed to be measurable.
(10.16) Theorem Let (У\Е,//) be a measure space, and let f be a nonnegative,
simple measurable function defined on a measurable set E. If f takes
values vr, . . . , vN on disjoint Ei9 ... , EN, then
I* fdg = £ vjfi(Ej).
JE
Proof. Since f is measurable, each Ej is measurable by (10.13)(iii).
Clearly, JE f dp > £ Vjp(Ej). On the other hand, consider any decomposi-
tion E — |J Ak of E into a finite number of disjoint measurable sets, and let
wk = infAk f. If Ak n Ej is not empty, then wk < Vj. Therefore, by the addi-
tivity of //,
£ и’ХА) = £ £ wkg(Ak n Ej)
J к
£ Vj £ g(Ak n Ej) = £ VjH(Ej).
j к
Taking the supremum over all such decompositions, we obtain JE f dp <
£ Vjp(Ej), which completes the proof.
Note that the last theorem holds even if some of the Vj are + oo.
(10.17) Theorem Let be a measure space, and let f and д be mea-
surable functions defined on a set E el
(i) IfG < f < д on E, then f dg < fEg dg.
(ii) If f > 0 on E and g(E) = 0, then JE f dg = 0.
Proof. Both parts follow immediately from the definition (10.15). For
part (ii), note that p(Ej) — 0 whenever Ej с E and Ej e L. Hence, each term
of the sum in (10.15) is zero.
In order to investigate further the properties of the integral, we need the
next two lemmas. In these and the results which follow, the measure space
(У ,£,;/) is fixed.
(10.20)
immeasurable Functions; integration
16^
(10.18) Lemma
(i) If f and g are nonnegative, simple measurable functions on E, and
if c is a nonnegative constant, then J£ (/ 4- g) dg — J£ f dp 4-
J£ g dg and j£ cf dg = c fE f dg.
(ii) If f is a nonnegative, simple measurable function on E, and E =
Er и E2 is the union of two disjoint measurable sets, then J£ f dp =
R f dg + fE2 fdg.
The proof of the first part of (i) is like the first part of the proof of (5.14).
The second part of (i) follows immediately from (10.16). The details and the
proof of (ii) are left as an exercise.
(10.19) Lemma Let fk, к — 1,2,..., and g be nonnegative, simple mea-
surable functions defined on a set E el If fk / and lim^^ fk>g
on E, then
lim 1 dp > g dp.
k~* co JE Je
Proof Suppose that g takes values v19 . . . , vm on disjoint sets Et, ...,
Em. By (10.18)(ii), it is enough to show that
lim fk dp > g dp for each j.
oo JEj JEj
We thus reduce the proof to the case when g is constant on E, that is, g =
v > 0 on E. If v = 0, the result is obvious. Suppose then that 0 < v < 4- oo,
and let 0 < 8 < v and Ak = {хе E : fk(x) > v — e}, к = 1,2, ... . Since
fk we have Ak / E, so that p(E). Moreover,
A dg > fk dg > (v - e)g(Ak).
JE jAk
Therefore, lim^^ J£ fk dp > (v — s)p(E). Letting 8 -> 0 and observing that
vp(E) = J£ g dp, we obtain the desired result.
The next theorem is helpful in deriving properties of J f dp for arbitrary
nonnegative f from those for simple f.
(10.20) Theorem Let {fk} be a sequence of nonnegative, simple measurable
functions defined on EeL If fk /* f on E, then J£ fk dp -> J£ f dp.
Proof. Clearly, lim^^ J£ fk dp < J£ f dp. To show the opposite inequal-
ity, consider a partition E = |J Ej of E into a finite number of disjoint mea-
surable sets Ej, and let Vj = infE.f and a = X The function g
defined by g = X Vj%Ej *s nonnegative and measurable, and J£ g dp = a.
Since linifc.^ fk > g, we have limk„>00 J£ fk dp > a by (10.19). Taking the
supremum of such a over all partitions of E, we obtain the desired result.
170
Abstract Integration
(10.21)
As a corollary, we have the following theorem.
(10.21) Theorem Let f and g be nonnegative measurable functions defined on
EgL, and let c be a nonnegative constant. Then
(i) Je(/+ g)dg = \Efdg + д ф and JE cf du = c\Efdg.
(ii) If E = Е{ и E2, where E{ and E2 are disjoint and measurable,
then \Efdg = J£, f dg + jE2 f dg.
Proof By (10.13)(iv), choose simple measurable fk and gk such that
0 < fk/f and 0 < gk Z g. Then fk + gk is simple and measurable, and
0 < fk + gk / f + g. Therefore, by (10.20) and (10.18),
(f + в) dg = lim (fk + gk) dg = lim ( dg + gk dg I
Je oo Je &->oo \ Je Je J
= fdg + g dg.
JE JE
This proves the first part of (i); the other parts are proved similarly.
If f is any real-valued measurable function defined on a measurable set
E, we define its integral with respect to g by
(10.22) f /(x) dg(x) = { fdg=[ f+dg-[ f~ dg,
JE JE JE JE
provided not both integrals on the right are + oo.
We say that f is integrable with respect to g, or g-integrable, over E if
Je f dg exists and is finite. When this is the case, we write f e L{E’,dg) or
/еОД/z).
It is immediate from (10.22) and (10.17) that JE f dg = 0 if g(E} — 0,
and that JE fdg < J£ g dg if f < g on E and both integrals exist. The
familiar properties of the Lebesgue integral are shared by J£ f dg\ some of
them are listed in the following theorem.
(10.23) Theorem
(i) IJe f dg\ < |/| dg; furthermore, f e L(E’,dg) if and only if
\f\eL(E;dg).
(ii) If |/| < |#| a.e.(p) in E, and if g e L(E*,dg), then f e L(E',dg)
and J£ |/| dg < J£ |^| dg.
(iii) Iffe LfEfig), then f is finite a.e.(g) in E.
(iv) If f = g a.efg) in E and if JE f dg exists, then J£ g dg exists and
\Egdg = jEfdg.
(v) If Je f dg exists and c is a cbnstant, then $Ecf dg exists and
\Ecf dg = c J/; f dg.
(vi) If f,g e L(E;dg), then f + g e L{E;dg) and f£ (/ + g) dg =
$Efdg + $Egdg.
(vii) Iff>Q and m < g < M on E, then
m \ f dg < \ fg dg < M \ fdg.
Je Je Je
Proof. The proofs are similar to those for Lebesgue integrals. As ex-
amples, we will prove (iii) and (iv). For (iii), suppose that f e L(E\dg). Then,
by (i), |/| 6 L(E-,dg). Let Z = {x e E : \ f(x)\ = + oo}. Then for any positive
integer k,
kg(Z) < [ |/| dg < f |/| dg.
Jz Je
Since f g L(E',dg), it follows that g(Z) = 0, which proves (iii).
For (iv), since both f+ = g+ and f~=g~ a.e. in E, we may assume
that f > 0. Let Ex = {x e E : /(x) / g(x)}. Since /ДЕ)) = 0, (10.21)(ii)
implies that
fdg = fdg = g dg = g dg,
Je Je-Ei Je—Ei Je
as asserted.
(10.24) Theorem If {fk} is a sequence of nonnegative measurable functions on
E, then
f (ЕЛ) = E f fk Ф-
JE JE
Proof. Let f = E"=1 fk- Since f > E”=i Л, the integral of f over E
majorizes Je fk f°r any m- Hence, the left side above majorizes the
right. To show the opposite inequality, let {f(kj)} be a sequence of non-
negative, simple measurable functions increasing to fk. Let Sj = = i JY-
Then Sj is nonnegative and simple, and Sj /. We will show that Sj / f.
Clearly, limy^^ Sj < f. On the other hand, for any m,
m rn
lim Sj > lim E Л° = E fk-
j-> <x> j-r oo к = 1 к = 1
Therefore, lim7^ro Sj > f. It follows from (10.20) that \ESj d/л -> f d/л.
Since fk > Sj, we obtain
QO Г J Г ' p p
E fk dg = bm E fk dg > lim dg = fdg.
k=t je j-* oo к = i Je j-> co Je Je
This proves the desired inequality, and the theorem follows.
The next three results are essentially corollaries of (10.24).
(10.25) Theorem If J£ f dp exists, and if E — [j Ek is a countable union of
disjoint measurable sets Ek, then
f fdp = £ f fdp.
Je
Proof Suppose first that f > 0. Let fk = fyEk on E, so that fk is mea-
surable and nonnegative, and f = £ fk. By (10.24),
f fdp = X ( fk dp- = X f fdp-
je Je jEfc
For arbitrary measurable f, the existence of J£ f dp implies that of
J£fc f dp; in fact, the integrals of f+ and f~ over any Ek are majorized by
those over E. Moreover, by the case already considered,
I* /+ dp = £ 1* /+ dp, I" f~ dp = X [ f~ dp.
JE jEk JE JEk
Since at least one of these sums is finite, the conclusion follows by subtrac-
tion. [Cf. theorem (5.24).]
(10.26) Theorem If fk are measurable and 0 < fk/fonE, then J£ fkdp ->
Ь fdp.
Proof. If J£ fkdp = + oo for some к, the result is obvious. We may there-
fore assume that each fkGL(dp)-. Write f = fx + £“=2 (A — Л-i)- Since
each term on the right is nonnegative, we obtain from (10.24) and (10.23)(vi)
that
\ fdp = /1 dp + X A dp - j dp = lim dp.
je je jt=2\jE Je / k-> 00 Je
(10.27) Theorem (Monotone Convergence Theorem) Let {fk} and f be mea-
surable functions on E.
(i) Suppose that fk/f a.e. (p) on E. If there exists ф e L(E;dp) such
that fk> ф on Efor all k, then J£ fk dp -^Cj£ f dp.
(ii) Suppose that fk\f a.e. on E. If there exists ф e L(E;dp) such
that fk < ф on E for all k, then J£ fk dp -> J£ f dp.
Proof. The proof of (i) follows by applying (10.26) to the functions
fk — ф. The details are as in the proof of (5.32). Part (ii) follows by applying
(i) to the functions — fk.
(10.28) Theorem (Uniform Convergence Theorem) Suppose that fke L(E;dp),
к = 1,2,..., and that {fk} converges uniformly to f on E, p(E) <
-F co. Then f g L(E;dp) and J£ fk dp -> J£ f dp.
ieasuraoie runcnons; miegraxion
1 /3
The proof is the same as for Lebesgue measure [see (5.33)], and is omitted.
Fatou’s lemma and the Lebesgue dominated convergence theorem are true
for abstract measures. They are stated below without proof; the proofs are
like those of (5.34) and (5.36).
(10.29) Theorem (Fatou’s Lemma) Let {fk} be a sequence of measurable
functions on E. If there exists ф e L(E\d[i) such that fk > ф on E for
all k, then
(liminffk) dp < liminf fk dp.
Je jt-> oo k-*oo Je
The case ф = 0 (that is, fk > 0) is of special importance. In this case,
we also have the following useful corollary.
(10.30) Corollary LeE{fk} and f be nonnegative measurable functions on E
such that fk -> f a.e. (д) in E. If f£ /k dp < M for all k, then
jEfdp < M.
(10.31) Theorem (Lebesgue Dominated Convergence Theorem) Let f, {fk}
and ф be measurable functions on E such that \fk\ < ф a.e. (//) on E and
ф e L(E\dp). Then
(i) J£(liminfHoo fk) dp < liminf^ fk dp < limsup^ M fkdp <
(limsup^ x dp.
(i>) V fk~> f a.e. (p) in E, then fk dp -> f dp.
(10.32) Corollary (Bounded Convergence Theorem) Let {fk} and f be mea-
surable functions on E such that fk —> f a.e. in E. If p(E) < + co and
there is a constant M such that \fk\ < M a.e. in Et then JE fk dp
fEf dp.
We conclude our brief study of integration with respect to abstract mea-
sures by defining Lp(E\dp) == Lp(E,\dp\ 0 < p < oo, to be the collection of
all measurable real or complex-valued f such that \f\p dp < + oo. We set
/ f \
ll/ll, = ll/ll,= l/r (0 < p < «)•
\ Je /
When p = oo, Lx(E-,dp) is the collection of all measurable f such that
II/II «> < +°o, where
ll/ll« = Н/Ноо.ЕЛд = ess sup I f\ = inf {a : p(x e E : |/(x)| > a) = 0}.
E
Observe that lp is Lp(If^,dp) when У7 is the set of integers, E is the set
of all subsets of If, and p(E) is the number of elements of E.
For 1 < p < oo, Holder’s and Minkowski’s inequalities hold:
Ш11 < \\f+g\\p< ll/ll, + [\g\\P.
1 /4
Abstract Integration
(10.33)
Moreover, Lp is a Banach space with norm || • ||p. In general, Lp is not separa-
ble (see Exercise 9). However, if L2 is separable, we can define orthogonality,
linear independence, completeness, Fourier coefficients, and Fourier series as
usual, obtaining Bessel’s inequality and Parseval’s formula, as well as the
usual result relating L2 and I2.
3. Absolutely Continuous and Singular Set Functions and Measures
We now turn out attention from the familiar results above to some new
ones arising naturally in the context of abstract measure spaces. Thus, let
(У,Х,д) be a measure space, and let ф be an additive set function on L.
If E g E, then ф is said to be absolutely continuous on E with respect to p if
ф{А) = 0 for every measurable A cz E with д(Л) = 0. Note that this defini-
tion has a somewhat different pattern from that given for Lebesgue measure
(see p. 99). However, we shall obtain in (10.34) a reformulation of the
present definition in terms of the old one.
On the other hand, ф is said to be singular on E with respect to p if there is
a set Z cz E such that p{Z) = 0 and ф{А) = 0 for every measurable A cz
E — Z. Thus, ф is singular if it is supported on a set of д-measure zero,
so that E splits into the union of two sets, Z and E — Z, one with д-measure
zero and the other with the property that ф is zero on each nje^surable
subset of it.
As examples, note that if f g L{E;dp), then the function ф{А) = f dp
is absolutely continuous on E with respect to p. If Z is any set with p{Z) = 0
and ф is any additive set function, then the function ф{А) = ф{А n Z) is
singular on E with respect to p.
We list several simple properties of such set functions in the next theorem.
(10.33) Theorem
(i) If ф is both absolutely continuous and singular on E with respect
to p, then ф{А) = 0 for every measurable A cz E.
(ii) If both ф and ф are absolutely continuous {singular) on E with
respect to p, then so are ф + ф and сф, where c is any real constant.
(iii) ф is absolutely continuous {singular) on E with respect to p if and
only if its variations V and V are, or, equivalently, if and only if
its total variation V is.
(iv) If {фк} is a sequence of additive set functions which are absolutely
continuous {singular) on E with respect to p, and if ф{А) =
lim^ фк{А) exists for every measurable A cz E, then ф is abso-
lutely continuous {singular) on E with respect to p.
Proof For part (i), suppose that ф is absolutely continuous and singular
on E. Let Z be a subset of E with д-measure zero such that ф{Н) = 0 if
H is measurable and H cz E — Z. If A is any measurable subset of E, then
ф(А) = ф(Л n Z) -h ф(А — Z). Since ф is absolutely continuous and
p(A n Z) = 0, we have ф(А n Z) = 0. Moreover, since A — Z cz E — Z
and ф is singular, we have ф(А — Z) = 0. Hence, ф(А) — 0, and part (i) is
proved. The proofs of parts (ii)-(iv) are left as exercises.
The next two theorems give alternate characterizations of absolutely con-
tinuous and singular set functions.
(10.34) Theorem An additive set function ф is absolutely continuous on E
with respect to p if and only if given 8 > 0, there exists b > 0 such that
|ф(Л)| < 8 for any measurable A cz E with p(A) < b.
Proof The sufficiency of the condition is immediate since if //(Л) = 0,
then p(A) < b for all b > 0, so that |</>(Я)| < e for all e. Consequently,
ф(А) = 0. For the converse, suppose that ф is absolutely continuous, but that
there is an 8 > 0 for which no b > 0 gives the desired result. Then, taking
b = 2~k for к = 1, 2, ... , there would exist measurable Ak cz E with
p(A^ < 2~k and |</>(A)I Let = limsup Ak. Then, for any m,
/co \ 00
11(A) < /z U Л < £ 2-\
\fc=m / k=m
sothat/i(z() = 0. Therefore, ф(А) = 0. Assuming for the moment that ф > 0,
we obtain from (10.2) that ф(А) = ^(limsup Ak) > limsup ф(Ак) > в. This
contradiction establishes the result in case ф > 0. For the general case, the
variation V of an absolutely continuous ф is absolutely continuous [by
(10.33)(iii)] and nonnegative. Since | ф(А)\ < V(A), the theorem follows.
(10.35) Theorem An additive set function ф is singular on E with respect to p
if and only if given 8 > 0, there is a measurable subset EQ of E such that
p(E0) < 8 and V(E - Eq; ф) < s. (
Proof If ф is singular, there exists Z cz E with p(Z) — 0 such that
V(E — Z; ф) = 0. Taking Eo = Z, we obtain the necessity of the condition.
To prove its sufficiency, choose for each к = 1, 2, ... a measurable Ek cz E
with /z(£k) <2~k and V(E — Ek; ф) < 2~k. Let Z = limsup Ek. Since
Z cz (j£°=m Ek for every m, it follows as usual that p(Z) = 0. Moreover, by
(Ю.2),
V(E — Z; ф) = V(E — limsup Ek; ф) = F(liminf (E — Ek); ф)
< liminf V(E - Ek\ ф) = 0.,
Hence, ф is singular with respect to p, which completes the proof.
Let ф be the indefinite integral of an f e L(E\dp): ф(Л) = f dp for mea-
surable A cz E. Letting P = {x e E : /(x) > 0}, we see that ф(А) > 0 for
any measurable A cz P, and that ф(А) < 0 for any measurable A cz E — P,
It follows that У(Е;ф) = ф(Р) and У(Е;ф) = — ф(Е — P). Moreover, since
P = {x e E : f(x) = /+(x)}, we have
7(Р;ф) = f f+ dp, У(Е;ф) = f Г dp.
JE JE
For an arbitrary ф, we have the following basic result.
(10.36) Theorem (Hahn Decomposition) Let E be a measurable set and let
ф be an additive set function defined on the measurable subsets A of E.
Then there is a measurable P cz E such that ф(А) > 0 for A cz P and
ф(А) <0 for A cz E - P. Equivalently, У(Р;ф) = V(E - Р;ф) = 0.
Hence,
У(Е;ф) = У(Р;ф) = ф(Р),
У(Е;ф) = V(E -Р;ф)= -ф(Е - Р).
Proof Write У(А) = У(А;ф) and У(А) = У(А;ф) for A cz Е. For each
positive integer к, choose a measurable Ak cz E such that ф(Ак) > У(Е) —
2Tk. Then У(Ак) > У(Е) ~ 2~к. Since У is additive, У(Е - Ак) = У(Е) -
У(Ак) < 2~к. Moreover, by the Jordan decomposition, У(Ак) — У(Ак) =
ф(Ак) > У(Е) — 2“*, so that У(Ак) < 2~к. Let Р = liminf Ак. Since У is
nonnegative, (10.2) implies that У(Р) < liminf У(Ак) = 0. Also,
V(E - P) = V(E - liminf Ak) = /(limsup (E - Ak)) < V( Q (£ - Л))
\k = m J
for any m. Therefore, for any m,
oo oo
VIE - P) < X V(E - Ak) < £ 2~k,
k = m k=m
which gives V(E — P) = 0 and completes the proof.
In the next theorem, we use the Hahn decomposition to split E into sets
where ф is comparable to p. In doing so, we assume that ф is nonnegative
and p is finite; thus, we are in fact dealing with two finite measures.
(10.37) Theorem Let ф be a nonnegative additive set function defined on the
measurable subsets of a measurable set E, and let p be a measure with
p(E) < + oo. Then given a > 0, there is a decomposition E = Z и
((Jk°= i ^k) °f E into disjoint measurable sets such that
(i) p(Z) = 0.
(ii) a(k — 1)^(Л) < ф(А) < akp(A) for measurable A cz Ek, к =
1,2,....
Proof We may assume that a = 1 by considering ф/а. For each positive
integer k, let фк(А) = ф(А) — kp(A) for measurable A cz E. Since ф and /r
are finite and additive, фк is an additive set function. By the Hahn decom-
position, there is a set Pk cz E such that фк(А) > 0 if A cz Pk and фк(А) < 0
if A cz E — Pk. Thus, ф(Л) > k[i(A) if A cz Pk and ф(А) < кц(А) if A cz:
E- Pk-
Now, let Qk = Em ^ог к = h 2, . . . , and observe that Pk cz Qk
and Qk \. We will show that ф(А) > кц(А) if Л cz Qk, and ф(А) < к/л(А)
if A cz E — Qk. To see this, write
Qk — EkU (Pk+ 1 “ Pк) U (Pk + 2 ~~ Pk+ 1 — Pk) U
and note that the terms on the right side are disjoint. Hence, if Л cz Qk,
we may write A = Am, where the Am are disjoint and Am cz Pm, by
simply intersecting A with each such term of Qk. Then
00 oo 00
Ф(А) = E «AGQ S: E тц(Ат) > k^ fi(Am) = kfi(A),
m = k m = k m=k
so that ф(А) > k^(A) for A cz Qk, as claimed. On the other hand, if A cz
E — Qk, then A cz E — Pk9 so that ф(А) < кц(А). This proves the assertion
above.
We can now give the decomposition of E. Let Z = t Qk = limsup Pk9
and write
E = Z и (E - Qt) u (Q. - Q2) u (02 - g3) u • • •
= Z и Et и E2 u E3 u • • •.
The terms in this decomposition are disjoint. If A cz E1(= E — Qt), then
ф(А) < /z(Z) by what was shown above, and ф(А) > 0 by hypothesis. For
к > 2, we have Ek = Qk-i — Qk = Qk-1 n (E — Q^. Hence, if к > 2 and
A cz Ek, then ф(А) > (к — 1)X^) due to the fact that A c. Qk-i ', also,
ф(А) < кц(А) due to A cz E — Qk. Finally, since Z cz Qk for all k, we have
ф(& > kn(Z) for all k. Since ф is finite, it follows that /z(Z) = 0, which
completes the proof.
To give some idea of the significance of the last result, write A =
(A n Z) u [|Jk G4 и Ek)] for measurable A cz E. Then
ф(А) = ф(А n Z) + X Ф(А Ek).
The set function а(Л) = ф(А n Z) is singular with respect to ц. By (ii) of the
theorem, ф is absolutely continuous with respect to p. on each Ek. Hence, the
set function a defined by
а(Л) = ф(А) - а(Л) = £ Ф(А Ек)
is absolutely continuous with respect to /z since if д(Л) = 0, then д(Л n Ek) =
0 and ф(А n Ek) ~ 0 for all k. Note also that (ii) can be written
178
Abstract Integration
(10.3^)
a(k — 1) i du < ф(А) < ak | ф
Ja ‘ Ja
for measurable A C Ek.
We will use these ideas to write any set function as the sum of an abso-
lutely continuous part, which will be an indefinite integral, and a singular
part. This decomposition, which is of major importance, is stated in the
following theorem. We assume that the measure д defined on the measurable
subsets of E is a-finite, that is, that E can be written as a countable union of
measurable sets with finite д-measure.
(10.38) Theorem (Lebesgue Decomposition) Let ф be an additive set function
on the measurable subsets of a measurable set E, and let p be a a-finite
measure on E. Then there is a unique decomposition
ф(А) = a(A) 4- a(A) for measurable A cz E,
where a and a are additive set functions, a is absolutely continuous,
and о is singular with respect to p. These functions are
a(A) = f dp, a(A) = ф(А n Z)
Ja
for appropriate f e L(E',dp) and Z with p(Z) = 0. Moreover, if ф > 0,
then f > 0.
Proof Assuming that such a decomposition exists, we will show it is
unique. If ф = 4- is another decomposition of ф into absolutely con-
tinuous and singular parts, then a — ax = — a, which (being both abso-
lutely continuous and singular with respect to p) must vanish identically.
Hence a = and a =
To show that the decomposition exists, first assume that ф > 0 and
p(E) <4-oo. Taking a = 2~m, m = 1, 2, . . . , in (10.37), we may write E as
a disjoint union E = Z(m) u ((Jk E(kmf where
p(Z(m)) = 0, 2~~m(k - l)p(A) < ф(А) < 2~mkp(A), A cz
Given m, k, m', and k', let p = 2~m(k - 1), 7 = 2~mk, P' = 2~m' (kf -1)
and yf = 2~m k’. If the intervals [P,y] and [/?',/] are disjoint, we will show
that the set A = Ejf^ n Ejf} has д-measure zero. In fact, we have both
Pp(A) < ф(А) < yp(A) and p'p(A) < ф(А) < y'p(A).
If, for example, у < /У, the inequalities P'p(A) < ф(А) < yp(A) imply that
p(A) = 0. A similar argument applies if y' < p. Fixing m and k, and setting
m' = m 4- 1, we see that there are at most four values of k' such that EkM
intersects E^ in a set of positive д-measure: namely, k' = 2k — 2, 2k — 1,
2k, 2k 4- 1. Hence,
i u.vu /
£'<”> c и E%t}> и £^+1) и и Y^\
where ^(Y^) = 0.
Let
z = (U z(m)) и (U y<m)),
m k,m
so that ji(Z) = 0, and define functions {fm}m=i on £by fm(x) = 2~m(k — 1)
if x e E^m) — Z and fm(x) = 0 if x e Z. Therefore, if x g E^m) — Z, fm(x) =
2~m(k — 1) and fm+i(x) takes one of the four values 2~m~1J, j = 2fc — 3,
2k — 2,2k — 1, 2k. Hence, \fm(x) - /m+1(x)| < 2~m if x g - Z, and
so also if x g E. It follows that {/w} converges uniformly on £ to a limit /.
Since fm > 0, also f > 0.
Since E is the disjoint union Z u [(Jfc (Efcmy — Z)] and ф is absolutely
continuous on each Е^п\
ф(А) = ф(А n Z) + £ <№
= ф(А n Z) + X Ф(А n E^)
к
for A с E. Therefore,
ф(А n Z) + £ 2-m(k - l)/z(Z n £<"’) Ф<А)
к
< ф(А n Z)+ £ 2~ткц(А n
к
which can be rewritten
ф(А л Z) + [ fm du < ф(А) < ф(А n Z) + f fmdfi + 2~тц(А).
JA JA
Since ju(A) is finite, we obtain from the uniform convergence theorem that
Ja fm dfi U fdV- Therefore,
ф{А) = ф(А r>Z) + ^fdn,
which is the theorem in case ф > 0 and ц(Е) < 4-oo.
If ф > 0 and ja(E) = Ч-oo, then E can still be written as a disjoint union
E = (J Ej with [f(Ej) <4-oo, since E is ст-finite. Hence, there exist Zj c E^
^(Zj) = 0, and nonnegative on Ej such that
ф(А n Ej) = ф(А n ZD 4- f fj dii (A cz E).
JAdEj
Letting Z = \J Zj and f = fj%Ep we obtain p(Z) = 0, f > 0, and
ф(А) = £ Ф(А n Ej) = 1Ф(Ап Zj)
+ E f fjdk = <ХЛ П Z) + [ fdn.
JXnEj jA
180
Abstract Integration
(10.39)
Of course, f is integrable since ф is finite. The proof is now complete if
ф > 0.
For an arbitrary ф, apply the decomposition to each of V and K, and
subtract the results. By the Jordan decomposition, we obtain ф(Л) =
f dp + <т(Л), where f e L(Efip) and a is singular with respect to д. It
remains to show that there is a set Z, ^(Z) = 0, such that сг(Л) = ф(А n Z).
Let Z be the set of д-measure zero corresponding to a in the definition of a
singular set function. Then a(A n Z) = a(A) and JAnZ f dp = 0. Hence,
replacing A by A n Z in the formula for ф above, we obtain a(A) =
ф(А n Z). This completes the proof. For a result concerning the uniqueness
of /, see Exercise 6.
We have already noted that the indefinite integral of an integrable function
is absolutely continuous. The following fundamental result gives a converse:
namely, in a а-finite space, the only absolutely continuous set functions are
indefinite integrals.
(10.39) Theorem (Radon-Nikodym) Let ф be an additive set function on the
measurable subsets of a measurable E, and let pbe a a-finite measure on
E. If ф is absolutely continuous with respect to д, there exists a unique
f g L(E;dp) such that
ф(А) = f fdn
JA
for every measurable A E E.
Proof The result follows from the Lebesgue decomposition. In fact,
ф(А) = f dp + ф(А n Z) for appropriate f g LfEfip) and Z with
д^) = 0. Since ф is absolutely continuous, we have ф(А n Z) = 0, so that
ф(А) = f dp. For the uniqueness of /, see Exercise 6.
If v and д are two measures defined on the same family of measurable
sets, we say that v is absolutely continuous with respect to p on a measurable
set E if v(Z) = 0 for every A с E with д(Л) = 0. If v is finite, (10.34) implies
that a necessary and sufficient condition for v to be absolutely continuous
with respect to д is that given s > 0, there exist d > 0 such that v(Z) < e if
p(A) < 5. The necessity of this condition may fail if v is not finite; see
Exercise 12.
We say that v and д are mutually singular on E if E can be written as a
disjoint union, E = Er u E2, of two measurable sets with vfEfi = p{E^ =
0. The reader can check that the following analogue of (10.35) is valid:
Two measures v and д are mutually singular on E if and only if given s > 0,
there are disjoint measurable E1 and E2 with E = EY u E2 and v(Efi < s,
№2) < e-
We also note that if v and p are mutually singular on E and if д g L(E\dv),
(10.40) Abso 2iy Continuous bet t-unciions <anu kVieci&ui
fi W 8
then the set function g dv is singular with respect to g. To see this, write
E = Er и E2, where Er and E2 are disjoint with v{Ex) = g^E^ — 0. Setting
Z = E2, wehave/i(Z) = Oandv(X) = 0 for every measurable Л cz E — Z =
E\. Hence, jj g dv = 0 for such A, which proves the assertion.
We have the following analogue of the Lebesgue decomposition.
(10.40) Theorem Let v and g be two ofinite measures defined on the mea-
surable subsets of a measurable E, Then there is a unique, nonnegative
measurable f on E and a unique measure a on the measurable subsets
of E such that о and g are mutually singular on E and г(Л) = f dg 4-
o(A) for every measurable A cz E. Moreover,
g dv = gf dg + g do
jA JA JA
whenever g dv exists.
Before giving the proof, we add several remarks. First, f dg is an
absolutely continuous measure with respect to g since f > 0. Next, if E ~
Z и (E — Z), where g(Z) = o(E — Z) = 0, then о has the form
а(Л) = г(Л n Z),
as can be seen by replacing A by A n Z in the decomposition of v. Note also
that if v(E) is finite, then f e L{E',dg). Finally, if g e L(E;dv), then g e LfEfio).
In this case, the second formula in the theorem implies that fg e L(E\dg)
and expresses the Lebesgue decomposition of g dv with respect to g.
Proof If v(E) < 4-oo, the Lebesgue decomposition implies that v(^) =
f dg 4- o(A) for measurable A cz E, where f > 0, f e L(E',dg), and о and
g are mutually singular. If v(E) = 4-oo, then since v is n-finite, we have
E = (J Ej with Ej disjoint and v(Efi < 4- oo. Choose Zj cz Ej and fj on Ej
such that g(Zfi = 0, fj > 0 and
v(A n Ej) = fj dg + v(A n Zfi for measurable A <=. E.
jA^Ej
Let Z = |J Zj and / = £ fj%Ej- Then g(Z) — 0, and adding over j, we have
v(A) =| f dg 4- v(A n Z) = | f dg 4- o(A),
JA Ja
as claimed. The proof of the uniqueness of f and a is left as an exercise.
If g is the characteristic function of a measurable set B, the formula
in question, namely, g dv = gfdg + g do, reduces to у(Л n B) =
J^nB f dg + n(/l n B). Hence, the formula is valid for any simple measurable
g, and therefore, by the monotone convergence theorem, for any measurable
g > 0. Now let g be any measurable function for which g dv exists. Then
о
182
Abstract Integration
(10.41)
at least one of JA g + dv and JA g dv is finite, and the formula for g follows
by subtracting those for g+ and g~~.
(10.41) Corollary Let v and g be two а-finite measures defined on the mea-
surable subsets of a measurable E.
(i) Then v is absolutely continuous with respect to g on E if and only if
there is a nonnegative measurable f such that v(A) = JA f dg for
every measurable A cz E. In this case,
g dv = \ gf dg
Ja Ja
for any measurable g and A cz E for which JA g dv exists.
(ii) Let g e L(E',dv). Then JA g dv = ]*л gf dg for some nonnegative f
and all measurable A cz E if and only if JA g dv is absolutely con-
tinuous with respect to g.
Proof Part (i) follows from (10.40) since a = 0 if and only if v is abso-
lutely continuous with respect to g. Let у(Л) = f dg 4- a(A) be the decom-
position given by (10.40). Part (ii) follows from the fact that the formula
Ja 9 dv = J4 gf dg 4- JA g da is the Lebesgue decomposition of J4 g dv.
4. The Dual Space of Lp
If В is a Banach space (or, more generally, a normed linear space) over the
real numbers, a real-valued linear functional I on В is by definition a real-
valued function 1(f), f e B, which satisfies
l(fi +f2) = /(/1) + /(Л), /(«/) = od(f), -к < a < 4-oo.
A linear functional / is said to be bounded if there is a constant c such that
\l(f)\ < c\\f\\ for all f e B. A bounded linear functional I is continuous with
respect to the norm in B, by which we mean that if \\f — fk\\ -> 0 as к oo,
then l(fk) -> 1(f), since \l(f) - l(fk)\ = \l(f - fk)\ < c\\f - fk\\ 0.
The norm ||/|| of a bounded linear functional / is defined as
(Ю.42) ||/|| = sup |/(/)|.
Since //У/ll has norm 1 for any / / 0, and since I is linear, we have ||/|| =
sup |/(/)|/||/||. .
The collection of all bounded linear functionals on В is called the dual
space Bf of B. We shall consider the case when В = Lp = Lp(E;dg), and for
simplicity restrict our attention to real-valued functions. Our goal is to show
that if 1 < p < oo and g is сг-finite, then the dual space (Lp)f of Lp can be
identified in a natural way with Lp'. The main tool in doing so is the Radon-
Nikodym theorem. The first result is the following.
(10.44)
I ne uuai opctue ui
(10.43) Theorem Let 1 < p < oo, 1/p + 1/p' = 1. If g e Lp'(E',dp), then the
formula
1(f) = f f9
Je
defines a bounded linear functional I e\Lp(E',dp)}f. Moreover, ||/|| <
\\g\\P-
Proof This follows immediately from Holder’s inequality and the linear
properties of the integral. We have
IV)I =
fgdp <
Je
IK'll/llp,
so that ||/|| < \\g\\p>.
The theorem shows that with each g e Lp' we can associate a bounded
linear functional, 1(f) = JE fg dp, on Lp. The correspondence between g and
I is unique (see Exercise 6) and defines an embedding of Lp' in (Lp)'. We now
give the characterization of (Lp)r, 1 < p < oo.
(10.44) Theorem Let 1 < p < co, 1/p + 1/p' = 1, and let p be a-finite. If
I e [Lp(E',dp)}', there is a unique g e Lp'(E‘,dp) such that
i(f) = [ fg dp.
je
Moreover, ||/|| = ||tj||p,, so that the correspondence between I and g
defines an isometry between (Lp)r and Lp .
Proof Suppose first that p(E) < 4-oo. Let I e (Lp)' and write ||/|| = c.
Define a set function ф on the measurable sets A с E by
Ф(А) = l(xA).
Note that ф is finite; in fact, \ф(А)\ < с||хл||р = cp(A)i/p. Clearly, ф is finitely
additive. To show that it is countably additive, suppose that A = Ak,
Ak disjoint. Write A = ((j£=i Лк) (U?=™+i Ак) = A' и A". Then
ф(А) = ф(А’) + ф(А") = f ф(Ак) + ф(А").
к=1
Since \ф(А")\ < ср(А")1/р, ф(А") tends to zero as m -> oo. Hence, ф(А) =
1 ф(Ак), which shows that ф is countably additive. The fact that |0(Я)| <
cp(A)l/p implies that ф is absolutely continuous with respect to p.
By the Radon-Nikodym theorem, there is a g e Ll(E\dp) such that ф(А) =
J/t g dp for measurable A с E. This means that /(/л) = %Ag dp, so that
1(f) = j£ fg dp for any simple measurable f. To show the same formula holds
for any^/eLp, we first claim that g e Lp' and \\д\\р> < c. If p > 1, choose
HUbitciGi Ниедrauori
simple functions hk with 0 < hk / |g|₽'. Let {gj be the simple functions
defined by
Qk = hl!p sign g.
Then Ш|р = ||M}/₽, and
f gkg dy = /(#*) < с|Ы|р = c\\hk\\{lp.
JE
Since gkg = h£/p]g] > h£/p+1/p' = hk, we obtain ||йк||, < с||й*||}/р. We may
assume that \\hk\\ t 0 for large k. (Otherwise, g would be zero a.e., and there
would be nothing to prove.) Hence, dividing both sides of the last inequality
by |IMi/p, we have ||/zj|i/p' < c, so that \\д\\р> < c by the monotone con-
vergence theorem. This proves the claim when p > 1. The case p = 1 is left
as an exercise.
To show that /(/) = fE fg dp for any f e Lp, choose simple fk converging
to f in Lp norm (see Exercise 8). Then /(/fc) -» /(/), and fE fkg dp -> J£ fg dp
by Holder’s inequality:
fk9 dp -
IE
fg dp
JE
< f \fk-f\\9\dy <
JE
The fact that the formula holds for /k- thus implies that it holds for f by
passing to the limit.
To complete the proof for the case p(E) < 4- oo, it remains to show that
\\g||p, = c and that the correspondence between I and g is unique. However,
we already know that ||g||p, < c, and the opposite inequality follows from
(10.43). For the uniqueness of the correspondence, see Exercise 6.
If p(E) = +oo, then since p is а-finite, there exist Ej / E with p(Ej) <
4- oo. Let I e [Lp(E)]f. Since the restriction of / to Lp(Ef) is a bounded linear
functional, there is a unique gjG Lp\Ej), И^-Цр^ < ||/||, such that
1(f) = [ fdj dy
for every f in Lp which vanishes outside Ej. For such /, the fact that Ej c Ej+l
also gives
1(f) = fgj+i dy = fgJ+l dy.
JEJ+i JEj
Therefore, gj+l = gj a.e. in Ej. We may assume that gj+i = gj everywhere
in Ej. Define g(x) by g(x) = gfx) if x e Ej. Then g is measurable and it
follows that \\g\\p, < ||/||. If/e Lp(Ef then
l(fxEj) = [ fgj dy = f fg dy.
JEj Jej
Since fxEj converges in Lp to f and f£j fg dg -> fg dg (note that fg e L1
by Holder’s inequality), we obtain 1(f) = fg dg in the limit. Therefore, by
(10.43), ||/|| < ||#||P', so that ||/|| = ||^||p', and the proof is complete.
We remark that not every bounded linear functional on Z00 can be
represented 1(f) = J fg dg for some g e L1; an example is indicated in
Exercise 18.
5. Relative Differentiation of Measures
Lebesgue’s differentiation theorem (7.11) states that if f is locally integrable
in R", then
lim 1п 7м| I ЯУ) = Лх) a-e-
Й-Ю ICJxvOI jQx(h)
where Qx(h) is the cube with center x and edge length h. We will now study
an analogue of this result for other measures on R". Specifically, if g and v
are two а-finite measures on the Borel subsets of R", we will study the
existence of
lim
h-*O
v(gx(A))
and its relation to the Lebesgue decomposition of v with respect to g.
We will follow the method used to prove Lebesgue’s differentiation theo-
rem. To do this, we must find a replacement for Vitali’s lemma: the simple
form of Vitali’s lemma (see (7.4)) relies heavily on the fact that expanding a
cube concentrically by a factor (say 5) only enlarges its Lebesgue measure
proportionally, whereas no such relation may hold for general measures.
In order to bypass this difficulty, we shall present a covering lemma which is
purely geometric in nature, that is, which makes no mention of measure.
We consider only cubes whose edges are parallel to the coordinate axes,
and write Q = Qx for those with center x. We say that a family К of cubes
has bounded overlaps if there is a constant c such that every x e R" belongs to
at most c cubes from K. Thus, К has bounded overlaps if and only if
E Xq(x) c (x 6 R").
QeK
(10.45) Theorem (Besicovitch Covering Lemma) Let E be a bounded subset
‘ ofRn, and let К be a family of cubes covering E which contains a cube
Qx with center x for each хе E. Then there exist points {xfc} in E such
that
(D e c u eXfc.
(ii) {Qxk} has bounded overlaps.
nuiliavi iiiicyjaiiuit
t u.Hru ;
Moreover, the constant c for which %Qxk < c can be chosen to depend
only on n.
In order to prove this, it will be convenient to first prove the following
lemma.
(10.46) Lemma Let i be a sequence of cubes with centers {xk} such that
if j < k, then xk ф Qj and |<2k| < 2|Qj\. Then {Qk} has bounded over-
laps, and the constant c for which %Qk — c can be chosen to depend
only on n.
Proof We will consider only n = 2; the case n > 2 is similar and left
as an exercise. Let Qkm, m = 1, 2,. . ., be those Qk which contain the origin
and whose centers are in the first quadrant, and let hm denote the edge length
of Qkm. Then Qki covers at least the region
A = {(AK) • 0 < x < 0 < у < |/q}.
Hence, no Qkm can have its center outside the set {(x,y) : 0 < x < hx,
0 < у < hr}; otherwise, we would have hm > 2hr for some m, so that |£)km| >
4|£>kl|, a contradiction. Therefore, the center of each Qkm, m > 2, must lie
in one of the regions А, В, C, or D indicated below:
The center cannot be in A since that would contradict the assumption that
the center of Qkm, m > 2, does not lie in Qkl. If it lies in B, then since Qkm
contains 0, it covers B, and so there is at most one Qkm with center in B.
Similar arguments hold for C and D. Applying the same reasoning to each
quadrant, we see that there are at most 16 cubes in {Qk} which contain the
origin. By translation, the same holds at any point of the plane.
Proof of Besicovitch's lemma. Let
<*i = sup{l6xl :xgE}.
If cq = -boo, there are arbitrarily large Q*, and since E is bounded, we
simply choose one which contains E. If < + oo, write Er = E and choose
Xj g Ег with |2X1I > ai/2. Let
E2 E1l — a2 SUP {ISJ * X e ^2}-
If a2 / 0, choose x2 e E2 with |gX2| > a2/2. Proceed in this way, obtaining
at the £th stage
Ek = Ek-r - QXk_t = E - (J Qx., ak = sup {|gx| : xeEk},
J= 1
Xk 6 Ek, |0J > at/2.
The process can be continued as long as ak > 0.
Since xkG£k, we have xk e for all j < k. Therefore, |£)Xfc| < ak <
Uj < 2|gx.| if j < k. It follows that {QXk} satisfies the hypothesis of (10.46),
and so has bounded overlaps. It remains only to show that E c |J QXk.
If some ako = 0, then Eko is empty, and E is contained in the union of the
0Xk, к < k0 — 1. If no ak = 0, there are infinitely many QXk. Since ak \
and ak/2 < |gxJ < ak, it follows that either |gxJ -> 0 or that there exists
<5 > 0 such that |(2XJ > <5 for all k. The second possibility cannot arise;
otherwise, E would not be bounded since xk g E but xk is not in any Q with
j < k. Hence, |6XJ -» 0, or equivalently, ak 0. If x g E - (J gXfc, then
xeEk for all k. Therefore, |gx| < ak for all k, which means that |gx| = 0.
This shows that E — |J QXk is actually empty, and completes the proof.
A Borel measure /z on R” (i.e., a measure on the Borel subsets of Rrt) is
called regular if
g(E) = inf (g(G) : G о E, G open}
for every Borel set E. From now on, we will consider two regular Borel
measures д and v on Rn which are finite on the bounded Borel sets. If Qx(h)
denotes the cube with center x and edge length /1, we will also assume that
(i) /z(6x(/?)) > 0 for x g R”, h > 0;
(ii) sets of the form
f v(2x(A)) If v v(Sx(/2)) )
<x : sup z^zy<< > <x : limsup z_ ZZ44 > a>, etc.,
I k>oP(2xW) J I HO J
are measurable.
Assumption (ii) is made for simplicity and is not necessary (see Exercise 17
of Chapter 11). Also, the assumption that /z and v are regular is redundant
(see (11.24) and the remarks following it).
We will use the fact that the class of continuous functions with compact
support is dense in L(dg); i.e., given fe L(dg), there exist continuous gk with
compact support such that J \f ~~ gk\ dg -> 0. (See Exercise 27.)
Note that when g is Lebesgue measure and v(E) = f£ |/| dx for a locally
integrable*/, then sup/l>0 v(Q^(h))/g(Qx(h)) is the Hardy-Littlewood maximal
188
Abstract Integration
(10.47)
function of f [see (7.5)]. In the lemma below, we estimate the size of this
expression for general /z and v. The results we will prove on differentiation are
corollaries of this estimate.
(10.47) Lemma Let p and v satisfy the stated conditions. Then there is a
constant c depending only on n such that.
(a) /z{x e R" : sup„>0 [г(<2х(/г))//г(2х(/г))] > a} < (c/a)v(R").
(b) ц{хеЕ : limsup^0 [v(Qx(h))/p(Qx(h))] > a} < (c/cc)v(E)
for any Borel set E c: R” and any a > 0.
Proof (a) Fix a > 0, and let
XCRB-5UPV(&(A))
If В is any bounded Borel set and x e 5 n B, there is a cube gx with center x
such that v(gx)//z(2x) > a. Using Besicovitch’s lemma, select {QXk} and c
such that v(2Xfc) > a^(2xj, S r> В c (J gX)t and £ < c. We then have
№ n B)< HU Qv) E KQ.k) < v(eXfc),
E v(2xfc) = E f Л < c [ dv = cv(U О
JuQxk JUQxfc
Therefore, p(S n B) < cv(|J so that p(S n B) < cv(R")/a. Letting
В/Rn, we obtain ^(5) < cv(R")/a, as desired.
(b) Fix a > 0, and let
x e E : limsup
й-»0
*(&(*))
p(2x(A))
a
If v(E) = Too, there is nothing to prove. Otherwise, choose an open set
G E with v(G) < v(E) + e, and let В be a bounded Borel set. If x e T n B,
there is a cube Qx such that Qx c G and v(2x)/X2x) > a- As above, there
exists {QXk}, QXk <= G, such that /z(T n B) < cv((J Therefore,
p(T n B) < cv(G)/a < c[v(F) + e]/a. The result now follows by first letting
e —> 0 and then letting В / R”.
The first result about differentiation of measures is the following.
(10.48) Theorem Let v and p satisfy the stated conditions. If v and p are
mutually singular, then
lim
h->0
v(6x(/Q)
л(2х(/0)
0 a.e. (д).
(10.50) Relative Differentiation of Measures 189
Proof. Since v and p are mutually singular, there is a set Z with
v(R" - Z) = p(Z) = 0. Let £ = R" - Z, and consider the sets
y(QxW) I
д(2х(А)) “j ’
v(6xW) J
Х2х(Л)) J •
Ta = < x e E : limsup
I h-*0
(a > 0),
T = < x e E : limsup
I h-*0
By (10.47)(b), we have д(Л) - cv(E)/a, = 0. Since T is the union of the Tak
for any sequence afc -► 0, it also has д-measure zero, and the result follows.
(10.49) Theorem Let p satisfy the stated conditions, and let f be a Borel
measurable function which is integrable (du) over every bounded Borel
set in R". Then
lim „(rilhXx I ae-
h-*0 Д(Сх(")) JQx(h)
Proof Assume first that f g L(Rn;dp). For any integrable g, we have
йо» ~/(x) я(Сх(А)) L,z -gi dg
+ хёЖ)к(*/ dfl ~ /(x)'
If g is also continuous, the last term on the right converges to |#(x) - /(x)|
as h -► 0. Hence, letting £(x) denote the limsup as h -► 0 of the term on the
left, we obtain
L(x) < sup I \f~ g\ d/г + |g(x) -/(x)|.
Л>0 MU/xVW jQx(h)
Therefore, the set St where L(x) > a, a > 0, is contained in the union of the
two sets where the corresponding terms on the right side of the last inequality
exceed je. From (10.47) and Tchebyshev’s inequality, we obtain
n(St) < cQe)”1 [ I/- g| du + (le)’1 f \f- ф.
Jr* Jr*
As noted earlier, g can be chosen such that fRn \f — g\ dp is arbitrarily small.
Hence, д(5е) = 0 for every a > 0, and the result follows.
The case when f ф L(Rn;dp) is left as an exercise [cf. theorem (7.11)].
Combining the last two theorems, we obtain the main result.
(10.50) Corollary Let v and p satisfy the stated conditions. If v(£) —
190
Abstract Integration
(10.50)
Je f dp 4- o(E) is the decomposition of v into parts which are absolutely
continuous and singular with respect to p, then
™KQx(hJ) /(x)a-e’Gi)-
Exercises
1. Prove theorem (10.13).
2. A measure space (<7\E,/z) is said to be complete if E contains all subsets of sets
with measure zero; i.e., (5/1,//) is complete if Ye E whenever Y c Z, Ze E,
and /z(Z) = 0. In this case, show that if /is measurable and д = f a.e., then д
is also measurable. Is this true if (5/E,/z) is not complete?
3. Prove Egorov’s theorem (10.14).
4. If (^,E,/z) is a measure space, and if / and {/} are measurable and finite a.e.
in a measurable set E, then {fk} is said to converge in p-measure on E to
limit f if
lim p{x e E : |/(x) — «/*(-*)! > £} = 0, e > 0.
Formulate and prove analogues of theorems (4.21)-(4.23).
5. Complete the proof of lemma (10.18).
6. (a) If /i,/2 G L(dp) and $Ej\ dp = fEf2 dp for all measurable E, show that
/i = fi a.e. (/z).
(b) Prove the uniqueness of / and a in (10.40).
(c) Let p be a-finite, and let /t,/2 eLp\dp\ 1/p + 1/p' = 1, 1 < p < oo. If
f f\9 dp = f f2g dp for all g e Lp(dp\ show that ft = f2 a.e.
7. Prove the integral convergence results in (10.27)—(10.31).
8. Show that for 1 < p < oo, the class of simple functions vanishing outside sets
of finite measure is dense in Lp(dp).
9. The symmetric difference of two sets Er and E2 is defined as
E1AE2-(E1 -E2)v(E2 -EJ.
Let (У7^,^) be a measure space, and identify Et and E2 if p(Er A E2) = 0.
Show that E is a metric space with distance d{Ey,E2) = p(Et AE2), and that
if p is finite Lp(ff/L,p) is separable if and only if E is, 1 < p < °°.
10. If ф is a set function whose Jordan decomposition is ф = V — JZ, define
( ЕЛф= f fdV- f fdVt
JE JE JE
provided not both integrals on the right are infinite with the same sign. If V
is the total variation of ф on E, and if |/| < M, prove that |fEfd(/)\ < MV.
11. Prove parts (ii)—(iv) of (10.33).
Exercises
191
12. Give an example of a pair of measures v and p such that v is absolutely con-
tinuous with respect to //, but given e > 0, there is no <5 > 0 such that г(Л) < a
for every A with p(A) < 3. [Thus, the analogue for measures of (10.34) may
fail.]
Prove the analogue of (10.35) for mutually singular measures v and p.
13. Show that the set P of the Hahn decomposition is unique up to null sets.
[By a null set for ф, we mean a set W such that ф(А) = 0 for every measurable
A c AC]
14. Complete the proof of (10.44) for p = 1.
15. (Converse of Holder’s inequality) If p is <7-finite and 1 < p < oo, show that
ll/lb = su₽ IJV0M
where the supremum is taken over all bounded g which vanish outside sets of
finite measure, and for which ||^||p. < 1 and J fg dp exists. [If 1 < p < oo and
||/||p <4-00, this can be deduced from (10.44).]
16. Consider a convolution operator Tf(x) = JR»/(y)X(x — y) dy with К > 0.
If \\Tf\\p < MI\f\|p, 1 <p < oo, for fe Lp(W,dx), show that\\Tf\\p, < M11f ||p,,
1/P + Up' = 1- [Use Exercise 15 to write ||Г/||Р, = supllff|IpSi |]> (Tf)g tZx|,
and note that
|к„(Г/)(х)0(х) dx = jR„(T^)(-y)/(y) dy,
where ^(x) = g(—x).]
17. Let p be ст-finite and define <Lp(dp) to be the class of camp/ex-valued f with
J \f\p dp < 4-oo. Let / be a cwnp/ex-valued bounded linear functional on
£Lp(dp). If 1 < p < oo, show that there is a function g g ^P'(dp) such that
> /(/) = J fg dp. (Here, as usual, we define J h dp — f hi dp + i J h2 dp if h =
hi + ih2 with hi and h2 real-valued.) (Hint: Reduce to the real case.)
18. Give an example to show that (L00)' # L1. (Consider £00 [—1,1] with Lebesgue
measure, and let # be the subspace of continuous functions on [—1,1] with
the sup norm. Define 1(f) = f(Q) for f e <%. Then / is a bounded linear functional
on so by the Hahn-Banach theorem (see, e.g., p. 62 of N. Dunford and J. T.
Schwartz, Linear Operators, Part 1,4th printing, Interscience, New York, 1967.),
I has an extension 7g(£°°)'. If there were a function geL{[- 1,1] such that
/(/) = ^_ifg dx for /gjL°°[—1,1], then we would have /(0) = \Lifgdx for
f e 4L Show that this implies that g = 0 a.e., so that I = 0. The functional I
is called the “^-function.”)
19. Complete the proof of (10.49).
20. Under the hypothesis of (10.49), prove that
L(/ol/(y) -/(x)1 <My) = ° a'e-U
21. Derive an analogue of the Bcsicovitch covering lemma (10.45) for the case of
two dimensions (x,y) when the squares С(Х)У) are replaced by rectangles R(x>yfh)
102
Abstract integration
whose x and у dimensions are h and A2, respectively. Use this result to prove
that under the hypothesis of (10.49),
„(Р 1 (hx\ L ,kJdfi ->f(x,y) a.e. (/z) as h — 0.
Д\-*Чх,у)оШ ^(x,y)(^)
22. Let p be a finite measure on A, let /be measurable and bounded on A, and
let ф be convex in an interval containing the range of /. Prove that
JhfdpX <SA ф(Щр
Ф\\ас1ц)~ fAdp ‘
[This is Jensen’s inequality for measures. See (7.44).]
23. A sequence {фк} of set functions is said to be uniformly absolutely continuous
with respect to a measure p if given e > 0, there exists 6 > 0 such that if E
satisfies p{E) < b, then |^fc(£)| < s for all k. If {fk} is a sequence of integrable
functions on a finite measure space (y\S,/z) which converges pointwise a.e. to
an integrable /, show that fk f in L(dp) norm if and only if the indefinite
integrals of the fk are uniformly absolutely continuous with respect to /z.
24. Let (У7,^,^) be a сг-finite measure space, and let /be S-measurable and integrable
over У7. Let So be a сг-algebra satisfying So с S. Of course, / may not be
S0-measurable. Show that there is a unique function /0 which is S0-measurable
such that j* fg dp = J fQg dp for every S0-measurable g for which the integrals
are finite. The function /0 is called the conditional expectation of / with respect
to So, denoted /0 = Д/^о). [Apply the Radon-Nikodym theorem to the set
function ф{Е) — jEf dp, Ее So.]
25. Using the notation of the preceding exercise, prove the following:
(a) E(af 4- M^o) = fl£(/|So) + ^(^|^o), a,b constants.
(b) E(/|S0) > 0 if/> 0.
(с) £(/^|S0) = ^E(/|S0) if g is S0-measurable.
(d) If Si c So c S, then £X/|Si) = E(E{f |S0)|Si).
26. {Hardy's inequality) Let /> 0 on (0,oo), 1 < p < oo, dp(x) = x* dx and
dv(x) = x“+p dx on (0,oo). Prove there exists a constant c independent of /
such that
(i) Jo (£/(0 df dn(x) < cjo fPMdv(.x) (a < —1)
(ii) (£ f{t)df d/j(x) < c £ fp(x) dv(x) (a > —1).
[For (i), {\of{t)dt)p < cxp~n~1 \of{t)tt’dt by Holder’s inequality, provided
p — tj — 1 > 0. Multiply both sides by xa, integrate over (0,oo), change the
order of integration, and observe that an appropriate rj exists since a < — 1.]
27. If p is a regular Borel measure on R”, show that the class of continuous func-
tions with compact support is dense in Lp(dp), 1 < p < oo. [By Exercise 8,
it is enough to approximate where E is a Borel set with finite measure.
Given c > 0, there exist open G and closed F'with F с E c G and p(G — F) <
c. Now use Urysohn’s lemma: if Fr and F2 are disjoint closed sets in R", there
is a continuous / on R" with 0 < / < 1, / = 1 on Ft, f = 0 on F2.}
Chapter 11
Outer Measure; Measure
1. Constructing Measures from Outer Measures
A function Г = Г(Л) which is defined for every subset A of a space is
called an outer measure if it satisfies the following:
(i) Г(Л) > О, Г(0) = 0.
(ii) Г(Л) < Г(Л2) if A cz a2.
(iii) F((J Ak) < HA) f°r апУ countable collection of sets {Лл}.
For example, ordinary Lebesgue outer measure is an outer measure on the
subsets of R”. Some other concrete examples will be constructed later in the
chapter.
As with Lebesgue outer measure, it is possible to use any outer measure to
introduce a class of measurable sets and a corresponding measure. In doing
so, we base the definition of measurability on Caratheodory’s theorem
(3.30). Thus, given an outer measure Г, we say that a subset E of is Г-
measurable, or simply measurable, if
(ИЛ) Г(Л) = Г(Я n E) + Г(Л - E)
for every A cz . Equivalently, E is measurable if and only if
Г(ЛГ u A2) = Ц^) + Г(Л2) whenever At cz E, A2 cz — E.
As a simple example, we will show that any set Z with T(Z) — 0 is
measurable. In fact, for such Z and any Л с У, property (ii) gives
Г(Л n Z) + T(T - Z) < T(Z) + Г(Л) = Г(Л).
But by (iii), the opposite inequality Г(Т) < Г(Х n Z) 4- Г(Л - Z) is always
true, and the measurability of Z follows.
. If E is a measurable set, then Г(£) is called its Г-measure, or simply its
measure. The terminology is justified by the following theorem.
193
194
Outer Measure; Measure
(11.2)
(11.2) Theorem Let Г be an outer measure on the subsets of SP.
(i) The family of Г-measurable subsets of У7 forms a o-algebra.
(ii) If {Ek} is a countable collection of disjoint measurable sets, then
T((J Ek) = £ Г(Ек). More generally, for any A, measurable or not,
Г(А r.[jEk) = l Г(А n £*) and Г(Л) = £ Г(Л n EJ + Г(А - J EJ.
Proof Let {Ek} be a collection of disjoint measurable sets, let H =
Ek and If = = 1,2,.... We first claim that
Г(Л) = f Г(Л n EJ + Г(А - HJ.
k=l
The proof will be by induction on j. If/ = 1, the formula follows from the
measurability of Er. Assuming that the formula holds for j — 1, we have
Г(Л) - Г(Л n Ej) + Г(Л - Ej)
= Г(Я n Ej) + f Г((Л - EJ n Ек) + Г((Л - EJ - Hj_ J.
k=l
Since the Ek are disjoint, (A — Ej) cs Ek ~ A cs Ek for к < j — 1. Hence,
since (A — Ej) — Hj_r = A — Hj, we obtain Г(Л) = П/l n Ek) +
Г(Л - Hj), as required. This proves the claim.
Since Hj cz H, we have Г(Л — Hj) > Г(Л — H). Using this fact in the
formula above and letting j -> oo, it follows that
Г(Л) > E ГЦ n Ej) + Г(Л - H) > Г(Л n H) + Г(Л - Я).
к=1
However, we also have Г(Л) < Г(А n H) + Г(Л - H). Therefore, H is
measurable and Г(у4) = ^Г-i Г(Л n Ek) 4- Г(Л — H). Replacing A by
A n H in this equation, we obtain Г(Л n H) = £*°=1 ГЦ n Ek), and the
proof of (ii) is complete.
Note we have also shown that a countable union of disjoint measurable
sets is measurable. To prove (i), it remains to show that a countable union of
arbitrary measurable sets is measurable, and that the complement of a
measurable set is measurable. To prove these facts, we shall use the following
lemma.
(11.3) Lemma If Er and E2 are measurable, then so is Er — E2.
Proof We will show that Г(Л u В) = Г(А) + T(B) whenever A cz
Ex — E2 and В cz C(El — E2). Since В = (В n E2) о (В — E2), we have
A u В = [A u (B — E2)J u [Z? n E2}. Hence, since A u (B — E2) cz CE2
and В n E2 cz E2, it follows from the measurability of E2 that Г(Л u B) =
Г(Л о {В — Ejj) + Г(В n E2). However, A cz Er and В — E2 cz
(11.4) Co. .ructing Measures from Outer Measures 195
C(Er — E2) — E2 CEr. Therefore, since Er is measurable, Г(Л u
(В — E2)) — Г(Л) + Г(В — E2). Combining equalities and using the mea-
surability of E2, we obtain
Г(Л и В) = Г(Л) + Г(В - Е2) + Г(В п Е2) = Г(Л) + Г(В),
which proves the lemma.
Returning now to the proof of (11.2), we see from the lemma and the fact
that the whole space У’ is measurable that the complement of a measurable
set is measurable. Moreover, since Er и E2 = — Ef), it follows that
Er u E2 is measurable if Er and E2 are. Therefore, any finite union of
measurable sets is measurable. Now, let {Ek} be a countable collection of
measurable sets. If Hj = Ek, then
0 Ek = H, и [0 (HJ+l - hE
k=l Lj=l
and since the Hj are measurable and increasing, the terms on the right are
measurable and disjoint. Thus, by the case already proved, it follows that
! Ek is measurable. This completes the proof of the theorem.
According to (11.2), an outer measure Г is a measure on the сг-algebra of
Г-measurable sets, and so enjoys the usual properties of measures. We also
have the following result.
(11.4) Corollary Let Г be an outer measure on If, let {Ek} be a collection of
measurable sets, and let A be any set.
(i) If Ek then Г(А n lim Ek) ~ lim^^ Г (A n Ek); if Ek \ and if
Г(Л n Ekfi is finite for some k0, then Г(А n lim Ek) = lim^^
Г(Л n Ekf
(ii) Г(Л n liminf^) < liminf*^ Г(Л n Ek); if Г(Л n Ek) is
finite for some k0, then Г(Л n limsup Ek) > limsup^^ Г(А n Ek).
Proof We will prove the first statements in (i) and (ii); the proofs of the
second statements are left as exercises. Let Ek / and be measurable. To
prove the first part of (i), we may assume that Г(Л n Ef) is finite for each k\
otherwise, the result is clear. The sets Ex, E2 — E19 . . . , Ek+l — Ek, ... , are
disjoint and measurable. Since
lim Ek = (J Ek = Ex и (j {Ek+i - Ek) ,
к = 1 U = 1
it follows from (11.2) that
Г(А n lim Ek) = Г(А n EJ + £ T(A n (Ek+l - EJ).
k=l
Moreover, since Ek and Ek+1 — Ek are disjoint and measurable with finite
measures, we have Г(Л n (Efc+1 — Efc)) = Г(Л n Ek+1) — Г(Л n Ek). There-
fore,
00
Г(Л n lim Ek) = Г(Л n + X [Г(Л n Ek+l) - Г(Л n Ek)]
k=l
= lim Г(Л n Ek+1),
k-+ 00
which proves the first part of (i).
For the first part of (ii), let {Ek} be measurable and define Xj = ПГ=;
j = 1,2,.... Then Xj / liminf Ek, so that by (i), Г(А n liminf Ek) —
lim,^ Г(А n Xj). But since A n Xj c A n Ey, we have lim,-^ Г(Л n Xj) <
liminf^ F(d n ЕД and the result follows.
2. Metric Outer Measures
Now let us introduce a new assumption concerning the underlying space У:
namely, that it is a metric space with metric d. The distance between two sets
A± and A 2 is then defined by
d(Al,A2) = inf {d(x,y) : x e Л15 у e A2}>
as in Euclidean space (see p. 5). An outer measure Г on is called a
metric outer measure, or an outer measure in the sense of Caratheodory, if
T^i u A2) = T(Tj) + T(T2) whenever d(A19A2) > 0.
For example, by (3.16) of Chapter 3, Lebesgue outer measure satisfies this
condition.
Since У' is a metric space, it has the topology induced by its metric. Thus,
a set G in У' is said to be open if for every x e G, there is a <5 > 0 such that
the “ball” {y : d(x,y) < 5} lies in G. A closed set is by definition the com-
plement of an open set, and & denotes the сг-algebra of Borel subsets of У7;
i.e., £3 is the smallest сг-algebra containing all the open (closed) subsets of ST.
(11.5) Theorem Let Г be a metric outer measure on a metric space . Then
every Borel subset of is Г-measurable.
Proof Since the collection of Г-measurable sets is a сг-algebra, it is enough
to prove that every closed set is Г-measurable. To prove this, we will use the
following fact.
(11.6) Lemma Let Г be a metric outer measure on a space with metric d.
Let A be any set contained in an open set G, and let Ak = {xeA :
d(x,CG) > (!//<)}, /<=1,2,.... Then lim^ Г(А) = Г(Л).
Proof. Since G is open, we have Ak / A. Clearly, Г(ЛЛ) < Г(Л).
I о /
To prove the opposite inequality, let Dk = Ak+1 — Ak, к = 1,2,.... Then
d(Dk+i,Ak) > [(1/fc) - (\/(k + 1))] > 0 and
A = AkvDkuDk+1 о •••, Г(Л) < Г(Л*) + T(Dt) + T(Z>t+1) + ••••
If £ Г(Т>7) < +co, then Г(£>7) tends to zero as к -> co, and it follows
that Г(Л) < Г(Лк), as desired.
If Z r(Dj) — 4-go, then at least one of £ r(D2j) and is
infinite. We can therefore choose N so that T(DN) 4- T(Z)N_2) 4- T(Z)N_4) 4-
• • • is arbitrarily large. However, the fact that Dk_1 cz Ak implies that the
distance between Dk+i and Dk} is positive. Therefore,
T(Z>N u DN^.2 Av-4 u • • •) ” F(Av) + Г(Р]у_2) 4- r(Z)Y_4) + • • •.
Since An+i contains DN и DN 2 и £>N_4 u • • •, it follows that lim Г(Лк) =
4- co, and the lemma is proved.
Proof of (1/.5). Let F be any closed set. It is enough to show that
Г(Л о В) '= Г(А) 4- Г(В) for A cz CF, В cz F. If Ak = {x e A : d(x,F) >
(l/Л)}, then d(Ak,B) > (1/fc), so that Г(Л и В) = Г(А) 4- Г(Б). Therefore,
Г(Л и В) > Г(Лк) 4- Г(В). Letting к -> oo, it follows from the lemma that
Г(Л и В) > Г(Л) 4- T(F). Since the opposite inequality is also true, the
theorem is proved.
If У is a metric space, the notions of upper and lower semicontinuity of
functions can be defined just as in R". For example, a function f defined near
a point x0 is said to be upper semicontinuous at xQ if
limsup/(x) < /(x0).
x-*x0
Here, of course, the notation x -> x0 means that d(x,x0) 0. The results of
Theorem (4.14) are valid for metric spaces; for example f is use at every
point of У if and only if {f > a} is closed for every a. We thus obtain the
following fact.
(11.7) Corollary Let Г be a metric outer measure on У. Then every semi-
continuous function on is Г-measurable.
Proof. Suppose for example that f is upper semicontinuous on У. Then
{x : /(x) > a} is closed for every a, and so is Г-measurable by (11.5). Hence,
f is Г-measurable. If f is lower semicontinuous, then —/is upper semi-
continuous, and the corollary follows.
3. Lebesgue-Stieltjes Measure
In this section and the next, we will consider two specific examples of
Caratheodory outer measures. The first of these, known as Lebesgue-Stieltjes
outer measure, elucidates the connection between measures and monotone
functions. The situation is relatively simple for measures on R1 and monotone
functions of a single variable, and we shall restrict our attention to this case.
Extensions to higher dimensions are possible, but more complicated.
To construct Lebesgue-Stieltjes outer measure, consider any fixed function
f which is finite and monotone increasing on (— oo, + oo). For each half-open
finite interval of the form (a,b], let
X(a,b] = A/W]) =f(b) — f(a).
Note that 2 > 0 since f is increasing. If A is a nonempty subset of R1, let
А*(Л) = Л*(Л) = тГр(Ш
where the inf is taken over all countable collections {(ak,bk]} such that
A cz |J (ak,bk]. Further, define A*(0) = 0.
(11.8) Theorem A* is a Caratheodory outer measure on R1.
Proof. We have A* > 0 and A*(0) = 0. First, we will show that if
Ax cz A2, then A*^) < A*(T2). This is obvious if either Ax = 0 or
Л*(Л2) = +oo. In any other case, choose such that A2 cz
and £ < A*(T2) + e. Then Ar cz (J (ufc,&fc], so that A*(/4j) <
£ 2(алД]. Therefore, A*(A J < A*(^2) + and the result follows by letting
£ —> 0.
To show that A* is subadditive, let {Я7}}°=1 be a collection of nonempty
subsets of R1 and let A = |J Aj. We may assume that A*(Aj) < Too for
each J. Choose {(aJkibJk]} such that
Aj c U ШИ, £ Kai,bft < A*(/Q + 82"<
к к
Since A c (Jm (akM]> we have
л*(л) <ЕЖ^]<Елчл-) +
j,k j
It follows that Л*(Л) < £ Л*(Л7), and therefore that A* is an outer measure.
To show that A* is a Caratheodory outer measure, observe that if a —
a0 < ar < *' • < aN ~ Z>, then
Ца,Ь\ = f(b) — f(a) = f [/(at) -Ж-i)] = £ X{ak^,ak].
k=l k=l
It follows that in defining A*, we can always work with arbitrarily short
intervals (ak,bk]. Hence, if A{ and A2 satisfy d(At,A2) > 0, then given £ > 0,
we can choose {[ak,bk]} such that each (ak,bk] has length less than dfA^Af)
and
Ai v; A2 c (J (ak,bk], £ Mak,bk] < \*(Ai и Л2) + e.
(11.10)
Lebesgue-Stieltjes Measure
199
Thus, {(ak,bk]} splits into two coverings, one of and the other of A2.
Therefore, A*(^i) + А*(Л2) < £ so that since e is arbitrary,
Л*^) + Л*(^2) ~ E^'(A1 u A2). But the opposite inequality is always
true, which completes the proof.
A* is called the Lebesgue-Stieltjes outer measure corresponding to /, and
its restriction to those sets which are Aj-measurable is cafied the Lebesgue-
Stieltjes measure corresponding to f, and denoted or simply A.
We leave it as an exercise to show that the Lebesgue-Stieltjes outer mea-
sure A* corresponding to /(x) = x coincides with ordinary Lebesgue outer
measure. Hence, by Caratheodory’s theorem (3.30), a set if A*-measurable
if and only if it is Lebesgue measurable.
An outer measure Г defined on the subsets of a set У7 is said to be regular
if for every A cz there is a Г-measurable set E such that A cz E and
Г(^4) = Г(Г). The next theorem shows that any Lebesgue-Stieltjes outer
measure is regular; in fact, it shows that any set in R1 can be included in a
Borel set with the same Lebesgue-Stieltjes outer measure. Of course, Borel
sets are A*-measurabIe by (11.5).
(11.9) Theorem Let A* be a Lebesgue-Stieltjes outer measure. If A is a sub-
set u/R1, there is a Borel set В containing A such that А*(Л) = A(/?).
Proof Given j ~ 1,2,..., choose {(ak,bk]} such that
A c (J (4,41 E 44,4] < Л*(Л) + i
к к J
Let Bj = (aJk,bJk] and В = Q Bj. Then A cz В and В is a Borel set.
Moreover,
Л(Б7.) < E 44,bi] < Л*(Л) + j.
к J
Since В cz Bj. it follows that A(B) < A*(^4) + (1/j), so that A(B) < Л*(Л).
But the opposite inequality is also true since A cz B, and the theorem follows.
If p is a finite Borel measure on R1, define
/д(х) = /х((— °o,x]), — °o < x < +oo.
Note that Д is monotone increasing and that p((a,b]) = ffjb) — ffa). It is
natural to ask if the Lebesgue-Stieltjes measure induced by Д agrees with p
as a Borel measure. An affirmative answer would mean that every finite Borel
measure is a Lebesgue-Stieltjes measure. We shall see later [(11.22)] that this
is actually the case, and that the continuity from the right of Д (see Exercise 2)
plays a role.
(11.10) Theorem If f is an increasing function which is continuous from the
riglp. then its Lebesgue-Stieltjes measure A satisfies
i i.iuj
* Nb» f fi - - %*- %p4 w* U« ft W
Л((аД) = /(&) - /(a).
In particular, A({a}) = /(a) — /(a — ).
Proof. Since (а,Ъ\ covers itself, we always have A((#,Z>]) < /(Z>) — f(a).
To show the opposite inequality, suppose that (a,b\ c (J (ak,b^. Given
g > 0, use the right continuity of f to choose {bk} with
bk<K, Ж)>Ж)~г2“А
If a' satisfies a < a' < b, then [a',b] is covered by the (ак,Ь$, and therefore,
there is a finite N such that [tz',5] c (J£= i (aM)- By discarding any un-
necessary (ak,bk) and reindexing the rest, we may assume that ak+i < bk for
к = 1, . . . , N — 1. Also, ai < a' and b < bfN, so that f(at) < f(a') and i
f(b) < f(b'N). We have |
E > X A(ak,bt] = £ №) -/(«,)] j
к k=l k=l .
= ЛЫ - f(aj + "f (Ж) - Ж+ >)]. S
k= 1 .
Now,
f(bN) - Ж) = [f(bN) - f(b'N)] + [Ж) - /(a,)]
> _е + [Л6)_Ж)].
Also, since f(bk) — f(ak+l) > 0 for к = 1,.. ., N — 1,
N-l N-l N-l
E №) -Ж+1)] = E [Ш -f(b'k)] + E (Ж) -Ж+1)]
k=l k=l k=l
00
> E (-e2'/[) -£-
k=l
Combining estimates, we obtain
WJ > -28 + [f(b) -/(«')].
к
Letting e -> 0 and a' -> a, we have X(ak,bk] > f(b) — f(a). Hence,
Л((я,£]) > f(b) — f(a), and the first statement of the theorem follows. The
second statement is proved by applying the first to the intervals (a — (1/k), a],
к = 1,2,..., which decrease to {a}.
Let g be a Borel measurable function defined on R1, and let Az be a
Lebesgue-Stieltjes measure. Then the integral f g d\f is called the Lebesgue-
Stieltjes integral of д with respect to Ay.f The next theorem gives a relation
between f д d\f and the usual Riemann-Stieltjes integral J д df.
fin some other texts, any integral $ f dp. of the kind considered in Chapter 10 is called a
Lebesgue-Stieltjes integral. We shall use the terminology only when p is a Lebesgue-
Stieltjes measure.
(11.11) Theorem Let f be an increasing function which is right continuous on
[a,b], and let д be a bounded Borel measurable function on [a,b]. If the
Riemann-Stieltjes integral \bag df exists, then
Г rb
g dAf = g df
J(a,b] J a
Proof Let Г = {Xj} be a partition of \a,b\, and let and Mj be the inf
and sup of g in respectively. Let
LT = Z mjlfixj) - f(Xj_,)]. Ur = £ Mj[f(xj) - f(Xj_,)]
denote the corresponding lower and upper Riemann-Stieltjes sums. Define
functions gr and g2 by setting gr = m} in (x;_ 15x7] and g2 = Mj in (xy_
Since f is right continuous, it follows from (11.10) that
gi dK = Lr, g2dA = Ur.
J(a,bl J(a,bl
Therefore, since gt < g < g2, we obtain Lr < J(a b] g dK < Ur. However,
as [Г| 0, both Lr and Ur converge to fb g df by (2.29). This completes the
proof.
We remark in passing that a right continuous function f of bounded
variation can be written f = fx — f2, where fr and f2 are right continuous,
bounded, and increasing. If Ai and Л2 are the Lebesgue-Stieltjes measures
corresponding to ft and f2, consider the Borel set function Ф = At — Л2,
and define
pj dФ = dAt — <7A2.
If Jb g df and jb g df2 exist and are finite, it then follows from (11.11) and
(2.16) that
Г Cb
g dФ = g df
J(a,b] J a
4. Hausdorff Measure
Our second example of a Caratheodory outer measure is Hausdorff outer
measure. To define it, fix a > 0, and let A be any subset of R". Given a > 0,
let
= infX^r,
к
where <5(Лк) denotes the diameter of Ak, and the inf is taken over all countable
collections {Ak} such that A cz (J Ak and b(Ak) < £ for all k. We may always
assume that the Ak in a given covering are disjoint and that A — |J Ak.
If s' < e, each covering of A by sets with diameters less than s' is also such
a cover for 8. Hence, as 8 decreases, the collection of coverings decreases, and
consequently H^(A) increases. Define
Ha(A) = lim 77<‘>(Л).
£->0
(11.12) Theorem For a > 0, Ha is a Caratheodory outer measure on Rn.
Proof. Clearly, Ha > 0 and Яа(0) = 0. If Ar cz A2, then any covering
of A2 is also one of A 1? so that HfXAi) < Я<е)(Л2). Letting 8 -> 0, we obtain
HfAx) < Ha(A2). To show that Яа is subadditive, let A = (J Ak9 and choose
a cover of Ak for each k. The union of these is a cover of A, and it is easy to
show that H^\A) < £ H<f(Ak) < £ HfAf). Letting e -> 0, we get НЛ(А) <
HfA^. The details of this argument and the proof that Яа is a Caratheodory
outer measure are left as an exercise.
Яа is called Hausdorff outer measure of dimension a on R", and the corre-
sponding measure is called Hausdorff measure of dimension a, and also
denoted Яа. It has the following basic property.
(11.13) Theorem
(i) If HfA) <+oo, then H^A) = 0 for /? > a.
(ii) If Hf A) > 0, then HfA) = 4- oo for /3 < a.
Proof Statements (i) and (ii) are equivalent. To prove (i), let A ~ |J Ak9
d(Ak) < 8. If fi > a, then
h^(a) <£<W < ^-аЕ^(Л)а-
Therefore, H^(A) < 8p~aH^f(A)- Letting 8 -» 0, we obtain Hfi(A) = 0 if
HfA) < 4-аз.
The next theorem shows that Hausdorff outer measure is regular, by
showing that any set in R" can be included in a Borel set with the same
Hausdorff outer measure.
(11.14) Theorem Given A cz R” and a > 0, there is a set В of type Gd con-
taining A such that HfA) = Ha(B).
Proof Given 8 > 0, choose {Лк} such that A = |J Ak9 d(Ak) < 8/2 and
£ 8(Ak)a < H^XA) 4- a < Яа(Л) 4- £.
Enclose Ak in an open set Gk with ^(Gf) <(14- е)<5(Лк); this can be done by
letting Gk = {x : d(x,Ak) < 8d(Af)/2}. Let G = Q Gk. Then G is open and
A cz G. Since <5(Ck) <(14- 8)8/2 < 8 for 0 <: s < 1, we have
G^(G) < £ d(Gky < (1 + £y £ (5(Aky
< (1 + + 8].
Now let e -> 0 through a sequence {ej, and let Gj be the corresponding open
sets. If В = Q Gj, then В is of type Gd and A с B. Also, since В c Gj for
each j, we have
H^\B) < (1 + е)а[Яа(Л) + 8]
for e = &j. Letting j -> oo, we obtain HfB) < Ha(A). Since the opposite
inequality is clearly true, the result follows.
If A is a subset of R1 and A = (J Ak with d(Ak) < e, then <5(Лк) = |7fe|,
where Ik is the smallest interval containing Ak. Hence, in the one-dimensional
case,
H^(A) = inf £ 141“ (n=l),
where the Iks are intervals with lengths less than £ such that A c |J Ik.
If а = 1, it follows that Я/Л) is the usual Lebesgue outer measure of A.
In R”, n > 1, Hn is not the same as Lebesgue outer measure (see Exercise 10).
Nevertheless, there is a simple relation between the two, which is a corollary
of the next lemma.
Let
я;<‘>(Л) = inf £<$(&)“,
where {Qk} is any collection of cubes with edges parallel to the axes such that
A cz |J Qk and <5((2fc) < £. Also, let
Я'(Л) = lim
£->0
Thus, Щ is defined in the same way that Ha is, except that cubes are used
instead of arbitrary sets.
(11.15) Lemma There is a constant c depending only on n and a such that
HfA) < H&A) < cHfA), A c R".
Proof. Since every covering of A by cubes is a covering of A, we obtain
HfA) < H'fA). Any set with diameter 5, say, is contained in a cube with
edge length 25, and so with diameter 2^nd. Now let A = (J Ak, b(Ak) < e.
Select cubes Qk Ak with — 2j«<5(y4fc). Then
Zd(Akf = (2^ny^5(Qkf > {2^Н^^\АУ
Therefore, Я<£)(Л) > (2j/i)“^(2>£)(^). Letting 8 -> 0, we obtain HfA) >
(2yJn)~aHa(A), which completes the proof.
(11.16) Theorem
(i) There are positive constants cr and c2 depending only on the dimen-
; sion n such that ctHn(A) < |A\e < c2Hn(A) for A cz R”.
(ii) If a > и, then Ha(A) — 0 for every A cz R”.
Proof For a cube Q, d(ff)n is proportional to the volume of Q \ <5(2)” =
nn/2\Q\. Hence, due to the additivity of volume,
Я'(Л) = inf X <5ШП, A c R",
where {Qk} is any collection of cubes covering M, without restriction on the
size of the diameters. Therefore, H'n(A) = nn/2\A\e, A c R”. Part (i) now
follows from the fact that H'n and Hn are comparable [(11.15)].
For part (ii), if Hn(A) is finite, then HfA) = 0 for a > n by (11.13). If
Я„(Л) = + oo, write A = (J (A n Qj), where the Qj are disjoint (partly open)
cubes. Since \A n Q.L is finite, so is Hn(A n Q}). Hence, H'(A nO,-) = 0
for a > n. Therefore, HfA) < £ HfA n Qj) = 0.
It is natural to ask if HfA) is comparable to the expression
inf£<5(A)‘>
where the inf is taken over all coverings {Лк} of A, without any requirement
on the size of the diameters. It is not difficult to see that the answer in general
is no. In fact, this expression is finite for any bounded A, as is easily seen
from covering A by itself. On the other hand, it is clear from (11.13) and
(11.16) that if \A\e > 0, then HfA) = +oo for a < n.
However, in case a = n, HfA) is comparable to the expression above.
To see this, it is enough by (11.15) to prove that H'n(A) is comparable to it.
But since H'fA) = inf £ <5(gfc)", where {Qk} is any collection of cubes cover-
ing A, this follows from the fact that inf £ d(Qk)n and inf £ d(Ak)n are com-
parable [cf. (11.15)].
We remark in passing that Hausdorff measure is particularly useful in
measuring sets with Lebesgue measure zero since these may have positive
Hausdorff measure for some a < n. For example, it can be shown that the
Cantor set in [0,1] has Hausdorff measure of dimension log 2/log 3 equal to 1.
5. The Carath£odory-Hahn Extension Theorem
In this section, we will settle the question which arose in the discussion
preceding theorem (11.10). We recall the situation. Let ц be a finite Borel
measure on R1, and let A be the Lebesgue-Stieltjes measure induced by the
function f(x) = p((—oo,x]). Since f is continuous from the right, we have
by (11.10) that
A((^]) = X(^]) [ = Ж-Ж].
The point in question is whether this implies that д and A agree on every
Borel set in R1. More generally, we may ask if two Borel measures can be
finite and equal on every (a,b] without being identical. It is worthwhile to
consider this question in a still more general context, which we now present.
Let If be a fixed set. By an algebra sd of subsets of If9 we mean a non-
empty collection of subsets of Sf which is closed under the operations of
taking complements and finite unions; that is, j/ is an algebra if it satisfies
the following:
(i) If Л gj/, then СЛ(= If — A) e^.
(ii) If A19. . ., An gjI, then (J^=1 Ak erf.
What distinguishes an algebra from a <r-algebra is that an algebra is only
; closed under finite unions. It follows from the definition that an algebra is
also closed under finite intersections and differences (relative complements),
and that both the empty set 0 and the whole space If belong to it.
The collection of finite intervals (a,b\ on the line is clearly not an algebra.
However, we can generate an algebra from it by adjoining 0, R1, and all
intervals of the form (—00,a] and (Z>, + oo), as well as all possible finite dis-
joint unions of these and the intervals (a,b]. This algebra will be called the
algebra generated by the intervals (a9b\.
By a measure A on an algebras^, we mean a function A which is defined
on the elements of j/ and which satisfies
; (i) A(A) > 0, A(0) - 0.
(ii) Adjfc°=i A) = Efc°=i^(A) whenever {Лк} is a countable collection of
disjoint sets in j/ whose union also belongs toj/.
; A measure A on j/ is called a-finite (with respect to j/) if У' can be written
$ У = U Sfc with SfcGj/ and 2(Sft) < +00. For example, any Lebesgue-
i Stieltjes measure A on the line is a u-finite measure on the algebra generated
I by the intervals (a9b\.
r Using the ideas behind the construction of Lebesgue-Stieltjes outer
measure, we can construct an outer measure Л* from A. Thus, let A be a
I measure on an algebra sd of subsets of У. For any subset A of define
’ (11.17) Л*(Л) = inf £ ад,
f where the infimum is taken over all countable collections {Лл} such that
f A cz (J Ak, Ak esd. It is always possible to find such a covering of A since У
itself belongs to j/. The fact that stf is an algebra allows us to assume without
loss of generality that the sets Ak covering A are disjoint.
(11.18) Theorem Let Abe a measure on an algebra stf, and let A* be defined by
Д (11.17). Then A* is an outer measure.
' • The proof is similar to the first part of the proof of (11.8), and is left as an
! exercise.
(11.19) Theorem Let A be a measure on an algebra^, and let A* be the cor-
responding outer measure. If A csf then A*(A) = A(A) and A is mea-
ly, surable with respect to Л*.
Proof. Let A ed. Clearly, A* (A) < A(A). On the other hand, given
disjoint Ak ed with A cz (J Ak, let Ak = Ak n A. Then Ак Esd and A is the
disjoint union of the Ak. Hence, A(A) — £ A(Ak). Since Ak cz Ak, it follows
that A(A) < £ А(Лк). Therefore, A(A) < A*G4), and the proof of the first
part of the theorem is complete.
For the second part, let dej/. To show that A is measurable, we must
show that
A*(£) = A*(E n A) + A*(E - A) for every E cz <f.
Since A* is subadditive, the right side majorizes the left. To show the opposite
inequality, we may assume that A*(£) is finite. Given e > 0, choose {Ek}
such that Ekesf E cz (J Ek and £ A(Ffc) < A*(£) + e. Since Ek and A are
in j/ and (Ek n A) u (Ek — A) = Ek, we have A(Ek n A) + A(Ek — A) —
A(Ek). Hence,
X Л(Ек n А) + £ Л(Ек - A) < Л*(£) + s.
Therefore, since E n A cz (J (Ek n A) and E — A cz (J (Ek — A), it follows
from the definition of A* that
A*(E n A) + A*(E — A) < A*{E) + e.
Letting £ —> 0, we obtain the desired inequality, which completes the proof.
Let A be a measure on an algebra j/, and let /z be a measure on a tr-algebra
E which contains л/. Then д is said to be an extension of A to IL if д(Л) = A(A)
for every A E^d. If A* is the outer measure generated by A and ifja/* denotes
the сг-algebra of A*-measurable sets, it follows from the last theorem that
A* is an extension of A to j/*. This proves the first part of the following
theorem, which is the main result of this section.
(11.20) Theorem {Caratheodory-Hahn Extension Theorem) Let A be a mea-
sure on an algebra sd, let A* be the corresponding outer measure, and let
sd* be the a-algebra of A*-measurable sets. Then
(i) the restriction of A* tosd*is an extension of A;
(ii) if A is a-finite with respect to sd, and if E is any o-algebra with
jZ cz E cz j/*, then A* is the only measure on E which is an exten-
sion of A.
Proof. As we have already observed, (i) follows from (11.19). To prove
(ii), which states the uniqueness of the extension, let p be any measure on E,
j/ cz E cz j/*, which agrees with A on jZ. Given a set E e E, consider any
countable collection {Л*} such that E cz (J Ak and each Ak Esd. Then
p(E) < /<(U Ak) < £ ц(Ак) = X
Therefore, by the definition of A*, we have p(E) < A*(E). To show that
equality holds, first suppose that there exists a set Л with E cz A and
Л(А) < +00. Applying what has just been proved to A — E (which belongs
to L), we obtain p(A — E) < A*(A — E). However,
p(E) + p(A - E) = M(J) = Л*(Л) = Л*(£) + А*(Л - £).
Since all these terms are finite (due to the fact that А(Л) is finite), it follows
that //(£) = A*(£) in this case.
In the general case, since A is rr-finite, there exist disjoint such
that = (J SA and A(Sfc) < + oo. We may apply the result above to each
£ n Sfc (which is a subset of SA), obtaining /i(£ n SA) = A*(£ n SA). By
adding over k, we obtain /a(^) = A*(£), which completes the proof.
As a corollary, we can answer the questions raised at the beginning of
this section.
(11.21) Corollary Let p and v be two Borel measures on R1 which are finite
and equal on every half-open interval (#,/>], —oo < a <b < + oo.
Then p(B) — v(B) for every Borel set В cz R1.
Proof Such p and v must agree on the algebra generated by the and
are ст-finite with respect to this algebra. Since the smallest ст-algebra contain-
ing all (afi\ is the Borel a-algebra, it follows from (11.20) that p and v are
identical.
Although we have confined ourselves to n = 1, we note that analogous
results can be formulated in higher dimensions. See Exercise 18.
Now let p be any finite Borel measure on R *, and define /д by /д(х) =
oo,x]). Clearly, ц((а,Ь]) = f^b) - f^a). Moreover, if A denotes the
Lebesgue-Stieltjes measure constructed from /д, then since /д is continuous
from the right, it follows from (11.10) that A((a,Z>]) == /M(Z>) — ffa). There7
fore, by the corollary above, p and A are identical as Borel measures, and we
easily obtain
(11.22) Corollary The class of finite Borel measures on R1 is identical with
the class of Lebesgue-Stieltjes measures induced by bounded increasing
functions that are continuous from the right.
See also Exercise 4.
Let p be a Borel measure on R1 which is finite on every («,&], — oo <
a < b < + oo (equivalently, p is finite on every bounded Borel set). Consider
the restriction of p to the algebra j/ generated by the (afi], and let p* be the
corresponding outer measure. The smallest сг-algebra containing j/ is the
Borel sets @L Thus,j2/ c cz j/*. Since p and p* are measures on which
agree on j/, it follows that p = p* on In particular, if В e we see from
the definition of p* [see (11.17)] that
ц(В) = inf {£ ц(Ак) :B c (J Ak, Ake j/}.
Each is a countable union of disjoint (a,b]. Hence, we obtain the
formula
(11.23) = UKy, Be®.
We recall (see p. 187) that a Borel measure /1 is said to be regular if
/л(В) = inf {jw(G) : В c G, G open}, Beffl.
(11.24) Theorem If fils a Borel measure on R1 which is finite on every bounded
Borel set, then д is regular.
Proof This is a corollary of (11.23). Given a Borel set В and s > 0, find
a cover {(ak9bk]} of В such that
E 1*(аМ < ц(В) + e.
Since ft is finite on intervals, it follows from (10.11) that ц(а,Ь] = lim£^0 +
/i(a, b 4- e). Hence, by slightly enlarging each (ak,bk], we see that there is an
open set G, G = |J (ak, bk 4- ek) for sufficiently small efc, containing В such
that
fi(G) < X pfat, bk + efc) < /x(5) + 2e.
This completes the proof.
In this theorem, as in (11.21), we have limited ourselves to n = 1. For
n > 1, see Exercise 18. In particular, note that the assumption on p. 187,
Chapter 10, concerning the regularity of д and v is redundant.
Exercises
1. Prove the second statements in both parts of (11.4).
2. Let ytz be a finite Borel measure on R1, and define = X(~°°>*])> — 00 < x
<4-00. Show that is monotone increasing, = ffb) — ffa\ is
continuous from the right, and limx_, _ ffx) = 0.
3. Let /be monotone increasing on R1.
(a) Show that Az is finite if and only if /is bounded.
(b) Let /be bounded and right continuous, let // = Af and let /denote the
function defined in Exercise 2. Show that / and / differ by a constant
Thus, if we make the additional assumption that limx^ _ f(x) = 0, then / = /.
4. If we identify two functions on R1 which differ by a constant, prove that there
is a one-to-one correspondence between the class of finite Borel measures on
R1 and the class of bounded increasing functions that are continuous from
the right.
5. Let/be monotone increasing and right continuous on R1 .
(a) Show that A^ is absolutely continuous with respect to Lebesgue measure if
and only if / is absolutely continuous. (By absolutely continuous on R1, we
mean absolutely continuous on every compact interval.
(b) If Af is absolutely continuous with respect to Lebesgue measure, show that
its Radon-Nikodym derivative equals (dfldx).
6. Prove that the Lebesgue-Stieltjes outer measure constructed from f(x) = x is
the same as Lebesgue outer measure.
7. If f is monotone increasing and right continuous on R1, show that AJ/4)
= Xj(A where A J is defined in the same way as AJ except that we use open
intervals (ak,bk)-
8. If /is monotone increasing and right continuous, derive formulas for A/([<7,Z>])
and Af((a,b)).
9. Complete the proof of theorem (11.12).
10. Show that in R", n > 1, Hausdorff outer measure Hn is not identical to Lebesgue
outer measure. (For example, let n = 2, and write A — (J Ak, b(Ak) < e.
Enclose Ak in a circle Ck with the same diameter, and show that 2 b(Ak)2
> (4M)|^|e. Thus, H</(A > (4/л)|Л|е.)
11. If A is a subset of R", define the Hausdorff dimension of A as follows: If Ha(A)
= 0 for all a > 0, let dim A = 0; otherwise, let
dim A = sup {a : Ha(A) = + oo}.
(a) Show that Ha{A) = 0 if a > dim Л, and that Ha{A) = d-oo if a < dim A.
Show that in R" we have dim A < n.
(b) If dimA = d for each Ak in a countable collection {Ak}, show that
dim (|J Ak) = d. Hence, show that every countable set has Hausdorff
dimension 0.
12. Let Г be an outer measure on «Z, and let Г' denote Г restricted to the Г-
measurable sets. Since Г' is a measure on an algebra, it induces an outer measure
Г*. Show that Г*(Л) > Г(Л) for А с У’, and that equality holds for a given
A if and only if there is a Г-measurable set E such that A с E and Г(£) = Г(Л).
Thus, Г = Г* if Г is regular.
13. Let 2 be a measure on an algebra л/, and let 2* be the corresponding outer
measure. Given A, show that there is a set H of the form (J7 AkJ such that
Ak j e A с H and 2*(/4) = 2*(77). Thus, every outer measure which is
induced by a measure on an algebra is regular.
14. Prove theorem (11.18).
15. (a) Show that the intersection of a family of algebras is an algebra.
(b) A collection of subsets of tf Is called a subalgebra if it is closed under
finite intersections and if the complement of any set in # is the union of a
finite number of disjoint sets in #. Give an example of a subalgebra. Show that
a subalgebra # generates an algebra by adding 0, and all finite disjoint
unions of sets of
16. If g is a finite Borel measure on R1, show that g(B) = sup g(F) for every Borel
set B, where F denotes any closed subset of B.
17. Show that the conclusions of theorems (10.48) and (10.49) of Chapter 10,
and therefore also the conclusion of (10.50), remain true without the assump-
tion (ii) stated before (10.47). [Show that without this assumption, the conclu-
sions of (10.47) are true with // replaced by //*; for example,
v(QxWK
x g E : sup > a
<Ж]
“ a
18. Derive analogues of (11.21) and (11.24) in R", n > 1. (Use partly open n-
dimensional intervals in place of the intervals
Chapter 12
A Few Facts from Harmonic Analysis
1. Trigonometric Fourier Series
Lebesgue measure and integration have been decisive in the development of
many branches of analysis and are applied there in ever greater degree. But,
conversely, some of the applications have had considerable impact on the
theory of integration. In this chapter, we will consider one topic where this
interdependence has been particularly fruitful: harmonic analysis (see p. 214).
We begin by describing some elementary notions and facts; the concept of
an orthogonal system, and in particular of the trigonometric system, is
basic here.
The notion of an orthogonal system, defined generally on a subset E of
positive measure in R”, was introduced in Chapter 8, and we refer the reader
to that place for the definitions and properties of general orthogonal systems,
restating only a few facts here.
A system of complex-valued functions {фа(х)}9 all belonging to L2(E), is
called orthogonal over E if
(Фа’ФрУ I ФаФр
JE
The second condition means that фа 0. If {фа,фа} = 1 for all a, the
orthogonal system is called normal, or orthonormal. If {фа} is orthogonal,
the system {</>a/||<^a||2} is orthonormal. Thus, by merely multiplying the
functions of an orthogonal system by suitable constants, we can normalize it,
and formulas for orthonormal systems can be easily and automatically
extended to general, not necessarily normal, orthogonal systems. On the
other hand, certain important orthogonal systems very naturally appear,
often for historical reasons, in a nonnormalized form, and because of this,
it may be desirable not to insist on the normality of the system under con-
sideration. Let us, therefore, briefly restate the definitions in this somewhat
more general set-up.
= 0 a
> 0 a =
211
212 A hew t-acts from Harmonic Analy^ (12.1)
Since orthogonal systems are countable, we may index them by integers.
Let $i(x), ф2(х),... be an orthogonal system on E cz R”. Thus,
Given any (complex-valued) f e L2(E), we call the numbers
T” J f$k
лк JE
the Fourier coefficients of /, and the series S[f] = £ скфк(х) the Fourier
series of /, with respect to {фк}. As before, we write
f~
If we set
(12.1) = Ак1/2фк, dk = = Г ж,
Je
then {фк} is orthonormal over E, and {dk} is the sequence of Fourier coeffi-
cients of f with respect to {фк}. Clearly,
(12.2) dkt//k = скфк.
This set of formulas enables us to rewrite relations for orthonormal systems
in forms valid for general orthogonal systems. Thus, Bessel’s inequality
£ \dk\2 < J£ |/|2 and Parseval’s formula £ |dk|2 = |/|2 take the forms
(12.3) £Ш|2 < f l/l2- £ Ш12 = f 1Л2-
Je
The notion of completeness of an orthogonal system (“the vanishing of all
the Fourier coefficients implies the vanishing of the function”) remains un-
changed in the general case, and as in the case of normalized systems, the
validity of the second formula (12.3) is a necessary and sufficient condition
for the completeness of {фк}.
Let sn denote the «th partial sum of 5[/]. As a corollary of the correspond-
ing result for orthonormal systems, we see that the equation £ 2л|ск|2 =
|/|2 is equivalent to
[|/-^|2-0.
Je
Thus, if an orthogonal system is complete, thd Fourier series of every / e L2(E)
converges to /, convergence being understood in the metric L2. Of course,
this says nothing about the pointwise convergence of £ скфк(х). On the other
hand, it holds for any rearrangement of the terms of £ скФк since the orthogo-
i uui ici oci ic;b
гц/л
к — m
e.*xeImx dx =
'e
nality and completeness of a system are not affected by a permutation of the
functions within the system.
We shall now consider a special orthogonal system, the trigonometric
system. This name is given to the system of functions
егкх = cos kx + i sin kx (k — 0, + 1, ±2, . . .).
These functions are all periodic, with period 2л, and it is immediate that
they form an orthogonal system over any interval Q = (a, a + 2л) of length
2л, since if к and m are distinct integers, then
““ ' a+2n
= 0.
a
The system is not orthonormal since lk = |е1кл|2 dx = 2л for all k. Thus,
with any f 6 L(Q), we may associate its Fourier coefficients
1 f _______ 1 f
(12.4) ck = — I f(f)eikt dt = — f(t)e~ikt dt (к = 0, ± 1, ±2,. . .),
2л Jq 2л Jq
and its Fourier series
+ 00
(12.5)
— oo
In what follows, this series will be designated by S[/], and its coefficients by
Cklf]-
Observe that if two functions ф and ф are orthogonal over a set E and if
j£ |<A|2 — f£ |(A|2, then the pair ф ± ф is also orthogonal over E, as seen from
the equation
[ (Ф + -’/')=[ И12 - I* Ш2 = 0-
Je Je Je
Applying this to the pairs e±lkx (k = 1, 2,. . .), we see that the functions
ifcx | „~ikx ikx ____________________ p-ikx
(12.6) i, . . ., +2 g—, 2/ , • • • (k = 1, 2, . . .),
or what is the same thing, the functions
(12.7) 1, cos x, sin x, ... , cos kx, sin kx,. . .
are orthogonal over any interval Q of length 2л. Using the form (12.6), we
find that the numbers for (12.7) are
|л, Л, Л, . . . .
Thus, any f g L(Q) can be developed into a new Fourier series
(12.8) f ~ |a0 + X tak cos kx + bk sin kx),
k=l
where
(12.9)
«о = (I*)’1 f V = ; f f,
Jq n Je
ak = - I f(t) cos ktdt, bk = - I /(?) sin kt dt.
я Je
The numbers ak and bk are easily expressible in terms of the coefficients ck
of (12.4):
(12.10) |u0 = c0) ak = ck + c_k, bk = i(ck - c_k).
Hence,
X ckeikx = c0 + X ckeikx + X c-k^ikx
k= —n k—1 k—1
n
— 2aQ + E (ak C0S kx + bk sin kx),
fc=l
and the Mh partial sum of the series in (12.8) turns out to be the zzth symmetric
partial sum of E ckelkx. The numbers ak = ak[f] and bk = bk[f] are called the
Fourier cosine and sine coefficients of /, respectively.
To sum up, we may consider the trigonometric system in two forms.
One consists of the functions elkx (k = 0, +1, ±2, . . .), and the Fourier
series has the form E-S cke'kx> where the ck are given by (12.4). The other
consists of the functions (12.7), the Fourier series is (12.8), and the coefficients
ak and bk are given by (12.9). The partial sums of (12.8) are the symmetric
partial sums of (12.5). In both cases, the terms of the Fourier series are
harmonic oscillations, and for this reason, the study of 5[/] is called the
harmonic analysis of f.
Each form of the trigonometric system has its advantages. For example,
if f is real-valued, then the numbers ak and bk are real, while the ck have the
property c_k = ck, Note also that if Q — { — it, it) and /is an even function,
i.e., if/(—x) = /(x), then
2 P
ak — - I /(Z) cos kt dt, bk = 0,
л J о
and if / is an odd function, i.e., if /( — x) = — f{x), then
2 P
ak — 0, bk = - I //) sin kt dt,
71J о
Thus, if /is even, (12.8) reduces to the cosine series + Efc°=i ak cos kx,
and if / is odd, to the sine series t bk sin kx.
Since the terms of a (trigonometric) Fourier series are periodic with
4.IUj
inyuiiuiheiiiu ruunei oeues
£. i Э
period 2л, if we expect to represent a function / by its Fourier series, we may
assume from the start that f is defined everywhere (or almost everywhere)
on the real axis and is periodic with period 2л. This amounts to considering
the function as defined on the circumference of the unit circle. We do not
distinguish between points which are congruent mod 2л. By an integrable
function, we shall mean a function integrable over a period. Similarly for Lp
and other classes of functions. In what follows, periodic will mean of period 2 л.
In the preceding chapters, we proved a number of theorems about func-
tions in Z/(Rn), and in particular, in //(R1). Usually these results have
analogues for periodic functions, where integrals over R1 are replaced by
integrals over a period. The proofs are usually in essence identical with those
for R1 (or are merely corollaries of the results for R1), and may be left as
exercises.
We would like to stress one point. The definition of a general orthogonal
system presupposes that the functions in the system are of class L2. This makes
it possible to define Fourier coefficients for any f eL1. If f is not in L2,
it may be impossible to define its Fourier coefficients with respect to certain
orthogonal systems. The situation is different for special orthogonal systems.
For example, in the case of the functions {егкх}, which are bounded, the
coefficients ck are defined for any f which is merely integrable over Q, and in
particular, for any f e LP(Q\ 1 < p < oo. Thus, the trigonometric system is
richer in properties than general orthogonal systems.
Some simple developments are important for the general theory of
Fourier series. We consider two here and refer the reader to Exercise 5 for
others.
Example (a). Let f be periodic and equal to |(л — x) for 0 < x < 2л, with
/(0) = /(2л) = 0. Since/is odd, its Fourier series is a sine series, and integra-
tion by parts gives
, 2 ГЧ A • А А Ч71 1 Г J, A 1
bk = - - (л — X) sm kx dx = - 7 — 7 cos kx dx] —
TrJ02v 7 n\k kj0 J к
Thus,
sin kx __ 1 L?, eikx
where E' denotes
Example (b). Let f be periodic and equal in (—л,л) to the characteristic
function of the interval ( — hfi), 0 < h < л. Then / is even, and if к Ф 0, its
cosine coefficient is
2 Г;’ 7 1 2 sin kh
ak = - cos kxdx =---------;—.
л J о л к
4»
z t о
a rew racts тгс-m Harmonic Analyst
(12.11)
Since aQ = 2h/n, we obtain
7U+ E
л (2
sin kh
—г-,— cos kx
kh
h
n
kh
Series of the form
E ckeikx,
00
2^0 + E (ak cos kx + bk sin kx),
k=l
whether they are Fourier series or not, are called trigonometric series. In
defining the convergence of E-S cke'kx> we usually consider the limit,
ordinary or generalized, of the symmetric partial sums and it is im-
mediate that
n n
E скегкх = + E (ak cos kx + bk sin kx),
— n k—1
where
2^0 c0, @k ^k "h c~_k, bk = i(ck c_k).
A finite sum T = ckelkx is called a trigonometric polynomial of order n,
andif|c_n| 4- |си| / 0, Th strictly of order n. If Tis of order «and vanishes at
more than 2n distinct points (i.e., distinct mod 2n), then it vanishes identi-
cally, i.e., all the ck are 0. For Teinx = ckel{k+n}x is a (power) polynomial
in z = elx of degree < 2n, and if it has more than 2n zeros, then it vanishes
identically.
If the numbers ak and bk are real, the trigonometric series
1 00
S = ~ a0 4- E (ak cos kx 4- bk sin kx)
k=l
is the real part of the power series
1 00
O«o + E - ibk)zk
z fc=l
on the unit circle z = eix. The imaginary part is then the series
00
(12.11) E (ak si*1 kx — bk cos kx)
k=l
(with vanishing constant term). If S is writtenjn the complex form E^2 ске'кх>
it is easy to see that (12.11) is
+ 00
(12.12) EC-^ignW^
— 00
i'i«sorerns aooui rounei voetticients 217
(where, by convention, sign 0 = 0). The series (12.11), or (12.12), is said to be
conjugate to S'. A series conjugate to a trigonometric series S is denoted by S.
If S' has constant term 0, then
§ = -S'.
It is natural to study the properties of S[/] simultaneously with those of S[/].
One more remark. Properties of functions in Lp(Rn), and in particular in
L^R1), are important for the theory of Fourier integrals, which for non-
periodic functions play the same role as Fourier series in the periodic case.
Though the two theories run largely parallel, we shall limit ourselves to
Fourier series since our primary aim is to show the role that Lebesgue
integration plays in problems of representability of functions, and both the
results and techniques of Fourier series are sufficiently indicative of the
situation.
2. Theorems about Fourier Coefficients
(12.13) Theorem If a periodic f is the indefinite integral of its derivative
ff (i.e., if f is periodic and absolutely continuous), and if f ~ cke'kx,
then
+ oo
f ~ £ ск(гкУкх.
— 00
In symbols,
S[f] = S'[f],
where S'[/] denotes the result of the termwise differentiation of S[/]-
Proof It is clear that /' is also periodic and that its constant term equals
f2n
(2/r)-1 f’ dx = (InYWn) - /(0)] = 0.
J 0
If к / 0, integrating by parts and observing that the integrated term is zero,
we have
(2tt)~1 I* f'e~lkx dx = (2n)~lik f fe~tkx dx = ikck,
Jo Jo
which proves the theorem.
By repeated application of this result, we see that if a periodic f is the
/?7th indefinite integral of an integrable function /(m), then
= 5<-V] = (ik)mckeikx.
(12.14) Theorem If f is periodic, f ~ £ скегкх, and if F is the indefinite
218 A rew racis irom narmoniu мислуыУ
integral of /, then F(x) — cQx is periodic and
F(x)~ cox~ C0 + £'^eikx,
where Co is a suitable constant (depending on the choice of the arbitrary
constant of integration in F), andlL' denotes ^k*o-
Proof Let G(x) = F(x) — cQx. The periodicity of G follows from the
equation
G(x + 2zr) — G(x) = f f dt — c0(2tt) = 2лс0 — 2itc0 = 0.
J X
Since G is also absolutely continuous, (12.13) gives
S'[G] = 5[G'] = c0] = £ ckeikx.
k*0
Hence, S[G] is obtained by termwise integration of ckelkx, which leads to
the result, Co being the constant term of S[F — cox].
For the trigonometric system, we certainly have BessePs inequality (see
(12.3))
+ 00 1 г2л
Bl’sJ 1Л2*-
- 00 27Г J 0
but actually, as we shall see, we also have ParsevaPs formula
+ oo 1 f2n
(12.15) £ kJ2 = x- |/|2 dx
- 00 271 J 0
for every f e L2. We know (see (8.31)) that this is a corollary of the following
(12.16) Theorem The trigonometric system is complete. More precisely, if all
the Fourier coefficients of an integrable f are zero, then f = 0 a.e.
Proof Assume first that f is continuous and real-valued, with all ck = 0.
If f =£ 0, then \ f\ attains a nonzero maximum M at some point x0. Suppose,
e.g., that f(xQ) = M > 0. Let b > 0 be so small that f(x) > in the
interval I = (x0 — <5, x0 + <5). Consider the trigonometric polynomial
t(x) — 1 + cos (x — x0) — cos b.
It is strictly greater than 1 inside I and does not exceed 1 in absolute value
elsewhere. The hypothesis that all the Fourier coefficients of f are 0 implies
that fT dx = 0 for any trigonometric polynomial T, and in particular,
Г ftN dx - 0 (N - 1,2, . . .).
— я
ineuiciiia auuui ruuiiui uuciHUienib
z. i»
We claim that this is impossible for N large enough. The absolute value of
the part of the last integral extended over the complement of Z is < 2л • M• 1N
= 2лМ. If Г is the middle half of /, then t(x) > в > 1 in Г, so that
f ft* dx > f ft* dx > |Af-|Z'|0N-> +oo.
ji jr
Collecting results, we see that ft* dx -» +oo; this contradiction shows
that f = 0.
If J is continuous but not real-valued, the hypothesis f q* fe~ikx dx = 0
for all к implies that \2f fe~lkx dx = 0 for all k. By adding and subtracting
the last two equations, we see that both the real and imaginary parts of f
have all their Fourier coefficients equal to 0, and so vanish identically.
Finally, if f is merely integrable, the hypothesis cQ = 0 implies that the
function F(x) = Jq f dt is periodic, and by (12.14), for a suitable Co, the
Fourier coefficients of the continuous function F — Co are all 0. Hence,
F — CQ = 0, F is a constant, and f = F' = 0 a.e. This completes the proof
of the theorem. Another proof is given on p. 241.
An immediate corollary is the following
(12.17) Theorem ParsevaFs formula (12.15) holds for any f e L2.
Parseval’s formula can be written in more general forms, which are,
however, corollaries of (12.15). Thus, besides f ~ ^cketkxeL2, consider
another function # ~ £ dkelkx e L2. Then, by an argument like that on p. 140,
i Г2тг 4- co
(12.18) z- fg dx = £ ckdk.
J 0 - oo
This reduces to (12.15) in the special case f = g.
The completeness of the trigonometric system also gives the following two
theorems.
(12.19) Theorem If the Fourier series of a continuous f converges uniformly,
then the sum of the series is f.
For let g be the sum of the uniformly convergent series 5[/]. The Fourier
coefficients of g can be obtained by multiplying 5[/] by e~ikx and integrating
the result termwise. Thus, ck[g] = ck[f] for all k, so that f = g.
(12.20) Theorem If a periodic f is the integral of a function in L2, then S[/]
converges absolutely and uniformly. In particular, the Fourier series of
a continuously differentiable function converges uniformly to the func-
tion.
For let f be the integral of д e L2, д ~ ckeikx (c0 = 0). Then £[/] =
Co + Скегкх, Ck = cffk, к / 0. We have £ Ы2 < +00 by Bessel’s
inequality, so that £ |Ck| < + oo by Schwarz’s inequality. This completes
the proof.
The theorem which follows is of basic importance.
(12.21) Theorem {Riemann-Lebesgue) The Fourier coefficients ck of any
integrable f tend to 0 as к -> ± oo. Hence, also ak,bk -> 0 as к -> + oo.
Proof First, we note the obvious but important inequality
1 Г2л
Iq[/H Jo 1/1 dx-
We will give two proofs of the theorem.
(a) (See also Exercise 15 of Chapter 8.) If f e Z2, then ck -> 0 as a corol-
lary of Bessel’s inequality (p. 218). If f e Z, write f = д + h, where д e L2
and |/z| < s. (This decomposition can be made in various ways’ we may,
e.g., take M large enough and define A to be f wherever |/| > M and 0
elsewhere; clearly, |g| < M, and so д e Z2.) Then
ck[f] = Qfo] + ck[h].
Since q[#] -> 0 and |сл[Л]| < (2я)“1 \h\ < s/2n for all k, the relation
ck[f] 0 follows.
(b) Observe that
1 Г271 1
q[/] = 5" f{x)e~lkx dx = — fix + у-1е“г7с(х+(я/Л)) dx
271 J 0 271 J — (tc/Jc) \ .
= -ST-1 f\x + -Де~'кх dx.
2тг Jo \ kJ
Taking the semi-sum of the first and third integrals, we obtain
i Л2я 1 Г / \ “
Cklf] = 2л J 0 2[/W - f\x + k)\e~ikX dx’
1 f2nl
|ct[/]|<2- у
Z7T Jq 2
fW - fix +
\ KJ
dx.
However, we know that the last integral tends to 0 as к ± oo [cf. theorem
(8.19); the analogous result for periodic functions is left to the reader]. This
completes the proof.
Given any continuous periodic /, the expression
sup |/(x + h) -/(x)| (<5>0)
is called the modulus of continuity of /, and denoted by m(<5) or ocrffff) (cf.,
Chapter 1, Exercise 17). If/is in Lp, 1 < p < oo, the expression
UGUIGHiS (WUUl ruunci kz U b 4 ilG i bfl'lS
ZZi
sup
|й|<<5
1/P
is called the pth modulus of continuity of /, and denoted ct>p(<5). Clearly,
mp(<5) < co((5) and, as is easily seen from Holder’s inequality,
03p(S) < mq(d) if p < q.
We know that if f gLp, then cop(d;f) -► 0 with 5 [theorem (8.19)]. The last
inequality in proof (b) above gives
(12.22) |Q[/]I
These two inequalities contain the Riemann-Lebesgue theorem in a sharp
form since they quantitatively estimate the magnitude of the Fourier coeffi-
cients of f in terms of various moduli of continuity.
The estimates (12.22) are also useful for families of functions. The
following special case deserves a separate mention. A continuous periodic
f is said to satisfy a Lipschitz condition of order a, 0 < a < 1, if m(<5;/) =
(9(<5a). or equivalently, if there is a finite constant M independent of x, h
such t. \at
l/(x + A) < M\h\*.
(12.23) Theorem If f satisfies a Lipschitz condition of order a, 0 < a < 1,
then
1^[/]| = W“).
If a = 1, the stronger estimate
kk[/]| =
is valid.
Proof The first part follows from the first inequality (12.22). If a = 1,
then f is absolutely continuous [see (7.26)] and so equals the indefinite
integral of its derivative Since f is bounded (and so is in L2), its Fourier
coefficients tend to zero. Hence, the coefficients of f are 6>(1/|A;|) by (12.14).
(12.24) Theorem If a periodic f is of bounded variation over a period, then
|Q[/]| = ^(l/I^D- More precisely,
Ы < F/2<|,
where V is the total variation of f over a closed interval of periodicity.
Proof. Integrating by parts and taking account of the periodicity of /,
we have
= | e~tkxf(x) dx = ~ | e~tkx df(xf
J — n lk J —it
where the last integral is a Riemann-Stieltjes integral. Hence,
2я|^[/]| < 1АГ1 Г l#(*)l = И'1 И.
J —It
It must be stressed that V is the total variation over a closed interval of
periodicity.
3. Convergence of S[/] and S[/]
We shall now briefly discuss the problem of pointwise convergence of 5[/],
treating side-by-side the parallel problem for S[/]. Among many existing re-
sults, we will consider only the simplest. Without loss of generality, we may
restrict our attention to real-valued f.
We begin by computing the partial sums of S[f] and §[/]. If ak and bk
denote the cosine and sine coefficients of /, then the «th partial sum of
^[/] is
1 n
s„(x) = x a0 + £ (ak cos kx + bk sin kx)
k=l
1 Г ” f 1 Г
= T" /(0 dt + Y \cos kx- - /(z) cos kt dt + sin kx
2^J-it k = l L TCJ-я
1 Г )
•- I f(t) sin kt dt>
n J -n J
= I I" f(t) t + £ cos k(x - t) dt = 1 f f(t)Dn(x - t) dt,
n J -n L2 4=1 J Я J -я
say, where
1 n
Dn(f) = Э + X C0S kt-
2 k=l
The trigonometric polynomial Dn is called the «th Dirichlet kernel. Similarly,
the «th partial sum of S[/] is
n 1 Гя Г n 'J
sn(x) — £ (ak sin kx — bk cos kx) = - f(t)< £ sin k(x — t)> dt
k=l 71 J -it u=i J
= f(t)Dn(x - t) dt,
convergence от г>|/j and 5[/]
223
where
Dn(t) = X sin kt
k = l
is the «th conjugate Dirichlet kernel. Notice that Dn and Dn are even and odd
functions of t, respectively, and that
(12.25) i Г Dn(t) dt = 1, t Г D„(t) dt = 0.
71 J “Л 71 J ~1t
Moreover, Dn and Dn are respectively the real and imaginary parts of
1 » 1 « i
z k=l z k=l z
zn — Z
z — 1
(z = e;t),
and an elementary computation gives
(12.26) =
Z, bill 21
cos У — cos (n 4- |)r
2 sin |Z
A quicker, though somewhat artificial, method of obtaining the first formula
is to multiply Dn(f) termwise by 2 sin |Z, replace the products 2 sin cos kt
by sin (к + I)/ — sin (k — |)t, and make use of cancellation of terms. Like-
wise for Dn(t).
Given a function f and a fixed point x, let us consider the expressions
Фх(0 = |[/(x + t) + f(x - t)], = i(f(x + t) - fix - /)]
as functions of t. They are called the even and odd parts of f at the point x,
respectively. Clearly,
f(x + t) = фх(Г) + фх(Г).
It turns out that the behaviors of фх(г) and i/<x(z) near t = 0 are decisive for
the behaviors of £[/] and S[/], as the case may be, at the point x.
Returning to the formula for ^(x) and making use of the even character
of Dn(tf we can write
1 ГЯ i p
*„(*) =Z f(tWx-t)dt=-\ f(f)Dn(t - x) dt
71 J -7t 71 J -It
If71 2 Г711
= - f(x + t)D„(t)dt =- -[f(x+t)+f(x-t)]Dn(t)dt
71 J -it 71 J 0 z
2 Г
= - ожол.
The first formula (12.25) immediately gives
2 Г
- f{x) = - [O) - dt
Tt Jo
_ 2 p sin (n + l)r
7tJ0^x() 2sin|Z dt'
It will be convenient to modify this formula slightly by replacing n by n — 1
and taking the semi-sum of the two formulas. Writing
(12.27) s*(x) = Ил(х) + Vi(*)] = “ ъ(ап cos nx + bn sin nx)>
we obtain
2 fn sin nt
s*(x) - f(x) = - [фх(0 - f(x)] ~ . n ,
тс j о - ran yt
The right side here is the «th Fourier sine coefficient of the odd function
[<Ax(O - /(*)H cot it,
and if this function happens to be integrable near t — 0, the Riemann-
Lebesgue theorem immediately gives s„(x) — fix) -» 0. Hence, making use
of the fact that an,bn -> 0, we obtain from (12.27) the following basic result.
(12.28) Theorem (Dini's Test) If the integral
cotlzrfz
о
is finite, then 5[/] converges at the point x to the value fix).
Since only small values of t matter here, and since for small t we have
| cot ~ t~i, Dini’s condition can be restated in the form
ri^)-/WL/z
-------------< 4-oo,
Jo *
or what is the same thing,
+ о + я* - <> - w>c ,
Jo *
The following special case is useful. Suppose that f has a jump discon-
tinuity at x, so that the one-sided limits fix-у), fix — ) exist. Since changing f
at a single point does not affect S[/], we may"assume that
/(*) - M/(x + ) +/(*-)],
in which case we say that f has a regular discontinuity at x. Condition (12.29)
к *
convergence от j and 5[/j
225
is then certainly satisfied if both
,'l/(x + O-/(x+)K
----------------at < +
о *
p|/(x-Z)-/(x-)|
----------------dt < 4- oo
Jo *
Thus, a corollary of Dini’s test is that if both f(x + ) and fix —) exist, and if
both of the last two integrals are finite, then S[/] converges at the point x to
the value
/(x+)+/(x~)
2
There is a result analogous to (12.28) for S[/], and we will be brief here.
Using the formula for s„ and the odd character of D„, we have
s-.w - i Г /(« + .)«.(» at --П\,(0^-~“5,(" + ^ a.,
Ч-n 71 J Q SHI 21
4 «»(•*) + s»-iW 2 Г , , 2 Г” , z,4 cosm
s* (x) =---------Z------= - - «AxCOi cot dt + - iAx(t) y-r-j- dt,
2 7Г J 0 7C J о 2 Sin it
provided that
Г dt
(12.30) ||ДХ(Г)| — < + co.
JO I
Under this hypothesis, the last term in the preceding equation tends to zero
by the Riemann-Lebesgue theorem, and we obtain
(12.31) Theorem Under the hypothesis (12.30), the series S[/] converges at
the point x to the sum
2
71
l/^(Z)| COt У
1 r/(x + - 0 ,
_ ---- ----- -----at
7t J 0 2 tan |z
We denote the last integral, if it exists, by fix):
(12.32)
Л*) = -
1 Г f(x + t) - f(x - t)
- | -----—-----:-------at.
7i J 0 2 tan |Z
This function is called the conjugate function of f, and is intimately connected
with the behavior of S[/]. We shall study the existence and properties of f
in detail later.
Observe that condition (12.30) is of a nature completely different from
(12.29); the former precludes the possibility that f may have a jump at x.
See Exercise 15.
The proofs of (12.28) and (12.31) are based on the Riemann-Lebesgue
theorem and give convergence results only at individual points. They cannot
give uniform convergence in an interval without additional and rather
226
A Few Facts from Harmonic Analysis
( IZ.J4)
strong assumptions. We consider one such assumption which, though very
restrictive, leads to an important result.
(12.33) Theorem If f = Q in an interval (a,b), then 5[/] and S[/] converge
uniformly in every smaller interval (a + e, b — e). The sum of 5[/] is 0.
Proof The pointwise convergence in (a,b) is a corollary of (12.28) and
(12.31), and it is only the question of uniformity which requires additional
comment. We, will consider only £[/]; the argument for £[/] is similar. Fix
e > 0. From (12.27), we deduce that
4 ^(Xo) + J„_i(x0) 1
5"(x°) =---------2--------= я
i г
= ~ f(xo + 0/(0 sin nt dt,
Я J -n
sinnZ
/(х» +
Xq e (a — e, b + e),
where yff) is periodic, equals |cot|r for e < |Z| < я and is arbitrary for
|Z| < 8. Suppose % is defined so that it is continuous everywhere. Write
f(xQ + /)/(/) — treating t as the variable and x0 as a parameter, and
consider the modulus of continuity of gXo(t) in the metric L1. If we show that
mi(<5,#Xo) tends to 0 with 5 uniformly for x0 e (a 4- e, b — e), then the
Fourier coefficients of gxft) will also tend to 0 uniformly for such x0 [see the
second formula (12.22)], and the theorem will follow. Now, for h > 0,
Г2л f2n
ItfxoO + A) - 0XO(?)| dt = |/(x0 + t + A)/(Z + h) ~/(x0 + t)x(t)\ dt
Jo Jo
C2n
£ l/(*o + t + h) - f(x0 + 01 lx(t + A)| dt
J о
f*2n
+ IZOo + t)\\x(t + h) - X(0l dt.
J о
The last integral clearly tends to 0 with A, uniformly in x0, since maxf |%(r 4- h)
— y(t)| -» 0 as h 0. If M = max |/|, the preceding integral is majorized by
M Г |/(x0 + t + A) -f(x0 + Z)| dt = M Г \f(t + A) -/(01 dt,
Jo Jo
a quantity independent of x0 and tending to 0. This completes the proof.
Two trigonometric series Tx and T2 are said to be equiconvergent at a
point xQ if their difference — T2 converges to 0 at x0. If — T2 merely
converges, but not necessarily to 0, then Tx and T2 are said to be equicon-
vergent in the wider sense at xQ. Each of two equiconvergent series may be
individually divergent, but the character of divergence is so similar that
divergence cancels out in — T2,
(12.34) Theorem Let and f2 be two periodic functions which are equal in an
(12.35)
Divergence of Fourier Series
227
interval (a,b). Then Sf/J and 5[/2] are uniformly equiconvergent in
every subinterval (a 4- s, b — e); S[/J and S[/2] are uniformly equi-
convergent in the wider sense in every (a + e, b — g).
This is a corollary of (12.33), since, for example, — 5[/2] = 5[/]
where f (= fr — f2) vanishes in (afi).
Thus, if we change the values of f in an arbitrary way outside an interval
(a,b), we do not affect the behavior of 5[/] in (a 4- a, b — e). Likewise for
S[/], although in this case, if the series converges, the value of the sum may
change. Therefore, the convergence or divergence of S[/] and S[/] at a point
x0 is a local property, i.e., depends only on the behavior of f near x0.
4. Divergence of Fourier Series
(12.35) Theorem There exists a continuous periodic f such that 5[/] diverges
(more specifically, the partial sums of 5[/] are unbounded) at some
point.
Proof. Let 1 < m < n and consider the polynomials
z s СО5Ж cos (m 4- l)x cos (m 4- n — l)x
- — + Lr + + ~H—L
cos (m 4- n 4- l)x cos (m 4- n 4- 2)x cos (m 4- 2ri)x
. _ ... - .
We will show that all these polynomials are uniformly bounded, but that
their partial sums are not. To prove the first statement, we need the fact
that the partial sums of the series
sin kx
Г ’
jt=i K
which we considered on p. 215, are uniformly bounded. This is an ele-
mentary fact which can be proved in many ways [see, e.g., (12.50(c))], but here
we take it for granted. Thus, since
" cos (m 4- n — k)x — cos (m 4- n 4- k)x
2m,nW = L -------------------Г--------------------
fc = l K
„ . , ” sin kx
= 2 sm (m 4- n)x ) j —-—,
к=1 к
we obtain |0nij/J(x)| < C, where C is independent of m and n. On the other
hand, when x = 0, the partial sum
z ч cos mx cos (m 4- n — l)x
2,L.W = -------—j---------
has the value 1 + (1/2) + • • • + (l/и), which is of order log n.
228
A Few Facts from Harmonic Analysis
(12.36)
Now select integers mk and nk such that
mk + 2nk < mk+l (k = 1,2, . . .),
and take a series of positive numbers ak such that £ a* < 4- oo, afc log nk ->
4- oo. (We will make the construction in a moment.) The series
k = l
then converges uniformly to a continuous function f. In view of the inequality
above relating mk and mk+1, the polynomials Qmk,nk do not overlap. Hence,
the last series can be written as a single trigonometric series, whose coeffi-
cients (because of uniform convergence) are the Fourier coefficients of f.
Thus, this series, unbracketed, is S[f]. But 5[/] has unbounded partial sums
at x = 0 since a single block of terms, namely, akQmk,nk(x)> *s °f order ak log nk
at x = 0.
It is easy to verify that if we set
= 5*3, nk — 2mk = 2(5P), ak — 1/k1 2,
then all the conditions required above are fulfilled. This completes the proof.
We leave it to the reader to check that if we choose
mk = 5k2, nk = 2mk, ak = 1/k2
in the construction above, we get a continuous f whose partial sums are
divergent but bounded at x = 0.
Theorem (12.35) asserts that the partial sums of S[/] can be unbounded
even if f is continuous. It is of interest to know “how unbounded” they can
be. From the formula
1 Cn
s„(x,f) = - fix + t)Dn(t) dt,
J —71
we see that if |/| < 1, then
ip 2 f”
WxJ)|<- |Р„(7)|Л = -
n J -Я 71 J 0
uniformly in x. The right side here is called the /7th Lebesgue constant, and will
be denoted by Ln. Note that Ln is actually the value of ^(0,/) for a specific /,
namely, f(t) = sign Dn(t).
(12.36) Theorem We have
2 p 4
Ln = - I |Pn(OI — -jlogn as n -> oo,
(12.37)
Divergence of Fourier Series
229
Proof. Write
2 p
Ln = - |Z>„WI dt =
11J о
2 p / 1\ _
= 2 I sm ( n + “)/—+-
dt 2 p
* Я Jo
2 sin dt
sin ( n + - и
2 sin У t)
71 J о
t
1
Since the expression in brackets is bounded for 0 < t < я, and
|sin (n + %)t\ < 1, the last integral is majorized by an absolute constant.
The change of variable (n 4- |)r = и shows that the preceding term equals
2 p(«+(l/2))n|sin w|
- -------du.
n J о и
We may disregard the parts of this integral extended over (0,я) and
(mt, (n 4- |)я), since the integrand is bounded. In view of the periodicity of
sin u, what remains can be written
2 |sin wl 2 p . 1 1 \
- ------du = - I sin w >, ----------r— du.
Я J n U Я J 0 \k=l u + K7t J
For 0 < и < я, the sum in brackets is contained between я-1 ОДО
and я-1 ^=i ОДО, and so is strictly of order я-1 log n. If we now note
that sin и du = 2, and collect estimates, we obtain Ln ~ (4/я2) log n.
(12.37) Theorem Iff is integrable, then at each point xQ of continuity of f,
sn(*o J) = o(\og ri).
The estimate is uniform over every closed interval of continuity of f.
Proof. We will prove only the first statement, leaving the second to the
reader. Suppose, as we may, that x0 = 0, f(xf) = 0. Because of our results
about localization [see (12.34)], we may assume that f vanishes outside an
arbitrarily small fixed interval (—5,5). Then
i p
- /т(рл
71 j
k„(0)l =
< sup 1/(01 • dt.
111 <, & J —n
Since the sup here is small with 5 and the integral is of order log«, the
assertion follows.
5. Summability of Sequences and Series
Theorem (12.35) shows that even continuous functions, when developed into
Fourier series, may not be representable by those series in terms of pointwise
convergence. The situation can be remedied by considering generalized sums
of the series. This topic is vast and basic for analysis, and we will study only a
few facts important for the theory of Fourier series.
230 A Few Facts from Harmonic Analysis (12.38)
Consider a fixed doubly infinite matrix of numbers (real or complex):
a00 a01 •’• a0n
a10 au • • • а1и
(^) !
’ * %mn '• • •
Given an infinite sequence of numbers sQ, slf . . . , sn,. . . , we transform it
into a sequence cr0, . . . , ... by means of the formulas
= am0^0 + ’ + W. + ’ ’ ’ («1 = 0, 1, 2, . . .),
assuming that the series defining converges for each m. We may ask what
conditions on (^) will guarantee that whenever {5Л} converges to a finite
limit s, lim om also exists and equals s. An answer is given by the following
theorem.
(12.38) Theorem Suppose that {JI) satisfies the following three conditions:
(i) |amJ < A {for all m, with A independent of m).
(ii) lim^-i-oo <xmn) 1*
(iii) lim^^ amn = 0 for each n.
Then for any sequence converging to a finite limit s, lim am exists
and equals s.
Proof First of all, since is bounded, (i) implies that am exists for
each m. Next, write sn = s + sn, where 0. Then
&т У^ &n) У^ "b У^
n n n
We have s amn -> 5 by (ii), and it remains only to show that the expression
Pm ttmnSn
n
tends to 0 as m -> 00. Given 3 > 0, split pm into two sums,
Pm Уу ~b У2 Pm ~b Pm’
n<;no n>no
say, where nQ is so large that |ел| < 3 for n > nQ. By (i),
\Pm\ < I KJW < E K„l<5 < Ad.
n>no n>no
On the other hand, p'm consists of a fixed number of terms each of which
tends to 0 as m -> co. Hence, |p'J < A3 for m large enough. Combining
estimates, we see that pm -> 0, which completes the proof.
It is useful to note that if 5 = 0, then condition (ii) is not required in the
(12.40)
Summability of Sequences and Series
proof (and so in the statement of the theorem) above. It is also irzrrezH^
from the proof that if {^} depends on a parameter, and if sn tenfs Ezffxrrr
to s, then am tends uniformly to s too.
If (7m -> s, we shall say that the sequence {5Л}—or the series whose puzu.
sums are the sn—is summable to limit (sum) s by means of the L/L ir
simply is summable (Л) to 5.
The matrix (Л) is called positive if ашп > 0 for all mji. Confirioz □ r
then a corollary of (ii). For positive (Л), theorem (12.38) holds if s — zr-
we leave the proof to the reader.
Two methods of summability are of special significance for Fours:
series.
(a) The method of the arithmetic mean. Given sQ, s15. . ., . - - -no-
si der the arithmetic means cr0, 04, . . ., . defined by
If -> 5- (— oo < s < + 00), then om -> s. This is clearly a special case of
(12.38); the matrix is positive.
It is useful to note that if the sn are the partial sums of a series
then
I
f
I
I
I
_ Sq + 4- ••• 4- sm ~ Uq 4- (uQ 4- Uj) + • • • 4-Q/q 4- 4- •'' ~
~ m+l m+l
1 m
Thus,
(12.39) <Tm = £ ( 1-------гт)мъ Sm “ = ---Уг X kUk-
k=0 \ m + 1/ m+l k=0
(b) The method of Abel. Given a series u0 + ur + • • • + u„ + - - -
consider the power series
/(r) = X unrn, 0 < r < 1,
n = 0
assuming that it converges for 0 < r < 1. If f(r) -> 5 as r -> 1, we say that
£ un is Abel summable (or А-summable) to sum s. The method can also be
applied to sequences since any sequence {.?„} can be written as a series
50 + GT “ 50) + (s2 - 5i) + • • •.
Let us now see the relation of Abel summability to the general scheme.
We claim that for 0 < r < 1, the formula
(12.40) f unrn = (1 - r) £ snT (s„ = u0 + • • • + u„),
n-0 n~0
232
A Few Facts from Harmonic Analysis
(12.41)
is valid assuming only that one of the two series which appear is convergent.
If the right side converges, it equals
oo oo 00 OO
E v" - X v"+1 = E v" - E
n = 0 n=0 n=0 n=l
= Jo + E = E unr”-
n= 1 n = 0
Conversely, if £“=0 converges for some r, 0 < r < 1, its Cauchy product
with the absolutely convergent series r” = (1 — r)""1 converges to sum
00 / n \ 00 oo
E ( E Ukrk-rn~k\ = e (“o + «1 + • • • + un)rn = E v”-
n~0 \fc = 0 / n~0 и = 0
This proves (12.40). Now, if {rm} is any sequence tending to 1, 0 < rm < 1,
then the positive numbers
^mn 0 ^m)^m
satisfy conditions (i), (ii), (iii) of (12.38). We leave the verification to the
reader.
(12.41) Theorem {Abel) If un converges to sum s, — oo < s < +oo,
then it is A-summable to s.
Proof Suppose first that s is finite. Applying (12.40), we have to show
that (1 — r) ^n=Q Vn as r -> 1. It is enough to prove that this relation
holds for any sequence r = rm, m = 0, 1,. . . , where 0 < rnm < 1, rm -> 1.
This is a corollary of (12.38) since the numbers = (1 ~ rm)r” satisfy
(i), (ii), (iii). The matrix a,nzj is positive, and so the proof holds for s — ± oo,
the only prerequisite being that the series £ unrn converges for 0 < r < 1.
We may also consider the power series
OO
/(z) = E
n = 0
where z is a complex variable lying in the unit disc: z = relX, 0 < r < 1.
If f{z) tends to a limit л* as z tends nontangentially to 1, i.e., as z -» 1 in such
a way that
y—< c <+oo (И < 1),
i - kl
then ^„°°=o un is said to be nontangentiallyjAbel summable to sum s. The last
inequality means that, in approaching 1, z remains between two chords of the
unit circle through z = 1.
(12.42) Theorem {Abel-Stolz) If £Г=о un converges to a finite sum s, then it
is nontangentially Abel summable to s.
(12.44)
jmmability of Sequences and Series
233
Proof, The proof is identical to that of Abel’s theorem, except that now
we use the formula £ unzn = (1 - z)£s„zn and consider any sequence
{zm} tending to 1 from the interior of the unit disc. The matrix amn is now
(1 — zm)z", conditions (ii) and (iii) are satisfied as before, and (i) takes the
form
H , r
I - kJ -
(12.43) Theorem IfYffv un is summable by the method of the arithmetic mean
to sum s, then it is A-summable to s. If in addition, s is finite, then
^«=0 un is nontangentially A-summable to s.
Proof Suppose that s is finite. By hypothesis,
+ h + ’ * ’ + sn
n + 1
Write 50 + + • • • + = /„. Applying formula (12.40) twice, we have
OO 00 00 OO
£ w„r" = (1 - r) £ snrn = (1 - r)2 £ tnrn = (1 - r)2 £ (n +
n=0 л=0 n = 0 и=0
Again, it is enough to consider any sequence rm 1, 0 < rm < 1. We then
have to apply (12.38) with matrix
= (1 - rm)2(n +1) r",
and we easily verify that (amn) satisfies conditions (i), (ii), (iii). The proof of
the rest of the theorem is left to the reader.
While convergence of a series implies summability A, the converse is
generally false: for example, X^°=o(“”O” diverges, but is Л-summable to
sum | since £^°=0 ("0” = VC1 + r) 2 as r If one makes addi-
tional assumptions on the terms of the series, however, the converse will hold.
The following result is both elementary and useful.
(12.44) Theorem {Tauber) If^ un is A-summable to sum s, — co < s < 4- co,
and if un = o(\/n) as n -> 00, then un converges to sum s.
Proof Write un = £„/«, n > 1, where en -> 0. Let rm be a sequence tending
to 1 which we shall determine in a moment. Then sm — f(rm) is a transforma-
tion of the sequence an:
m p 00 p co
= E 4 - E ^4 = E
n = 1 tL n^l a /1=1
where
= zO ~ if n - if n> m-
® n n
If we verify conditions (i) and (iii) of (12.38), then the fact that en -> 0 will
give sm — f(rm) -> 0, and so also sm -> s. Condition (iii) is obvious for any
{rm} 1. As for (i), observing that
1 - rn = (1 - r)(l + • • • + r""1) < (1 - r>,
we have
| 00
< ™(1 - r.) + „ + !
" m(1 - r-) + ™ + I 1 - r.'
Hence, if we choose rm = 1 — (I//??), then |amn| < 2. Thus, condition
(i) holds, and the theorem follows.
If^un is summable by the method of the arithmetic mean and nun -+ 0,
then £ un converges. Of course, this is a corollary of (12.43) and (12.44),
but a direct proof is on the surface: By (12.39),
1 m
~ am — I i /L
m + 1 k = 0
and the assumption kuk -> 0 clearly implies sm — am 0. Thus, if atn -+ s,
then also sm s. Actually, this argument shows that if kuk 0, then whether
{c^} converges or not, the difference sm — tends to 0, i.e., the behavior of
imitates that of {am}. The same argument shows that if {сгш} is a bounded
sequence, and |wn| < Afn for n = 1,2,..., then the sequence is bounded.
The result which follows lies deeper.
(12.45) Theorem (Hardy) If^ un is summable by the method of the arithmetic
mean to a finite sum s and if
A
|w„| < - («=1,2,.. .),
/Лея £ un converges to s.
Proof Consider the expressions (which we shall call the delayed arithmetic
means)
_ ‘S'n+l + ^ + 2 + -Z • + $n + k
°n’k ~ к
They are easily expressible in terms of the an:
L' J L' J
__ Go + • • • + $п+к) ~~ Go + • • *_+ sn) __ n + к + 1 n
~ к к an + k ~kan
_n, . к+1
j^\^n + k &n) + ~j^ @n + k'
It is clear that if kn is any sequence of integers such that n/k„ is bounded as
n ->• co, then o-„ -» s implies that а„Лп -> s. Using the definition of <r„ fc, we
also deduce that
_ <, _j_ (Jn+] ~ ‘ ' + (Sn + k — ^n)
°n,k = Sn+ ---------------£----------------
1 k
= Sn + k 1Лк ~J + iX + y
KJ=1
Hence, assuming as we may that A = 1,
k к
“ zL lW« + jl — w I 1 •
j=l П -f- 1
Let к = kn — [ея], where e > 0 is arbitrarily small and fixed, and [x] desig-
nates the integral part of x. Then njkn is bounded, and so onkn -» a1. But by
taking kn = [етз] in the last estimate, we obtain
limsup |алЛп - sj < e.
n-> OO
Hence, sn —► s, and the proof is complete.
We remark in passing that the conclusion of Hardy’s theorem is true if
the assumption of summability by the method of the arithmetic mean is
replaced by Л-summability (theorem of Littlewood; see A. Zygmund, Tri-
gonometric Series, vol. 1, 2nd ed., Cambridge University Press, Cambridge,
1968, p. 81).
6. Summability of S[/] and 5[/] by
the Method of the Arithmetic Mean
Given a periodic /, we denote by 5n(x) = sn(x^f) the partial sums of 5[/]
and by <Jn(x) = an(x,f) their arithmetic means. Thus (see p. 222),
$„(*) = " f /^„(х - t)dt = t f ftx + t)Dn(t) dt,
71 J -я
f f(t)Kn(x - t)dt = i f f(x + t)Kn(t) dt,
71 J -П 71 J -It
where
К"^ = n + 1 = (h + 1)2 siHr Д8Ш(7 +
Multiplying the last sum termwise by 2 sin and using the equation
2 sin (j 4- %)t sin У — cos jt — cos (j 4- l)f, we get
r(f\ l-cos(« + l)z 2 /sin [(и 4- l)r/2]\2
t } ~ + 1)(2 sin 1Z)2 n+1^ 2siniz y-
The trigonometric polynomial Kn(t) is called the nth Fejer kernel. An analo-
gous kernel was considered in Chapter 9, (9.11), for nonperiodic functions.
The formula
<L.(V) = ;;f - t)dt = if f(x — t)Kn(t)dt
is a periodic version of the notion of convolving a function and a kernel.
Some of the facts we will prove below are similar to ones we have already had
in Chapter 9, but rather than connecting the present case with those results,
we shall give brief direct proofs of the theorems we need.
Using the formula Dj(t) = | 4- = i cos/m, the Fejer kernel can be
written [see (12.39)]
1”/ m \ 1 n ( \m\ \
™ = i+Z cosm/ = i z
2 m = l \ П 4- I/ 2т=-л\ П -f- 1/
It has the following properties:
(a) Kn(t) > 0; ^(-0 = O).
(Ь) (1/я)^_яО)Л= i.
(c) Kn(t) < (n + l)/2; Kn(t) < A/(n 4- l)r2 (0 < t < я; A an absolute con-
stant).
Here, (a) and (b) are obvious from the various formulas above for Kn. The
first part of (c) follows from the formula Kn(t) = (n 4- I)-1 £"=0 Д/0) an<^
the obvious estimate \Dj\ < j 4- The second part follows from (12.46) if
we note that sin и > (2/л)и for 0 < и < |я.
From the second inequality (c), we immediately deduce
(c') f J ^«(C dt 0 as n -> oo for any fixed <5, 0 < <5 < я.
These properties of Kn lead to the following basic result, which is related
to (9.9) in Chapter 9.
(12.47) Theorem (Fejer) Let f be integrable and periodic. Then
<?„(*) -vw
at each point of continuity of /, and the convergence is uniform over
every closed interval of continuity. In particular, aH(x) tends to f(x)
uniformly everywhere if f is continuous everywhere. If f has a jump
L' J ° U J
Z4/
discontinuity at х0, then
«МЛо) -+ i[/(*o + ) +/(*<>-)]•
Proof. Suppose f is continuous on a closed interval I == [a,b] (which may
reduce to a point). Given e > 0, we can find <5 so that \f(x + t) — /(x)| < e
for x e I, |Z| < b. Using (b) above, we can write
- f(x) = j_ I/O + 0 - f(x)]Kn(t) dt
If If
~ _ I Jr | + ftp
say. Clearly, if x в I, then
If 8 f®
|a„| < - I eKn(t) dt < - \ Kn(t) dt = e.
71 J|t|<<5
Let M = max | f\ in I. Then, for x g Z,
|/?„|<;f (l/(x + ')l + M)Kn(t)dt
71 <.n
<;[max7C„(0]f [|/(x + i)l + M] dt.
The last integral is majorized by J* я \f(x + 01 dt + 2nM = f*J/(z)| dt +
2пM, and by (c), the factor preceding it tends to 0. Hence \fin\ -> 0 uniformly
for x g I, and |an| + \fin\ < 2e for n large enough and x g I. This proves the
first part of (12.47).
The proof of the second part is similar. We may assume that /(x0) =:
Wo+) + /(x0-)]. Then
<’„(*0) -/Uo) = ~ I [/(*o + 0 + f(x0 - t) -f(x0 + ) - f(x0-)]K„(t) dt,
71 J 0
1 г
K(*o) -/(Xo)l - l/(*O + 0 - f(x0 +)\K„(t) dt
71 J 0
+ [ l/(^o - t) ~ f(xQ-)\Kn(t) dt = a„ + bn.
71 J 0
To show, for example, that an 0, write Jg = -4- and use the fact that
in (0,<5) the difference |/(x0 + t) — /(x0 + )| is small, while in (^,я) we have
max Kn{t) tending to zero.
The following result, although it is simple, deserves a statement.
(12.48) Theorem
(a) Let f be periodic and integrable. If f(x) < В for all x, then also
<ти(%) < В. If f{x) > A, then crn(x) > A. If |/(x)| < M, then
|сг„(х)| < M.
(b) If f(x) -> + oo as x —> xQ, then crn(x0) ± oo as n oo.
We leave the proofs to the reader.
The following two results are corollaries of Fejer’s theorem.
(12.49) Theorem Let f be periodic and integrable, f ~ £ скегкх, and let F
be the indefinite integral of f. Then the series in the formula
F(x) - cox ~ Co + X'^,7tx
[see (12.14)] converges uniformly.
The series on the right is the Fourier series of a continuous function, and
since its terms are o(l/|&|) uniformly in x, the difference between its partial
sums and the arithmetic means of its partial sums tends uniformly to 0
[see the discussion following the proof of (12.44)].
(12.50) Theorem {Dirichlet-Jordan) If f is periodic and of bounded variation,
then
(a) 5[/] con verges to f(x) at each point of continuity of f, and to
![/(* + ) + al each point of discontinuity.
(b) The convergence of S[/] is uniform over every closed interval of
continuity of f.
(c) The partial sums of 5[/] are uniformly bounded.
Parts (a) and (b) follow immediately from Fejer’s theorem if one uses (12.45)
and the fact that the Fourier coefficients of a function of bounded variation
are <2(1/|A:|) [see (12.24)]. For (c), use (12.48) and the remark before (12.45).
Perhaps it is of interest to observe here that the classical theorem of
Weierstrass about the uniform approximability of functions which are con-
tinuous in finite closed intervals by power polynomials can be easily deduced
from Fejer’s theorem. For suppose that f{x) is continuous for a < x < b.
The formula x = %{a + b) + %{b — d)t establishes a mapping between the
intervals a < x < b and — 1 < t < 4-1, and every f(x) continuous in [tz,6]
becomes a g{t) continuous in [— 1,4- 1]. If we approximate g(t) by polynomials
in t, we at the same time approximate f(x) by polynomials in x. Hence, we
may assume from the start that f{x) is defined and continuous in [-1,4-1]-
Write x = cos 6. The function h(e) = f (cos 0) is then defined and continuous
in [0,я], and if we extend it to [—л,я] by the condition of evenness, and after
that to (— oo, 4- oo) by periodicity, then h{0) dan be approximated arbitrarily
closely and uniformly on [0,л] by cosine polynomials
T{ff) = £ аЛ cos кв,
e.g., the arithmetic means of S[A], these being cosine polynomials since h is
even. It is easy to see that cos k9 is a power polynomial of degree к in x =
cos 0: for к = 0, 1, this is obvious, and for general к it follows by induction
from the formula cos k0 + cos (k - 2)0 = 2 cos 0 cos (k - 1)0. Thus, the
polynomials T(0) above are power polynomials P (cos 0) in cos 0, and the
approximability of Л(0) by T(0) is the same thing as the approximability of
f(x) by P(x).
We shall now consider the arithmetic means of S[/] when f is merely
integrable. In Chapter 7, we introduced the notion of a Lebesgue point of a
function in R", but here we are only interested in the case n = 1. We recall
the definition. A point x0 is a Lebesgue point for a locally integrable f if
1 P
2AJ J/(X° + О-/Ы1Л-0 (A->0),
and we proved that almost all points have this property.
Simultaneously with S[/], we shall also consider £[/], for f merely
integrable. For a > 0, we write
fax) = - Ц f(x + Г)
2 tan
1 fV(x + t)-f(x- t) .
71 J e 2 tan
If limfi_0/e(x) exists, we will denote it /(x) and call it the function conjugate
to ft
f(x) = lim
e-*0
1 fV(* +
- -----———------dt
7i J e 2 tan у
lim
1Г /(x + t) Л
2 tan
о
We came across this function on p. 225 in connection with Dini’s criterion.
Occasionally, one also uses the notation
/(•x) =
-P.V.
Г /(x + Q
J 2 tan it
where P.V. stands for principal value, indicating that the integral, which as a
Lebesgue integral is generally divergent at t — 0, is given a new meaning by
first removing a symmetric neighborhood around t = 0 and then making
that neighborhood shrink to 0. Formally, / is the convolution of f and
| cot although the latter is not an integrable function. We will study the
existence of / later.
The arithmetic means of £[/] will be denoted by a„(x) = an(x,/). From
the formula on p. 222 for sn, we obtain
1 f*
5„(x) = - - fix + t)K„{t) dt,
л J
where
Of course, Do = 0.
(12.51) Theorem {Lebesgue) Suppose that f is periodic and integrable. Then
at every Lebesgue point x0 of f {in particular, for almost every xQ),
(i) t7„(x0) -» /(x0).
(ii) °fx0} - /1/п(х0) -> 0.
{n oo)
Proof We will use the estimates
(12.52) Kft) < n, Kn(t) < L {n > 1, 0 < t < n\
which are just variants of (c), p. 236. If we have to use both estimates, then
clearly the first is preferable for t < 1/n and the second for t > 1/n. The
proof that follows is basically a repetition of the argument for (9.13),
Chapter 9.
Let xQ be a Lebesgue point of/ (see p. 108). Assuming as we may that
/(x0) = 0, and letting '
ф(0 = |/(x0 + 01 + l/(*o - 01, ’/'(O = [ Ф(и) du,
J о
the condition that x0 is a Lebesgue point takes the form ^{hfjh -» 0 as
h 0. The formula for crn(x0) gives
k„(*o)l [ <X0-^»(0 dt = f + t f = a„ + /?„,
ftjo 71 J 0 71 J 1/n
say. Clearly,
p/n
an < | (j){t)n dt =
J о
•KW n
1/n
Next, using the second
estimate for Kn and integrating by parts, we have
Pn
2A
n
Ф(0 h
t3 dt'
The integrated term tends to 0 as n oo. As for the last term, we will show
that it also tends to 0. Given any s > 0, take <5 so small that < s if
0 < t < 8. Then
dt.
1 Г л 1 Г £t
— I л < — I Tbdt 4“
«Ji/n t n}l{ne
The first term on the right is majorized by (е/и) t ~2 dt = e, while the last
term clearly tends to 0. Collecting results, we conclude that o-„(xo) -> 0. This
proves (i).
To prove (ii), we need estimates for Kn — [\/(n 4- 1)] &j- The ob-
vious inequality |5y| < j shows that
(12.53) |O)I < n.
On the other hand, from the formula
~ 4 1 1
D/О = 2 cot 2*
cos (j 4- j)t
2 sin
(see (12.26)), we find that
O) ~ |cot|z =
sin (n 4- |)z
(n 4- 1)(2 sin |/)2’
which shows that
(12.54)
~ x 1 1
K„(Z) - 2 cot 2?
(I?| < я, n — 1, 2, . . .).
The estimates (12.53) and (12.54) are analogues of (12.52), and they easily
lead to (ii). We write
f(x0 +
Г1 1
^n(x0) -fi/niXo)
= -d ( f(x0 + t)Kn(t) dt + t f
71 J - я 11 J:
= -“ I f(xa +t)K„(t)dt+^_\ f(x0 + t) ^cot~t - Kn(t) dt,
J|tJ<l/n 71 J1 /n <. 111 <.n L2 2
2^2
and use the estimates (12.53) and (12.54) in the last two integrals, respectively.
An argument identical to that for an and pn in the preceding proof shows that
these integrals tend to 0. This completes the proof.
We remark that part (i) of Lebesgue’s theorem leads to a new proof of the
completeness of the trigonometric system [see (12.16)]. For if all the Fourier
coefficients of f are 0, then an(x,f) vanishes identically.
Part (ii) of Lebesgue’s theorem shows that lim <rn(x0) exists at every
Lebesgue point of f at which the conjugate function
7(*o) = limj£(*o)
£->0
exists, lhe converse is also true, though it requires an additional argument.
Let l/(n + 1) < e < 1/n. Then
1 f1/n 1 1
lJE(*o) - 7i/„(*o)l ;£ - l/(*o + t) -f(x0 - t)\-cot^tdt
7rJl/(n+l) Z Z
1 f1/n dt
(12.55) <- |/(Xo + Z)_/(Xo_-Z)|
^Jl/Cn+l) 1
n + 1 f1/"
< —F ’ l/(*o + 0 ~/(*o - 01 dt -> 0
л Jo
in view of the Lebesgue point condition. Hence, we obtain
(12.56) Theorem At every Lebesgue point x0 of an integrable f, the existence
of J(x0) is equivalent to the summability of S[/] by the method of the
arithmetic mean, andf(x^ = lim an(x0,/).
Suppose now that f is not only integrable but also in L2. If f ~ £ скегкх,
this means that £ |q|2 < + oo. Observing that
s[/l = E ck^eikx, ч = — i sign к
[see (12.12)], we see by the Riesz-Fisher theorem (8.30) that there is a function
д e L2 such that S[/] = 5[#] and
1 C2k
2tc Jo = £ lCfcE^2-
Therefore,
l9l2<EW2
1 f
=4
i.e., ||#||2 < ||/||2. Since an(x,f) = an(x,g) and lim an(x,g) exists and equals
g a.e. by (12.51),/exists and equals g a.e. by (12.56), and we have proved the
following result.
(12.57) Theorem If f is periodic and in L2, then the conjugate function
j(x) = - - —------—Vs------------ dt ~ — hm
л Jo 2tan|Z
exists almost everywhere and is in L2. Moreover, ||/||2 < ||/||2 (more
precisely, ||J||2 = ||/||2 if c0 = 0) andS[f] = 5[/].
The existence a.e. of Jis a remarkable result which shows that the odd part
of/,
= i[/(x + 0 -/(x - 0],
has special properties which are not immediate consequences of the theory
of integration. Observing that | cot — (1//) is bounded for 0 < t < n, we
deduce from (12.57) that if f e L2, then the integral
[V(* + 0-/(x-0 P
---------------------= I
Jo £~>0 J £
exists almost everywhere, a result which is not, obvious even for continuous /.
That J exists almost everywhere for f merely integrable will be proved
later [see (12.67)].
In the theorems which follow, we will use the notation
/ p \ 1/p
ll/ll, = ( |/(х)|р dx) , l<p<co; ll/ll „ = esssup |/(x)|
(although sometimes it may be convenient to modify the definition of ||/||p
by inserting a numerical factor ;e.g., by writing ||/||p = [(l/2n) l/l2’ с/х]1/р).
(12.58) Theorem If feL”, then
(i) KU, < ll/llp, 1 < P <
(ii) II/ - <b.llp -> 0, 1 < p < oo.
Proof. The theorem and its proof are repetitions of (9.1) and (9.6) of
Chapter 9. Ifp = oo, (i) isjust (12.48)(a). If 1 < p < co andp' is the exponent
conjugate to p, we have
k„(x)i <If /(/Жри-
я J -It
;Г l/(t)IX(.v -
J ~7t
tfKln/p'(x - t)dt
t) dt
i/рГ i
71
Kn(x — t) dt
i/pf
by Holder’s inequality, and
|<7„(Х)|₽<;Г \f(t)\pKn(x—t)dt,
an inequality which clearly also holds for p = 1. Integrating both sides over
— 7i < x < 7Г, and interchanging the order of integration on the right, we
obtain (i).
Part (ii) is proved similarly. We write
1 Г
k,(x)-/(x)l <- |/(X + t) — f(x)\Kn(t) dt
“ J -Я
< If +
J “Я
1Ф) - /(х)Г < 1 Г \f(x + t) — f(x)\p Kn(t) dt.
7^ J —7t
1/РГ1
%
Kn(t)dt
1/p'
Integrating both sides over — n < x < n and interchanging the order of
integration on the right, we obtain
1 f71
K-/||'<-
where
<^(0=Г l/(* + 0 - f(*)\P dx.
J — n
Clearly, ф is a bounded function, and we know that it tends to 0 with t.
Hence, by Fejer’s theorem (12.47),
if*
Я J -я
and (ii) follows.
We conclude this section by considering the maximal function
<T*(x) = = sup |ff„(x,/)|.
n^O
In view of (12.51), cr*(x) is finite almost everywhere. It has properties not
unlike those of the Hardy-Littlewood maximal function /* considered in
Chapters 7 and 9 and which are easily deducible from those of /*. (Using
similar symbols, cr* and/*, for different notions should not cause confusion.)
First, we consider an adaptation of the definition of /* to the case of periodic
functions. For periodic /, it is natural to set
1 p
(12.59) /*(x) = sup |/(x + 01 dt.
Q<h^K^n J _h
Clearly, /* is also periodic.
(12.60) Theorem Let f be periodic and integrable. Then
\\f*\\P < cp\\f\\p, 1<р<ъ,
|{x : 0 < X < 2n;f*(x) > a}| < ll/lli, a > 0.
These inequalities are analogues (actually, corollaries) of (9.16) and (7.9).
For let g(x) be defined as equal to f(x) in ( — я,Зя) and to 0 elsewhere. Then,
in (0,2я), the maximal function /* just defined is majorized by the Hardy-
Littlewood maximal function of g, and the norms of g in ( —oo, + oo) are
majorized by multiples of the corresponding fiorms of / in (0,2я).
The first part of the following result is an analogue of (9.17) of Chapter 9.
(12.61) Theorem Let f be periodic and integrable. Then there is an absolute
constant c such that
(i) a*(x,/) < c/*(x).
(ii) sup„ \a„(x,f) -Ji/n(x)| < cf*(x).
Proof. The proof can be based on either Fubini’s theorem [see the proof
of (9.17)] or on the formula for integration by parts. We choose the second
approach since it follows the same line as the proof of (12.51), but is actually
easier since we do not have to consider Lebesgue points. Using the notation
and proof of (12.51), we have
K(*,/)l £ 1 I* (l/(* + 01 + l/(* - t)\)Kn(t) dt
71 J 0
1 f1/” 1
“ z I z = an +
71 Jo 71 J 1/n
where
1 Г1/п А Г71 (Mt)
v-n <X0 dt, Pn <-~\ —2-dt
71 JO m J 1/n 1
and </>(0 = \f(x + 01 + \f(x ~ 01- Let j/(t) = Jo ф(и) du. The inequality
(i^(/)/20 < /*(x) shows that an < /*(x). If we integrate the integral majoriz-
ing fin by parts so as to introduce if/(f) and again use the inequality (il/(t)/2t) <
f*(x), we obtain < Л/*(х), and (i) follows. The proof of (ii) is left to the
reader.
The following result is a corollary of (12.61) and completes (12.57).
It will be useful later.
(12.62) Theorem If f is periodic and in L2, then the maximal conjugate
function
fax) = sup |Л(л)|
0 < E<,n
is also in L2 and
IIAll < All/ll2 (A independent of/).
We put the asterisk as a subscript here to avoid confusion with (/)*, the
Hardy-Littlewood maximal function of f which also appears in the proof
below.
Proof Let 1/(t7 + 1) < £ < 1/n. The inequalities in (12.55) give
1Л(О - Л/„(х)|
n + 1 r1/n
— J W + 01 + \f(x - 01} dt < f*(x).
0
Combining this with (12.61)(ii), we find successively (with different A’s at
different places) that
1Л001 < sup KCOI + Af*(x) = sup + ДР(х),
l7*(x)l < ^((7)*(x) + /*(%)) (by (12.61)(i)),
117*112 < А(.\\?\\г + IIZII2X
Il7*ll2 £ 4|/||2.
7. Summability of S[/] by Abel Means
Given a periodic and integrable /,
1 00
f ~ 5 a0 + £ (an cos nx + bn sin nx),
z n=l
let
j 00
f{r,x) = -<20 + £ cos nx 4- bn sin nx)rn, 0 < r < 1,
denote its Abel means. Since summability by arithmetic means implies Abel
summability, the results of the preceding section immediately lead to results
about Л-summability of S[/]. For example, we have the relation
(12.63)
f(r,x0) ->/(*o) (r -> 1)
at every Lebesgue point of /, and so almost everywhere. In particular, the
last relation holds at each point of continuity of /, and uniformly over every
closed interval of continuity.
However, an independent discussion of Abel summability has some
merits, if only for the following two reasons: (a) the relation (12.63) holds
at points which need not be Lebesgue points; (b) instead of (12.63), we may
consider the more general relation
/(r,x) -+f(x0)
as (r,x) tends to (l,x0) C-е., as reix tends to elXo) not only radially but also
along more general curves, e.g., nontangentially (see p. 232).
We compute /(r,x). Using the formulas for an and we have
Ж*) = тГ [ Я0 dt + £ r"
J -n Л= 1
1
COS nx -
It
f(t) cos nt dt
1 Г
+ sm nx- f(t) sin nt dt
1
п
"1 oo
ЯО ч + Xr”cos п(х ~ О
_z 1
dt
- j /(z)P(r, x — t) dt = - j f{x 4- t)P(r,t) dt,
where
1 00
P(r,t) = - + £ rn cos nt
n = l
is called the periodic Poisson kernel. The function f(r,x) is called the Poisson
integral of f. All the formal operations above (like the interchange of the
order of summation and integration) are easily justifiable.
We can write P(r,t) in a finite form by observing that | 4- j rn cos nt
is the real part of the series
__ 1 1 4- z
~ 21 - z
1 2
- + z 4- z2 4- • •
(z = relt, 0 < r < 1),
1 — r2
and a simple computation shows that
1 1 - r2 _ jl__________________________
2 1 — 2r cos t 4- r2 2 (1 — r)2 4- 4r sin2 *
This may be compared to the nonperiodic version of the Poisson kernel
discussed on p. 151. The Poisson kernel has all the properties of the Fejer
kernel, but is also much smoother. We list the properties:
(a) P(r,t) > 0; P(r, — f) = P(r,t).
(b) (l/K)^nP(r,t)dt== 1.
(c) P(r,t) < 1/(1 — r); P(r,t) < A(1 — r)/t2 (| < r < 1, |Z| < л, A an ab-
solute constant).
Properties (a) and (b) are obvious. The first part of (c) is a corollary of
ч 1 2 1 1 + r 1
P(r,t) < - 4- r 4- r2 4- • • • = T---< -----,
v 2 2 1 — r 1 — r
and the second part follows from
( л = 1 a - rX* + < 1 -r
' 2 (1 — r)2 4- 4r sin2 ~ 4r sin2 |Z
The estimates (c) are analogues of the corresponding estimates for the Fejer
kernel Kn(f) if we identify (1 — r) and l/лг. Thus, results for the Fejer means
have analogues for Abel means, and the proofs are basically the same. We
shall, however, not dwell on this point and shall limit ourselves to several
results of a somewhat different nature.
Given a periodic and integrable /, we shall systematically denote by F
its indefinite integral; F need not be periodic. Besides the ordinary derivative
of F at x,
+ A) - F(x)
F'(x) = hm —--------£------
h^O n
we shall also consider the symmetric derivative
F's(x) = lim
h->0 +
F(x + h) — F(x — h)
2h
Clearly, the existence of F'(x) implies that of F's(x) and Ffs(x) = Ff(x).
The converse is not true, however, as shown by the simple example F(x) =
|x| at x = 0. Using the notions of the even and odd parts of a function
introduced on p. 223, we see that F’s is the ordinary derivative at 0 of the odd
part of F at the point x (x is fixed, differentiation is with respect to h, at
h = 0). Also, since
r 1 (hf(x+t)+f(x-t)
F's(x) = hm т --------------5----------dt,
h^0+ n JO z
Ffx) is the ordinary derivative at 0 of the integral of the even part of f at the
point x.
(12.64) Theorem Let f be periodic and integrable, and let F be the integral
of f. Then
(i) at each point x0 where F's(xq) exists, finite or infinite, £[/] is Abel
summable to the value F'(x0) J
(ii) at each point x0 where F has an ordinary and finite derivative
F'(x0), the Poisson integral f(r,x) tends to F'(xq) as (r,x) tends to
(l,x0) nontangentially.
Proof, (i) Suppose, as we may, that x0 = 0. Write
Ф(1) = IL/XO + /(- OL <K0 = Г </•(«) du.
J 0
We have
2 2 C3
/(r,0) = - </>(t)P(r,t) dt = - </>(Z)P(r,t) dt + o(l)
tt J 0 71 J 0
J
for any fixed <5, 0 < <5 < я, in view of the second part of property (c) for
P(r,t). Integration by parts shows that the last integral equals
2 Га
-- $(t)P'(r,t) dt + о(1),
71 J О
where
гл/z ч d . (1 — r2)r sin t
P (r,t) = -j-P(r,t) = — 7;------------------7-5.
v 7 dt v 7 (1 — 2r cos t 4- r2)2
Since — P' > 0 in (0,л), if i/<(0A is contained between m and M in (0,<5), then
the last integral is contained between m and M multiplied by
— [ tP'(r,t) dt = - [ P(r,t) dt 4- o(l)
2 p
= - P(r,t) dt 4- o(l) = 1 4- o(l).
я J 0
Collecting results, we see that the limsup and liminf of J(r,0) as r -» 1 are
contained between m and M. This gives (i) when F'(0) is either finite or
infinite.
(ii) Assume again that x0 = 0. Since the result is obvious when f is
constant, we may also assume that F'(0) = 0. Suppose that (r,x) -► (1,0)
nontangentially, that is, r -> 1 and x -» 0 in such a way that
(12.65)
1*1
1 — r
< C.
Given 8 > 0, take d so small that \F(ti)/u\ < 8 for |m| < 2<5. Write
1 p 1 P
f(r,x} = - f(x 4- t)P(r,t) dt = - f(x 4- t)P(r,t) dt 4- o(l)
71 J-n
1 c5
= — I F(x 4- /)P'(r,0 dt 4- o(l),
л J -<5
using integration by parts and property (c) of P. If |x| < d then |x 4- /| < 2<5
in the last integral, and the integral itself is majorized in absolute value by
p p
(12.66) £ (\x\ + \t\)\Pr(r,t)\dt = 28\x\\ \P'(r,t)\dt
J -5 Jo
p
4- 2s t\P\r,t)\ dt.
J 0
Since
r<5 j
IP'POI dt = P(r,0) - P(r,a) < P(r,0) < -----------
Jo 1 — r
and
p Г<5 Г<5
Z|P'(r,Z)| dt = — tP'(r,f) dt = — [zP(r,Z)]o 4- P(r,t)dt
Jo Jo Jo
< | P(r,t) dt = |я,
J о
condition (12.65) implies that the right side of (12.66) is less than a fixed
multiple of s. Hence, f(r,x) tends to 0 under the hypothesis (12.65). This
completes the proof of (ii).
8. Existence of /
In this section, we prove the following basic result.
(12.67) Theorem If f is periodic and integrable, then the conjugate function
J(x) = lim Je(x) = lim| —- f + О- <уД
v £->o £->o I ti JE<|d^u 2 tan У J
exists almost everywhere, and is in weak L1: for a > 0
|{x : |x| < n, |/(x)| > a}| <
where c is independent of f and a.
Remark: If f is integrable,/ need not be, as the following example shows.
Let f be any periodic integrable function, nonnegative in (0,|я) and zero
elsewhere in ( — n,n). Then for — < x < 0,
r 1Г dt -1P" ft dt
~ n J 2 tan |(x — t) л J 0 Z 2 tan |(x — t)
1 [‘"‘я dt 1 Г|х| dt
~ ti J о 2 tan |(x — t) ti J 0 2 tan %(t — xf
1 Г'х' dt 1 C|xl
“ я J 0 2 tan ±(t + |x|) - я2 tan |x| J 0 dt’
Now choosing /(z) = (Hog2 tf"1 (= (/Wz)[log (1/z)]”1) for 0 < t < | and
/(z) = 0 for | < Z < ti, we obtain fix) > c[|x| log (l/|x|)]-1 near x — 0.
Clearly, f e L but / ф L.
The lemma that follows is essential for the proof of (12.67).
(12.68) Decomposition Lemma Let Q be a finite interval in R1 and suppose
that f g L(6), f > 0. Then for any a satisfying
(l“”
there is a sequence of nonoverlapping intervals Ql9 Q2, . . . contained in
Q such that
/ < 2a (A; = 1, 2, . . .);
iVfcl JQk
(ii) almost everywhere in P = Q — (J Qk, we have f(x) < a;
(iii) IlMl /•
a Ju<2k a JQ
Proof. We split Q in half, obtaining 2 subintervals Q of equal length.
For each Q', there are only two possibilities: either I2'|-1 JQ'/ < a or
I21-1 / > a. Since |2'| = i\Q\, the hypothesis (12.69) implies that
|2'|-1 Jq' f 2a. Thus, for each Q'9 we have either
12'Г1 f or a< le'R1 f f<2a.
JQ' JQ'
If Q satisfies the first condition, we call it an interval of the first kind—
otherwise, of the second kind.
We save any Q' of the second kind. If Q is of the first kind, we may repeat
the previous argument by splitting Qr into 2 equal parts Q". For each Q",
we again have either
I2T1 f /< a or a < I2T1 f f< 2a.
Jq" Jq"
Saving those of the second kind, we repeat the procedure for each g" of the
first kind, and so on.
Let Qlf Q2, . . . , Qk, • • • be the sequence of all the intervals of the second
kind in the procedure above. Clearly, the Qk are nonoverlapping and satisfy
condition (i). Also, each x g P = Q — (J Qk belongs to a sequence of inter-
vals {Q} with |g| tending to 0 such that \Q\~l f < a. Since the ratio
12Г1 Jo f tends to f(x) almost everywhere in P, (ii) follows. Finally, writing
the first inequality (i) in the form a|£?J < J2fc /, and summing over k, we
deduce (iii).
Remarks:
(1) Lemma (12.68) holds for periodic functions of period 2я considered on
the circumference of the unit circle, and a > (2я)“! f. The proof is
identical with that above.
(2) Lemma (12.68) is valid for Q = R1, f g L^R1), and any a > 0. More-
over, it has an analogue in Rn, n > 1, where Q and the Qk are taken to be
«-dimensional cubes with edges parallel to the coordinate axes. The proofs
are left to the reader.
Proof of (12.67). Assume first that the periodic function f eL is > 0.
Fix any a > (2л)”1 f and apply remark (1) above. We then obtain a
sequence of nonoverlapping arcs Qx, Q2, ... on the circumference Q of the
unit circle such that
(12.70) a<i7n[ f - 2a> f - a ae-in p = Q ~ U Qk-
ILfkl jQk
Make a decomposition
f = 9 +
where g is defined as equal to f in P and as I2fe|-1 f2k f on each arc Qk.
Hence, h equals 0 in P and f — IftJ"1 $Qkf on each Qk. Using (12.70),
we have
(a) 0 < g < a a.e. in P, 0 < g < 2a in each Qk;
(12.71) (b) h = 0 in P, \Qkh = Q-,
(с) |Л| <f+ \вкГ^вк/</ + 2a in each Qk.
In particular, since g is bounded a.e., and so is in L2, g(x) exists a.e. and
II0II2 < № by (12.57).
The study of h requires the theorem (6.17) of Marcinkiewicz. That
theorem was stated for functions defined on intervals of R1, while here we
need it for functions defined on the circumference of the unit circle. Clearly,
these two situations are identical. We restate the theorem.
(12.72) Theorem Let F be a closed subset of the unit circumference Q, and
let G — Q — F. Let d{x) be the {circular) distance of the point x e Q
from F. Then, for each A > 0, the integral
, r <>'<>> . Г i'O-) .
dy “ .1 “y
is finite almost everywhere in F. Moreover,
j Mk(x)dx < 22~1|G|.
We shall need the result only in case A — 1.
We shall now study the existence of h = lim hE, using properties (12.7 l)(b),
(c) of h. Let Q* denote the interior of the arc Qk expanded concentrically
twice, and let tk denote the center of Qk and dk its length. Let g* = |J Q*
and let P* be the complement of g* in Q ; P* is closed. We will consider only
points x e P*. ,
Fix x e P*, and consider the equation
hfx) = - j A(0cot^(x - t) dt.
Since h = 0 outside |J Qk, the last integral is a sum
(i) of integrals extended over those Qk which are totally outside (x — s,
x 4- s),
(ii) of at most two integrals extended over portions of those Qk which contain
the points x + s.
We shall investigate these two cases separately, beginning with (ii). The
interval Qk which, say, contains x 4- s is distant from x by < s and at the
same time by > idk (since x 6 P*). Hence, %dk < s, and the integral under
consideration is majorized in absolute value by
। fx + c + dfc j px + 3c । C3c
i-----h Л < — I \h(t)\ dt < — Ih(x 4- 01 dt.
7Cjx + e |x - И K8 Jx + e Зб J 0 71
A similar estimate is valid for the interval Qk containing x — s, and using
definition (12.59), we see that the contribution of (ii) is majorized by 2й*(х).
(We have just presupposed implicitly that s < f тс, but since we are primarily
interested in small e, this is an unimportant restriction.)
We also notice that h = 0 in P* so that by Lebesgue’s theorem on the
differentiability of integrals, the contribution of (ii) tends to 0 with e at
almost every x e P*.
Consider now any integral from (i). Since h — 0 [see (12.71)(b)], it can
be written in the form
if
71 jQk 2L
h(t)^ cot-(x — t) — cot^(x - h) dt
(tk is the midpoint of Qk). Let ^к(х) denote the last integral with the integrand
replaced by its absolute value. Denoting absolute constants by A, we have
-If 1ЛГН11 lsM(/-4)l ,,
7Г Jet |ЯИЯ 2 |sin - t) sin i(x - ^1
(12.73)
< Adk I |Л(Г)| ;-----------------------r
к 7I|X- r||x- rt|
Adk
~ (.X- tk)2
\h(t)[dt,
tak
where to arrive at the last term, we have used the fact that |/ — /J < ^dk and
|x — tk\ > dk (recall that x ф Q* since x e P*), so that
(12.74)
1 |x - t\ 3
2~ \x - tk\~ 2
(x e P*).
Using the first inequality in (12.7l)(c) and (12.70), we also have
(12.75)
•Ф
Г |A| <2f /<4а|6к|.
'<2k jQk
If S(f) denotes the distance from t to P*, then <5(0 > 1*4 f°r *e Qk- Collecting
results and using (12.73)—(12.75), we get the final estimates
yfe(x) < -—- 4a|gk| < A<x Г —
(x~4)2 k(x-O2
Z Лй < Яа £ I" , ^,<2 dt
for x g P* and the intervals Qk which are entirely contained in the com-
plement of (x — s, x + e).
The last sum is majorized by
a quantity which is finite almost everywhere in P* by (12.72) with 2=1.
Hence, in view of our observation that the integrals in (ii) above tend to 0
almost everywhere in P*, h = lim he exists almost everywhere in P* and
(12.76) |h(x)| < Act [ z ^^ <2 dt (a.e. in P*).
J<2 Vх ~ И
Since / = g 4- /?, and g exists almost everywhere, / exists almost every-
where in P*, i.e., everywhere in Q with the exception of a set of measure
at most
2 Г
(12.77) ie*i<2£iai<- z
Taking a arbitrarily large, we see that / exists almost everywhere.
The assumption f > 0 can be dropped by considering the decomposition
f=f+ - Г-
We still have to prove the weak integrability of /. This can be deduced
from the estimates above, and we shall be brief. We may again assume that
f > 0. Fix any а > (2л)-1 /, and consider the decomposition f =
g + h corresponding to this a. Then
{x : |/| > a} c {x : |£| > |a} и {x : |/i| > |a] = 5 и T,
say. Recall that 0 < g < 2а; we also have JQ g = JQ f since fQ Л = 0 by
(12.71)(b). Hence, by Tchebyshev’s inequality,
4 Г 2 8 Г _8f
~''<xJ<2d aJe
To estimate |T|, let I\ = Tn P*, T2 = Tn g*.Thus, |T| = |TJ + |T2|,
and by (12.77),
2 f
|T2| < |g*| <- /.
a Je
On T1; |/j(x)| is majorized a.e. by [see (12.76)]
a“ L <Нп?L (rn?л
say, and if |h(x)| is to be > |a there, then necessarily M(x) > 1/(2A). But
M(x) is the integral Мл(х) of (12.72) corresponding to 2 = 1, G = g*,
F — P*. Therefore, by the estimate given in (12.72) and (12.77),
Г 4 Г
M < 212*1 < - \ f
Jp* a JQ
and by Tchebyshev’s inequality, the subset of P* where M(x) > 1/(2A)
has measure at most 2A JP* M < 8Ta-1 f- Thus, ITJ < 8Лос-1 f, and
collecting estimates we have
|{x > a}| <^11/11! (c = 10 + 8Л).
This was proved for a > (2л)-1 f but is trivially true for smaller a since
our set certainly has measure < 2л, while 2л < a-1 / for such a.
This completes the proof of Theorem (12.67). The theorem is strengthened
by the following result.
(12.78) Theorem If f e L, then the maximal conjugate function
A(*) = sup |X(x)|
0 <&<3n
is in weak if, i.e.,
|{x:A(x)> a}| <^|/||, (a > 0),
where c is independent of f and a.
Proof The proof is practically the same as that of (12.67). Leaving the
details for the reader to fill, we argue as follows. We fix a > (2л)-1 /,
and make the previous decomposition f = д + h, so that /*<$* + h*.
By (12.62), H#*||2 < and the estimates we had for h also hold for h*.
The restriction s < |л that was imposed in the argument is unimportant since
if e > then |/7j»| < c f2 f for all x, so that supc>;t/3 |йс(х)| < c \Q f.
Therefore, the set where sup^K/3 |/?£(^)l > « is empty if a > c$Qf, while
if a < c f, its measure is < 2тс < 2ш-1 JQ f.
9. Properties of f for f e Lp, 1 < p < oo
(12.79) Theorem If f g Lp,\ < p < co, thenf e Lp and S[f] — 5[/]. Moreover,
(12.80) (a) ||/||p < Лр||/||р and (b) ||Д||Р < Лр||/||р.
The constant Ap depends only on p and is bounded for p away from 1
and co.
This is the central theorem of the section. Its proof is long and has to be
split into several parts. Of course, (b) implies (a). The theorem is false for
p = 1 [see the remark after the statement of (12.67)], and (12.67) is a sub-
stitute for this case. See also Exercise 21. The theorem is also false for/? = oo:
see Exercises 19 and 20.
(12.81) Lemma If feLp, 1 < p < 2, thenf*eLp and(Y2.W) holds.
Proof For p — 2, this is Theorem (12.62). For p = 1, J* is in weak L1 by
theorem (12.67). The lemma will be obtained from these extreme cases by an
interpolation argument similar to the one we used in the proof of the Hardy-
Littlewood maximal theorem (9.16).
Let f g Lp, 1 < p <2. For each fixed a > 0, we make the decomposition
f = g + h with
#(*) = f(x) wherever |/| < a, g(x) = 0 otherwise,
h(x) — /(*) wherever |/| > a, h(x) — 0 otherwise.
Clearly, g g L2, h g Ll,f* < g* + h*. Let w(a) = \{f* > a}| be the distribu-
tion function of/*. We have
a>(f) < \{g* > + \{h* > Ml-
By (12.67),
|{h+ > Ml < (M"1 f \h\ = Aa~l f |/|,
and by Tchebyshev’s inequality and (12.62),
\{g* > Ml < (M~2 f gi
JQ
< Aa~2 f g2 = Ла~2 f f2.
We have [see (5.51) and Exercise 16 of Chapter 5]
\\f*\\pP
ap dco(a) = p \ ap 1 co(d) da.
0 Jo
Using the estimates above, the last integral is majorized by
Ap (* a^fa"1 | |/| dx) Ja 4- Ap | ap-1(a~2 f f2 dx) Ja.
JO \ J{|/|>a} J Jo \ J{|r|^a} J
Interchanging the order of integration, we can write this as
Ap\ 1/0)1 ( 2 Ja ) dx 4- Ap\ f\x)[ ap 3 da] dx
Je \Jo J Jq \Ji/(x)| /
= [ 1Я*)1РЛг: + [ l/O)lp Jxk
LP ~~ 1 JQ z P Jq J
due to the fact that 1 < p < 2. It follows that (12.80)(b), and so also (12.80(a),
holds with
(12.82) App = Apl-±- + y/-l (1 < p < 2).
LP 1 z P)
It is not surprising that Ap becomes infinite as p -> 1 since as we have
observed, the integrability of f does not imply that of /. On the other hand,
in view of the validity of (12.80)(b) for p = 2, it is natural to expect that
there is a better estimate for Ap near p — 2 than (12.82) (which becomes
infinite as p -> 2). This we shall see below.
Meanwhile, note that the exponent 2 plays a rather accidental role in
Lemma (12.81). In fact, if we knew (12.80)(b) for any p0 > 1, then the
argument above would give it for 1 < p < p0, with Ap bounded for p away
from p = 1 and p = pQ. The only change necessary in the proof is replacing
the exponent 2 in the argument for g by p0. Moreover, if instead of (12.80)(b)
we only knew (12.80)(a) for some p0, then by applying the same argument to
J instead of Д, we would obtain (12.80)(a) for 1 < p < p0, with Ap bounded
for p away from 1 and pQ. We will use this idea below.
Inequality (12.80)(b) for any p implies
(12.83) ИЛИ, < Лр||/||р
for the same p and all s > 0. From the fact that ||7e||p equals sup^ |?ед| for
all g with ||^||p, < 1 ((1/p) 4- (1/p') = 1), and the easily verifiable formula
f Лз = - [
JQ Jq
(apply Fubini’s theorem), it follows that if (12.83) holds for any p, 1 < p <
oo, then it also holds for the conjugate exponent pf, and Apf = Ap (see also
Exercise 16, Chapter 10). But we proved (12.83) for 1 < p < 2. Hence, it
also holds for 2 < p < co, and an observation analogous to the one made
above (this time for /£ rather than f) shows that the Ap in (12.83) remains
bounded for p away from 1 and oo.
I
Using (12.82), we thus see that the Ap in (12.83) satisfies the inequality
(12.84) Ap < (1 < P < 2),
and so also (since Ap = Ap,)
(12.85) Ap < Ap for p > 2.
Since/£(x) ->/(x) as e -> 0 for any integrable /, (12.83) leads to the basic j
inequality ||7llP < Лр||/||р, 1 < p < oo, where Ap satisfies the two estimates I
above. This proves (12.80)(a); we still have to prove (12.80)(b) for 2 < p < oo ,
and the formula S[f] = S[/] for f e Lp, p > 1.
Let us show that if f e Lp, p > I, then S[/] = Sf/]. We may assume that
p < 2. In view of (12.56), lim <j„(x) exists and equals J(x) almost everywhere
for f merely integrable. Next, by (12.61)(ii), |a„(x)| is majorized by f*(x) + i
cf*(x), an integrable function in our case since f e Lp, 1 < p < 2. Hence,
by the dominated convergence theorem, each Fourier coefficient of an tends i
to the corresponding Fourier coefficient of J. But it also tends to the cor- |
responding coefficient of <S[/]. Thus, S[/] = 5[/].
To complete the proof of (12.79), it remains to show that (12.80)(b)
holds for 1 < p < oo (we have shown it only for 1 < p < 2), with Ap
bounded for p away from 1 and oo. It is immediate [see the proof of (12.62)]
that if 1/(и + 1) < £ < l/и, then |J£(x) — 7i/„(x)l is majorized by /*(x).
Hence, by (12.61)(ii), ?
7*(*) < cf*(x) + sup |(f„(x)|. I
n
By (12.61)(i) applied to S[/] = 5(7],
sup Itf/x,/)! = a*(xj) < c(J)*.
n
Hence,
7* <•(/* + (7)*), j
IIAllp £ c(||/*||p + 11(7)*IIp) ;
< ucp(||/||p + ||J||P), S
where cp is the constant of the Hardy-Littlewood maximal theorem. In view
of (12.80)(a), the inequality ||Д||Р < 2?p||/||p is thus established for all p,
1 < p < oo, with Bp = + Ap], Ap being the constant in (12.80)(a).
Combining the estimates (12.84) and (12.85) for Ap with the estimate on j
p. 156 for cp, it follows that Bp is bounded for p away from 1 and oo. In J
particular, since cp remains bounded as p —> oo, it follows that Bp is 0(p) as
p -> oo. To estimate Bp as p -> 1, it is best to use the estimate derived in the j
fnpjjHuanui» о-к
proof of (12.81) [see (12.82)]; thus, Bp = O(l/(p — 1)) as p -» 1, so that Bp
satisfies the same sorts of estimates as Ap. This completes the proof of (12.79).
We have the following important corollary.
(12.86) Corollary Let f e Lp, 1 < p < oo. Then fE converges in LP norm to f.
This is an immediate consequence of the dominated convergence theorem
since fE converges pointwise almost everywhere to/and |/j < /* e Lp.
10. Application of Conjugate Functions to Partial Sums of S[/]
The behavior of the partial sums sfx) = sn(x,f) is a much more delicate
topic than the behavior of the arithmetic means. We will consider only the
question of the convergence of sn to f in the metric LP. The main tool here is a
connection between the partial sums and conjugate functions.
Instead of sn, it will be convenient to consider the expressions [see (12.27)]
(12.87) s„ (x) =----------------= sn(x) — -(an cos nx + bn sm nx).
We have
5„*(x) = - P ° + Dn~l(X °
7l ’
I
2
sinn(x - t)
•^ 2 tan i(x — t)
I P f(t) cos nt
= sm nx-- ——г;---------г dt -
я J 2 tan |(x — Z)
dt
I fK f(t) sin nt
cos nx • - -L----------- dt.
тс J 2 tan |(x — t)
The last decomposition was purely formal, but from Section 8, we know that
the cofactors of sin nx and cos nx exist almost everywhere in the principal
value sense (and represent the conjugate functions of / cos nt and / sin nt,
respectively).
(12.88) Theorem (M. Riesz) If f e LP, I < p < oo, then
0) INIp < c\\f\\p, \\f-sn\\p->0,
(ii) НУр < c||/llP, ИУ-Ур^О,
where c depends only on p.
Proof. It is enough to prove (i), which implies (ii) since sn(x,f) = sn(x,f)
and IlfH, < Ap\\f\\p, I < p < oo. The first part of (i) with s„ instead of sn is
an immediate corollary of the last formula for s„ and the inequality ||/||p <
Лр||/||р: letting gn = f cos nt, hn — f sin nt, we have
^*(x) = sin nxgfx) — cos nxhfx) (a.e.),
K*llp < Ш!р + \\h„wp < лр(|Ш1р + UMp) < ЛрИЛр
»»
260
A hew i-acts rrom harmonic analyst
I 4,00;
for 1 < p < co. Since, by (12.87), ||s* - 5„||p < A||/|| t < Л||/11Р, we obtain
IKIlp < Ap\\f\\p.
The relation \\f — s„||p -> 0 is obvious for functions which have a con-
tinuous derivative [see (12.20)], and since such functions are dense in Lp,
also follows in the general case.
Exercises
1. Prove the following versions of theorems (12.13) and (12.14).
(a) If/ - ia0 + Sk°= i (flk cos kx + bk sin kx), and / is the indefinite integral
of /', then /' — i k(bk cos kx — ak sin kx).
(b) If / — ia0 + 2?= i (ak cos kx 4- bk sin kx), and F is the indefinite integral
of /, then
1 1 00 1
F(x) — aox - 5 Aq 4- S T (ak sin kx — bk cos kx).
z z fc= 1K
2. Prove that if f(x) — £ ckeikx, then f(x + a) — S ckeikaeikx.
3. If/is real or complex-valued and/ — + Ел°= i &k cos kx + bk sin kx), show
that
П2“ и2 = 5 l«ol2 + S (№ + №)•
ftjo z k=l
4. Deduce from (12.18) that if fg eL2,f~ 2 ckeikx, g - S dkeikx, then
4 - t) dt =2 ckdkeik*.
Thus, a Fourier coefficient of the convolution of /and g equals the product of
the corresponding coefficients of / and g.
5. Prove the following.
(a) If /is periodic and equal to sign x in (—n,n), then
r, x 4 1 . . sin 3x . sin 5x . 1
/(x)~-|sinx+-3— + — + •••)•
(b) Let 0 < h < %n, and let /be the “triangular” function defined as follows:
/ is periodic, even, continuous, /(0) = 1, f(x) — 0 for 2/z < x < n, f is
linear in (0,2Л). Then
(c) Let g be periodic and equal to | log [1/|2 sin |x|] in (—n,n). Then
cos kx
9 ~ fc?l к J'
[For (b), the coefficients can be computed directly, or by using Exercise 4
and example (b), p. 215. Observe that the convolution of two characteristic
functions of intervals is a “triangular” function. For (c), one may either
integrate by parts in the formula for the cosine coefficients of g, or consider
the real part of the series
00 yfc 1
^r10gT^’ z = re<x’
for r < 1, and then let r —> 1.]
6. Using the formula for the Fourier series of 4(tc — x) given in Example (a),
p. 215, prove the formulas
v -LX v X —X
n^in2 6’ ^«4 90’
00 1
S ~2k — n2kBk (k = 1, 2,..., Bk rational).
n= i n
7. Prove that each of the systems
(a) J, cos x, cos 2x,. .., cos kx,...
(b) sin x, sin 2x,. . ., sin kx,. ..
is orthogonal and complete over (0,л).
8. Let {^(x)} and {^(y)} be two orthogonal systems, the first over a set A c Rm,
the second over a set В c R”. Then the (double) system
"лк(х>У) = ^(х)^к(у)
is orthogonal over the Cartesian product С = A X В c Rm+". If both
and {ipk} are complete, so is {coJtk}.
Generalize this to the case of more orthogonal systems.
9. Let (ki,ki,... ,kn) be all lattice points in the space Rn (i.e., all distinct points
with integral coordinates). Then the system
> exp i(k±X! + k2x2 + • • • + knxn)
is orthogonal and complete over any cube in R" with edge 2zr.
10. If / ~ S ckeikx e L2 and g ~ £ dkeikx e L2, then h = fg is integrable, and
if h ~ 2 Cneinx, then
+ 00
G = ckdn—k,
k = - oo
where the series on the right converges absolutely. [For n = 0, this means
(1/2я) ffifg = S Ckd_ki which is a variant of (12.18).] See also Exercises 17
and 18.
11. Let f ~ 2 ckeikx gL2. For each n, let
v 1 __ v 1
Уп — 2л С к n 2j cn-k 7, •
кФп П — К *
Show that {/„} g I2 and, more precisely, that £ |yn|2 < тг22 кл|2- The numbers
yn are discrete (and formal) analogues of the integral f [/(z)/(x — /)] dt. (The
numbers yn are the Fourier coefficients of the function fg, where g ~
elkx/k = i(n — x)forO < x < 2n [see Example (a), p. 215], and we have
С2тс р2тс
[ \f9\2 <^2 n 1/p.)
Jo Jo
12. Let/and g be periodic,/eLp, geLp', 1 < p < oo, 1/p 4- 1/p' = 1. Consider
the product hx(f) = f(x + t)g(t) as a function of t. Show that the «th Fourier
coefficient of hx(t) tends to 0 as n 00, uniformly in x. [Show that the L1-
modulus of continuity of h tends to 0 uniformly in x; apply (12.22).]
13. Let f be periodic and integrable. Show that for the partial sum sn we have
the formula
(a) sn(x) = (1/zr) flnf(x + r)[(sin nt)/t] dt + £n(x), where cn{x) tends to 0 uni-
formly in x as n —* 00. Likewise,
(b) sn(x) = (l/я) Jl„/(x + /)[(! — cos nt)/t] dt + where gn(x) tends to 0
uniformly in x. [For (a), except for an error which is <?(1) uniformly in x,
we have
= f(.X + t)
Л J -Л
sin nt .
2 tan it ’
and since J cot it — 1/t is bounded in (—я,я), the result follows easily
from Exercise 12 with p = 1, p' = 00.)
14. Let Ln = 2л:"1 Jo |5„(/)| dt, Dn denoting the conjugate Dirichlet kernel.
Prove the following analogue of (12.36):
r 2.
Ln - - log n.
15. Show that if x is a point of continuity of an integrable / then sn(x) = o(log ri).
If f has a jump discontinuity at x and the value of the jump is d = /(%+)
— f(x—), show that sn(x) ex —{d/n) log/?. [The second statement follows from
the first by considering (if, for example, x = 0) the function /?(/) = i(n — t)
of Example (a), p. 215, whose conjugate function h satisfies h = — g, where g
is the function of Exercise 5(c).]
16. Prove the following generalization of (12.38). Suppose that liminf sn = s
and limsup sn = s are finite. Then both liminf an and limsup an are contained
between the numbers
i(s + J) ± A i(s — s),
where A is the same as in condition (i) of (12.38).
17. Prove the following extension of Exercise 10. If f ~ fckeikxELp, g ~
S dkeikx gLp', 1 < p < 00, 1/p + 1/p' = 1 (thus also 1 < p' < 00), then the
Fourier coefficients Cn of the (integrable) function h = fg are given by
00 _ м
Cn = X] ckdn~k = lim
k=-oo М-* + ооЛ=— M
Thus, the Cn are the same as in Exercise 10, but the series representing the Cn
are no longer claimed to be absolutely convergent, and we must consider the
limits of their symmetric partial sums. [The proof is parallel to that of Exercise
10. We write
1 f2zt 1 Г2я 1 г2я
як +
where = sM(x,f), observe that the first integral on the right is majorized by
Jo71 \ f — apply Holder’s inequality, and use the fact that ||/ — *0
by (12.88).]
18. The result of Exercise 17 is valid when p = 1, p' = oo (or when p = oo,;/ = 1),
but the series defining the Cn must be taken in the sense of the (symmetric)
first arithmetic means:
Cn = lim 2 Cki_JLL/„_k.
M-> + oofc=-M \ ~r 1/
19. We know that theorem (12.79) is false for p = 1. Show that it is also false for
p = oo, i.e., that the conjugate function of a bounded function need not be
bounded. [Consider, for example, the two series Sk°=i (sin kx)jk ~ Цл — x),
0 < x < 2л, and S?=i (cos kx)/k ~ | log (1/|2 sin %x\), — n < x <n (see
Exercise 5).]
20. There is a substitute result for theorem (12.79) in casep = oo. Let/be a periodic
function with |/| < 1. Then there are absolute constants 2,// > 0 such that
J eAlfl < p.
[Write
еЛ|/| _ A|/| - 1 = g М/
' n = 2 •
and integrate termwise. Use (12.80)(a), (12.85) and the fact that nn < c"/z!]
21. Suppose that /is a periodic function for which
J-„ 1^1,og+ 71 < +°°’
where log+ stands for the positive part of log. This clearly implies that /6 L1.
Show that feLr and that there are absolute constants A and B such that
Г 171 < а Г |/| log+ |/| dx + B.
J —n *> — it
[Assume as we may that |/| > 2, say, and write
w(a) = |{x : |x| < л, |/(x)| > a}|. Then
(•It ~ foo Г2 foo
Lj/|=Jo ^Mo+L-
For the first integral of the right, use the fact that w(a) < 2л, and for the second,
use an argument like that in the proof of (12.81.)
Notation
X + у addition [«,/>]
xy dot product {a,b)
|x| absolute value
x+ positive part [a,*)J
x~ negative part
/'(*+) limit from the right
/(*-) limit from the left К
—► converges to
converges in measure 0
to sup
increases to decreases to inf
{**} sequence limsup
set of x satisfying .. .
x e E x an element of E liminf
хфЕ x not an element of E ess sup
Ei u E21 union ess inf
im j L, L1)
к Lp
Etc\E2\ cm j к intersection L™ lp
Ei -E2 difference, relative Rl
complement Rn
Ei c E2 subset
CE complement C
E closure
О E interior Uf
diam E diameter $Efdfi
6(E) diameter
3(x) distance function
|E| measure
|£|e outermeasure II-II
closed interval
open interval
partly open intervals
countable intersection
of open sets
countable union of
closed sets
empty set
supremum, least upper
bound
infimum, greatest lower
bound
limit superior
limit inferior
essential supremum
essential infimum
classes of functions
classes of sequences
real number system
//-dimensional
Euclidean space
Complex number
system
Lebesgue integral in Rn
Lebesgue integral in a
measure space
Riemann-Stieltjes
integral
norm
265
П-Bp I/norm /*g convolution Xe characteristic function a.e. almost everywhere (5,S,/z) measure space о(ф(хУ) | order of magnitude <2(^(x))J relations И(Е) _ 1 K(jE’), F(E)? variations V[f',a,b} ’ S[/] Fourier series 5[/J conjugate Fourier series use upper semicontinuous Isc lower semicontinuous Rr Riemann-Stieltjes sum 5r sum of increments fr moduli of continuity <Чг,е(я) distribution function /* Hardy-Littlewood maximal function НЛ(А) Hausdorff measure Л(Л), Л*(Л) Lebesgue-Stieltjes measure, outer measure v(I) volume
Index
A
Abel’s theorem (on summability), 232
Abel-Stolz theorem, 232
Abel summability (A-summability),
231, 246
nontangential, 232, 248
Absolutely continuous function of a
real variable, 115, 116
Absolutely continuous measure, 180,
182
Absolutely continuous set function,
99, 174, 180
Absolute value, 3
Abstract measure and integration,
161
Additive set function, 98, 162
Algebra of sets, 205
cr-Algebra of sets, 39, 161
Almost everywhere (a.e.), 52, 167
Approximation of the identity, 123,
148
Arithmetic means, 231, 235
delayed, 234
Axiom of choice (Zermelo), 46
В
Banach space, 131
Basis for L2, 137
Besicovitch covering lemma, 185
Bessel’s inequality, 138, 212, 218
Bolzano-Weierstrass theorem, 13
Borel o-algebra, 40, 196
Borel measurable function, 50
Borel measure, 187
Borel set, 40, 196
Boundary of a set, 5
Bounded convergence theorem, 76,
173
Bounded function, 9, 126
Bounded linear functional, 182
Bounded overlaps, 185
Bounded set, 5
Bounded variation, function of, 15,
113, 117, 221, 238
C
Cantor-Lebesgue function, 35, 116
Cantor set, 35
Caratheodory-Hahn extension
theorem, 206
Caratheodory outer measure, 196
Caratheodory’s theorem, 43
Cauchy-Schwarz inequality, 3
Cauchy sequence, 4, 60, 132
Change of variable, 124
Characteristic function, 54
Choice, axiom of, 46
Closed interval, 7
Closed set, 5
267
Closure of a set, 5
Compact set, 8
Compact support, 9
Complement
relative, 2
of a set, 2
Complete (metric space, normed
space), 4, 131, 132
Complete measure space, 190
Complete orthogonal system, 137,
139, 212, 218
Complex-valued measurable function,
61, 125
Conditional expectation, 192
Conjugate Dirichlet kernel, 223
Conjugate exponents, 127
Conjugate function, 225, 239, 242,
245, 250
Conjugate series, 217
Continuity
absolute, 99, 115, 116, 174, 180
in Lp, 134
at a point, 10
relative to a set, 10
uniform, 11
Continuous function (of a real
variable), 10
Continuous set function, 99
Convergence
of Fourier series, 139, 222, 259
in measure, 59, 190
in norm, 131
of a sequence of points, 3
of a sequence of sets, 165
weak, 144
Convergence theorem
Bounded, 76, 173
Lebesgue dominated, 71, 76, 173
Monotone, 66, 75, 172
Uniform, 75, 172
Converse of Holder’s inequality, 128,
191
Convex function, 118
Convolution, 93, 145
Convolution operator, 146
Countably additive family of sets, 162
Cover of a set, 8
in the Vitali sense, 109
Cube in R", 7
Curve, 21, 22
Cylinder set, 65
D
6-function, 191
Decomposition
of a function of bounded variation,
117
Hahn, 176
Jordan, 165
Lebesgue, 178
lemma for functions, 250
of a measure, 181
of a set function, 165, 178
Decreasing sequence of sets, 2
De Morgan’s laws, 2
Dense, 4, 149
Density, point of, 107
Derivates, 111
Derivative
of a convolution, 146
of a function of bounded variation,
ИЗ
of the indefinite integral, 100
of measures, 185
of a monotone function, 111
Diameter of a set, 5, 99
Difference of two sets, 2
Differentiability of the indefinite
integral, 100
Differentiability of a monotone
function, 111
Differentiation of measures, relative,
185
Dini number, 111
Dini’s test, 224
Dirichlet function, 16, 84
Dirichlet-Jordan theorem, 238
Dirichlet kernel, 222
Dirichlet problem, 152
Discontinuity of the first or second
kind, 10, 19
Discrete measure space, 162
Dispersion, point of, 107
Distance
in a Banach space, 131
between two points, 3
between two sets, 5, 196
from a point to a set, 95
Distribution function, 77
Divergence of Fourier series, 227
Dominated convergence theorem, 71,
76, 173
Dot product, 2
Dual space of a normed linear space,
182
Dyadic cubes, 8
E
Egorov’s theorem, 57, 167
Empty set, 2
Equiconvergent (series), 226
in the wider sense, 226
Equidistributed (equimeasurable)
functions, 79
Essentially bounded, 126
ess inf (essential infimum), 143
ess sup (essential supremum), 125
Euclidean space, 1
Even function, 214
Even part of a function, 223
Extension of a measure on an
algebra, 206
Exterior measure, 33
F
6
Fatou’s lemma, 70, 75, 173
Fejer kernel, 152, 236
Fejer’s theorem, 236
Finite (-valued) function, 9
Fourier coefficient, 137, 212, 217
Fourier integral, 217
Fourier series, 137, 212
Fourier transform, 97
Fubini’s lemma, 113 y
Fubini’s theorem, 87
Function
absolutely continuous, 115, 116
bounded, 9, 126
of bounded variation, 15, 113, 117,
' 221, 238
Cantor-Lebesgue, 35, 116
characteristic, 54
conjugate, 225, 239, 250
continuous, 10
convex, 118
integrable, 72, 170
in Z?, 82, 125, 173
Lipschitz, 17, 95, 115
maximal, 104, 155, 244
measurable, 50, 167
monotone, 9, 111
semicontinuous, 55, 197
set, 98, 162
simple, 54
singular, 116
step, 23
uniformly continuous, 11
Functional, bounded linear, 182
G
G.,6
Gauss-Weierstrass kernel, 152
Gram-Schmidt process, 136
Graph
of a curve, 21
of a function, 64
Greatest lower bound, 4
H
Hahn decomposition, 176
Hardy-Littlewood maximal function,
104, 155, 244
Hardy’s inequality, 192
Hardy’s theorem, 234
Harmonic analysis, 214
Harmonic function, 151
Harmonic oscillation, 214
Hausdorff dimension, 209
Hausdorff measure, 202
Hausdorff outer measure, 202
Heine-Borel theorem, 8
Hilbert space, 141
Holder’s inequality, 128, 131, 173
converse of, 128, 191
I
Image, 11
Improper integral, 84
Increasing sequence of sets, 2
Indefinite integral, 98
Indicator function, 54
inf (infimum), 4
Infinite dimensional Hilbert space,
141
Inner measure, 48
Inner product, 2, 135, 141
Integrable function, 72, 170
Integral
in an abstract space, 167, 170
of a complex-valued function, 76,
135
improper, 84
Lebesgue, 64, 72
Lebesgue-Stieltjes, 200
principal value, 239
Riemann, 12, 85
Riemann-Stieltjes, 23, 76
Integration by parts, 26, 118
Interior point, 5
Interior of a set, 5
Intersection of sets, 2
Interval in R1, R", 7
open, 7
partly open, 7
Inverse image, 51
Isolated point, 3
Isometric, 140
Isometry, 140
J
Jensen’s inequality, 119, 192
Jensen’s integral inequality, 122
Jordan decomposition, 165
Jordan’s theorem, 19
Jump discontinuity, 10
К
Kernel, 146
Dirichlet, 222
conjugate, 223
Fejer, 152, 236
Gauss-Weierstrass, 152
Poisson, 151, 247
L
L, L\ 72, 170
L2, 135
L00, 126
Lp classes, 82, 125, 173
lp classes, 130
Least upper bound, 4
Lebesgue constant, 228
Lebesgue decomposition, 178, 181
Lebesgue differentiation theorem, 100
Lebesgue dominated convergence
theorem, 71, 76, 173
Lebesgue integrable function, 72, 170
Lebesgue integral, 64, 72, 167, 170
Lebesgue measurable function, 50
Lebesgue measurable set, 37
Lebesgue measure, 37
Lebesgue outer measure, 33
Lebesgue point, 108, 153, 239
Lebesgue-Stieltjes integral, 200
Lebesgue-Stieltjes measure and outer
measure, 199
Lebesgue theorem on summability,
240
Length
of a curve, 21, 22
of a vector, 3
liminf
of a function, 10
of a sequence of real numbers, 4
of a sequence of sets, 2
Limit point of a sequence or a set, 3
Limit from the right or left, 10
limsup
of a function, 10
of a sequence of real numbers, 4
of a sequence of sets, 2
Linear functional, 182
Linear isometry, 140
Linearly independent functions, 136
Lipschitz condition, 16, 115
of order a, 221
Lipschitz transformation, 44
Locally integrable function, 107
Lower Riemann-Stieltjes sum, 27
Lower semicontinuous, 55, 197
Lower variation of a set function, 164
Lusin’s theorem, 58
M
Marcinkiewicz integral, 95, 157, 252
Marcinkiewicz’s theorem, 95, 157,
252
max (maximum), 4
Maximal function, Hardy-Littlewood,
104, 155, 244
Mean value theorem for integrals, 28
Measurable function, 50, 167
complex-valued, 61, 125
Measurable set, 37, 162, 193
Measure
abstract, 162, 193
Borel, 187
[Measure]
Hausdorff, 202
inner, 48 <
Lebesgue, 37
Lebesgue-Stieltjes, 199
outer, 33, 193
Measure on an algebra, 205
Measure on a cr-algebra, 162
Measure of a set, 37, 193
Measure space, 162
Measure zero, 167
Metric, 132
Metric outer measure, 196
Metric space, 2, 132
min (minimum), 4
Minkowski’s inequality
for integrals, 129, 173
for series, 131
Minkowski’s integral inequality, 143
Modulus of continuity, 14, 220
in L\ 221
Monotone convergence theorem, 66,
75, 172
Monotone decreasing or increasing, 9
Mutually singular measures, 180
N
n-dimensional cube, 7
n-dimensional Euclidean space, 1
n-dimensional interval, 7
Negative part of x, 18
Negative variation of a function, 18
Nonmeasurable set, 46
Nonoverlapping intervals, 7, 8
Nontangential Abel summability,
232, 248
Norm in a Banach space, 131
Norm of a bounded linear functional,
182
Norm ini/, 132
Norm of a partition, 12, 16
Normed linear space, 131
Null set, 191
о
R
о(ф(х))9 150
О(^(х)), 150
Odd function, 214
Odd part of a function, 223
Open ball, 5
Open cover, 8
Open interval, 7
Open set, 5
Order of a trigonometric polynomial,
216
Origin, 1
Orthogonal set of functions, 135, 211,
213
Orthonormal set of functions, 135, 211
Outer measure
Carathdodory, 196
general, 193
Hausdorff, 202
Lebesgue, 33
Lebesgue-Stieltjes, 199
metric, 196
regular, 199
P
Parallelepiped, 7
Parallelogram law, 144
Parseval’s formula, 138, 212, 219
Partition, 15
Partly open interval, 6
Perfect set, 6
Point of density, 107
Point of dispersion, 107
Point of the Lebesgue set, 108
Poisson integral, 151, 247
Poisson kernel, 151, 247
Positive part of x, 18
Positive summability matrix, 231
Positive variation of a function, 18
Principal value integral, 239
Property 57
Property ^,88
th modulus of continuity, 221
Radon-Nikodym theorem, 180
Real-valued function, 9
Rectifiable curve, 21
Refinement of a partition, 17
Region under /, 64
Regular Borel measure, 187, 208
Regular discontinuity, 224
Regular outer measure, 199
Regular shrinking of sets, 108
Relation
of convergence in measure and
pointwise convergence, 59, 60
of Riemann and Lebesgue integrals,
83
of Riemann-Stieltjes and Lebesgue
integrals, 76
of Riemann-Stieltjes and Lebesgue-
Stieltjes integrals, 201
of two definitions of the Riemann-
Stieltjes integral, 29, 30
Relative complement of a set, 2
Relatively closed (open) set, 6
Removable discontinuity, 10
Riemann integral, 12, 85
Riemann-Lebesgue lemma, 144, 220
Riemann-Stieltjes integral, 23
Riemann-Stieltjes sums, 23, 27
(F.) Riesz-Fischer theorem, 139
(M.) Riesz’s theorem, 259
S
cr-algebra, 39, 161
smallest containing 40
tr-finite measure, 178
on an algebra, 205
E-measurable, 162
E-measurable function, 167
Schur’s lemma, '160
Schwarz’s inequality, 3, 128, 141
Semicontinuous function, 55, 197
Separable, 4, 132, 174
Sequence of points, 3
Sequence of sets, 2, 165
Set
Borel, 40, 196
Cantor, 35
closed, 5
compact, 8
measurable, 37, 162, 193
nonmeasurable, 46
open, 5
perfect, 6
of type As, Aa, Gs, Ga, 6
Set function, 98, 162
absolutely continuous, 99, 174
decomposition of, 165, 178
singular, 174
variations of a, 164
Shrink regularly, 108
Sign x, 129
Simple function, 54
Simple Vitali lemma, 102
Singular function of a real variable,
116
Singular measure, 180
Singular set function, 174
Smallest сг-algebra containing <%, 40
Space
Banach, 131
Hilbert, 141
Lp, 125, 173
measure, 162
metric, 2, 132
normed linear, 131
(Span,137
Step function, 23
Strictly monotone function, 9
Subalgebra, 209
Subcover, 8
Sublinear operator, 160
Summability
Abel, 231, 246
by arithmetic means, 231, 235
nontangential Abel, 232, 248
of a sequence, 231
of a series, 231
Support of a function, 9
Supporting line, 122
sup (supremum), 4
Symmetric derivative, 248
Symmetric difference, 190
T
Tauber’s theorem, 233
Tchebyshev’s inequality, 68, 82
Tietze extension theorem, 14
Tonelli’s theorem, 92
Total variation of a set function, 164
Transformation, 10
Lipschitz, 44
Translation invariant, 48
Triadic (ternary) expansion, 47
Triangle inequality, 3
Trigonometric Fourier series, 211
Trigonometric polynomial, 216
Trigonometric series, 216
Trigonometric system, 213
Type A6, Aa, Fa, G6, 6
. U
Uniform convergence theorem, 75,
172
Uniformly absolutely continuous,
124, 192
Uniformly bounded, 9
Uniformly continuous, 11
Union, 2
Upper Riemann-Stieltjes sum, 27
Upper semicontinuous, 55, 197
Upper variation of a set function, 164
Urysohn’s lemma, 192
V
Variation, bounded, 15
Variations
of a function, 15, 18
of a set function, 164
Vector, 1
274
Vitali covering Iemma, 109
simple form of, 102
Vitali’s theorem, 46
Volume of an interval
(parallelepiped), 7
W
Weak convergence, 144
Weak £(R"), 105, 250
Weierstrass approximation theorem,
238
Index
Y
Young’s convolution theorem, 146
Young’s inequality, 127
Z
Zermelo’s axiom, 46