AN. ·INTRODUCTION TO LEBESGUE INTEGRATION AND FOURIER SERIES by
HOWARD WILCOX Professor of Mathematics Wellesley College and DAVID L. MYERS Professor of Mathematics Wellesley College
ROBERT E. KRIEGER PUBLISHING COMPANY HUNTINGTON. NEW YORK 1178
Original edition 1978 Printed and Published by ROBERT E. KRIEGER PUBLISHING COMPANY, INC. 645 NEW YORK AVENUE HUNTINGTON, NEW YORK 11743 Copyright o 1978 by ROBERT E. KRIEGER PUBLISHING COMPANY, INC.
All rights reserved. No reproduction in any form of this book (except for brief quotation in critical articles or reviews), may be made without written authorization from the publisher.
Printed in the United States of America
Library of Congre11 Cataloging In Publication Data Wilcox, Howard J. An introduction to Lebesgue integration and Fourier series. Bibliography: P. 1. Integrals, Generalized. 2. Fourier series. I. Myers, David L., joint author. II. Title. 77-12013 515'.43 QA312.W52 ISBN Q-88275-614-1
Conten ts
Chapter 1
The Riemann Integral
1 Defmition of the Riemann Integral 2 Properties of the Riemann Integral 3 Examples
4 Drawbacks of the Riemann Integral 5 Exercises Chapter 2 Measurable Sets
6 7 8 9
Introduction Outer Measure Measurable Sets Exercises
Chapter 3 Properties of Measurable Sets 1 0 Countable Additivity 1 1 Summary *12 Borel Sets and the Cantor Set *13 Necessary and Sufficient Conditions for a Set to be Measurable 14 Lebesgue Measure for Bounded Sets *15 Lebesgue Measure for Unbounded Sets 16 Exercises
1
1 6 7 8 9 13
13 17 21 26 29
29 32 33 36 38 40 41 iii
iv CONTE NTS
Chapter 4 Measurable Functions
17 18 19 20
Definition of Measurable Functions Preservation of Measurability for Functions Simple Functions Exercises
Chapter 5 The Lebesgue Integral
21 22 23 24 25 26
The Lebesgue Integral for Bounded Measurable Functions Simple Functions Integrability of Bounded Measurable Functions Elementary Properties of the Integral for Bounded Functions The Lebesgue Integral for Unbounded Functions Exercises
Chapter 6 Convergence and The Lebesgue Integral
47
47 50 53 55 59
59 60 62 65 68 73 77
27 28 *29 *30
77 Examples Convergence Theorems 78 A Necessary and Sufficient Condition for Riemann Integrability 82 Egorofrs and Lusin's Theorems and an Alternative Proof of the Lebesgue Dominated Convergence Theorem 85 31 Exercises 88
Chapter 7 Function Spaces and t2
32 Linear Spaces 33 The Spacel2 34 Exercises
93
93 101 108
Chapter 8 The 1.2 Theory of Fourier Series
35 36 37 38
Definition and Examples Elementary Properties !.2 Convergence of Fourier Series Exercises
113 118 120 1 27
CONTENTS
Chapter 9 Pointwise Convergence of Fourier Series
39 An Application: Vibrating Strings 40 Some Bad Examples and Good Theorems
41 More Convergence Theorems 42 Exercises
v
133
1 33 137 140 144 147
Appendix Logic and Sets Open and Closed Sets Bounded Sets of Real Numbers Countable and Uncountable Sets (and discussion of the Axiom of Choice) Real Functions Real Sequences Sequences of Functions
147 148 148 149 151 153 153
Bibliography
155
Index
157
Prefa ce
This book arose out of our desire to present an introduction to Lebesgue Integration and Fourier Series in the second semester of our real variables course at Wellesley College. We found that most undergraduate texts do not cover these topics, or do so only in a cursory way. Graduate texts, we felt, lack motivation and depend on a level of sophistication not attained by most undergraduates. We feel this text could be used for a course lasting one semester or less (there are several optional sections, marked with an asterix, which could easily be omitted). We assume knowledge of advanced calculus, in cluding the notions of compactness, continuity, uniform convergence aad Riemann integration (i.e., a usual one-semester undergraduate course in advanced calculus). Therefore, the book would be suitable for advanced undergraduates and beginning graduate students. It is our intention throughout the book to motivate what we are doing. Goals of the theory are kept before the reader, and each step of the development is justified by reference to them. For example, the in adequacies of the Riemann integral are pointed out, and each new accom plishment of the Lebesgue theory is measured against the goal of over coming these difficulties. In addition, each new concept is related to concepts already in the student's repertoire, whenever this is possible. The Lebesgue integral is defmed in terms precisely analogous to the Riemann sum definition of the Riemann integral. The primary difference is that in Lebesgue's approach, "Lebesgue sums" are formed relative to an arbitrary
PR EF ACE
partition of an interval containing the range of a bounded function, in contrast to Riemann's partitioning of the domain. The formation of "Lebesgue sums" leads naturally to the goal of defining the measure of an arbitrary set. Outer measure is defined in the classical way, and is shown to lack countable additivity on the collection of all subsets of [0, 1] . This leads to the restriction of attention to measurable sets, and hence to measurable functions. The theory is pursued through the usual convergence results, which overcome some of the de ficiencies of the Riemann theory. This is followed by a discussion of linear spaces and £2 in particular. This leads in a natural way to the £2 theory of Fourier series. Finally, pointwise convergence of Fourier series is discussed.
CHAPTER
1
The Riemann Integ ral
1.
Definition of the Riemann Integral The problem of fmding the area of a plane region bounded by vertical lines x and x the horizontal line y 0, and the graph of the non-negative function y f(x), is a very old one (although, of course, it has not always been stated in this terminology). The Greeks had a method which they applied successfully to simple cases such as f(x) = x2 • This "method of exhaustion" consisted essentially in approx· imating the area by figures whose areas were known already-such as rectangles and triangles. Then an appropriate limit was taken to obtain the result.
= b,
=a
=
=
In the seventeenth century, Newton and Leibnitz independently found an easy method for solving the problem. The area is given by where is an antiderivative of f. This is the familiar Funda mental Theorem of Calculus; it reduced the problem of fmding areas to that of finding antiderivatives.
F(b) - F(a),
F
Eventually mathematicians began to worry about functions not haYing antiderivatives. When that happened, they were forced to return again to the basic problem of area. At the same time, it became clear that a more precise formulation of the problem was necessary. Exactly what is area, anyway? Or, more generally, how can J:f(x)dx be defmed rigorously for as wide a class of functions as possible?
2
LEBES G U E I NTEGR AT I O N A N D FOUR I ER S ER I ES
In the middle of the nineteenth century, Cauchy and Riemann put the theory of integration on a finn footing. They described-at least theo retically-how to carry out� program of the Greeks for any function{. The result is the definition of what is now called the Riemann integral of f. This is the integral studied in standard calculus courses.
[a,b ) is a finite sequence (X o.X�> . . . ,xn) such that a = Xo < x 1 < . . . < Xn = b . The norm (or width, or mesh) of P, denoted IIPII , is defmed by IIPII = max (x,-x,_ 1 ). l
1 .1 Definition: A partition P of a closed interval
That is, IIPII is the length of the longest of the subintervals
[x2 ,x3] ,
·
·
·
,
[Xn - l.Xn]
[x0,xd,
·
(x0, ... .Xn) be a partition of [a,b] , and let f be [ a,b] . For each i = 1 , . . . ,n, let x1 * be an arbitrary point [x1_ 1 ,x;] . Then any sum of the fonn n R(f,P) = I f(x ,*)(x1-x,_ 1 ) I= I
1 .2 Definition : Let P = defined on in the interval
is called a Riemann sum off relative to P. Notice that R (/,P) is not completely determined by f and P; it de pends also on the choice of the elements For a non-negative function f, R (f,P) is the sum of the areas of rectangles approximating the area under the graph of {(see diagram).
x1*.
x,•
"•
THE R E I MA N N I NTEG RA L
3
Now we take a limit of our approximating areas.
1 .3 Definition : A function f is Riemann integrable on [a,b] if there is a real number R such that for any e > 0, there exists a 8 >0 such that for any partition P of [a,b] satisfying IIPII < 8, and for any Riemann sumR{f,P) of {relative toP, we have IR(fP , )-R I< e. We can rewrite this in logical shorthand as: f is integrable on [a,b] if there is a number R such that (Ve>0)(38 > O)(VP)[IIPII < 8 -+ IR{f,P)- R I< e ), where we must remember the meanings of P and R(fP , ). What is this number R? First of all, there can be at most one number R which satisfies the condition. For, suppose thatR andR' both worked; then we could take e = (1/2) IR - R' 1 . As the reader can verify (Exercise 5.2), this would entail the existence of a Riemann sum R(fP , ) satisfying both IR{f,P) - R I < e and IR(f.P) - R' I < e. An application of the triangle inequality yields a contradiction. Since at most one number R can satisfy our defmition, and it is evidently the "limit" of the Riemann sums, we define it, if it exists, to be the Riemann integral off on [a,b] , denoted J:f(x)dx. (For { non-negative, we also define this number to be the area we have described above.) The fact that the defmition of integrability is a kind of limit will be emphasized if we write down the usual defmition of a functionH having a limit as x approaches 0:
there is a real numberb such that (Ve>0}(38>0)(Vx+ O)[lxl < 8 -+ IH(x)-bl< e). where we understand that x is restricted to the domain of H. Since we write lim H(x) =bin this case, we might write x.. o
lim R(f,P) =t f(x)dx
IIIPit+O
a
for the integral. But we must keep in mind that this is only an abbrevia tion of the (necessarily) involved e -8 definition� The difference between the two types of limits, arising mainly from the fact that R(f,P) is not a function simply ofiiPII , whereasH is a function ofx, should also be noted.
4 LEBES G U E I NTE G R A T I O N A N D F O U R I E R S E R I ES
The defmition of the Riemann integral given above has a drawback which it shares with all limit defmitions-to prove that a particular function is integrable, you must first know the value of R = J:f(x)dx. The defini tion itself gives no direct means of finding R. If we restrict ourselves for the moment to non-negative functions, more information about the area R can be obtained by considering rectangles lying entirely below the graph of {, and rectangles whose tops lie above the graph of f. Then it would seem reasonable that the area R should lie between the area of the inner rectangles and the area of the outer rectangles. In fact, it turns out that f is integrable if and only if inner and outer rectangles can be '· --·----..
--I
I I I I I I
.- -- ---------1 1 I I I I I I I I
----------- ----- ---------;
I I I I I I I
l
found whose total areas are arbitrarily close to each other. This is the content of Theorem 1 .6 below. Note that this theorem holds for any function (not necessarily non-negative). Before we state the theorem, we need to introduce some useful nota tion. Note that the tops of the outer rectangles defme a function, except for some ambiguity at the shared boundaries between two rectangles (and similarly for the inner rectangles). Furthermore, such a function has a particularly nice form. 1 .4 Definition : A function g, defined on [a,b 1, is a step function if there is a partition P = ,x,.) such that g is constant on each open sub interval ,g(x,.) are for i = 1 , ... ,n. (The values of irrelevant.)
(x0,x 1 , (x1_ 1 ,x1),
•
•
•
g(x0),
•
•
•
Notice that a step function on [a,b 1 has finitely many values. Fur thermore, a step function is Riemann integrable, and the value of its in tegral is the obvious area given in the following proposition.
TH E R E I M A N N I NT E G R A L
5
Any step function g on [a,b 1 is Riemann integrable. Fur thermore, if g(x) = c1 for x E (x1_1, x1) , where (x0, , x ) is a partition of [a,b 1 , then
1 .5 Proposition:
•
·b
•
•
n
n
I g(x)dx = 1I= 1 cl.,x1- x1_1 ) tJ
(Notice that the values of g at x0,
•
•
•
,Xn
have no effect on the integral.)
See Exercise 5.5. Now we are ready for the theorem. Proof:
D
function f, defmed on [a,b 1 , is Riemann integrable on [a,b 1 if and only if for every e > 0, there are step functionsfl and /2 such that for all x E [a,b] , and
1 .6 Theorem: A
Proof: See Exercise 5 1 1 Note that iff is Riemann integrable, then {is bounded (Exercise 5 .9). D Since the condition in the theorem is equivalent to integrability, we will use it and the definition interchangeably. We also have the following expressions for J: J(x)dx. .
1 .7 Corollary:
.
Iff is Riemann integrable on [a,b 1 , then
6
LEBESG U E I NTEGRAT I O N A N D F O UR I ER SER I ES
2. Properti• of the Riemann I ntegral
We state here for future reference some of the fundamental prop erties of the Riemann integral. Proofs of these can be found in any standard text on advanced calculus or analysis. Iff and g are Riemann integrable on [a,b] , so are cf (for any real number c), and/ + g. Furthennore,
2.1 Theorem (Linearity) : 'b
and
b
fa cf(x)dx =c fc f(x)dx,
If a < c < b , then /is integrable on only iff is integrable on both [a,c] and [c,b] . Furthennore,
2.2 Theorem (Additivity):
[a,b]
if and
If f(x) < g(x) for all x E [a,b] , and iff and g are Riemann integrable on [a,b] , then
2.3 Theorem (Monotonicity) :
If there are real constants m and M such that m < f(x) <M for all x E [a,b] , then
2.4 Corollary :
m (b - a) <
L f(x)dx <M(b - a). b
Any function which is continuous on grable on [a,b] .
2.5 Theorem :
2.6 Theorem :
on [a,b].
[a,b]
is Riemann in
Any function which is monotone on [a,b] is Riemann integrable
T H E R I EMA N N I NTE G R A L
7
Using Theorem 2.2 and Exercise 5 .5, we can extend .the last two existence theorems to bounded piecewise continuous or bounded piecewise monotone functions. Finally, the Fundamental Theorem of Calculus. Let U be an open interval containing [a,b] . If /is continuous on U and F is an antiderivative of f on U, then
2.7 Theorem :
ef(x)dx = F(b) - F(a).
3. Examples
We have stated in the last section that any piecewise continuous or piecewise monotone function is Riemann integrable. We will now present two important examples of functions which are bounded but are neither piecewise continuous nor piecewise monotone. One will tum out to be Riemann integrable, and the other will not. With any set of real numbers A , we associate a function called the characteristic function ofA, defmed by if x E A x..t (x) = 0 if xEA .
3.1 Definition :
x..t.
l1
Let Q be the set of rational numbers. Then XQ is bounded, but is neither piecewise continuous nor piecewise monotone, and it is not Riemann integrable on [0.1 ] . In fact, if /1 and /2 are step functions such that fJ < XQ < /2 , then there is an irrational x in each subinterval on which /1 is constant, so that for that x,
3.2 Example:
Therefore, except at fmitely many points (the points of the partition for we have fl(x) < 0, so that f01 !1 (x)dx < 0 by Corollary 2.4 and Exercise 5.5. On the other hand, there is a rational number x in each interval on which h is constant, so that h(x) > xa(x) = Hence f�h(x)dx >
/1 ),
1.
1,
8
LEBESG U E I NT EGRATI O N A N D F OUR I ER SER I ES
1,1!1
and J� fl(x)dx (x)dx ;> 1 cannot be made arbitrarily small , as required for integra«tility, by Theorem -
1.6.
It is one of the advantages of the Lebesgue theory, to be introduced soon, that XQ will be Legesgue integrable. We will fmd its Lebesgue in tegral on [0,1) to equal 0.
pfq, qpfq p q {1/q =p/q
3.3 Example: Now we present a function which is neither piecewise monotone nor piecewise continuous, but which surprisingly enough is Riemann in tegrable. Let us agree for the purposes of this example to write non-zero rational numbers only in the form where ¢ 0 and and are in tegers having no factors in common (that is, is written in "lowest terms''). Then define ifx
g(x) =
1
ifx =0
0
ifx is irrational.
Then g is continuous at each irrational and discontinuous at each rational (Exercise 5.23). Thus g is not quite as discontinuous as XQ• which is dis continuous at every point (Exercise 5.22). As part of our study of the Lebesgue integral, we will see just how discontinuous a Riemann integrable function can be (see Theorem 29.2). It is not too difficult to prove at this point that g is Riemann integrable on (0,1) (Exercise 5.24). 4. D rawbacks of the Riemann I ntegral
The Riemann integral is adequate for most practical applications. The functions we usually encounter are piecewise continuous and very often have nice antiderivatives as well. However, in advanced theoretical investigations of functions of a real variable, we would like to have an integral with certain properties that the Riemann integral lacks. First of all, there are certain easily described functions such as XQ which have no Riemann integral at all . We will greatly expand the range of integrable functions when we defme the Lebesgue integral. In addition, the Riemann integral lacks certain desirable limit prop erties. In particular, the class of Riemann integrable functions is not closed under pointwise limits, even for bounded monotone sequences.
TH E R I E M A N N I NTEGRAL
4.1
9
Example: Let Q n [0, 1 ] be enumerated without duplications by , 3 , . . . . This is possible of course because Q is countable. Then q1oq2q defme, for n = 1 ,2, . . . , andx E [0, 1 ], /,.(x) =
11
if x =q1 or x =q2 or . . . or x==q,.
0 otherwise.
The sequence {{,.} is monotone increasing and is uniformly bounded (in fact 0 < f,.(x) < 1 for all n and x). Each /11 is Riemann integrable, since /11 has exactly n discontinuities, hence is piecewise continuous. Finally, {{,.} converges pointwise to XQ • a non-Riemann integrable function. The Lebesgue integral not only overcomes many of these difficulties inherent in the Riemann integral, but-like any great mathematical idea its study has also generated concepts and techniques which are extremely valuable to amthematicians doing research in diverse areas.
5. Exarci•
5.1
Given /(x) = x3 on [ 0, 1 ] and the partition P four different Riemann sums R(f,P).
5.2
Prove that the number R
5.3
Let f(x) 1 .3 that
5.4
5.5
f
(0,1/8,1/3 ,2/3 ,1),
find
in Definition 1.3 must be unique, if it exists.
c for all x E [a,b ] and some real number c. Show by Definition is Riemann integrable on [a,b ] , and J:f(x ) dx = c(b -a).
Let g(x) = O for x + I ,g( l ) = I . Show from Definition 1 .3 thatf; g(x)dx = 0. (Hint : given a partition P = (x0,x�o . . . ,x11 ) of [ 0,2] , x = 1 is in at m ost two subintervals [x1_ 1 ,x1] and [x,,xi+ 1] . Thus show R(g,P) < 2c5.) (a) (b)
5.6
==
==
Prove that if f is Riemann integrable on [a,b ] , c E [a,b ] , and g(x) = f(x) for all x + c, then g is Riemann integrable on [a,b ] , and fabf(x) dx = f bg(x ) dx. (Hint: See Exercise 5.4.) Repeat (a) in the case where g(x) = f(x) except at finitely m any points c�r ... ,ck in [a,b].
11
Let /(x) =
f01/(x)dx
=
5.7 Let o(x) =
0 for x 0.
1
1 0 -I
+
1 /n, n
if x > O if x = 0 if x < 0.
=
1 ,2,3 , . . .
, and let
/0/n)
=
1.
Show that
10
L E B ESGUE I NTEGRATION AN D FOUR I ER SER I ES
(a) (b) (c)
5.8
Show that f • o is a step function for any function f defined on any closed interval [a,b]. In contrast, find a function g such that o • g is not a step function on [ 0, 1 ] . Show that o • sin is a step function on every closed interval, and find f.!;1'"/8 o(sin x)dx.
If f and g are step functions on [a,b], show that f + g is a step function. Is fg a step function? If g: [a,b1 -+ [a,b], is f g a step function? •
5.9
Prove that if f is Riemann integrable on [a,b], then f is bounded, using only Definition 1 .3 . (The result follows easily from Theorem 1 .6, but is needed in its proof.) (Hint : Assume that f is unbounded. Then relative to any partition P, R(f.P) can be made arbitrarily large by making appro priate choices of x;* E [x;- 1,x1] .)
5.1 0 Show that, in contrast to Exercises 5.5 and 5.6, there is a Riemann in tegrable function f on [0, 1 ] and a non-Riemann integrable function g on [0, 1 ] such that f(x) = g(x) for all but countably many points in [ 0, 1 ] . (Hint : see Exercise 5.9.)
[a,b]. Let e, 6,P be as in Definition 1 .3 . Now if /1(x) = glb{/(x)jx E ( x,_ 1 ,x,)} for x E (x1_.,x,), then h (defined appropriately at x0 , , xn ) is a step function lying below f. Show that there is a point Xt* E [x;-l,x;] such that f(x,*) - fl(x) < e. Find h above f in a similar manner, and x,** E [x;-l,x;1 with f2(x)-f(x,**) < e. For the converse, let R = lub{(,11!1(x)dxlfl a step function and /1 < /}. Show that R = glb{f:f2(x)dxlf2 a step function and f < f2 }. Then given e > 0, obtain step functions /1
5.1 1 Prove Theorem 1 .6. (Hint : let / be Riemann integrable on .
.
•
5.1 2 The following is a popular alternative development of the Riem ann integral.
, xn > is a If f is a bounded function defined on [a,b1, and P = (x0 , partition of [a,b], let m; = g lb {f(x)j x E [x1_1,x1]}, M; = lub {f( x) j x .
n
•
•
n
E [x1_.,x1]}. ThenL({,P) = •=l .E m,{x;-x;-1), U({,P) = •= .ElM,{x;- x;-1 ). (a) (b) (c)
Let .C = {L(f,P)j P is a partition of [a,b]}, U = {U({,P) I P is a partition of [a,b]}. Show that .C is bounded above and U is bounded below. Define J1111[�x)dx = lubl, and J 11[(x)dx = glbU, called the 11 lower and upper mtegrals of {, respectively. Show thatf11f(x)dx <J11"f(x)dx. Prove that '/ is Riemann integrable (in the sense of Definition 1 .3 and Theorem 1 .6) if and only if J11f(x)dx = i"f(x)dx. (In the alterna tive development using uppe/ and lower �urns, f is said to be in tegrable if the upper and lower integrals are equal.)
THE R I EMANN I NTEG R A L 1 1
5 .13 Prove Theorem 2.1. ( Hint : use 1.6 and 1.7.) 5 .14 Prove Theorem 2.2. 5. 15 Prove Theorem 2.3. 5.16 Prove Corollary 2.4. 5.17 Prove Theorem 2.5. ( Hint: use 1.6 and uniform continuity.)
,xn ), 5.18 Prove Theorem 2.6. ( Hint: use 1.6. Given any partition P = (x0, if f is non-decreasing, then 1 et / 1 (x) = f(x1_1 ) for x E (x1 _ 1,x1) and {2(x) = {(Xt) for X E (X;-J,Xt).) .
•
•
5.19 Consider the function
f(x) = Is
l
sin ( l/x)
0
ifx #: 0
ifx = 0.
f piecewise continuous on [ -1 ,1 ) '! Is f piecewise monotone on
'!
[ -1, 1 )
5.20 Invent a function which is monotone on [ 0,1 ) , but is not piecewise continuous.
5.21 (a)
( b) (c)
Show that for any sets A and B, XA. nB = XA. XB. Find similar expressions for XA. us and XA. \B. Show that XA. + XB = XA.us + XA.nB·
5. 22 Show that
XQ is discontinuous at every point.
5. 23 Show that the function g in Example 3.3 is continuous at every irrational and discontinuous at every rational. ( Hint : for x irrational, given e > 0, there are only finitely many rationals pfq E [ x - 1,x + 1 ) with 1/q ;;a. e. (Why?) Let 6 > 0 be so small that no such rational is in [ x - 6,x + 6 ] . ) 5 .24 Prove that the function g of Example 3 .3 is Riemann integrable on [ 0,1) . What is the value of J01g(x)dx'! ( Hint : Given e > 0, let /1(x) = 0 for all x E [ 0,1 ) . If 1/n < e , then f2(x) = 1/n except at those finitely many rationals p/q E [ 0, 1 ) ( written in ''lowest terms") where q < n.) 5.25 Show that the class of Riemann integrable functions on [a, b ] is closed
under uniform limits (i.e. , if ifn} converges to f uniformly, and each In is Riemann integrable on [a,b ] , then f is Riemann integrable on [a,b ] ) Also show that in this case, f."f(x)dx = lim J.,." fn(x)dx. a .
n•• ..
CHAPTER
2
Measu rable Sets
6. Introduction In his publications beginning in 1 902, Henri Lebesgue presented one of the most exciting new ideas in the history of analysis. Some of his ideas had been anticipated by Borel and Cantor, but it was Lebesgue who fully developed the theory now known as Lebesgue measure and integration. His idea basically was to eliminate the deficiencies of the Riemann integral by starting with a partition of the range of {rather than a partition of the domain as in the Riemann integral.* For example, if
f is a
bounded non-negative function defmed on
(a,b] , and if M is a strict upper bound of the values of f(x), for x E [a,b] ,
then the area under the graph of f could be approximated in the following Mt!llaure and the Integral, Holden Day, San that we must break up not (a,b), but the bounded by the lower and upper bounds of f(x) in (ll, b). Let us do this
*In his own words (Henri Lebesgue, Francisco,
interval
1966,
(f.h
p.
180):
"It is clear
•
•
•
with the aid of numbers Yi differing among themselves by less than E. We are led to consider the values of f(;c) defined by Y t < f(x) < Yt- l · The corresponding values of
x form a set E1
• • • •
With some continuous functions it might consist of an inf'mity of
intervals. For an arbitrary function it might be very complicated. But this matters
little."
13
14
LEBESGUE I NTE GRATION A N D FOUR I ER SER I ES
distinctly non-Riemann fashion. Let (y 0y y , n ) be a partition of , 1, [O,M) along they-axis, and choose y1* E [y _. y , i], for i = I, .. . ,n. •
1
•
•
,4•M y.• JIJ
y,•
11•
y,•
,
.. , ..
,. aQ A
•
Defme
Et = {x E [a,bJIYi-1
In the end, of course, the Lebesgue measure will be the "limit" of such sums. Lebesgue himself points out an analogy which contrasts his method with Riemann's in the simplest terms. Given a row of coins consisting, in order, of : quarter, dime, penny, quarter, nickel, dime, dime, Riemann would determine the total value of money by adding
25 +
IO + I
+ 25 + 5 +
IO +
IO.
MEASUR A B L E SETS
15
Lebesgue, on the other hand, would first count, for each denomination, the number of coins with that value. His sum would be 25 • 2 + 10 • 3 + s • 1 + 1 • 1 . Of course, the result is the same. For non-step functions, however, it is not so evident that Lebesgue's method gives the same result as Riemann's. The difficulty with Lesbegue's procedure, as outlined above, is that it requires a systematic way of assigning '1ength" to subsets of the real .line. In the diagram above, for example, one must be able to arrive at a '1ength" (or measure) for the setE2 • lfE 2 turns out to be an interval or a union of intervals, there is no problem. But this need not be the case. Consider XQ , the characteristic function of the rationals, and any partition (y0, ••• ,yn) of an interval containing its range {0,1 }. At most two of the sets E1 = {x E [0,1 J iy1_1 < XQ(x) < y1} will be non-empty (namely, those for which Yt- 1 < 0
Let S be a collection of subsets of [a, b] which is closed under
countable unions, that is, if A �oA 2 , ••• E S, then nU= l A n E S. ...
(i)
A set function on S is a function which assigns to each set S E S a real number. (ii) A set functioniJ on S is called a measure if (a) O
n=
...
IJ(A) = n�l iJ
(countable additivity).
18
L EBESGUE I NTEGR AT I ON A N D FOUR I ER SER I ES
We will also require that subintervals of [a,b] be in S, and that the Lebesgue measure of an interval be equal to its ordinary length. Finally , we hope that every subset of [a,b] will be in S ; that is, we will be able to take the measure of any subset of [a,b] . However, we will see that this will be impossible if we wish to retain countable additivity (property (d) in the defmition). 6.2 Example:
�
Defme p.(S) for any subset of [0,1 ] by
p.(S)
=
1 if !2 es
1 0 if 2� s.
It is easy to verify that this defines a measure on [0,1) , but it is not con sistent with our concept of the length of an interval since for example p.([t.fD 1 and p.( [o.tD = o. =
Define p.(S) = 0 for every S C [0,1 ] . This is called the "trivial measure," and it is also inconsistent with lengths of intervals, although it obeys properties (a)- (d) of a measure.
6.3 Example:
Anyone familiar with the Riemann integral might try to define a measure by
6.4 Example:
p(A)
=
J; x_.(x)dx.
This is consistent with lengths of intervals, but as we have seen it fails to assign a measure to the set Q f"'' [0,1 ] . In fact, if S {A C [0,1] I XA is Riemann integrable}, then S is not closed under countable unions. =
The non-Riemann integrability of XQ results from the fact that for step functions /1 < XQ <12. we have
6.5 Example:
J; fl (x)dx < O < 1 < J; f2(x)dx.
(See Example 3.2.) We could instead consider only step functions f2 lying above XQ· Or, in general if A C [0,1] , let
p.(A) = glb
u; f2(x)dx l t2 a step function and f2 ;;.ox_. on [0,1] � .
M EASU RAB LE SETS
17
Clearly an equivalent expression is
ll(A) = glb{1 �1 (b1 - a,) IA c , ";11 (a1,b1)} . n
n
This function is called "the Jordan content of A ," and it is surely consistent with length for intervals. But it fails to be a measure; namely, p.(Q n [0,1 ] ) = 1 , p.((O, l)\Q) = 1 , but p.((O,l ]) = I, contradicting property (d) of Defmition 6.1 . (See Exercise 9.8.) Thus in this section we have seen the necessity of obtaining a con cept of measure which obeys certain basic properties and agrees with our concept of the length of an interval. We have also seen that some obvious attempts to produce such a measure result in failure. In the next section we shall chart a course which results in a degree of success toward this end.
7. Outer Measure For the remainder of this chapter, we will denote the unit interval by E. We will restrict our attention to subsets ofE in order to avoid unnecessary complications. After we have developed the theory of meas ure on subsets ofE, we will consider subsets of an arbitrary closed interval [a,b ) , and even unbounded sets.
[0,1 ]
We begin with the following natural definition of measure for in tervals, which will at the outset guarantee that our measure will be con sistent with length for intervals.
7.1 Definition : The outer measure of any interval (open, closed, half-open) inE with endpoints a < b will be the positive real number b -a. We will use the symbol m * to denote the outer measure function, so that m*([a,b] ) = b -a, for example. Note that it is possible to have A s; B CE, but m *(A) = m *(B) . This outer measure is our first attempt to defme a measure. We will need to refine our ideas slightly to succeed in fulfilling this goal. Now we extend our outer measure from intervals to arbitrary open subsets of E. This procedure depends on a general theorem about open sets of real numbers. Recall that any open set in 6t can be written as a union of open intervals. The following theorem says even more.
18
LE BESG U E I NTEG RATION AN D FOU R I E R SE R I ES
7.2 Theorem : Every non-empty open set G C IR can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals. Proof: Suppose first that
G is bounded. Since G is open, for each
x E G there is an open subinterval of G containing x. Let
bx = lub {yl(x,y) C G}, and ax =glb {zl(z,x) C G}. Let I"
= (ax, bx), called the component of x in G. Clearly x E I".
Now lx C G, for ifw E lx, say x <w < bx, then by defmition of bx, there is a number y such that w < y and (x,y) C G. Hence w E G. The case where ax <w < x is handled similarly. (What aboutw = x?) Also ax €1. G and bx €1. G (see ExercisP 9 . 1 0). Now we show that G can be expressed as the disjoint union of I"'s. Clearly G = x U Ix. Furthermore, for x,y E G, either I" and Iy are disjoint
EG
or they are identical. Indeed, suppose lx n ly ::/= 0. Then ay < bx and ax < by {draw pictures to verify). But since ax €1. G and ly C G, it follows that ax €1. ly; therefore ax < ar Similarly ay €1. lx, so that ay < ax· Thus ax = ar A similar argument shows that bx = by, so that lx = lr Hence any two intervals in the collection Ux l x E G} are equal or disjoint, and G is the union of a disjoint collection of open intervals of the form Ix. That this collection is countable follows from the fact that one can choose a rational number in each lx (using the axiom of choice), and the disjointness of the intervals guarantees that no duplication will occur in the choice of rationals. The subset of Q thus chosen is countable, and is in one-to-one correspondence with the collection {I" I x E G}. The proof that this representation of G is unique is left to the reader (Exercise 9 . 1 1 ). The case where G is unbounded is handled in the same way, except that G may equal (-oo,oo) , which is trivial, or some components may be of the form (-, b) or (a,oo).
0
Whenever we use open sets in the unit intervalE, we shall use "open" to mean open in the relative .topology ofE as a subspace of IR. That is, G is
19
MEAS U R A B L E SETS
open in E if and only if G = E intersected with some open subset of 6t Then it is clear that the theorem holds in the relative topology of E, in the sense that if G is open in E, there is a unique representation G =
�/;, r
where the It are disjoint intervals of the form (a1,b1) n E. For example,
G = U(a1,bi).
G C E,
we allow the possibility that some (a1,b1) may be of the form or [0, 1 ] .
[O,b), (a, 1 ] ,
We now use the theorem to extend our defmition of outer measure to open sets inE.
7.3 Definition : The outer measure m*(G) of an open set G C E is defined as the real number 1 (b1 - a1), where G = l.( (a1, b1) as in Theorem 7 .2. Since the theorem says that G is uniquely represented as a disjoint union of countably many intervals, this defmition is unambiguous. Also, -
the number
(b a1) exists since it is the 1l: =1 1 -
sum
of a series of positive
terms with bounded partial sums. Finally, the order in which the terms of the series appear is unimportant since convergence is absolute. Note that 1 (b1 - a1) may be a fmite sum if l.( (a1, b1) is a finite union. We next extend the defmition of outer measure to by approximating arbitrary sets with open sets.
7.4 Definition : The outer measure the real number glb
all subsets ofE,
m*(A) of any set A C E is defined to be
{m*(G)IA c G and G open inE}.
Clearly m*(A) exists for any A C E since the set {m*(G)IA C G and G open in E} is bounded below by 0 and thus has a finite greatest lower bound. It is also clear that if A is open in E, then the previous definition of m*(A) for open sets (Defmition 7.3) agrees with this defmition. In case A is a non-open interval, it is left as an exercise (Exercise 9.16) to show that this new defmition agrees with the earlier defmition (Definition 7 . 1 ) for intervals.
We will often use the following proposition, which is an immediate consequence of the definition of outer measure for arbitrary sets.
20
L E B ESG UE I NTEG R AT I ON A N D F OU R I E R SE R I ES
Given any e > 0 and any set A C E, there exists an open set G C E such that A C G and m*(G)<m*(A) + e.
7.5 Proposition:
It would appear that we have arrived at a consistent manner of assigning a measure to any subset of E and that we need only verify that the desired properties for a measure hold. Unfortunately the outer measure we have defined fails to be countably additive on the class of all subsets of E. We will present an important example to show this, but first we need the following easy result. Let A C E. Then for any x, m*(A) = m*(x + A), where x +A = { x + aIa E A } is called the translate of the set A by x. (Since we are restricted to subsets of E, we may need to translate modulo 1 ; for example, r-t. 1 1 +t = lt. 1 1 u [o.t] .) Proof: If A is an interval or a countable union of pairwise disjoint open intervals, the lemma is clearly true. Thus the lemma holds if A is any open set in E. For arbitrary A C E, m*(x +A)= glb {m*(G)Ix +A C G and G open in E} = g1b {m*(-x + G)IA C -x +G and -x +G open inE}
7.6 Lemma :
:>m*(A).
The proof that m*(A) ;> m*(x +A) is similar.
0
Outer measure m* is not countably additive on the class of·all subsets of E. That is, there exist pairwise disjoint sets V,. that E = U V,., but 1 = m*(E) :/1: � m*( V ) . To obtain V,. let E. = {x E E l x - a is rational} for each a E E. That is, E. consists of all members of E which differ from a by a rational. Note that each E. is countably infmite and that if E. n EfJ :1= t'. then E. = EfJ (the reader should verify this; see Exercise 9.21). Using the axiom of choice, let V be a set consisting of exactly one xa from each distinct set Ea. Thus, no two distinct members of V can
7.7 Example:
00
n= l
00
n= l
,.
MEAS U RA B LE S E TS
21
differ by a rational. Now let q.,q2 , be an enumeration of the rational numbers in E, and for each n, let Vn =qn + V. Claim 1 : Vn n Vm = 0 for n =I= m. Proof: Let y E Yn n Vm . Then there exist E V such that xa +qn = y = +qm . Then xa - must be rational, which means that xa = xp. Henceqn =qm and Yn = Ym · Claim 2: E = nU Yn · =l Proof: If x E E, then x E Ea for some a. But there is one repre sentative of Ea in V, say xa, so that x = xa + qn for some rational Qn· That is, E Yn. Claim 3: m* is not countably additive on E. Proof : Clearly I = m*(E) = m *( nU Yn). By the lemma (7.6), =l m*(V) m*( Vn) for all n = 1 ,2, . . . , since the Yn's are translates of V. Now by Claim I, the Yn's are disjoint, so if countable additivity were to hold, we would have I = nl:= l m*(Vn) = n=ll: m*(V) . That is, we would have infinitely many equal numbers m*(V) adding up to I. This is clearly impossible. 0 This example would suggest that we have failed in our attempt to define a measure. However, the amount of work necessary to produce the example, and the use of the axiom of choice, suggest that perhaps all is not lost if we are willing to settle for slightly less than our original goal. To ensure that the resulting Lebesgue Integral satisfies desired properties, we insist on retaining countable additivity as a key property for our measure. However, we are forced to relax our hope that every subset of E have a measure assigned to it. We therefore set out to fmd a large class of subsets of E on which m* will be countably additive. There are many ways to do this and not all methods result in the same class of sets. In section 8 we will present one method of obtaining such a class of sets (called Lebesgue measurable sets). •
•
•
Xa.XtJ
X fJ
XfJ
00
X
00
=
00
00
8. Measurable Sets
In order to produce a large class of sets on which m* will be count ably additive, we need the following definition.
22
8.1
L E B ES G U E I NTE G RATI ON A N D FO U R I E R SE R I ES
Definition : The inner measure m.(A) of any subset A be the number 1 m*(E\A). -
C E is defined to
It is clear that m.(A) exists and is non-negative for each A m*(E\A) exists and is between 0 and 1.
C E since
We are now in a position to give the definition of measurable set. We will eventually show that the class M of all measurable sets is a large enough collection of subsets of for our purposes, and that m* is a measure (in particular, m* is countably additive) on M.
E
CE
8.2 Definition : A set A is said to be (Lebesgue) measurable if m.(A) = m*(A). In this case, the measure of A, denoted m(A), is the number
m.(A) = m*(A).
8.3 Proposition : A set A
C E is measurable if and only if
m * (A) + m * (E\A.) = 1.
Proof : This follows immediately from the two preceding definitions. 0
CE
8.4 Corollary : A set A is measurable if and only if E\A, the complement of A in is measurable.
E,
E,
E
It is easy to show that f/J, and finite sets of points in are meas urable. Also, every interval in is measurable (Exercise 9.22), and measure is consistent with length for intervals. It is also true that open sets are measurable , but we postpone proof of this until we have shown countable additivity of m* on the class of measurable sets in Chapter Three.
E
We now present further examples of measurable sets, preceded by several lemmas which we will find very useful in proving countable additivity. J.5 Lemma: (Monotonicity of m* and m.). Given A
C B CE, we have
( I) m*(A) < m*(B), and
(2) m.(A) < m.(B).
(I)
Proof: follows easily from the definition of outer measure (see Definition 7.4 and Exercise 9 . 1 7).
M E ASUR AB LE SETS
For (2), note that.E\B CE\4, so that .t m*(E\B) < m*(E\B) by part (1). Thus m.(A ) = 1 m*(E\4) < 1 m*(E\B) = m.(B). -
-
23
0
(Subadditivity of m *). Suppose {A ,. In = 1 ,2, . . . } is a collection of subsets ofE. Then
8.6 Lemma:
m*(nu=1 A,.) < ni:=1 m*(A,.).
Proof:
We present the proof in several parts.
(1) Finite collections ofopen intervals.
It is obvious that if / 1 and /2 are open intervals in E, then m*(/ 1 U /2 ) < m*{I1 ) + m*(/2 ) . (In fact, equality holds if and only if / 1 n /2 0.) It follows that =
for any finite collection of open intervals. (See Exercise 9.24.) (2) Countable collections ofopen intervals.
Given nU= 1 I,. , write this open set as nU= 1 J,., where the J,.'s are disjoint open intervals (Theorem? .2). Now given e > 0, there is an N such that -
-
n =N:t+ 1 m*(J,.) < e. -
Thus,
(Why?)
m*(nU= 1 I,.) = n:t= 1 m*(J,.) < nN:t=1 m*(J,.) + e. •
•
For each n = 1 ,2, . . . ,N, there is an open interval K,. = (a,., b,.) such that .K,. = [a,.,b,.] C J,. and such that
24
L E B ESGUE I NT E G R AT I ON A N D F OU R I E R SE R I ES
m*(J,)<m*(K,) + e/N. Thus,
m*(U I,)< n=1 :E m*(K,) + 2e. n=1 oo
N
·
N_
Now ':! K, is a compact (closed, bounded) set contained the open n-1 .. N _ open cover U I,. By the defmition of compactness, U K, is contained n=1 n=1 ,. N ,. the union of some fmite subcollection U I,1, so that U K, U I,1. 1=1 1=1 n=1 ,. N ,. Therefore, :E m*(K,) < m*(U I, ) < :E m*Q, ) by part (1). Therefore 1 n=1 j=1 /=1 1 we have
in
in
C
..
..
m*(n=l U I,)
Since e was arbitrary, the lemma holds for countable collections of open intervals.
(3) Countable collections ofarbitrary sets. Let {A, In = 1 ,2, . . . } be a collection of arbitrary subsets ofE. Given e > 0, for each n, there exists an open set G, such that A, G, and
(Proposition 7 .5).
m*(G,)<m*(A �) + e/2" But G, has a unique representation open intervals, and by definition,
C
'fin,k as a union of pairwise disj oint
m*(G,) = fm*Q,,")< m*(A,) + e/2" . Therefore,
..
"and U A, C n= U1 UI, k n=1
00
•
-
I
0G
00
UI, ")< n=1 m*(U A,)<m*(U :E :Em*Q, :E m*(A,)+e. ")< n=1 k n=1 n=1 k •
•
Since e was arbitrary, the lemma follows.
.. .. (Note that :E m*(A,) may be infinite but m*( UA,) must be fmite n=1 n=I .. since
U A, C E.)
n=1
0
8.7 Lemma: For any set A
MEAS U R A B LE SETS
25
E, so that m*(A U(E\.4)) 1 . But
by
C E, m.(A)<m*(A) .
Proof: Clearly A U (E\.4)
==
Lemma8.6,
==
m*(A U (E\.4))<m*(A) + m*(E\.4). Therefore,
m*(A) ;> 1 -m*(E\.4) m.(A). ==
8.8 Example: Any set has measure 0 since
A
with outer measure
m*(A)
==
0
0
is measurable and
m*(A) ;> m.(A) ;> 0 implies m*(A) = m.(A ) 0. (1) The empty set 0 and any finite set of points {p 1 , ,p11} CE are measurable sets with measure 0. We need only show that m*({p., . . . ,p11}) 0, but this is true since for any e > 0, the open set G =(pi -e/n, Pt + e/n)U (p2 -e/n, p2 + e/n)U .. . 0
==
==
•
.
•
==
U (p11- ejn,
Pn + e/n )
has outer measure< 2e. By definition,
m*({p., . . . p11})< m*(G)< 2e, {p.,p2 ,
so that
}
C E is measurable (2) Evezy countably infmite set and has measure 0. (In particular, n £) 0.) Again we show that the outer measure is 0. Indeed, by subadditivity (Lemma 8.6), and by part (1),
m(Q
m*({p 1 ,
•
•
•
==
}< 1=1:1 m*({p1}) 00
•
•
•
==
0.
This example raises the question: are there any sets of measure 0 which are not countable? We will see that the answer is yes the next chapter.
(3) If B C measure (Lemma measureO.
A C E and m*(A) 8.5); m*(B) 0,
0, then by monotonicity of outer so that B is measurable, and has
==
==
in
26
L E BESG UE I NT E G R A T I O N AND F OU R I E R S E R I ES
8.9 Example: Recall from Corollary 8.4 that the complement E\A. of any measurable set A is measurable. It follows in this case that the measure of E\A. is 1 m*(A). -
(1) The complement of any countable subset of E is measurable and has measure 1 . For example, the set of irrational numbers in E is measurable, with measure 1 . (2) Complements of intervals are measurable, since intervals are measurable. In subsequent sections we will show that all open (and hence all closed) subsets of E are measurable. We will also prove that outer measure m* is a measure, satisfying countable additivity, on the class of all measura ble sets, and we will produce an example of a set of measure 0 which has uncountably many points.
9. Exercises
9.1 9.2 9.3 9.4
Let f(x) = x 2 on [ O, l ] , y0 = O , J' t = i , y 2 = i · Y3 = l , y 4 = 3 . Et . E2 , E3, E4 .
Find
A, B E S, and A n B = (3,
then
Prove that if 1J. is a measure on S and
IJ.(A U B) = p.(A ) + p.(B).
Prove that if IJ. is a measure on s and {x} e s for p.({ x}) = p.({y }) for all x, y E [a, b ] , then p.(Q) = 0.
every
X
E [a, b ] , and
Let S be a collection of subsets of [a, b ] which is closed under countable unions and under complements in [a, b ] ( that is, if A E S, then
[a, b ] \.4 ES).
(a)
(b) (c)
Prove that S is closed under countable intersections. Prove that if A, B E S, then A \B E S. Show that if 1J. is a measure on S , then
(Hint for (c) :
p.(A U B) = p.(A )+ p.(B) - p.(A n B). A U B = (A\B) U (B\.4 ) U (A n B) disjointly.)
9.5
Prove that the set function 1J. of Exampl e 6.2 is a measure.
9.6
Prove that the "trivial measure" of Example 6 . 3 is indeed a measure,
9.7
Show that the collection S of Example 6 .4 is not closed under countable unions. Show however that 1J. satisfies (a), (b), (c) of Definition 6. 1 , and that 1J. is consistent with lengths of intervals.
M E ASU R A B LE SETS
9.8
(a) (b) (c) ( d)
Show that the two expressions for ll in Example 6.5 are equivalent. Show that "Jordan content" is consistent with lengths of intervals. Show that IJ.(Q n 0 , 1 ] ) = 1 and IJ.( 0, 1 ) \Q) = 1 . Prove or disprove each of the defining properties of measure for Jordan content.
[
[
/ .
..
9.9
27
Find the component of 1r in the open set U (n, n+ 1 n) n=l
9.10 Prove that a" � G, in the proof of Theorem 7.2.
7.2 ; if G
9.1 1 Prove uniqueness in Theorem
= UI,. = U J" , where I,., J" are ,. k open intervals and I,. n Im = 9, Jk n 12 = 9 for all n, m, k, 2, then show that for every n there is a k such that I,. = Jk , and for every k there is an n such that J" = I,. . ( Hint : if x E I,. , show that /,. = Ix .>
9. 1 2 Find m *(G), where G = (0, 1 ]\{ 1 /n j n = 1 , 2 ,
. . . }.
9.13 Do there exist open subsets G � o G 2 of E such that G 1 :I= G2 but m *( G 1 ) = m *( G 2 )?
9.14 Prove or disprove that if G 1
<,i
G2 C E are open, then m*(G t ) < m *(G2)·
9.15 Prove that m * is countably additive on the class of open subsets of E. 9.16 (a)
Show that m* is consistently defined for closed intervals by Defmi tion 7.4 ; that is, if [ a,b ] C E, show that
I
b - a = glb { m *(G> £a,b ] C G and G open in E}. (b)
Is there an open set G C E with
[ 1 /2,3/4]
C G and m*(G) =
t?
9. 1 7 Show that if A C B C E, then m *(A ) < m *(B). (Definition 7.4) 9.1 8 If A C
[ O,t J
t l]
and B C <
.
.
show that m *(A U B) = m *(A )
+
m *(B).
9.1 9 Given A C E, show that m *({ l - x l x E A }) = m *(A ). ( Hint : see proof of Lemma 7.6. Give details.)
9.20 Describe E 1/2 and Ev'i _ 1 in Example 7. 7. 9.21 In Example 7. 7 , prove that i f Ea n Ep :1: t/>, then Ea = Ep. 9.22 Prove from Definition 9.23 (a) (b)
9.24 (a)
8.2 that all subintervals of E are measurable.
+
If A C E is measurable, show that x A is measurable. If A C E is measurable, show that { I - x x E A } is measurable. Show that m *(I 1 U I2 ) < m *(/ 1 ) in E.
l
+ m *(I2 ) for I 1 , I2 open intervals
28
LEBESG UE I NTEG RATION A N D FOUR I E R SE R I ES
(b)
�l m*(Ik ) for open intervals /k in E. Show that m *( kV - 1 111) < k=
9.25 Referring to Lemma 8.6 , give an example of a collection {A n } n :'l of sub sets of E such that
m *(
.. U
n= l
..
A n ) < l: m*(A n >· n= l
9.26 Prove that m *(A ) = glb { n1; l m *(Jn > l ln open intervals and A C nU In} . = =1 (Note that the Jn need not be disjoint. Use Lemma 8.6.)
+
+
m *(B) = m .(A ) m .(B), prove that A is measurable and B is measurable. ( Hint : Lemma 8 . 7 . )
9.27 If m *(A )
9.28 Example 8 . 8( 2) implies the surprising result that for any e > 0, the dense
set Q () E is contained in an open set with outer measure less than e. Write down an expression for such an open set G. (Hint : let Q rt E = { q 1 , q , }. 2 .. Let G = nU In , where q n E In for each n, and In is a specified open =1 interval. See the proof of Lemma 8.6(3) and Example 8 . 8( 1 ) for an idea of what In should be.) .
•
.
9.29 Show that m* is not (finitely) additive on the collection of all subsets of
E. (See Lemma 8.6 and Example 7. 7 .)
CHAPTER
3
Properties of Measurab le Sets
10.
Countable Additivity In this section we will show that outer measure m* is countably additive when restricted to the class of measurable subsets of E. We first need two lemmas.
10.1 Lemma: lf G 1 and G2 are open subsets ofE, then
Proof: If G 1 and G2 are intervals, the Lemma is obvious. If G 1 and G2 are both finite unions of open intervals, the proof of the Lemma is contained in Exercise 1 6.1 at the end of the Chapter.
. 11 Fu1 ao
For G 1 and G2 arbitrary open subsets of E, we know that G 1 ao
and G2
=
U J , where the /; are disjoint open intervals and the 11 are dis· /= 1 1 joint open intervals. Given e 0, there is an integer such that =
ao
l': m*(/;)
i=N+ 1
1
ao
=
N
l': m *(J1) < e (Why?). < e and j=N+ 1
Define G 1 G2 "
>
ao
U 11.
j=N+ 1 m*(G 1 U G2)
=
N
U 1 /; and G2 1
i= Then G 1 U G2
<
m*(G 1 '
U
= =
G2')
0Jl 11. Also define G 1 j= 1 G 1 ' U G2'
+ 2e
U
11
ao
U
/; and i=.f' + 1 G t" U G2 ' so that =
by subadditivity (Lemma 8.6). 29
30
L E B ESGUE I NTEG RATI ON A N D F OU R I E R SE R I ES
Similarly G1 n G2 = (G1' U G' t) n'(G2' U G{') C (G1' n G2') U G ." U G; ' . Thus m*(G1 n G 2 ) <m*(G1 n G2 ) + 2e. Now' m*(G 1) + m*(G 2 ) ;;;.' . m*(G.") + m*(G ;) = m*(G1' U G;) ' ' + m*(G1 n G 2 ) since G1 and G2 are finite unions of open intervals. Thus m*(G1) + m*(G2) ;> m*(G1 U G2 ) - 2e + m*(G1 n G2 ) - 2e from above. Since e was arbitrary, the Lemma follows. 0 If A 1 and A 2 are subsets of E, then (1) m*(A 1) + m*(A 2 ) > m*(A 1 U A 2 ) + m*(A 1 n A 2 ) and (2) m.(A 1) + m.(A 2 ) < m.(A 1 UA2) + m.(A 1 n A2). Proof : Given e > 0, there exist open sets G1, G 2 C E such that G1 :> A 1, G 2 :> A 2 , and m*(Gi) < m*(A1) + e for i = 1 ,2. Thus <. m*(G1 U G2) 1 <.1 m*(G1) + m*(G2 ) - m*(G I n G 2 ) m*(A 1 U A 2 ) a.s 0. <. m*(A 1) + e + m*(A 2 ) + e - m*(A 1 n A 2 ). Since this is true for all e > 0, the result in (1) follows. To prove part (2), use E\.4 1 and E\.4 2 in placeof A 1 and A 2 above (Exercise 16.2). 0
10.2 Lemma:
If A 1 and A 2 are measurable subsets ofE, then A 1 U A 2 and A 1 n A 2 are measurable. Proof: By the Lemma, m*(A 1) + m*(A 2 ) > m*(A 1 U A 2 ) + m*(A 1 n A 2) 8.7;> m .(A 1 U A 2 ) + m.(A 1 n A 2 ) > m.(A 1 ) + m.(A 2 ). But m*(A1) = m.(A1) by hypothesis, so the inequalities become equal signs. Now m*(A 1 U A2) + m*(A 1 n A 2 ) = m.(A 1 U A 2 ) + m.(A 1 n A 2 ) together with Lemma 8.7 imply that m*(A 1 U A 2) = m.(A 1 U A 2 ) and m*(A 1 nA 2 ) = m.(A 1 n A 2 ). (Exercise 9.27.) 0
10.3 Corollary:
If A 1 n A 2 = � for sets A ., A 2 C E, then m .(A l U A2) > m *(A I ) + m .(A 2). Proof: Since m*(�) = 0, this is obvious from the Lemm a..
10.4 Corollary:
0
If {A1}j: 1 are pairwise disjoint measurable subsets ofE, then m.(l=UlAi) > i=l:l m.(A;).
1 0.5 Corollary: ...
...
P ROPE RTI ES OF M E AS U R A B L E SETS
31
"" N N N, m.( u A -) > m.( u A;) > l: m.(A;) 1 i= 1 1 i= 8. 5 i= . by the previOus Corollary extended to finite unions by induction. Letting N we obtain m.(1=1 P A1) > 1i:=1 m (A 1·) . 0 *
Proof : For any integer
I
-+Oo,
We are now in a position to prove the major theorem of this section. Once we know that is countably additive on the measurable subsets of E, then we will know that is a measure according to our earlier definition (Definition 6 . 1 .
m*
m*
)
{A;}i::..1 be a pairwise disjoint Then U A; is measurable and i=1
10.6 Theorem : (Countable Additivity). Let
.. .. m*(1�1 A1) = 1�1 m*(A 1).
collection of measurable sets in E.
A; is measurable, m*(A;) = m.(A;) for each i. Thus m*( � A;) < i: m*(A;) = i: m.(A;) < P A1). By Lemma 1= 1 8.6 i=1 i=1 1 0.5 m.(1=1 8.7, m. (� A;) < m*(� A;), so U A; is measurable and m*( U A1) i= 1 1- 1 .. 1- 1 i= 1 = 1� m*(A1). 0 Proof : Since each -
-
-
-
There are many useful results which follow easily from countable additivity. We list several as corollaries.
10.7 Corollary : All open and closed subsets ofE are measurable. Proof: Open sets are countable pairwise disjoint unions of open intervals, which are measurable. Closed sets are complements of open sets. 0
10.8 Corollary: There exist non-measurable sets. Proof: In Example 7 .7 we produced pairwise disjoint sets -
.. rr=l:1 m*(V,) would be
equal outer measure. Since
V, all of
mU=1 V, = E, if the V, were measurable sets,
1 . This is clearly impossible, so the V, 's are not measurable. That is, m.(V,) < m*( V,) for each of these sets. 0
A 1 C A 2 C A 3 C • • • are measurable subsets ofE, then U A;) = �im m*(A1). •�= ' A; is measurable and m*(i=1 1+•
10.9 Corollary: If -
-
32
L E B ESGUE I N TE G RATI O N A N D FOUR I E R SE R I ES
A t • A 2 \A t • A3 \A 2 , • • • . Each set A,. \A,._ t = A,. n (E\A,._ t ). Furl=UtA1 = A t U
0
A t :::> A 2 :::> A3 :::> • • • are measurable subsets ofE, then .n A; is measurable and m ( nt A1) = limm*(A;). r=t i+• r=
10.10 Corollary : If -
*
-
.
0
Proof: Exercise 1 6.7.
{A,}i= t
10.1 1 Corollary : Let be any (not necessarily disjoint) countable col lection of measurable sets inE. Then -
,� A; is measurable and (2) ntAt is measurable. i= Proof: (1} Write � A t as the union of pairwise disjoint sets �t n-t A t Un{A- t2 \A t } U (A3\ [A t U A 2 ]}U• • •U(A,.\( �Ut A ,] }U • • • . Each set A,.\[ i=Ut A;] is measurable by Corollary 1 0.3 and induction (Exercise 1 6.�.} . (1)
-
The result follows from countable additivity {Theorem 1 0.6}.
(2) Write l�t A; = E'< ";!t [E\A;] ) and use part {1}. -
-
,
0
11. Summary
Let us pause to take account of our accomplishments. Our original (ambitious) goal was to construct a set function defined for aU subsets of E 0,1 and satisfying
m =[ ] (1} O < m(A) < 1 for any A C E. (2} If A CB, then m(A) < m(B) for A,B CE . (3} m(�) = 0 and m(E) = 1 . (4} m( Ut A;) = . :Et m(A,) for pairwise disjoint subsets A; ofE. r= i= In addition we expected m to agree with our concept of length of an -
interval.
-
P R O PE R T I ES OF M EAS U R A B LE SETS
33
In Chapter 2, we built up a definition of the outer measure m * based on certain of these properties. However, we learned that m* was not countably additive (property 4 above) on the class of all subsets of E. Our solution was to retain m* but to apply it only to certain (measurable) subsets of E. In section 10 we saw that this procedure resulted in a class of sets on which m* obeys properties ( 1) - (4) of a measure. Let us state this fact formally .
1 1 .1 Theorem : Let M be the class of measurable subsets of E. Then M con tains all intervals in E and is closed under countable unions, countable intersections, and relative complements. Also, m* is a measure on M. That is, m* obeys properties (4) above when restricted to measurable sub sets of E. Also m* applied to any interval gives the length of the interval.
(I) -
NOTATION : For the remainder of the text we will use m(A) rather than
m*(A ) whenever A is a measurable set. 12. • • Borel Sets and the Cantor Set
The most difficult part of our development has been proving count able additivity. A great deal of this difficulty stemmed from having to verify that various sets were measurable according to our definition of measurable set. One might reasonably ask whether M or some other large class of sets on which m * is a measure could be more easily defined. We will present one such large class (the Borel Sets) and then show that the class M of measurable sets has many more sets than the class of Borel Sets. First we need the following concept.
12.1 Definition : Let S be a collection of subsets of E. Then S is called a countably-additive class of sets (or a a-algebra) if
(1) E E S. (2) A-E S ,. E\.4 E S.
(3) ;Y A ; E S whenever {A; } ;:, 1 c S. 1 The following result may easily be proved.
1 2.2 Proposition : If S is a countably-additive class of sets in E, then
(1)
� E S.
34
L E B ESGUE I NTEGRATION A N D F OU R I E R SE R I ES
(2)
er {A;};= t
1,n A 1 E S whenev ...
...
C
s.
(3) Finite unions and intersections of members of S are in S.
Proof: Exercise 16. 10.
0
Two examples of countably-additive classes are the class of all sub sets of E and the class of all measurable subsets of E (see Exercise 16. 1 1).
12.3 Theorem : Let A be any class of subsets of E. Then there exists a minimal countably-additive class S ::> A (Here "minimal" means that if W is any countably additive class containing A , then W ::> S). Proof: Let S be the intersection of all countably-additive classes containing A . Since the class of all subsets of E contains A and is a countably-additive class, this intersection exists. It is easily verified that S is a countably-additive class of sets con taining A (Exercise 16 . 13). Since S is the intersection of all such classes it is clearly minimal.
0
1 2.4 Definition : Consider the minimal countably-additive class 8 which con tains all open intervals in E. This class 8 is called the class of Borel Sets in E . The previous theorem assures us that 8 exists. It is clear that 8 contains a vast number of sets (all open sets, closed sets, intervals, count able unions and intersections of these, and many more). Since M is a countably-additive class containing the open intervals, we know that 8 C M . Could it be that 8 = M and that we could have obtained M i n this easier way? Or, even if M :1= B , are there enough additional sets in M to justify our previous labors? To answer these questions and provide one of the most famous examples in analysis, we present the Cantor Set.
12.5 Example : (The Cantor Set) Let
PROPERTI ES OF M EASU R A B LE SETS
c, :
! 0 ----�.� ,3----2/ ·3
35
...
____
c. . .
0
•
1/9
•
2/9
•
•
1/3
2/3
•
7/9
• 8/9
• I
That is, C,. results from removing the open middle thirds of each interval in Cn - 1 · 1 2.6 Definition : The Cantor Set
...
C = nn= l C,. .
1 2.7 Theorem : The Cantor Set is an uncountable measurable set with measure zero. Proof: Clearly C is closed since its complement is open . Thus C is measurable. Now m(C,.) =fm(C,. _ I) = f(fm(C,. _�) = • • • = (l.)"m (E) , and C is the intersection of a nested sequence of measurable sets t,., so by Corollary 1 0. 1 0 m(C} = lim C,. = lim (1.. 3 )" = 0. PI
+oo
II i
PI +oo
E
To show that C is uncountable, expand the numbers in in ternary form (i.e. base three). (See Exercise 1 6. 1 6.) Now C consists of the set of all points in E which have no 1 in their ternary expansion (e.g. C 1 = all numbers which can be written in ternary form with a 0 or a 2 in the first place after the decimal point-note that t = .02222 • • • and f = .2000 • • • here). But the set of all sequences of O's and 2's is uncountable by a familiar diagonal argument (Exercise 1 6.20}. 0
1 2.8 Corollary: Every subset of the Cantor set is measurable with measure zero. Proof: All subsets of sets of measure zero are measurable with measure zero by 8.5 and 8.8 . 0 For those with a knowledge of cardinal arithmetic, the above Corol � lary shows that there are at least 2c = 2 2 0 measurable sets in (The
E.
36
L E B ESG U E I NTEG RAT I O N A N O F OU R I E R SE R I ES
N
Cantor Set has 2 o points since it is in a one-to-one correspondence with E.) Now the collection of Borel Sets is generated by operations (unions, intersections, and complements) on countable collections of open intervals and on countable collections of the resulting sets etc. It follows from a rather involved argument using ordinal numbers that there are at most 2 N o = c Borel Sets. Thus the Cantor Set alone contains 2 2 N o measurable sets which are not Borel Sets. (See Exercise 20.1 2 for another approach.) We can now see that our method of obtaining a collection of sets on which is a measure yields a vast collection of measurable sets.
m*
Thus we have seen that there are many measurable sets which are not Borel sets. Nevertheless, the following result shows that in some sense every measurable set has a Borel set "almost" equal to it. 1 2.9 Theorem : Given a measurable set C E, there exists a Borel set C E = such that That is, differs from by a set of measure zero.
m(A) m(B).
Proof: Exercise
16.39.
B
A
B
A
0
In concluding section 12 we should point out that the complement of the Cantor Set is open, dense in E (i.e. every open interval in E meets it), and has measure 1 . Also, it is possible to construct a closed nowhere dense (i.e. containing no open interval of E) subset of E with measure > 1 - e where e > 0 is specified in advance (see Exercise 16.24). 1 3.** Necessary and Sufficient Conditions for a Set to be Measurable
There are several different ways to define the class M of measurable sets in E. One of these is called the Caratheodory Criterion. In essence it requires that for a set to be measurable, outer measure must be additive when and .E\A are intersected with any set in E.
A
A
X
A set A c E is m*(X) m*(X A) + m*(X n [E\.4] ) for
13.1 Theorem: (Caratheodory Criterion for Measurability) measurable if and only if = n
every subset
X C E.
Proof : Suppose A is measurable and X C E. < m*(X n A) + m*(X n [.E\A ] ) by subadditivity of m * . an
To prove the reverse inequality, let open set G � such that <
X
e
Clearly
m*(X)
> 0 b e given. Then there is
m(G) m*(X) + e . Thus m*(X n A) +
P R OPE R T I ES OF M E AS U R A B LE SETS
37
m*(X n [E\.4 ] ) < m(G n A ) + m (G n [E\.4] ) = m(G) < m*(X) + e since m is additive M disjoint measurable subsets of E. Since this inequality holds for every e > O,m*(X n A)+m*(X n [E\A. ] ) < m*(X).
0 The converse follows by letting X =E. One advantage of the Caratheodory Criterion is that it does not require the concept of inner measure. Thus it can be used to defme measure and measurability for subsets of tR . Recall that in Lemma 1 0. 1 we proved that if G1 and G2 are open in E, then m*(G 1) + m*(G2) > m*(G 1 U G2) + m*(G 1 n G2). The reverse inequality also holds (see Exercise 16.3) and we use this relationship to prove the following useful criterion for a set to be measurable.
subset A of E is measurable if and only if for any e > 0, there exist open sets G1 and G2 such that G 1 ::> A, G2 ::> E\.4 , and
13.2 Theorem : A
m(G 1 n G2)<e.
Proof: Let A be measurable and e > 0 be given. Then there exist open G. , G2 in E with G 1 ::> A, G2 ::> E\A , and m(G 1 ) < m*(A ) + e/2, and m(G2)<m*(E\.4)+e/2. Now m(G 1 n G2) =m(G 1 )+m(G2)-m(G 1 U G2)<m* (A) + e/2 + m* (E\A. )+e/2 -m(G1U G2). But G 1 U G2 = E, so m(G1 n G2)<m*(A) + m* (E\.4) - 1 +e. Since A is measurable, m*(A) + m*(E\.4) = 1 and so m(G1 n G2) < e. Conversely, given e > 0, suppose there exist open G 1 , G2 such that G 1 :> A, G2 ::> .E\A , and m(G 1 n G2)<e. Then m* (A)+m*(E\.4) < m(G 1 )+m(G2) =m(G 1 U G2 )
Thus m* (A )+ m*(E\A) < 1 . Clearly 1 =m*(E) < m*(A) + m*(E\A) by 0 subadditivity, so equality holds and A is measurable. set A C E is measurable i.f and only if for any e > 0, there exist open G ::> A and closed F C A such that m(G\F)<e.
13.3 Corollary: A
38
L E B ESGUE I NTEG RAT I ON A N D FOU R I E R SE R I ES
0
Proof: Let F = E\G 2 in the Theorem.
There is also a standard criterion for measurability involving the symmetric difference of two sets. We include this in Exercise 16.27.
1 4. Lebesgue Measure for Bounded Sets
It would be easy (and we leave it to the reader) to repeat our develop· ment of Lebesgue measure for subsets of [a,b] , for each a < b. The definitions and proofs will be nearly identical to those for subsets of E = [0, 1 ] . The most notable change is for inner measure :
m.(A) = (b - a) - m * ([a,b) \.4),
for A
C [a,b] .
So we will now assume that for each a < b , we have m*, m * defined on all subsets of [a,b] , we have the concept of measurability of a subset of [a,b] , and we have all the nice theorems developed in the last two chapters. We still need to achieve all of these results for arbitrary bounded sets. For, if A is a bounded set, it is contained in infinitely many closed intervals. It might turn out that A would be measurable as a subset of [a,b] and not measurable as a subset of [c,d] . Fortunately, this does not happen : m* (A), m.{A), and measurability of A do not change when we consider A as a subset of two different closed intervals. First of all , if G = y/1 is open and bounded, it is clear that the I
definition m*(G) = "i; m*(/;), where the 11 are pairwise disjoint open in· I
tervals, gives an unequivocal value, no matter what closed interval contains G. This is the basis of the following consistency result for the defmition of m*(A ) 14.1 Proposition : If A C [a,b] and A C [c,d] , then glb { m* (G) I G open in [a,b] and A C G} = glb { m * (G) I G open in [c,d] and A C G}. Proof : Both glb's are equal to glb { m * (G) I G open in iR and A See Exercise 16.28.
C G}. 0
The Proposition says that m*{A) is the same no matter what closed interval we take as containing it. The same is true of m.(A), as is shown in the next proposition.
P R O P E RT I ES OF MEAS U R A B L E SETS
39
14.2 Proposition : If A C [a,b] and A C [c,d] , then
(b - a) - m*( [a,b] \A ) = (d - c) - m*( [c,d] \A ). Proof: The result is easy if A = �. If A :/= �. then A C [a,b ] n [c,d] which is not empty. There are several ways this can happen-let us suppose that a < c < b = d. The remaining cases are left to the reader.
Let e > 0, and let G be open in [c,d] with ( [c,d] \A ) C G and m*(G) < m *( [c,d] \A) + e. Then H = G U [a,c) is open in [a,b] , and ( [a,b] \A) C H, so
m*( [a,b ] \A) < m *(G U [a,c)) = m*(G) + (c - a) < m*( [c,d] \A) + (c - a) + e. Since e was arbitrary, m*([a,b]\A) < m*([c,d]\A) + (c - a). On the other hand, if G ' is open in [a,b] , G ' ::) ( [a,b] \A), and m*(G ' ) < m * ([a,b] \A) + e, then [a,c) C G ', H ' = G '\ [a,c) is open in [c,d] , contains [c,d) \A , and
m*([c,d]\A ) < m*(H ' ) = m*(G ') - (c - a) < m*([a,b]\A) - (c - a) + e.
0
From now on we will spe ak freely of measurability, m*, and m * for arbitrary bounded sets, and will feel free to use all the results we developed for subsets of [ 0, 1 ) in this more generlll context. Of particular importance, of course, is countable additivity which now has the following form.
14.3 Theorem : Let {A;}; 1 be a pairwise disjoint collection of measurable .. .. sets with U A1 bounded. Then 1U A ; is measurable and
i= 1
=1
..
..
m*( .U A1) = i :E m*(A 1). r= 1 =1 In the remainder of the book we will use the numbers of theorems about measure on subsets of E to refer to analogous results about measure on bounded sets.
40
L E BESGUE I NT E G R A T I O N A N D F OU R I E R SE R I ES
1 6.** Lebesgue Measure for Unbounded Sets
If A is unbounded, then we cannot apply our previous construction directly to define m(A ), since m.(A ) = (b - a) - m*([a,b]\A) makes sense only if A C [a,b ]. On the other hand, many unbounded sets intui tively should be measurable and have finite measure. For example, let A = (O,f) U (I ,I+t ) u (2,2+-i" ) U (3,3+- 1� ) u • • • . Intuitively we ex1 pect m (A ) = f + -;} + i + 1 6 + • • • = I . How then should we define measurability and measure for unbounded sets? The simplest solution is to truncate our unbounded set A by inter secting it with [-n,n]. This intersection will be bounded, and if measur able, m (A n [-n,n]) will be a finite number. Since m(A n [-1 , 1 ]) < m(A n [-2,2])< , we are led to the defmition ·
·
·
m(A ) = lim m(A n [-n,n]). � + Y\
This limit is either +ao or a finite non-negative real number.
1 5.1 Definition : A set A (not necessarily bounded) is measurable if A n [-n,n] is measurable for every n = I ,2, • • • . In that case
m(A ) = lim"" m(A n [-n, n] ).
m+ Notice that this defmition is consistent with our previous concept of measurability of A and the value of m (A) for bounded sets A (Exercise 16.30). Notice also that m(A) = ao is a possibility. For example, IR is measurable and m(IR) = ao. Our extended measure has the nice properties we to, as exemplified by the following results.
are
accustomed
1 5.2 Proposition : If A C B are both measurable, then m(A)< m(B ) . Proof: Clearly A n [-n,n] C B n [-n,n] for every n.
0
15.3 Theorem : (Countable Additivity) If {A1}1": 1 is a collection of pairwise ...
disjoint measurable sets, then A
= r= .u1A 1 is measurable and
m( U1 A1) = :t1 m(A 1). 00
i=
...
i=
P R OPE RT I ES OF M E AS U R A B L E SETS
Proof: Clearly is measurable , so
41
A n [-n,n] = I=uI (A1 n [-n,n]). Each A1 n [-n,n]
A n [-n,n] is measurable .
m(•=.U1A 1) = 1=�1 m(A 1) is true if m(A 1) = some i. Therefore , let m(A1) < for all i. Then, since the sets Now the formula
oo
for
oo
are disjoint and measurable , countable additivity for bounded sets gives
n [-n,n]) = 1=�1 m(A; n [-n,n]) < 1=E1 m(A;) . Since every n ; letting n -+ 00 we obtain m (A) < 1 � m(A1) . 1
m(A
this is true for
=
On the other hand, if e
> 0 and if k is given , then there is an n such
n [-n,n]) + e/k for i = 1 ,2, • • • , k (Why?). Therefore m(A1)<m(A1 k k ... :t m(A1 < :t m(A1 n [-n,n]) + e < :t1 m(A1 n [-n,n]) + e ) 1 1 1= 1= 1= ... m(A n [-n,n]) + e < m(A ) + e. Since e and k were arbitrary, 1=:t1 m(A;) < m(A� that
=
0
16. Exercises 16.1 Prove Lemma 1 0. 1 for G . , G2 finite unions of disjoint open intervals. (Hint : see Exercise 5 .2 1 . Verify that is the ordinary Riemann integral.)
�f� Xo 1 (x) dx = m *(G 1 ), where this
16.2 Prove Lemma 1 0. 2( 2). 16.3 Prove that if A 1 and A 2 are measurable, then m *(A 1 )
+ m *(A 2 ) = m *(A 1 U
A 2)
+ m *(A 1
n A 2 ).
(Hint: 1 0.2, 1 0.3.)
16.4 Using Corollary 1 0.3 and induction, prove that finite unions and inter sections of measurable sets are measurable.
16.5 (a) (b)
Using Corollary I 0.3 , show that if A and B are measurable subsets of E, then A \B is measurable. Use Countable Additivity ( I 0.6) to show that ifB C A and B and A are measurable subsets of E, then m*(A \B) = m *(A ) - m*(B).
C E with m(A ) > 0, there are real numbers x and y E A such that x - y is rational. (Hint : see the construc tion in Example 7.7.)
16.6 Show that for any measurable set A
42
L E B ES G U E I NTEG R AT I ON A N D F O U R I E R SE R I ES
16.7 Prove Corollary
1 0. 1 0.
16.8 If m(A )
= 0 and {:E -+ E has bounded derivative, show that m(f(A )) = 0. (Hint : use the mean value theorem on each open interval of an open set G containing A . ) S C E, then there is a measurable set A C E such that S C A and m (A ) = m*(S). (Hint : take an intersection of a decreasing sequence of
1 6.9 If
approximating open sets.)
16.10 Prove Proposition
1 2. 2.
16.1 1 Verify that the following collections are countably additive classes of sets: (a) (b)
the collection of all subsets of E. the collection of all measurable subsets of E.
1 2. 1 is called "closure under countable unions. " Show that the class of measurable subsets of E is not closed under uncountable unions ; i.e. , there exists an uncountable collection of measur able subsets of E whose union is not measurable.
16.1 2 Part (3) of Definition
1 6. 1 3 Show that any arbitrary intersection of countably additive classes countably additive class.
16.14 (a) (b) (c) (d)
is
a
Show that every open set of E is a Borel set. Show that every closed set of E is a Borel set. Show that every half-open interval (a,b ] C E is a Borel set. Show that Q n E is a Borel set.
16.15 Show that if f:E -+ E is continuous and A C E is a Borel set, then r 1 (A ) = {x E Elf(x) E A } is a Borel set. ( J1int : show that the collection 1 of all sets A such that {- (A ) is a Borel set, is a countably additive class which contains all the open intervals.)
a be an integer bigger than 1 . An a'(lry expression for a number x E E is an expression of the form . a1a 2 a3a4 , where each 0 < a; < a, and · where x = 11; + 11! + 11: + . . . . The series on the right converges by II II • comparison with the convergent geometric series :E 11n. The decimal n= l ll (a = 1 0), binary (a = 2), and ternary (a = 3 ) expressions for a number in E
1 6. 1 6 Let
•
•
•
are special cases.
(a) (b) (c) ( d) (e)
Find the rational number whose 4-ary expression is .2222 . . . . Find the rational number whose S-ary expression is .2 1 20202020 . . . . Find the binary expression for Find the ternary expression for .! . Show that every number x E has an a-ary expression. (Hint : let
l
E
P ROP E RT I ES OF M E AS U R A B L E SETS
a1
= the greatest integer
43
b such that !a < x.
Then
a1 b a2 = the greatest integer b such that -a + 2 a < x, etc. Prove that the resulting series converges to x .)
(0
a1a2
a,. OOO
Show that for any a-ary expression of the form . , with "¢ 0, there is a different expression with the same value. (In ternary, for example, .02000 . . . = .01 222222 . . . . . )
a,.
•
•
•
•
•
•
= { x E El a binary expression for x has a 0 in every even position}. Show that B is m easurable and m(B) = 0. (Hint : write B as the intersection of a decreasing sequence of sets.)
1 6.17 Let B
16.18 Let f:C -+ E be defined by /( . a1a2a 3
�2!.. ; ;
, where the " " ) = expression on the left is ternary with no 1 's, and the expression on the right is binary. Prove that f is onto but not 1 - 1 . ( See Exercise 1 6 . 1 6(0.)
16.19 Prove that the function f in Exercise
•
•
•
•
•
•
1 6. 1 8 is continuous.
16.20 Prove that C is uncountable. (Hint : use the ternary characterization of C, and a modification of the usual diagonal argument used to show that
(0, 1 ) is uncountable. ) 1 6.21 (a) (b)
Prove that the Cantor set C is nowhere dense in E (contains no in terval of E.). Prove that every point of C is a limit point of C ; that is, for every E C and every € > 0, there is b E C with - b €. A closed set with this property is said to be perfect.
a
Ia
I<
A is nowhere dense in E (Exercise 16.2 1 ), then dense in E (i.e. , every point of A is a limit point of E\A ).
16.22 Show that if
16.23 Construct a perfect nowhere dense set D follow the construction of Example middle third at each step.)
E\A
C E such that m(D) = l (Hint:
1 2.5,
but throw away less than the
16.24 Show that for any € > 0, there is a perfect nowhere dense set Ce such that m(Ce) > 1 - E. 1 6.25 (a) ( b)
1
A2
is
A1
A2
CE
Prove that A U and are measurable, is measurable if using only the Caratheodory criterion (Theorem 1 3 . 1 ) and properties of m * . Repeat part (a), using the criterion of Theorem 1 3 .2.
44
L E B ESG U E I NT E G R A T I O N A N D F O U R I E R S E R I ES
C E, m,.. (A ) = lub { m*(F) I F is
16.26 Prove that for any A
1 6.27 Given A , B C E, define the symmetric difference A
closed and
F C A }.
tl B, by
A tl B = (A \B) U (B\A ). (a) (b) (c)
Prove that if A and B are measurable, so is A tl B. Prove that A is measurable if and only if for every e
> 0, there is a n finite union U It of open intervals such that m*(A tl U 11) < e. �· �· Prove that if A is measurable and m"'(A tl B) = 0, then B is measurable and m(A ) = m(B). n
16.28 Prove that for A glb
C [a,b ] ,
{ m*(G) I G open in [a,b ]
and A
C G}
= glb { m*(G) I G open in � and A C G}. A is a measurable set with positive measure, then A has a non-measurable subset. (Assume A is bounded if you need to. Follow the construction of Example 7.7. ( See also Corollary 1 0.8.) You may wish to try A = [a,b] first.)
16.29 Show that if
16.30 Show that if A is bounded, then A is measurable if and only if A
is measurable for every positive interger n , and in that case
n [-n,n ]
m(A ) = nlim m(A n [-n,n ] ). +• 1 6.31 Find m(Q), m(6l\Q),
m(Z) (where Z is the set of integers).
16.32 Prove that (even for unbounded sets) if A and B' are measurable, so are
A U B, A n B, and A\B.
16.33 Prove that if
A
and
B
are measurable (not necessarily bounded), then
m(A ) + m(B) = m(A U B) + m(A n B). 16.34 Prove that if
A t is
measurable for
are measurable, and
00
i = 1 ,2, . . .
00
, then
U A1
i=l
00
and
nA, i =l
00
m( 1U= 1 A1) < i1; m(A 1). =l
16.35 Prove Corollary
1 0.9
in the case of unbounded sets
A1, i = 1 ,2, . .
1 0. 1 0 is false for unbounded sets A1, i = 1 ,2, Where does the proof of Corollary 1 0. 1 0 break down in this case?
16.36 Show that Corollary
.
...
PRO PE RT I ES OF M E AS U R A BLE S ETS
45
16.37 Show that every compact set is measurable with finite measure. 16.38 Show that if A is bounded, m*(cA ) = Also show that
16.39 Prove Theorem
cA
l c lm*(A), where is measurable if A is measurable.
1 2.9.
cA
=
{cx l x E A }.
C HAPTER
4
Measurable Fu nctio ns
1 7. Definition of Measurable F unction At the beginning of Chapter 2, we indicated that our definition of the Lebesgue integral for l:[a,b] -+ � would require finding the measure of sets of the form E1 = {x E [a,bJ I Yt- 1 < l(x) < y1}. We have nearly accomplished this, but unfortunately we found that not all sets can suc cessfully be assigned a measure. We have shown that if we restrict our selves to measurable sets, then we do have a measure m. Therefore, we must require that sets of the form E1 be measurable. Since E1 must be measurable for arbitrary y1_ 1 , y1, we evidently must restrict ourselves to functions I for which {x E [a,b] lc < I< d} is measurable for every c < d. It turns ,out that we can relax this condition slightly and still get the same class of functions (see Proposition 1 7 .2). We would like to allow our functions more general types of domains than intervals of the form [a,b ], so with section 14 of Chapter 3 in mind we make the following convention. From now on A will always be a bounded measurable set.
17.1 Definition : Let A be a bounded measurable subset of �- Then f:A -+ � is measurable on A if {x E A lc < l(x)} is measurable for every real number c. 47
48
L E BESGU E I NT E GR AT I ON A N D F OU R I E R S E R I ES
in
Notice that the definition there are infinitely many sets to check for measurability, one for each real number We will delete the phrase "on if it is clear from context what the set is. Notice also that the function may be unbounded on
A"
f
c.
A.
A
Now we need to show that the class of measurable functions satisfies the requirements we discussed before the defmition.
A
17.2 Proposition: Let be bounded and measurable and following are equivalent :
f:A. -+ 6t Then the
(1) (2)
{x EA I c
(In other words, any of the conditions ( 1 ) - (8) could be taken as the defmition of measurability. After proving their equivalence, we will use the conditions interchangeably.)
,. (2). Surely c < f(x) if and only if for every natural n, c - 1 /n <{(x). Therefore {x EA lc
number
but the latter set is measurable by ( I) and preservation of measurability under countable intersections (Corollary 10.1 1 generalized to arbitrary bounded sets).
(2) ,. (3). It is clear that and the latter set is measurable by (2) and preservation of measurability under relative complements. (Corollary 8.4 and Corollary 10.3).
{x E A I /(x) < c} = A\{x EA i c
(3) ,. (4)
Exercise 20.1 .
(4) ,. (1)
Exercise 20.2.
M EAS U R A B L E F U NCT I O NS
49
We may now use ( 1 ) - (4) interchangeably in the remainder of the proof. (1) - (4) ,. (S). Notice that {x EA jc B if and only if B is measurable. In fact, t Hfc ;illo l. {x EA l:xs(x) > c} f}B if O < c < 1 . tA ifc < O. This means that XQ and XA \Q are measurable functions on A but X v is not measurable on [0,1] , where V is the non-measurable set in Example 7.7.
17.3 Example:
=
For every set A with m(A) > 0, there is a function f:A -+ � which is not measurable on A (see Exercise 20.4). Now we mention a characterization of measurability which is remi niscent of continuity. Recall that a function g:A -+ fl is continuous on A if and only if g- 1 (G) is open in A for every open set G C �- If we relax this condition to require only that g- 1 (G) be measurable, then we have measurability of g.
1 7A Example:
function /:A -+ fl is measurable if and only if r 1 (G) is measurable for every open set G C fl.
1 7.5 Theorem : A
50
L E B ESGUE I NT E G R AT I ON A N D FO U R I E R SE R I ES
A
=
Proof: Let f be measurable . Let G c It be open . If G = It, then f- 1 (G) is measurable . If G =I= It, then Theorem 7.2 yields G = Ul;,
i
11
where the are disjoint open intervals, at most two of which are infinite of the form (-a�!) or for real numbers. Since the order of the I;'s is irrelevant, let us suppose that I -•, c) and I 2 = and all the remaining I ;'s are of the form for real numbers and
(d,•)
c,d ( (d, •) (a1,b1) 1 a1 b1• Then r 1 (G) = Q f- 1 (11) , and by the preservation of measurability under unions (Coroll �i}J 10.1 1), it is sufficient to show that f- 1 (1;) is measurable for each i = 1 ,2,3 , • • • . Butf- 1 ((-•,c)) = {x EA lf(x) < c} ; /- 1((d,•)) = {x E A l f(x) > d} ; and f- 1 ((a1, b;)) = {x EA i a1 <{(x)
are all measurable since f is measurable .
For the converse see Exercise 20. 1 1 .
0
Another possible analogy with continuity would be to expect that f- 1 (B) should be measurable for every measurable set B. However there is a measurable function / for which this is not true. (See Exercise 20. 1 2.) Theorem 1 7.5 has a couple of immediate corollaries. 17.6 Corollary: If
f:A -+ It is continuous, then / is measurable. f:A -+ 6l is measurable and g:f(A) -+ It is continuous, then
17.7 Corollary : If g • f is measurable.
Proof: Exercise
0
20. 13.
Surprisingly , composition in the opposite order does not necessarily yield measurability. There is a measurable function g and a continuous function f such that g • f is not measurable. (See Exercise 20.12.) It follows that the composition of two measurable functions need not be measurable.
1 8. Preservation of Measurability for Functions
Measure theory provides a concept of "almost equality" for functions which is very useful.
almost
18.1 Definition : We will say that f equals g everywhere (abbreviated a.e.) on if the set where they differ has measure zero. That is, f = g (a.e.)
A
MEAS U R A BLE F U NCT I ONS
51
on A ifm({x EA lf(x) =Fg(x)}) = 0. Because subsets of sets of measure zero also have measure zero, this is equivalent to saying that there is a set B C A with m(B) = 0 and f(x) = g(x) for every x EA\B. We will extend this terminology to other circumstances as well. In general "almost everywhere" will mean "except on a set of measure zero." Thus, "f is continuous a.e. on A" means "there is a subset B C A such that m(B) = 0 and fis continuous on A \B." Now a simple but sometimes useful result. If f is measurable on A and f = g (a.e.) on A, then g is measurable on A.
18.2 Proposition :
Proof : Le t
B = {x EA lf(x) =Fg(x)}. Then m(B) = 0 and {x EA ig(x) > c} = {x EBig(x) > c} U {x EA\Big(x) > c} = {.x EB ig(x) > c}U {x EA\Bif(x) > c}. In this last union, the first set is a subset of B and hence is measurable (with measure 0). The second set is {x E A lf(x) > c} n (A \B), hence is
measurable.
0
Measurability is also preserved under many common manipulations of functions. If f is measurable on A, so are lfl, f , 1/[(if f(x) =F 0 for all x EA ), and .fl (iff(x) > 0 for all x EA).
18.3 Proposition :
Proof: This follows immediately from Corollary 1 7.7. For example 1!1 = g • f, where g(x) = lx I , a continuous function. 0
You should be able to extend the list in the above Proposition ad infinitum (see Exercise 20.14).
The same nice behavior also holds under algebraic combination of functions.
52
L E B ESG U E I NTEG R ATION A N D FOU R I E R SE R I ES
If f and g are measurable on A , so are f + g, fg, and f/g (for g non-zero).
18.4 Proposition :
Proof : A slick proof like that of the previous proposition is possible, but we would have to consider measure on sets in �2 • Directly, we note that for fJXed x E A ,
f(x) + g(x) > c if and only if f(x) > c -g(x) if and only if
there is a rational number q such that f(x) > q > c -g(x). Therefore, {x EA i f(x) + g(x) > c} = U {x E A if(x) > q and q > c - g(x)} qeQ
= U ({x E A if(x) > q} n {x E A ig(x) > c - q}) . qeQ
the sets involved are measurable since f and g are measurable and Q is countable.
All
For fg, notice that fg = i if + g)2 - t(f- g)2 • The result follows from the preceding. (How do you deal with the i?) Finally, f/g f(l /g).
D
=
1 8.5 Corollary:
measurable. Proof:
If f and g are measurable, then {x E A lf(x) > g(x)} is Exercise 20.21 .
D
Most important for us will be preservation of measurability under limits. The following results, in fact, form the underpinnings of the basic convergence theorems involving the Lebesgue Integral (see Chapter 6). If fn :A -+ � is measurable for each n 1 ,2, • • • and if f is the pointwise limit of {fn } (that is, f(x) = nlim + oofn(x) for all x E A), then f is measurable on A.
18.6 Theorem :
Proof:
=
We must carefully analyze the meaning of the limit. Let
x E A . Then, if f(x) > a, let p be a natural number such that f(x) > a + l/p. Then for large enough n, fn(x) > a + l/p. That is,
M E AS U R A B L E F UNCT I ONS
53
(Hp)(HN) (V n > N) lfn(x) > a + 1 /p) .
Conversely, if this condition is satisfied, then /(x) = lim /n(x) n +•
:> a + 1/p > a .
Therefore,
{x EA.I /(x) > a} = pu= l N=ul n=N+ n I {x EAI /n(x) > a + 1/p} ,
and this set is measurable.
0
Related to this result is the following. If Un l is a pointwise bounded sequence of functions (that is, {/n(x)ln = 1 ,2, • • • } is bounded for each x E A) each measurable on A , then f(x) = lub {/n Cx) l n = 1 ,2, • • • } is measurable on A . Similarly, glb {fn(x)l n = 1 ,2, • • • } is measurable on A .
18.7 Proposition :
Proof:
This follows from the equation
{x !f(x) > a } = n9 t {xlfn(x) > a} .
0
Under the conditions of the proposition, g(x) = nlim fn(x) +• and h(x) = lim fn(x) are measurable.
18.8 Corollary:
Proof: Use the relationship g(x) = lim (Iub UnCx)l n k +• a similar relationship for h(x).
> k}), and
0
1 9. Simple F unctions
In discussing the Lebesgue Integral in the next chapter, we will have need of some particularly uncomplicated measurable functions. These will play a role analogous to that of step functions in the Riemann theory. 19.1 Definition : A simple function
finitely many values.
g :A
19.2 Proposition : A function g :A -+ tR numbers b 1 , • • • , b n and pairwise
n n with A = i=ul B1 and g = i=l:l btXs ·· I
-+
tR
is a measurable function with
is simple if and only if there are real disjoint measurable sets B�o . . . ,Bn
54
L E B ESG U E I NTEG R AT I O N A N D FOU R I E R S E R I ES
Proof: If g is simple on A , let its distinct values be b I > · • • ,bn . Then by Exercise 20.4, B1 = {x E A lf(x) = b1} is measurable for each
n B1 n B1• = � for i =I= j and i=U1 B, = A . Therefore, g = 1� b{XBi • since if x E B;. XB1{x) = {� :� � ; : . so that 1 n g(x) = b1 = lJ1 b1 XB ,(X). i =
1 ,2, •
• •
,n.
Furthermore,
For the converse, see Exercise 20.27.
0
If g is simple, then any sum of the form in the Proposition will be called a representation of g. Clearly every simple function can have many different representations. (How many?) Simplicity is preserved by algebraic combinations.
1 9.3 Proposition: If f:A (for g non-zero).
-+
IR, g:A
-+
Proof: Exercise 20.30.
IR are
simple, so are f
+ g, fg, and ffg 0
Simplicity is not preserved under limits. However, the pointwise limit of a sequence of simple functions is measurable by Theorem 18.6. The converse i s also true.
19.4 Theorem : A function {:A -+ IR is measurable if and only iff is the point· wise limit of a sequence of simple functions on A . Proof: (•) . Theorem
18.6.
(�) . Let Pn be a partition of [-n,n] (on the y -axis) ob· tained by taking equal sub-intervals of length 1/n. (Thus
+ i/n for i = 1 ,2, • • • , 2n 2 .) 2 n2 Now let B; = { x E A IYi - 1 < f(x) < y,} , and let In = � Yt - 1 "XB i · i- 1
where Yt = - n
• Simp • 1e. Then 1I"n IS
Now given x0 E A , for each n big enough so that f(x0) E [-n,n] , Yt- 1 < f(x o) < Yi for some i = 1 ,2, • • • ,2n 2 • Thus x0 E Bi and 1/(x o) - /n(X o) l = lf(x o ) - Yt- 1 1 < lY; -Yt- 1 1 = 1/n.
MEAS U R A B L E F U NCT I O NS
So lim• /11(Xo) = f(xo) for all x0 E A .
55
0
n•
If f is bounded below (above), then it is possible to modify the construction in the proof of the theorem so that the sequence /11 is in creasing (decreasing). See Exercise 20.32. I f f is bounded above and below, we can do even better.
19.5 Corollary: A function f:A _. 6?. is bounded and measurable if and only if f is the uniform limit of a sequence of simple functions. Proof: If / is bounded , then for large enough n , f(x) E [-n, n] for all x E A . Therefore, the calculation in the proof of the Theorem shows that l/11(x) -f(x) I < 1 /n for all x E A . Unifonnity follows. For the converse, see Exercise 20.35.
0
20. Exercises
20.1
Prove ( 3)
'*
( 4) in Proposition 1 7. 2.
20.2
Prove ( 4)
=>
( 1 ) in Proposition 1 7. 2.
20.3
Prove the remainder of Proposition 1 7. 2.
20.4
(a) Prove that if f is measurable, then {x E A I c = f(x)} is measurable for each real number c. (b) Show that the converse of (a) is false if m(A ) > 0. (Hint : by Exercise 1 6 . 29, there is a non-measurable set B C A .)
= 0, then every function is measurable on A .
20.5
Prove that if m(A)
20.6
If B is any set, f:B _. 6?., and { x E B l f(x) real number c, show that B is measurable.
20.7
< c}
is measurable for every
Prove directly from Proposition 1 7.2 that each of the following functions is measurable on [ 0 , 1 ] . (a) f(x) = 3 for all x E [ 0, 1 ] . (b) f(x) = x for all x E [ 0, 1 ] . P /� if x E ( O, l ] (c) f(x) = l o 1f x = o.
20.8
If B C A. B measurable, f:A on B .
20.9
If f:A U B -. 6?. is measurable on on A U B, on A n B, and on A \B.
_.
6?. measurable on A , then f is measurable
A
and on B, prove that f is m easurable
56
L E B ESG U E I NT E G RATION A N D F OU R I E R SE R I ES
20.10 If f is measurable on A and m(B) = 0, show that f is measurable on
A U B.
1 -+ 6l and f- (G) is measura G C 6l, then f is measurable on A .
20.1 1 Prove the remainder of Theorem 1 7. 5 ; if f:A ble for every open set
20.1 2 Let C be the Cantor set (Example l 2. S and Exercise 1 6 .2 1 ). Let D C [ 0, 1 ]
be a nowhere dense measurable set with m(D) > 0 (Exercise 1 6. 22). Then there is a non-measurable set B C D (Exercise 1 6.29). At each stage of the construction of C and of D, a certain finite number of open intervals of [ 0, 1 ] are deleted (put into [ 0, 1 ] \C or [ 0, 1 ] \D). Let g map the in tervals put into [ 0, 1 ] \D at the nth stage linearly onto the intervals put into [ 0, 1 ] \c at the nth stage, for n = 1 ,2, . . . . ( See illustration.) Thus g is monotone and defined at every element of [ 0, 1 ] \D, m apping onto [ 0, 1 ] \C.
I
cl
0
I
I
D
a afler stqa 2
(a) Using the fact that g is increasing and that [ 0, 1 ] \D is dense in [ 0, 1 ] (Exercise 1 6. 22), prove that for every x0 E D,
l
and
lim g(x) = lub {g(x) x E [ 0 , 1 ] \D and x < xo } x +xo
lim g(x) = glb {g(x) J x E [ 0, 1 ] \D and x > xo}. x +xt (b) Use (a) to show that for x0 E D , lim g(x)< lim g(x). x +x0 x +xt (c) Show that if a = lim g(x) < lim+ g(x) = b for some x o E D, then x +x0 x +x0 C would contain the interval (a,b), contradicting the fact that C is nowhere dense. Therefore lim g(x) = lim g(x) = lim g(x), and x +x ; x +x 0 x +x� the function
MEAS U R A B L E F UNCT I O NS
fi(
x
)=
{
57
g(x) if X E ( 0, 1 ) \D ijm g(x') if x E D
x +x
is continuous and monotone. 1 (d) Let B o = /(B). Show that B0 is measurable but /- (80 ) = B is not, even though I is a measurable function. Since /- 1 (Bo) is not measurable, it is not a Borel set. By Exercise }6. 1 S , Bo is not a Borel set, even though it is measurable. (e) Using the fact that XB = XBo • /, show that composition in the opposite order in Corollary 1 7.7 is false.
20.13 Prove Corollary 1 7.7. 20.14 Show that if f is measurable, so is the function g defined by g(x) = eft.x >. 20.1 5 Show that if f is continuous a.e. on a compact set K , then for
every
C K such that f is bounded on A , and m(K\A ) < e. (Hint : modify the usual compactness argument for continuous functions.)
e > 0, there is a set A
20.16 (a) Prove or disprove : if I = g a.e. , and tinuous a.e. . (b) Prove or disprove the converse of (a).
g
is continuous, then f is con
20.1 7 Show that if f is continuous a.e. on a bounded measurable set A , then
I is measurable. (By Exercise 20. 1 6(b), you may not use Proposition 1 8. 2.)
20.18 Show that "equality a.e. " is an equivalence relation on the class of all functions on a bounded measurable set A .
20.19 Show directly from Proposition 1 7.2 that able if / is measurable.
VI, /2 , 1 /1 (/ ::1= 0) are measura
20.20 Prove or disprove: if 1/1 is measurable, so is /. 20.21 Prove Corollary 1 8. 5 . 20.22 Show that f i s measurable on A if and only i f /2 is measurable and {x E A l fi<x) > 0} is measurable.
20.23 If f and
g
measurable.
are measurable on A , show that {x
E A l t<x>
20.24 Let f and g be measurable on A , and define a function h by
h(x) =
{
:
/(x) x) if /(x) + g(x) ::1= 0 /(x) g(x) 0
if /(x)
+ g(x) = 0.
= g(x)} is
58
L E B ESG U E I NT E G R A T I O N A N D F O U R I E R SE R I ES
Show that
h is measurable on A .
20.25 Let /,. be measurable on bounded measurable set A , for n = 1 ,2, . . . . If B = {x E A I U,.(x)} converges} show that B is measurable. (Hint: look at the Cauchy criterion for convergence.)
20.26 Let /,. be measurable on A for n = 1 ,2, . . . . Let /,. on A . Show that for every real number c > 0,
f pointwise a.e.
1 :> c}) = 0. Note that m( n U { x E A l lf(x) - /,.(x) I :> c }) = 0.) k= l n = k l
lim m({ x E A lf(x) n +oo
(Hint :
-+
-
/,.(x)
n
20.27 Prove the remainder of Proposition 1 9. 2 ; if g = . E b;Xs,· , where the B; r= 1 are pairwise disjoint measurable sets with union A , then g is simple on A .
20.28 Show that every step function on false.
[a,b ]
is simple, but the converse is
20.29 Give an example of a simple function and two different representations of it.
m
20.30 Prove Proposition 1 9. 3 . Given representations f = . E b;Xs ; and n r= 1 g
= E
k= l
ck Xc k • find explicit representations of f
+ g, fg, f/g(g ::1= 0).
20.31 Suppose g is simple and f is any function. Under what conditions is f • g simple'? When is g · { simple'?
20.32 Prove that if f is measurable on A and f is bounded below, then / is the pointwise limit of an increasing sequence of simple functions. (Hint : at stage n , partition [ -n, n ] into 2n • 2" equal subintervals. Thus, in going from stage n to stage n 1 , each subinterval is split in half.)
+
20.33 Find an increasing sequence of simple functions on [ 0, 1 ] with limit /(x)
= x.
20.34 (a) Show that the simple function
Xc (where C is the Cantor set) is the pointwise limit of a decreasing sequence of step functions, but not of an in creasing sequence of step functions. (Hint : for the second part, you need the fact that C is uncountable and nowhere dense ( Exercise 1 6 .2 1 ).) (b) Show how to alter Xc to create a simple function which is not the pointwise limit of any monotone sequence of step functions.
20.35 Prove the remainder of Corollary 1 9. 5 ; if / is the uniform limit of simple functions, then f is bounded and measurable.
20.36 Prove that if f is monotone on a bounded measurable set A , then f is measurable on A .
CHAPTER
5
The Lebesgue Integ ral
21 . The Lebesgue I ntegral for Bounded Measurable Functions
f
Let be a bounded measurable function on a bounded measurable set C dt. We wish to defme the Lebesgue integral of f on As we said in Chpater 2, this involves partitioning the range of f, rather than the do main as for the Riemann integral. More precisely , we need to partition an interval [ll, u) such that ll < u for all This may be accom· plished by taking
A
A.
x EA .
ll =
glb {f(x)lx E A},
u=
1 + lub{f(x)lx EA}.
and
The evident arbitrariness of this choice will be eliminated as we proceed ; that is, we will se� that any choice of ll and u-as long a s i t satisfies the inequalities-will yield the same integral . Now for our approximation.
f
21 .1 Definition : If is bounded measurable function on a bounded measurable set C dt, ifP = is a partition of
A
(y0,y�r . . . ,yn) [ll,u} = [glb{f(x)lx EA}, 1 + lub {f(x) l x EA } } ,
59
60
LEBESGUE I NT E G R A T I O N A N D FOU R I E R SE R I ES
and if
Yt * E fy, _ . ,y1)
for i = 1 ,
. . . ,n,
then we will call
n L (/. P) = ��lt* • m({x EA ! Yt - 1
As indicated in Chapter 2, L (/. P) is an approximation to the area under the graph of f (iff is non-negative). It is the sum of areas of pseudo rectangles whose bases are not necessarily intervals, but rather are measura ble sets of the form A 1 = {x E A jy1_ 1 < f(x) < y1}. Note that the in equalities Yt- 1 < f(x) < Y; are chosen so that the sets A1 are pairwise disjoint and A = A 1 U A 2 U U A n . Now we take the limit of L(f.P ) as II P II -+ 0. .
.
•
21 .2 Definition: A bounded measurable function f:A. -+ � is Lebesgue in tegrable on A if there is a real number L such that for every e > 0, there exists a c5 > 0 such that if I IPI I < c5 , and if L(f.P) is a Lebesgue sum off relative to P, then IL(f.P) - L I < e. The reader should compare this with the definition of Riemann integrability (Defmition 1 .3). As for the Riemann case, we have unique ness: at most one number L can satisfy the definition. This unique L is called the Lebesgue integral off on A, and is denoted
f..t fdm. 22. Simple Functions
We would like to have a criterion of Lebesgue integrability which avoids mention of the limit L = f..tfdm, since the definition gives no clue as to how to fmd L . The relevant result for Riemann integrability (see Theorem 1 .6) involves approximating / above and below by step functions. The significance of step functions is that their Riemann integrals (Proposi tion 1 .5) are of the same form as Riemann sums. In our present situation we desire uncomplicated functions whose integrals are sums comparable to Lebesgue sums. It is not surprising that simple functions work (see section 1 9). Of course simple functions are bounded and measurable; they are also Lebesgue integrable.
THE L E B ESG UE I NTEG R A L
61
g is Lebesgue integrable on A . Furthermore, if g = .� b;XB; is any representation of g, then 1
22 .1 Proposition : If g:A
-+
n
R is simple, then
·-
fA gdm = t=1E b
·
I
•
m1B1). "
Proof: Note that we may assume without loss of generality that b; :1= b1 for i :1= j. {Otherwise , we could take unions of sets whose b;'s were the same. Since m(/11 U B1) = m(B;) + m(B1) for i :1= j, the result would be the same.) Now, given e > 0, let 0 < 6 < e/m(A) be so small that 6 < lb; - b1 l whenever i :l=j. I fP = (y0,y 1 , . . · Yk ) is a partition of [ll,u] (as in Definition 21 . 1 } such that II P II < 6 , then at most one b; can lie in each lYJ- 1 ·YJ). Therefore , for any
each set {x E A I.Y1 - 1 < g{x) < y1 } is either B1 (if b; E [y/_ 1 ,y1}}, or 0 {if no b; is in (y1_ 1 ,y1)). Therefore, eliminating the zero terms in which 0 occurs, there is a 1 - 1 correspondence between terms b1
•
m(B;)
Y/*
•
m(B;),
and terms
where
Thus
n
But, if b1 E [y1_ 1 ,y/}, then
so that
< 1=l:1 1y1 • - b1 I • m(B1).
62
L E B ESG U E I NT E G R A T I O N A N D F OU R I E R SE R I ES
n
Notice that the proposition implies that the sum i=l:l b1 • m(B1) is the same no matter what representation is chosen for g. lt is an important and useful fact that the Lebesgue integral, restricted to simple functions, satisfies certain nice properties expected of integrals. We will later extend these to other functions. 22.2 Proposition :
(1) (2) (3)
Iff and g are simple on A , then
fAfdm < fA gdm whenever {
Exercise 26.9.
D
If f is simple on A U B, where A and B are disjoint bounded measurable sets, then
22.3 Proposition :
Proof:
fA UB tdm = LA tdm + Jrs tdm.
Exercise 26.10.
D
23. Integrability of Bounded Measurable Functions
Now we state the criterion of Lebesgue integrability promised earlier. 23.1 Theorem : A bounded measurable
function { is Lebesgue integrable on a bounded measurable set A if and only if for every E > 0, there exist simple functions/1 and f2 such thatf1
L f2dm -JA !1 dm < e . Proof: lf /1 and /2 are simple and /1
THE L E B ESGUE I NTEG R A L
63
This set is non-empty and bounded above (verify!) By hypothesis, if e > 0, then there are simple functions{1 <.{<.{2 such that
L. f2dm -J.A. f1 dm < e/2.
For such simple functions, we have
(why?). Now let 6 = e/2m (A). All that remains to be proved is the following claim; the reader can verify that this will imply that f is Lebesgue integrable on A . Claim : I fP =
.
£ {1 dm - e/2 <. L (f,P) <. L /'].dm + e/2.
Proof of Claim : �
•
�
On A; = {x EA I Yt- 1 <.{(x)
and Therefore, Hence, on integrating over A 1 , by Proposition 22.2,
1 {1 dm
-
6m(A1) <.y1*m(A1) <. J..A. ; /'].dm + 6 m(A1).
Adding these inequalities, for i 22.3 and additivity of m : or
==
1 ,2, . . . , n , we obtain by Proposition
( ( l..t {1dm - 6 m (A) <. r=. �l y1*m(A 1) <. l..t hdm + 6m(A) ,
E {1 dm - e/2 <. L(f,P) <. £ {2dm - e/2.
To prove the converse, all we need to know is that f is measurable and bounded. In this case, if e > 0, let P {y0J11 , ,y,.) be any partition of [.IZ,u] such that II PII < 6 = e/m (A). Then if =
•
•
•
64
L E B ESG U E I NTEG R AT I O N A N D F OU R I E R SE R I ES
and if and
/1 = r=. l:l Yt - 1 XA t "
f2 = i=l;l yr. iVI v . r. ' "
then we have /1 < t <.f2, and The proof of the theorem yields several important corollaries. The first is a little surprising; it says that by restricting ourselves to measurable bounded functions in the definition of Lebesgue integrability, we have already eliminated aU non-Lebesgue integrable functions. 23.2 Corollary:
If f is bounded and measurable on a bounded measurable set
A , then f is Lebesgue integrable on A . Proof:
See the proof of the theorem.
0
Note that already we can see that many non-Riemann integrable functions are Lebesgue integrable. (Name some.) The next corollary is not so astounding, but it is of theoretical importance. In other treatments of the subject, it is often used as a definition for the Lebesgue integral. We will make use of it ourselves to extend our defmition to unbounded measurable functions (see section 25.) 23.3 Corollary:
A , then
Proof:
23A Corollary :
If f is bounded and measurable on a bounded measurable set
£ tdm = lub�£ !1 dm l t1 is simple and /1 < .rf = glb f £ f2dm l f2 is simple and f2 ;>.rt. See Exercise 26.1 2.
If m (A) = 0 and fis bounded and measurable on A , then
£ tdm = O.
0
T H E LE B ESG UE I NTEG R A L
Proof:
65
See Exercise 26.13.
0
24. Elementary Properties of the I ntegral for Bounded F unctions
The Lebesgue integral shares with the Riemann integral the nice properties stated in the following theorem. If f and g are bounded and measurable on the bounded meas urable set A , then:
24.1 Theorem :
(Monotonicity) If {<.g, then fAfdm <.fAgdm ; (2) If there are real numbers fl and u such that fl <.{<. u, then fl • m(A) <.fA fdm <. u • m (A ); (3) (Linearity) fA cfdm = cfAfdm for c real; (4) (Linearity) fA (f+ g)dm = fAfdm + fAgdm; (I)
(S)
lfAfdm I <. fA lfl dm.
Proof:
(1) If {1 <.{, then /1 <.g. Hence by Corollary 23 .3,
£ tdm = lub J£ t1dm l t1 <.{,{1 simple{ <. lub JL g 1dm l g 1 <.g. gl simpl� = L gdm.
(2) Use (1), noting that g(x) = fl and h(x) = u are simple functions, so that their integrals over A are easy to compute. (3) Follows from the identities (which the reader should verify)
(
c · lubl £ t.dm l t1 < t.f1 simple = lub J£ cfldm l t1 <.t.f1 simple /
= lub \ £ g1dm l g1 <. cf,g 1 simple l
Note that cf is measurable and bounded on A iff is.
(4) Recall that f + g is measurable and bounded on A , hence Lebesgue integrable. Let e > 0. By virtue of Corollary 23 .3, there are simple functions !1 <.{, g 1 <.g, such that and
Therefore,
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LE BESG U E I NTEG R AT I ON A N D F O U R I E R S E R I ES
Since e was arbitrary, we have
1 (f+ g)dm ;;. r fdm +1 gdm. •A
A
A
The reverse inequality is proved similarly, using the fact that
£tdm = glbJ£t2dm jt2 ;;.!. !2 simple f.
(Exercise 26.14.)
(5) Note that I ll is bounded and measurable. Also f < 1/1, and < Hence (5) follows from (1) and (3). 1/1. -t 0 Another fundamental property which the Riemann integral has is additivity on intervals:
Of course, for the Lebesgue integral we are no longer restricted to closed intervals, but may integrate over arbitrary bounded measurable sets. Our version of additivity might be
LA fdm + ·{.'B tdm = Jcr fdm,
where C = A U B disjointly (actually, the intersection could contain a set of measure 0, as we will see). It turns out that an even stronger kind of additivity, countable additivity, holds. But first the simpler property. 24.2 Theorem :
If A and B are disjoint bounded measurable sets, and
f:A U B -+ tR is bounded and measurable, then L fdm = £ fdm + £ fdm. Proof: Note that f is bounded and measurable on each of A, B, so UB
that the integrals all exist. Now let g1 be simple on A such that g 1 (x) < t(x) for all x E A , and let h 1 be simple on B such that h 1 (x) < f(x) for all x E B. Then the function /1 defined by ifx EA ifx E B, is simple on A U B, and /1 < f on A U B. Furthermore, by additivity for simple functions (Proposition 22.3),
·
TH E L E B ESG UE I NTE G R A L
67
£ K1 dm + k ht = �ftdm + kf1dm 2 �3L us ft dm 2�L u s fdm .
Therefore, taking lub's, we obtain
£ tam + fs tdm < £ u 8 fdm (Corollary 23.3). On the other hand, if {1 is simple and if {1 <. f on A U B, then /1 < t on A and on B separately, so that £ u s ft dm = £ ftdm + J8tt dm <. £tdm + J8tdm . Again taking lub's, we get f...t usfdm
24.3 Theorem :
=
r JAr fdm = r=.�1 JA ; fdm. Proof: We need to show that lf...t fdm - � f...t tfdm I becomes i=l U A1, finite additivity (Theorem arbitrarily small as n -+ 00• But if Bn = i=n+l 24.2) and induction give
I JAr fdm - 1=1JA.j � r fdm l 24.21J = I A.f fdm - fn tdm l i=l UA t
Given e > 0, there is a simple function {1 on A with 0 <. !1 <. Ill on A, such that f...t f1 dm > f...t lfldm - e/2 (Corollary 23 .3). It follows that for all n ,
J,Bn ft dm > J,'Bn l fl dm - e/2. (Exercise 26. 1 6). But {1 is bounded; say {1 (x) <. u for x E A . Hence by Theorem 24. 1 ,
68
LE BESG U E I NTEG R AT I O N A N D F OU R I E R SE R I ES
Since { B,.} is a decreasing sequence of measurable sets with empty inter section, nlim m(B,. ) = 0 (Corollary 10.10). Thus for large enough n , +• u • m(B,.) < e/2 . Therefore, ·
1; r tdm l < Jrs,. ltldm < Jrs,. t.dm + et2 1 JAr tdm - i=I)At
< u • m(B,. ) + e/2 < e.
0
If we define ll(B) = /8/dm for some bounded measurable function f:A -+ dl, where B C A , and A is bounded and measurable, then 1J. is a set function defined on all measurable subsets of A . It is countably additive by Theorem 24.3. If f"> 0, the ll is a measure (Defmition 6.1 and Exercise 26. 1 7). If f(x) = I for all x E A , then ll = m, the Lebesgue measure. As a direct consequence of additivity, we have a couple of important and easy results which say that sets of measure 0 don't matter in integra tion. If A is a bounded measurable set, B C A has m(B) = 0, and fis bounded and measurable on A , then fAfdm = fA 'VJ{dm.
24.4 Proposition :
Proof:
Exercise 26.18.
0
If f is bounded measurable on A , and g is measurable on A , and f = g a.e. on A , then fA fdm = fA gdm.
24.5 Proposition :
Proof: Strictly speaking, g may be unbounded, so that fA gdm is undefmed. In section 25, we will extend the definition of the integral to cover unbounded measurable functions. In the meantime, the previous proposition suggests how to deal with a function like g, which is "almost bounded." See Exercise 26.19 for details. 0
The last two propositions can be used to extend previous results to slightly more general situations. For example, in Theorem 24.2, if we assume m (A n B) = 0, rather than the stronger property of disjointness, then still /A us fdm = fA fdm + fs fdm. 25. The Lebesgue I ntegral for U nbounded Functions
The definition of the Lebesgue integral given in section 21 (Defini tion 21 .1) will not work for unbounded functions, since {/(x) jx E A }
THE L E B ESG UE I NTEG R A L
89
will not have both a finite lub and a finite glb. To extend the definition to unbounded functions, therefore, we must look further. We choose to take Corollary 23 .3 as our starting point. It says that for a bounded measurable function I on a bounded measurable set A ,
£ tam = lub l £ ttdm j ft simple andft < ll = glb l £ f2dm j12 simple and /2 > J'f·
We are going to take this property which holds for bounded measurable functions, and use it as the basis for a definition for unbounded measurable functions. However, there are difficulties. Namely, iff is unbounded above, there will be no simple functions 12 with 12 > f; and if f is unbounded below, there will be no simple functions ft with ft < f. That is, the sets in question could be empty, and the lub and glb may not make sense (may not exist as real numbers). The solution is to consider first only non-negative functions f. For such functions, there are always simple It with ft
non-negative.
Let A be bounded and measurable, f:A -+ � measurable and Define fAfdm = lub{fAftdm l ft simple andft <. /}.
Note that if {/Aft dm I ft simple and ft < /} is unbounded above, then the lub is not a real number. In that case, we take the lub to be -; that is, we say fAfdm = -.
Because of Corollary 23 .3 , this definition is consistent with the definition of the Lebesgue integral for bounded measurable functions. Perhaps it has occurred to you that there is no necessity to restrict the definition to measurable non-negative functions. The set {fAftdm l/1 simple and {1 < /} is non-empty in that case as well. The reason for the restriction is that measurability is required to prove some of the nice limit properties of the Lebesgue integral which appear in the next chapter. Now, to deal with an arbitrary measurable function f, we split it into positive and negative parts: let
f+ (x) = f_ (x) =
>O l �(x) if/(x) if/(x) < O. l --j'(x) if /(x) < 0 0
ifl(x) > O.
70
L E B ES G U E I NT E G R AT I O N A N D F O U R I E R SE R I ES
Then it is immediate that
f- = max { -{,0} = 21 ( 1 {1 - f) ;> 0, Therefore, we might expect JA.fdm = fA.f+dm - JA.f-dm. The only dif ficulty arises if fA.f+dm = fA.f-dm =00, since 00 cannot reasonably be given a value. - oo
Let A be a bounded measurable set, and f:A -+ R measurable. Then f is integrable on A if at least one of fA.f+dm or fA.f_dm is finite, and f is summable on A if both fA.f+dm and fA.!_dm are finite. In either case, define fA.fdm = fA.f+dm - fA.f_dm.
25.2 Definition :
Of course, in the definition, 00 - c is called and c - 00 is called for every (finite) real number c. Linearity of the Lebesgue integral for bounded functions (Theorem 24. 1) shows that this definition is still con sistent with the one for bounded functions. oo ,
-oo
As one would hope, the nice properties proved in section 24 for bounded functions carry over to this new situation. The proofs remain the same, except as indicated. All references to Corollary 23 .3 now are to Definition 25. 1 and 25.2, of course. Let us introduce the notation l(A) = { f:A -+ R I f is summable on A}, for A bounded and measurable. (Additivity) If A and B are disjoint, bounded measurable sets and{El(A U B), then {E l(A ) U l (B), and lA. us fdm =fA.fdm + !Jdm.
25.3 Theorem :
Proof:
and
By the proof of Theorem 24.2,
r
JA.
UB
f+dm = JrA. t.dm + r f+dm Js
( JA. us f_dm = Jsr f_dm + Jsr f_dm.
T H E L E B ESGUE I NT E G R A L
The result follows by subtraction (Definition 25 .2). 25A Theorem :
71
0
Let A be bounded and measurable,/,g E .t:(A). Then
(I) (Monotonicity) If{< g, then fAfdm
m(A) <£ {dm < u m(A); (Linearity) c{E l(A) for every real number c, and i cfdm = cL fdm: (Linearity) f + g E .C(A ), and fA (f+ g)dm = fAfdm + fAgdm; 1/I E .C(A) and lfAfdm l
(3) (4) (5)
•
•
Theorem 24.1 (1), and Definition 25.2.
(2) By ( l) of this theorem. Note that since f is bounded in this case, part (2) of this theorem is the same as part (2) of Theorem 24. 1 . (3) Apply the proof of Theorem 24.1 to f+ (5) Let B = {x E Alf(x) ;> 0}, and C by additivity (Theorem 25 .3),
=
andf_ . {x E A !f(x) < 0}. Then
JA l fldm = /, l fldm + J, l fldm £/+ dm + Jc f_dm = £ f+dm + £ f_dm, .B
C
=
and both of the latter integrals are finite by assumption (see Definition 25 .2). The remainder of the proof of this part is as in part (5) of Theorem 24.1 . (4) The proof of the corresponding part of Theorem 24.1 does not apply here, even for {;;;.. 0 and g ;;;.. 0, since we used
L fdm = glbl£ f2dmjf2 simple andfl ;;;.. tf. not valid for unbounded f. The simplest proof in our present
which is case involves using the Monotone Convergence Theorem of section 28. No circularity will result from using that theorem here.
72
L E B ESGUE I N TEG R AT I O N A N D FOU R I E R SE R I ES
Let us write h == f + g, and
F == {x EA j /(x) ;> 0},
F' == A\F;
G == {x EA j g(x) ;> 0};
G ' == A\G;
H == {x EA l h(x) ;> 0}; H ' ==A \H.
Note that A == (F n G) U (F n G ') U (F' n G) U (F' n G ') disjointly.We show the result holds on each of these intersections, then apply additivity (Theorem 25 .3).
(a) On F n G, by Exercise 20.32 there are non-decreasing sequences of simple functions {/,. } , {g,. } converging pointwise to /. g respectively. Then the sequence {/,. + g,.} is non-decreasing and converges pointwise to h == f + g. This Monotone Convergence Theorem says that lim {. /,.dm == ],•F nG fdm, n +•• F nG f lim J.FnG g,.dm == JFn gdm, G
n +•
and
lim
( hdm r (/,. + g,.)dm == Jpna
n +•JF n G
.
But since /,. and g,. are simple functions, Theorem 24.1 or Theorem 22.2 gives ( /,.dm + �( n G g,.dm. f.FnG (/,. + g,.)dm == �nG It follows easily that /p na h dm ==fFna fdm + fFna Kdm. (b) The result fp• na•h dm == /p• na •fdm + /p• n a•gdm follows by considering -/, -g, and -h on F' n G', and applying the technique of (a). . (c) On F n G ', consider the disjoint representation F n G ' == (F n G ' n H) U (F n G ' n H')
On F n G' n H, we have f;> 0, -g > 0, and h ;> O, so that/== h + (-g). Using the technique of (a), and part (3) of this theorem, we obtain { r h dm -J. fdm == .ftFna'nH JFna'nH Fna'nH gdm ·
THE L E B ESG UE I NTEG R A L
73
Similarly, on F n G ' n H', -g = -h + f, and -g, -h. /are all non-negative. Therefore our previous results give -L nG ' rlH' gdm = -�nG ' nH' h dm + k n G ' nH, fdm. Combining, we get fFflG 'fdm + fFnG•gdm =fFnG• hdm. (d) Similarly, fF'nG fdm + fF'nGgdm = fF'nG h dm. The sum of (a) - (d) gives the result; all the numbers (integrals) concerned are finite, ro f + g E l�� o (Countable Additivity) Let A be bounded and measurable, I E l(A), A = 1=U1A 1 , where the A; are pairwise disjoint measurable sets. Thenf.t /dm = 1�/.t,!dm.
25.5 Theorem :
00
By the proof of Theorem 24 .3 . f.t f+dm = 1=E1f.t tf+dm and f.t f-dm = 1=t1f.t tf- dm. Therefore, Proof:
00
L tdm = L t+dm - L t- dm = i=EU.tr 1 t+dm - i=E1}Ar t-dm t
r fdm. = r-.�1(' ( f+dm - JAr 1 f_ dm) = r-.�1JA; JA1
0
IfA is bounded and measurable, and B CA has measure O, and/E l(A ) , then f.t /dm = f.t\Bfdm. (2) If/E l(A) and g =fa.e. on A , then f.t fdm = f.t gdm . 0 The definition of the Lebesgue integral for unbounded functions may be approached in many ways other than the one we chose. You should be able, using the results of the next chapter especially, to prove that other defmitions you may encounter are equivalent to ours.
25.6 Corollaries: ( I)
26. Exercises
26.1
Pro e uniqueness of the Lebesgue integral. v
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L E B ES G U E I NT E G R AT I ON A N D FOU R I E R SE R I ES
26.2
Find two Lebesgue sums of f(x) = x 2 on ( -1 ,2] , relative to the partition (0, 1 /4 , 1 /2, 2/3 , 1 , 2, 4 , S) of [ ll, u] .
26.3
Show that if f is monotone on [a,b ] every Riemann sum of f is a Lebesgue sum. Is the converse true?
26.4
Where does boundedness of f come into Definition 2 1 . 1 ? Where does measurability of f come in?
26.5
Show directly from Definition 2 1 . 1 that if f is bounded and m(A ) = 0, then fA fdm = 0.
26.6
As an easy case of Proposition 22. 1 , show from the definition that if f(x) = c for x E A , then fA fdm = c • m(A ).
26.7
Show directlr from Definition 2 1 . 1 that if f(x) = x for x f1 o , 1 J fdm = 2·
26.8
Fmd f( o , 1 ] XQ dm .
26.9
Prove Proposition 22.2. (Hint : if f = write f =
n
:t .:t c1 Xc1 n01., etc.) k
:t Ct'Xc1 k
i= 1
and g =
E [ 0, 1 ] , then
n
.:t d ·XD -, then 1= 1 1 1
i= 1 J= 1
26.10 Prove Proposition 22. 3 . 26.1 1 Finish the proof o f Theorem 23. 1 ; show that the Claim implies that f is Lebesgue integrable on A .
26.12 Prove Corollary 23 . 3 . 26.13 Prove Corollary 23.4. 26.14 Prove the remainder of Theorem 24. 1 ( 4) ;fA fdm
-
+ fA gdm "> fA ({ + g)dm .
26.1 5 Let A = U A 1, where the A ; are disjoint and measurable and A i= 1
is
bounded. If f is measurable on each A t, show that { is measurable on A . Does the fact that f is bounded on each A i imply that f is bounded on A ?
< g is bounded and measurable on A , and B C A is measurable, show that fA gdm "> fsgdm . (b) If f < g are bounded and measurable, B C A is measurable, and fA fdm > fA gdm - e, then show that fs fdm > fsgdm - e without i e r ty of
26.16 (a) I f 0
using the l n a i
the integral.
26.17 If p.(B) = fs fdm , for ! "> 0 and bounded and measurable on a bounded
and measurable set A , show that 1J. is a measure on the measurable subsets of A .
TH E L E B ESG UE I N TE G R A L
75
26 .1 8 Prove Proposition 24.4 . 26. 1 9 Prove Proposition 24. 5. 26.20 (a) Prove that if f is bounded and measurable on A U B, where A and B are bounded and measurable, then
.ft u s fdm + .ft u s fdm .ft fdm + fs!dm. =
( b) If m(A
n B) = 0, show that fA u s fdm
= f..tfdm
+ fs fdm .
26.21 Prove that for any measurable function /, if m(A ) = 0,
then f.t fdm = 0.
26.22 Complete the proof of Theorem 2 5 .4 ( 1 ), ( 2) , (3), and (5). 26.23 (a) Suppose that (Hint: Let (b)
B
f ;>
0 and
f..tfdm
= { x E A l t(x)
f = 0 a.e. on A . E A j f(x) > 1 /n}. If
= 0. Show that
> 0}
=
n�l { x
m(B) > 0, obtain a contradiction.) Let f ;> g and f..tfdm = f..tgdm . Show that / = g a.e. on A .
f is measurable function such that fs fdm B C A , show that f = 0 a.e. on A .
26.24 If
f is measurable· on [a,b ] and I = 0 a. e. on [a,b ] .
26.25 If
= 0 for all measurable sets
fra.c J fdm =
0 for
all
c
< b , show that
26.26 Prove that if / is measurable on A and 1/1 <; g and g E l(A ) , then fE l(A ). 26.27 If
fE
l(A ) and
B
i s a measurable subset of A , show that
fE
.t(B).
26.28 If / E .t(A ) and g(x) = /(x) sin(ax), show that g E .t(A ).
f(x) = 1 /x is Lebesgue integrable on ( 0 , 1 ). Is it summable? Find f(o, l )fdm.
26.29 Determine whether
26.30 Find f(o,1 �dm if g(x) = x
_!.
2
26.31 Show that if f E .t(A ) and
26.32 If g E .t(A ) and A
::::>
E1
•
BEA
is measurable, then Xsf E l(A ) and
::::>
::::>
E2
..
•
•
•
, and
n=nl En
= 1,1, show that
76
L E B ESG U E I NTEG RAT I ON A N D FOU R I E R SE R I ES
E .t(A ) and e > 0, show that there exists a 6 > 0 such that whenever D C A and m (D) < 6, then l fD fdm I < e. (Hint : Let
26.33 If f
26.34
,
A,. = {x EA 1 1/(x) l < n}. Then find N such that I JA \A n fdm I < e/2. Let 6 = e/2N.) Let g, h E .t(A) and let a < g(x) < b for a.e. x EA. Show that there is a number c E [a,b ) such that fA g lh ldm = c(JA lh ldm).
26.35 In Riemann integration, the change of variable formula provides the following equation for g continuous: b
ia
g(u) du
= lce-a g(c - t) dt. -b
(Let t = c - u.)
A,
We can put this in Lebesgue terms as follows: given g summable on let c E where c is a fixed real number Let = g(c Prove that h is summable on and
B= h(x)
A = {c - x lx A }, x), x EB.
h+(x) = g+ (c - x) first, and let h 1 < h+ b e simple. Kl(x) = - x). Then K1 is simple and K 1 < K+ · Show that fAg 1 dm = fsh 1 dm (see Lemma 7.6 and Exercise 9.23). Conclude that fA gdm ;> fsh dm . Then prove the opposite inequality.) Let g E l [a,b ) , h(x) = f( a ,x ] gdm , for x E [a,b ) . Show that h is con ( Hint : consider Let h 1 (c
26.36
£ gdm = fs h dm .
B,
tinuous on
[a,b ) . (Hint:
consider first the case where g is simple.)
E l(A) D CA
26.37 Prove the following generalization of Exercise 26.39. If g and then for any e 0, there is a such that whenever is g measurable and we have JD gdm e. (Hint : Consider lg(x) < There exists an n such that fA \An gdm < e/2.)
;> 0, A,. = {x E A
> m(D) < 6, n}.
6>0
<
CHAPTER
6
Co n vergence and The Lebesgue Integ ral
27.
Examples
One of our reasons for being dissatisfied with the Riemann integral had to do with convergence properties. We are interested in results con cerning the integral of the limit bof a sequenceb of functions. In particular, under what conditions will nlim+oof fn(x)dx = L (nlim+oofn(x))dx? It is known on [a,bJ and all the fn and f are Riemann inte that if fn -+ f uniformly bfn(x)dx grable, then nlim f = f f(x)dx. (See Exercises 5 .25 and 3 1 .1 . ) +oo There are numerous examples in which fn -+ f pointwise on [a,b] , but nlim+oof bfn (x)dx + Lbf(x)dx. II
II
II
II
27.1 Example:
II
w
Let
{
when 1/2" � x � 1/2" - 1 fn(x) = 02"otherwise
for n = 1 ,2,3, • • • on [0,1 ] . Clearly fn -+ 0 pointwise, but ( 1 fn(x)dx = 1 + O = ( 1( lim fn(x))dx . nlim +ooJo Jo n + oo
It is tempting to suppose that the requirement of uniform boundedness l fn(x) l < M for all n, x) might remedy the situation, but the next example shows this is insufficient. (i.e.
77
78
LE B ESG U E I NTEG RATION A N D F O U R I E R SE R I ES
27.2 Example:
For each n = 1 ,2,3, • • • let
fn , k = X k k+ l J [VI • ¥
for k = 0, 1 , • • • , 2" 1 on [0,1 ] . Arrange the fn , k in a sequence {/p} first by order of n and then by k: /1 = /1 ,0 ; /, = /1 ,1 ;/] =f2 , o ; f4 =/2, 1 ; Is = /2,2 ; /6 = /2,3 ; {7 = !3 ,0; • • • Then +•fo1!p(x)dx = 0 but the sequence {/p} does not converge (show this). P Not even monotonicity and the existence of a limit help. Recall Example 4.1 of Chapter 1 in which a (strictly increasing) uniformly bounded sequence of Riemann integrable functions converged to a function which was not even Riemann integrable. We would hope that convergence properties for the Lebesgue integral might be better. However, this is not immediately clear. In Example 27.1 above, the {,. are all Lebesgue integrable (summable) and converge (point wise) to a Lebesgue integrable function f. Similarly, in Example 27.2 all the fn ,k are Lebesgue integrable (summable) and are uniformly bounded. However, they do not even converge abnost everywhere. Thus at first glance the Lebesgue integral does not seem to offer much of an advantage. However, it is worth noting that the difficulties presented by Ex ample 4.1 of Chapter 1 are overcome 'by the Lebesgue theory. In fact, ��f( o , 1 ) f (x)dm 0 = f( o , 1 J f(x)dm , now that the limit function XQ is Lebesgue integrable (summable). The advantage of the Lebesgue theory is that under much Jess stringent hypotheses than for the Riemann case, convergent sequences of integrable functions have integrable limits. -
·
,.
=
28. Convergence Theorems
There are two major convergence theorems involving the Lebesgue integral. They are the Monotone Convergence Theorem and the Lebesgue Dominated Convergence Theorem, and neither is true if we restrict our attention to Riemann integrable functions. Thus we will observe that our new Lebesgue theory actually offers an improvement over the Riemann theory with regard to convergence properties. It is this improvement which makes the Lebesgue theory valuable in many theoretical applications. We will use it in later chapters on Fourier analysis.
C O N VE R G E NCE A N D TH E L E BESG U E I NT E G R A L
79
To prove the Monotone Convergence Theorem we need the following Lemma. Let g be a non-negative measurable function on a bounded, measurable set A. If {A1 lT: 1 are measurable subsets ofA with
28.1 Lemma:
and if a is a real number satisfying a ;;a. fA gdm for all n = then a ;;.. fuA ,.gdm. Proof:
I ,2, •
• • .
Exercise 3 1 .3 .
0
(Monotone Convergence Theorem): Let A be a bounded meas urable subset of 6l and Un} be a sequence of measurable functions on A such that 0
28.2 Theorem
00
By the Lemma, L ;;.. cfAgdm and since this is true for all c E (0,1), L > f.tgdm. Therefore by the definition of the integral (25 . 1), L ;;.. fA fdm. 0
Since sets of measure zero have no effect on Lebesgue integrals, the following Corollary is immediate .
.-rbis theorem is true even under an extended concept of "function" which
anows infmite values. Under this interpretation, evel)' monotone increasing sequence of functions will converge to some function. If { "> 0 and f(x) = oo for x eS, then
m(B) = 0.
fsfdm = • if m (B) > O fsfdm and
=
0 if
80
L E BESG UE I NTEGRATI ON AN D FOU R I E R SE R I ES
If A and {/,.} are as in the Theorem and f(x) = nlim+oo /,.(x) almost everywhere, then nlim fA f,.dm = fA fdm . o +oo
28.3 Corollary :
Another Corollary follows immediately from the close relationship between sequences and series.
If {/,.}== • is a sequence of non-negative measurable functions on a bounded measurable set A , then � ,.�/ndm = J1 J.. f,.dm.
28.4 Corollary :
Proof: The partial sums of the series form a monotone increasing sequence. 0 The hypothesis that the /,. are non-negative seems a bit restrictive in the Monotone Convergence Theorem. What is really at stake is the desire to avoid the following kind of situation.
28.5 Example:
Let /,.(x) =
l
0 on (1 /n, 1 ] 1 /x on (0,1 /n] -
O at x = 0. Each /,. is integrable (not summable) with integral --. But lim /,.(x) n +• = 0 which obviously has integral 0. We can avoid this kind of situation and retain the Monotone Con vergence Theorem if we keep the /,. bounded below. (A stronger result in this direction is found in Exercise 3 1 .35 .) Let A be a bounded measurable set of real numbers and {/,.} a sequence ofmeasurable functions on A such that M
28.6 Corollary:
Proof: Use the sequence 0 < !1 - M < f2 - M < • • • in the Theorem. Then nlim+!.JA( (/,. -M)dm = J(A (lim n +•f,. -M)dm =lA fdm -lA Mdm. D The result follows.
CON V E R G E N CE A N D THE L E B E SG U E I NTEG R A L
81
Of course, if we have a decreasing sequence {gn } of measurable functions bounded above, then the Corollary can be applied to the se quence { -gn} to reach the conclusion nlim+• fA Kndm = JAgdm, where nlim+•Kn(X) = g(x) a.e. The other major result concerning integration term by term is calle d the Lebesgue Dominated Convergence Theorem. We will need a Lemma before proving it. Recall that nlim +•fn(x} = nlim+- {glbfk(x)jk;> n}. (see Appendix). Let A be a bounded measurable set of real numbers, If {fn} is a sequence of non-negative measurable functions on A , and f(x) = nlimfn(x) for every x E A , then fAfdm n}. Then each In is measurable on A by 18.7 and the Kn 's form a monotone increasing sequence of non-negative functions such that Kn(x) < fn(x) for each n. Now nlim lim fn(x), so by the Mono+ .. Kn(x) = f(x) by definition of n""+tone Convergence Theorem nlim +• fAKndm = fAfdm. Since £ fn(x)dm ;;;.£ Kn(x)dm for each n, nlim +• fafndm ;> fA fdm. See Exercise 3 1 . 1 0 for this last step.
28.7 Fatou's Lemma:
D
Fatou's Lemma holds if f(x) = nlim fn(x) almost everywhere +oo D Strict inequality may hold in Fatou's Lemma. See Exercise 3 1 . 1 1 for an example.
28.8 Corollary:
on A .
(Lebesgue Dominated Convergence Theorem): Let A be a bounded measurable subset oftR and let {fn } be a sequence of measurable functions on A such that nlim+oofn (x) = f(x) for every x E A . If there exists a function g E l(A) [i.e. g is summable] such that 1 /n(x)l < g(x) for
28.9 Theorem :
82
L E B ESGUE I NT E G R ATI ON A N D FOU R I E R SER I ES
n = 1 ,2,3 ,
o
o
o
. and all x EA , then nlim +• fAindm = fAidm
Proof: By Exercise 26.26, In and I are in .C(A). Now In + g ;;a. 0 on A so by Fatou's Lemma, fA (I + g) dm < nlim fA (In + g)dm or +oo fAidm < lim fAindm by linearity of the integral (25 .4). n""+-
It is also true that g - In ;;a. 0 on A so by Fatou's Lemma fA (g - /)dm < lim fA (g - ln)dm and - fAidm < lim [-fAindm ] .
Thus fA idm ;;a. nnm +• Llndm (see Exercise 3 1 . 1 2). So
lim 1A ln dm < nUm 1A ldm < nT+•iA lndm < 1A ldm .
Hence nlim +oo fAindm exists and is equal to fAidm.
D
The Lebesgue Dominated Convergence Theorem holds when and l!n(x) l
28. 1 0 Corollary :
=
If lln(x) I < c for n = 1 ,2,3 , • • and almost all x E A (i.e. {In } is a uniformly bounded sequence) and if In -+I almost everywhere on A , then nlim 0 +- fAindm = Lldm.
28.1 1 Corollary :
o
29.** A Necessary and Sufficient Condition for Riemann I ntegrability.
Armed with the convergence theorems of the previous section we can determine exactly which functions are Riemann Integrable. This question was far from solved by the existence theorems quoted in section 2. We know that continuous functions and monotone functions are Riemann integrable. Moreover, it is possible to alter a function at finitely many points without affecting integrability. Yet (see Example 3.3) there are Riemann integrable functions which are not even piecewise continuous (continuous at all but finitely many points). It turns out that the dis continuities must be restricted to a set of measure zero. In fact, a bounded function is Riemann integrable if and only if it is continuous almost everywhere. To prove this result, we need to relate step functions to continuity and to Riemann integrabilitY more intimately than we have heretofor.
CON V E R G E N C E AN D T H E L E B ESGU E I NTEG R A L
83
A function f: [a,b] -+ IR. is continuous a.e. if and only if there are sequences {gn} and {hn} of step functions such that
29.1 Lemma :
and nlim +ooKn(x) = f(x) = nlim +oo hn(X) a.e. on [a,b] . Proof: Assume the sequences {gn} and {hn } satisfy the given con ditions. The set {x E [a,b] I x is a point of discontinuity of some Kn or hn } is countable, hence has measure 0 . I fx ' E [a,b J satisfies
nlim +ooKn (x') f(x') nlim +oo h n(X1) =
=
and x' is not a point of discontinuity of any Kn or hn (hence for a.e. x'), then given e > 0, there is an n such that hn(x') -Kn(x') < e. Furthennore, since Kn and hn are step functions, there is an open interval containing x' in which hn(x) = hn(X1) and Kn(x) Kn(x'), so that in this interval, =
Kn(x) - hn(x') <.f(x) - f(x') <. hn(X) -Kn(x'), or
lf(x) - f(x') I <. l hn(x') -Kn(x') I < e. Conversely, let Pn be the regular partition of [a,b J into 2n equal subintervals. For each subinterval [y ;_ 1 ,y;] of Pn , let
Kn(x) glb{f(x')lx' E [yi - l •Yi)} =
and
hn(x) lub {f(x')l x' E [y, _ I IY;)} =
for x E [y;- t .Y;). Finally, let gn(b) hn(b) f(b) . l t is easy to verify that each Kn and h n is a step function, and since each subinterval of Pn is split into two equal subintervals in Pn + J, g 1 <.g2 < · • · < {<. • • • < h 2 < h t . =
=
If x' E [a,b] and x ' is not in any Pn , and f is continuous at x' (hence for a.e. x '), then for any e > 0, there is a � > 0 such that if l x - x' I < � , then f(x') - e
84
L E B ESGUE I NT E G R AT I O N A N D F OU R I E R S E R I ES
g(x') == glb{f(x) l x e fy; _ 1 ,y1)} ;> l(x') - e, and h n (x')
0
bounded function 1: [a,b] -+ ft is Riemann integrable if and only ifI is continuous a.e. on [a,b] . Proof: Suppose that I is Riemann integrable. Given n, Theorem 1 .6 8!Jarantees the existence of step functions Kn <.-I < hn such that f,11 hn(x)dx I11"Kn(x)dx < 1/n. Define step functionsgn and hn by in(X) == max {g l (x), . . . ,gn(x)}, hn (x) == min{h l (x), . . . ,hn (x)} . Then surely nlim+ J,g_ n (x)dx == J."l(x)dx = lim r-h n (x)dx.
29.2 Theorem : A
-
-
II
II
II
If g(x) == nlim+ooin(x), and h(x) = nlim +oohn(x), then g <. I < h, and by the Monotone Convergence Theorem, nlim +oo .J.[ll, b ) Kndm = J.[a, b l gdm < J.[ 11, b ) hdm = nlim+ooJ.[ a, b ) h,dm.
Also, fra.b ) Kndm = f:in(x)dx and J1 1 hndm = f:hn(x)dx, so that , J"l(x)dx = J.ia, b ) gdm = J.(ll, b ) hdm. Therefore, g = h a.e. (Exercise 26.23) and I == g = nlim +ooin a.e. Thus I is measurable and lis continuous a.e. by Lemma 29.1 . For the converse, suppose I is continuous almost everywhere. Let Kn . h n be as in Lemma 29.1 . Then by the Monotone Convergence Theorem (in particular, Corollary 28.6 and the discussion following it), nlim +oo}f[ a, b ) h ndm. +ooJ.[ll, b ) gndm = nlim Hence, given e > 0, there exists an n such that J"hn(x)dx - •(a"Kn(x)dx = J[a, b ) hndm -J·[ll, b )Kndm < e. 11 0 Riemann integrability follows from Theorem 1 .6. ,
II
CON VE R GE NCE A N D TH E L EB ESGUE I NTEG R A L
85
If f is Riemann integrable on [a,b ] , then f is Lebesgue in tegrable on [a,b] . and f: f(x)dx = f(a,b] fdm. Proof: In the proof of the theorem, we showed that f is measurable. (it is also measurable by virtue of being continuous a.e.). Since I is bounded, f is Lebesgue integrable. Since f = g a .e., b I( a, b] tdm = I.(11, b)gdm = nlim r i dm = nlim Lbin <x>dx = f. t<x)dx . +ooJra. b ] n 11 + • 11
29.3 Corollary :
0
The meaning of this corollary is that all the functions you could integrate in calculus can be integrated in the Lebesgue sense as well, using the same techniques. Furthermore, the Lebesgue values are the same as the Riemann values.
30. ** Egoroff's and Lusin's Theorems and an Alternative Proof
of the Lebesgue Dominated Convergence Theorem.
In this section we will prove two interesting and useful theorems. The fust, Egorofrs Theorem, says that a sequence {/n } of measurable functions which converges to a function f almost everywhere on a bounded measurable set A. is somehow close to being a uniformly convergent sequence. "Close to" means that for any e > 0 there is a measurable set A e C A. such that m (A.\Ae� < e and In -+ I uniformly on A.e . One can imagine some uses of such a result. In particular, convergence theorems involving the Lebesgue integral may be proved using Egorofrs Theorem since by Exercise 3 1 .1 uniform convergence allows integration term by term.
(Egoroff): Let A. be a bounded measurable set in Q . L et In and I be functions defined on A. such that each In is measurable and In -+ I almost everywhere on A. . Then given any e > 0, there exists a measurable subset A e C A. such that m(A.\Ae) < e and In -+ I uniformly on A.e.
30.1 Theorem
> 0 and positive integers m and n , define the set Em ,n = Qn {x l lfi (x) - f(x) I < 1/m}. Note that Em ,n is me�urable. If U i is the subset on which In -+ f, then clearly for any m, U C nU Em , n C A. . =l Proof:
Given
e
Now m(U) = m(A.) so m(n=Ul Em n ) = m(A.). But Em n C Em n + 1 for all m and n , so ..
•
'
'
88
LE BESGUE I NT E G RATI ON A N D F O U R I E R SE R I ES
nlim +• [m(A ) - m(Em ' n)J +• m(A\Em ' n) = nlim
= m(A ) - nlim m(Em ' n) = m (A ) m (nUl Em ' n) = O . +• = ..
-
Thus for each m , there exists an integer nm such that ..
Now let Ae = m'J 1Em, n m . Then Ae is measurable and < l; m(A -Em ' nm ) < ml;= l e/2m 8.6 m = l ..
..
= e.
We now claim that fn -+ f uniformly on Ae since given any m, there exists an nm such that for all n > nm , lfn(x) - f(x) l < 1 /m everywhere on Em . nm . But A e C Em . nm for evtlry m, so for all n > nm , lfn(x}-f(x) I < 1/m everywhere on Ae· But this is exactly uniform convergence. 0 We will now use Egorofrs Theorem to give another proof of the Lebesgue Dominated Convergence Theorem. Given measura ble functions fn converging to f on a bounded measurable set A with m lfn(x) l < g(x) for some g E .t(A ), then nlim .... . +oof....� f dm = f�{d
30.2 Corollary (Lebesgue Dominated Convergence Theorem): n
Proof: Clearly f is also measurable and 1!1 < g implies {E .t(A ). Defme disjoint measurable sets A k = {x lk - 1 < g(x) < k} for k = 1 ,2, · · · . Then A = �UIA k so that f"'"�gdm = �1;l f�"'" kgdm by Theorem 24.3 . Thus given e > 0, there is an N such that ..
J
U Ak
lr =N+ l
Thus
J lfn ldm ..
UAk lr•N+l-
..
gdm =
� J gdm < e/5 .
k=N + I A k
and
J lfldm ..
U.Ak lr = N + l
are each <e/5. This
takes care of the set on which fn and f are large. On k=UN lA k use Egoroffs Theorem to write kUN= l A k asB1 U B2 where m(B t ) < e/5N and
CO N V E R G E NCE AND T H E L E B ESG UE I NT E G R A L
87
J,B2 l/11 - fldm < e/ Sm(A)
for large enough n (by uniform convergence). Now for large enough n : IJ.A f11dm - Ll4 fdm I < Joo lf11 ldm + Joo UAk
II""N + l
UAk
lfldm
II =N + l
+ J,B 1 l/11 1dm +1B 1 lfldm +1B 2 1/11 - fldm < e.
0
This alternative proof of the Dominated Convergence Theorem shows the power of Egoroff's Theorem. The proof is in many ways more intuitive and natural than the one in section 29, but it requires the uniform con vergence provided by Egorofrs Theorem. We now proceed to Lusin's Theorem which says that a measurable function fon a bounded measurable set A is "nearly" continuous. "Nearly" here means given any e > 0, there is a subset C of A such that m�\C) < e and f is continuous on C.
If f is a measurable function on a bounded measurable set A , then given any e > 0, there exists a closed set Ce C A such that m� \Ce) < e and f is continuous on Ce . (That is, the restriction off to Ce is continuous.)
30.3 Theorem (Lusin) :
II
First, let f = k =l;l akXE..... be a simple function on A . Given e > 0, for each k there exists a closed set Ck C Ek such that Proof :
(see Corollary 1 3 .3). Now C = ku= l Ck is closed, m�\C) < e (Exercise 3 1 .23) and f is continuous on C since C is equal to the disjoint union of closed sets Ck and f is constant on each Ck· (Exercise 3 1 .24). Now let / be any measurable function. Since f = f+ -{_, we may /11 for a sequence assume f is non-negative. Since f is measurable, f = lim +ooo { /11 } of simple functions by Theorem 19 .4. By the first part of the proof, given e > 0, there exists for each n a closed set C11 such that m(A\C11) < e/211 + 1 and/11 is continuous on C11. II
II
88
L E B ESG U E I NTEG RATI O N AN D F O U R I E R SE R I ES ..
Let C0 = nn= l Cn . Then C0 is closed and
m(A\Co) = m(A\ n=nl Cn ) m(n=U l {A\Cn )) =
< n'i:= l m(A \Cn ) < en ='i:l l/2n + 1 = e/2.
8.6
We would like to conclude that f is continuous on C0, but we cannot since we need something like uniform convergence for this. However, by Egoroff's Theorem, there is a measurable subset C C C0 such that m(C0\C) < e/4 [and thus m{A\C) = m{A\C0 ) + m(C0\C) < te] , and
In -+{uniformly on C.
Now each fn is continuous on C C Co so that /, being the uniform limit of the fn , is continuous on C. If C is closed we are through. If not , take closed set F C C such that m(C\F) < e/4. o 31.
31.1 31 .2
Exercises
Show that i f In -+ I uniformly on a bounded measurable set A , and each In E .C{A ) , then iE .C(A ) , and nlim ,.. n dm = f.tldm. + • f.f n Consider the sequence of functions ln(x) = n2x( l - x 2 ) , x (a) Show that lim ln(x) = 0 for all x E [ 0, 1 ] . (b) Show that
n +• ��.J( o,l ] ln dm = oo. •
(c) What happens to the sequence
31.3
E [ 0, 1 ] .
K
n (x)
=
nx( l - x 2 )n , x
E [ 0, I ] ?
Prove Lemma 28. 1 . (Hint: use countable additivity of the integral (Theorem 25.5) to show that f.U A gdm = lim L gdm.)
n
n +• "' n
31 .4
Prove Corollary
3 1 .5
Use the Monotone Convergence Theorem to show that l(x) = 1 /x is not summable on (0,1 ). ( Hint : consider the sequence ln (x) = max {f(x), n}, X E (0, 1 ).)
31 .6
28.3 in detail.
Use the Monotone Convergence Theorem to find
g(x) = x- l.
3 1 .7
Suppose that
h
is a continuous function on
Then of course the
improper
where
but lim h(x) = +oo. x+ b fa h(x)dxb is defined by
[a,b),
Riemann integral
J(O,l)gdm,
CON VE R G E NCE A N D T H E L E B ESG U E I NTE G R A L
lim_l ' h (x) dx . Show that if this limit exists and is finite, then t+� a Lebesgue integrable on [a,b), and f h dm = �b h(x)dx.
h
89
is
(a, b )
31 .8 31 .9
Give an example to show that monotonicity in the Monotone Convergence Theorem is essential. n Let f E .C [ O, l ] and Kn(X) = x f(x) for x E [ 0, 1 ] , n = 1 , 2, . . . . that gn E l [ O, l ] for n = 1 ,2, . . . , and nlim J.10 '11 gn dm = 0. +•
31 .10 Suppose that
Sh ow
an ;;;;.. bn for all n . Show that nlim an ;;;;.. nlim bn . +• +•
31 .1 1 Show that strict inequality may hold in the conclusion of Fatou 's Lemma.
3 1 . 1 3 Where is the fact that g is
summable used in the proof of the Dominated
Convergence Theorem?
31 .14 Let fn ;;;;.. 0, {11
E l(A ), {11
-+
Kn (x)
= max{fi (x), . . . ,{11(x)}
for x E A , and assume that Prove that lim Lf11 dm = 0. II
31 . 1 5 Let {11 -+ {,
+oo .,.
O,a.e. on A . Define
.ftgn dm
lfn l
< M for some real number M, all n.
Show that
nlim +• L:.4 {11g dm = 1A fg dm . 31 . 1 6 Give an example of an
unboun ded
tegrable) which is continuous a. e. on
function (hence non-Riemann in [ 0, 1 ] .
31.17 Give an easy proof that the function integrable.
g
of Example 3.3 is Riemann
31 .18 Show that
X c , where C is the Cantor set (Example 1 2.5 and Exercises 1 6.20 and 1 6. 2 1 ) is Riemann integrable, but that X D (Exercise 1 6 .23 ) is not.
31 .19 If A if
C [a,b] is closed, show that XA A ) = 0.
m(
/, h E .C (A ) and JA f dm = .ft h dm . g E .C(A ) and ,.fttdm ,.ft g dm .
31 .20 Let
=
is Riemann integrable on If f <
g
[a,b]
on A , show that
90
LEBESG U E I NT E G R AT I O N A N D FO U R I E R SE R I ES
31 .21 Give an example to show that Egoroff's Theorem cannot be improved to
yield m(Ae) = 0.
31 .22 For each example in this chapter of an a.e. pointwise convergent se
quence of functions, find an A e as in Egoroff's Theorem.
,E,. be disjoint measurable bounded sets. Let C" measurable for k = 1 , . . . ,n. Show that ,. ,. ,. m( u E" \ u 1 C") = I: m(E" \c").
31 .23 Let E 1 ,
•
•
•
k= 1
C E" be
k= 1
k=
,C,. are disjoint closed sets and f is constant on ,. each C" , then f is continuous on U C". Is this true under any other k= l assumptions about the c"? (Hint: you must show that for every X E c", there is a 8 > 0 such that (x - 6 ,x + 6) n c1 = (J for i :/= k. Why does this prove the result?)
31 .24 Prove that if
C1 ,
•
•
•
31 .25 Let f = XQ on [ 0, 1 ] . Given
e
>
0,
find Ce as in Lusin's Theorem.
31 .26 Prove Lusin's Theorem for { non-negative, measurable, and bounded with
out using Egoroff's Theorem. (Hint: see Corollary 1 9.5.)
31 .27 If {: C -+ IR is continuous on the closed set C, show that f can be extended
to a continuous function on all of IR. (Hint: IR\C = U/;, where the /; are 1 disjoint open intervals.)
3 1 .27 to prove the following corollary of Lusin's Theorem : if f is measurable on a bounded measurable set A , and e > 0, then there is a continuous function g such that m({ x E A j {(x) :/= g(x)} < e.
31 .28 Use Exercise
31 .29 Prove the following converse of Exercise
3 1 .28: Suppose that A is a bounded measurable set and {:A -+ IR. If for every e > 0 there is a continuous function g such that m( { x E A I f(x) :/= g(x)}) < e, then f is measurable on A . (Hint : by taking e = 1 , 1 /2, 1 /3 , . . . , obtain a sequence {g,. } of continuous functions which converge pointwise a.e. to f.)
31 .30 Show that if f is a measurable function on A , then f is the a.e. pointwise
limit of a sequence of continuous functions. (Hint: see Exercise 3 1 .29 ..)
� fdm for A unbounded and measurable, f measurable on A . Restrict yourself to A = IR if you wish. What theorems follow from your definition?
31 .31 Discuss possible definitions of
31 .32 Find
!10,1 1 tdm , where (a) f(x) = x, and (b) f(x) = x3 + 2x.
31 .33 Let { h ,. } be an increasing sequence of functions in .C(A ), with h ,.
for all n and some h
E
CON VE R G E NCE A N D T H E L E BESGUE I NTEG R A L
91
h = nlim hn a.e. on A. (Hint: As in the proof of Theorem 29.2, let +.., g(x) = lim.., h n (x) , and show that g = h a.e. in A.) n +
A be bounded and measurable, f:A -+ R , and suppose that for all e > 0, there are simple functions g < t < h such that /A h dm - .L_gdm < e. Show that /is bounded and measurable. (See the proof of Theorem 29.2).
3 1 .34 Let
A be bounded and measurable and let Un l be a sequence of measura ble functions on A such that /1 < h < . . . , where LiJdm is finite. Prove that if /(x) = lim /n (x) for x E A , then lim Lfndm = Lfdm . +""
31 .35 Let
n +..,
n
CHAPT ER
7
Function Spaces and L2
32.
Linaar Spaces
Having developed the main points of the Lebesgue integration theory, we will devote the remainder of the book to theoretical applica tions in analysis. We will touch on only a few of the uses to which the theory has been put, but hopefully these will give some feeling for how important the Lebesgue integral has become in modern analysis. One of the most fruitful recent developments in analysis has been to apply the techniques and concepts of linear algebra to the study of functions. In particular, certain sets of functions have natural structures as vector spaces (in this context, they are called/unction spaces), and their vector space properties yield a great deal of information about the func tions themselves. We assume the reader has some familiarity with the basic notions of linear algebra. However, in this section we will review some of the definitions and results which will be most useful to us in the remainder of the book. real vector space (or linear space) is a non-empty set V with two operations. The first assigns to each 'P, w E V a unique element v + w e V. The second assigns to each a E 6l andv E V, a unique element avE V. These operations must satisfy the following properties:
32.1 Definition : A
94
L E B ESG UE I NT E G R A T I O N AND F O U R I E R SE R I ES
u + v = ii + u for all u, v e V; u + (v+ w) = (u + ii) + w for all u, v, w E V; there is an element 0 E V such that u + 0 = u for all u E V; (4} for each u E V, there is an element -'U E V such that u + (-u) = 0; (5) a(u + V) = au + avfor all a E dl, u, vE V; (6} (a + b}U = ail + bu for all a,b E dl, uE V; (7) (ab )u = a(bu) for all a, b e dl. u e V; (8} Iii' = u for all u E V. (I) (2) (3}
First, the prime example. (I} The set of all n-tuples of real numbers, denoted R", with addi tion and scalar multiplication defined respectively by
32.2 Examples:
The following examples are all function spaces. The operations are defined by (/+ g)(x) = f(x) + g(x), and (af)(x) = af(x). Most of the vector space properties are trivial to verify in each example below. The only remaining requirement in each case is to show that if f and g are in the set in question, so are f + g and af. In each case this is either trivial or an immediate consequence of elementary facts proved in calculus or in this book. (2) B(A.) = {gJ g:A -+ R, g bounded}, for any A C R. (3 } C(A) = {gJg:A -+ R,g continuous}, any A C R. (4) R([a,b] ) = {gJg: [a,b] -+ R,g Riemann summable} . (5) .C(A.) = {gJg:A -+ R, g Lebesgue integrable}, A bounded and measurable. (6} P(A ) = {g J g:A -+ R,g a polynomial}. (7} Pn(A.) = {gJ g E P, g of degree < n}. The concept of length of a vector is fundamental because it leads immediately to a definition of distance between two vectors. Distance is
F U NCTI ON SPACES A N D 1.2
95
particularly important in analysis, of course, since so much of analysis is concerned with approximations (as in limits, integrals, etc. ) . Based on the connection of IR3 with Euclidean geometry, length in .an is most reason ably taken to be Pythagorean length (length of (x1, . . . ,Xn) = .Jx 12 + . . . + xn2). In the function spaces, however, it is much less clear how best to defme this concept. real normed linear space is a vector space V (with addition and scalar multiplication denoted by u + v , a'U , respectively) such that for each u E v, there is a unique real number I ii n (called the norm or length of U) satisfying: (1) ll ii ll ;:;;.. 0 for all ii E V; (2) II iill = 0 if and only if ii = 0; (3) llaiill = Ia llliill for all a E IR, ii E V; (4) llii + vii <; ll iill + ll v ll for all u, v E V.
32.3 Definition : A
32.4 Examples:
(1) For (X J , . . . ,Xn) E IRn , II (X J , . . . ,Xn) ll = Vx 12 + x,'l + . . . . + Xn'J. · (2) For /E B(A), II III = lub{lf(x)l l x E..4 }. (3) For/E C([O,I] ), II III = f� lf(x) ldx. (4) For/E P1 ([0,1 ] ), where f(x) = a0 + a1x, }fo1 II II I = f2 (x)dx = Va o2 + aoal + }a t2 • Whenever a space has a norm, then a distance, or metric can be immediately defined by d(ii,v) = llii - vii. This has the properties, for all ii, v-. w e v, (I) d (ii, V) ;;;s. o; (2) d (ii, v) = O if and only ifii =v; (3) d(ii,V) = d(V,ii); (4) d(ii, W) <; d(ii,V) + d(V, W). So, a normed linear space is a metric space, and one can talk about limits, continuity, etc., for real valued functions defined on the space. We are most concerned with limits of sequences.
98 L E B ESG UE I NT E G RATION A N D F O U R I E R SE R I ES
If V is a normed linear space with norm II II , if v E V, and {vn } is a sequence of elements in V, then we say nlim+•Vn = v (in norm II I I) if for every e > 0, there is a positive integer N such that whenever n > N, then llv - vnll < e. In this case, we say {vn } converges to v. Note that nlim• •'Vn = v if and only if n +• llv -'Vnll = 0.
32.5 Definition:
32.6 Examples:
(I)
For (x i>
•
•
•
, Xn) E Rn , (x 1k, , Xnk) e .Rn , then k k klim+ • (xl , , Xn ) = (X J , , Xn) •
•
(2)
to us.
.
.
•
.
•
•
•
in the knorm of Example 32.4(1) if and only if for each i = 1 , . . . ,n, klim+•x1 = x1• For f, In E B(A), nlim+ ooIn = f in the norm of Example 32.4(2) if and only if Un} converges to f uniformly on A. The completeness of a normed linear space V will be of great interest
The sequence {vn } is a Cauchy sequence in the normed linear space V with norm II II , if for all e > 0, there is a positive integer N such that whenever m,n >N, llvn - vm ll < e. Thus a Cauchy sequence is one for which the elements eventually become arbitrarily close to one another. Of course, any convergent se· quence is a Cauchy sequence (Exercise 34.9), but only for certain spaces is the converse true.
32.7 Definition :
normed linear space V is complete in the norm II II if every Cauchy sequence converges (i.e., has a limit).
32.8 Definition : A
32.9 Examples:
(I ) (2) (3)
Rn is complete in the norm of Example 32.4(1). R(A) is complete with the norm of Example 32.4(2). (See Exercise 34. 1 1 .) C( (0,1 ] ) is not complete with the norm of Example 32.4(3). (See Exercise 34.1 2.)
F UNCTI O N SPACES A N D 1.2
(4)
97
"<: [0,1 ] ) is not complete with norm IIIII = J! 1/(x) l dx. (See Exercise 34. 1 3 .)
A complete normed linear space is called a BaMch space. In some normed linear spaces, the norm arises from an inner product. An inner product on a real linear space V is a real valued function ii • v, defined for all ii, v E V, such that: (1) ii • v = v · ii for all ii , v e V; (2) u · ii :> 0; and ii • ii = 0 if and only ifii = 0; (3) (ii + V) • w = ii · w + v • w, for all ii, v, w E V; (4) (aii) • v = a(ii • V) for all a E R, ii, v E V. It follows from the definition that ii • (v + W) = ii · v + ii • w, ii • (aV) = a(U • V), and 0 • v = 0, for all ii, v, w E V, a E 6t (Exercise 34. 14).
32.10 Definition :
32.1 1 Examples:
For (x 1 , ,xn ), (y 1 , ,yn) E Rn , the standard inner prod uct, also called the dot product, is defined by (1)
.
•
.
.
•
•
For /, g e C( [0, 1 ] ) , define I • g = f�f(x-.<x)dx = f1 0,1 lgdm. On 1'1( [0,1 ] ), define (a o + a 1 x) • (bo + b1x) = a ob o + 2aob l + 2a l b o + Sa l b l . Given an inner product ii • v, a norm is defined by ll iill � • ii. The properties of the norm are easy to verify (Exercise 34.1 7) except for the triangle inequality (Defmition 32.3(4)). For this we need the Cauchy inequality. First a simple lemma. (2) (3)
=
32.12 Lemma : u· u=
In an inner product space
1 , then Iii · v i < 1 .
V
with inner product
Proof: o < . (ii -•> = ii · ii - 2 (U • v> + v ii • v < 1. -(ii. • v) < 1,
Therefore,
Similarly,
•
ii
•
v,
if
• = 2(1 - ii • V).
so the result follows from
0
98
LE BESG U E I NTE G R AT I O N A N D F O U R I E R SE R I ES
32.13 Cauchy Inequality :
If V is an inner product space with inner product
ii • v, then for any ii, v E V,
Proof : If ii = 0 or v = 0, then equality holds trivially. Otherwise, consider the vectors -
-r
u=
U
.../;; • ii
It is easy to verify that ii' • Iii' • v ' I < 1 . But
and -v 1 =
-
II
.fi:i
ii' = v' . v' = 1 ,
.
so that by the lemma, 0
32.14 Corollary :
norm.
In an inner product space V, if ll iill = .Jii
·
ii, then II
II
is a
Proof: The defining properties (1) - (3) of Definition 32.3 are left to the reader (Exercise 34. 17) . For property (4), we calculate:
= ll iill 2 + 2(U • v> + llvll 2
< ll iill 2 + 2ll iill ll vll + ll vll 2 32 . 1 3 = (ll iill + ll vll )2 •
32. 1 5 Examples:
product:
(2)
(1)
0
The norm of Example 32 .4( 1) comes from the dot
The norm on P1([0,1] ) defined by lla0 + a1xll =
Jao2 + aoat + kal
m
F U NCT I O N SPA CES A N D t,2 99
(Example 32.4(4)) comes from the inner product f • g = I;f(x)g (x)dx. (3) The norm on C([0,1]) defmed by IIIl i = J� If(x) ldx (Example 32.4(3)) does not come from any inner product. Indeed,2 if we supposed there were an inner product such that f • f = (f� lf(x) ldx) , then we would have (I + x) •
2 (1 + x) =(10( 1 (I + x)dx}2 = ( 1 + 21 ) = 9/4.
But (1 + x) • (1 + x) = 1 • 1 + x • x + 2(1 • x) = 1 + i + 2(1 • x). Hence 1 • x = .!. Then (.!. - x) • (.! - x) = .!2 • .!.2 - 2(.!2 • x) + x • x = .!4 - .!2 + .!4 = 0, even2 " though2t - x is 2not the zero function (contradicting Defmition
32 . 1 0(2)) .
An inner product space that is complete in its norm is called a Hilbert space. The inner product is useful for defining angles in our space, in particular right angles. If V is an inner product space with inner product u • then the set {ul > u2 , } c V is an orthogonal set if for every i ::1= j, iii • u1 = 0. The set is orthonormal if it is orthogonal and for each i, ii1 • u1 = 1 , that is II u,�l = I .
32.16 Definition :
v,
•
•
•
(1) In fi3 , the set {(1 ,0,0),(0,1 ,0),(0,0,1)} is orthonormal relative to the dot product. (2) In C([0,1 ] ), the set consisting of /1 (x) = 1 , fl(x) = x - 1/2, [3 (x) = x 2 - x + 1 /6 ,f4 (x) 1= x3 - ix2 _:-'" tx - 2� is orthogonal relative to the inner product [ • g =/0 f(x)g(x)dx. (3) In C([-11', 11'] ), the set {-:;fi;, 1 v;1 . ..;;1 ..;;1 Sln" 2 ..,;;1 2 . . }
32.1 7 Example:
SID X,
COSX,
X,
COS X, .
is orthonormal relative to the inner product f g = � -w,rlgdm. (See Exercise 34.20.) ·
100
LE B ESG U E I NTEG RATION A N D FOU R I E R SE R I ES
You will recall from linear algebra that orthogonal sets are linearly independent (see any linear algebra book). Furthennore, if {v 1 ,v , , Vn } 2 is an orthononnal basis of then any element v E can be expressed as •
•
•
Y, Y v == l: (v • v1)v1 (see Exercise 34.21). The following is an important related 1= 1 idea which we will find useful later.
Let {v� r . . . ,vn } be an orthononnal set (not necessarily a basis) in an inner product space Consider the subspace W spanned by {v� r . . . ,'in } (i.e., W is the set of all linear combinations a 1v 1 + . . . + an'in , where a 1 , , an E 61). Then, given v E the vector
Y.
•
•
Y,
•
is an element of W, and is called the projection ofv on W. Note that if v E W, then vw == v.
Y,
32.1 8 Theorem : Given v E 'Vw is the vector in W which is close5t to v (in the sense of the nonn). That is, for any w e W, ll v - w ll ;;;;. llv 'Vw II . -
n Proof: Let w == :t a1v1 be an arbitrary element of W. Then i= 1 ll v - wll 2 == (v -
I: a,'V,) . (v - j=I:1 a,v. )
i= 1
== v • v == ll v ll 2 since
I
I: a (v • v·)1 - J=.I:1 a1(V • v1) + 1=I:1 /=I:1 a;a1(V1 • v1) 1= 1 1
- 21=I:1 a1 (V • "V,) + 1=I:1 al ,
- - ll II;
•
IIJ
and r.:
-
==
if i == i
0 if i =l=j n
-
r.: • v1). :t a \v • v1) == /=:t1 a1\v 1= 1 1
n
A similar calculation gives
F UNCT I O N SPACES A N D t2
1 01
Therefore
0
32.19 Corollary : Under the hypotheses of the theorem ,
llv - �E1 (ii • v1)v·ll 2 = llvll 2 - �E1 (v v1)2 ' I
·
and
Proof: From the proof of the theorem,
0 33. The Space !'-
Given a bounded measurable set A , we wish to define a particularly useful norm for certain functions f:A -+ 6!. This norm is a generalization of the Pythagorean norm in 6!" given by II +. . . + 2 Now we could represent the n-tuple (y l > . . . by a step function
(y1,
f: [O,n) -+ 6! defined by
•
Yn)ll = ..Jy 12 ,yn)
•
•
•
Yn
•
1 02
L E B ESGU E I NT E G R ATI ON A N D FO U R I E R S E R I ES
f(x) = Yi for x E [ i - l , i). Then
ll (yJ , . . . ,yn)ll = V)lr(O�t)J4"1dm. For a general function f:A norm
�
IR, we attempt to define by analogy the /,2
Of course, this will not be defined for every function ; since the norm should be finite for every f, we need f1 E /,(A ). Hence {2 must be meas urable. This does not imply f is measurable (see Exercise 34 .24), but for technical reasons we will require f to be measurable as well . 33.1 Definition : If A is bounded and measurable, let
l
l2 (A ) = t:A
�
IR j fis measurable and
£ {2dm < ao f.
That is, l2 (A ) is the set of measurable, square summable functions on A . = .J For { E l2 (A ), define
llilh
�f2dm. lh satisfies
We will verify that II properties (1), (3), and (4) of Defmition 32.3, so that it is almost a norm. Unfortunately (2) is false, since = 0 only implies that f = 0 a.e . on A . The way out of this dilemma is to replace our notion of "equality" with the modified notion of "equality almost everywhere." Therefore , we will consider two functions f,g E l2 (A ) to be identical if and only iff = g a.e . on A . Of course , this is a somewhat bizarre and informal procedure. It could be made rigorous by defining l2 (A) to be the set of equivalence classes-under the equivalence relation" = a.e." (see Exercise 34.25)-of measurable square summable functions on A . This procedure has the virtue of being logically unobjec tionable , but involves a notational nightmare. We will therefore follow the universal practice of identifying functions which are equal almost everywhere. Thus property (2) of Definition 32.3 holds for II The other properties will follow from the discussion below.
� {ldm
lh .
First , we should verify that l2 (A ) is a linear space (that is, a function space) under the ordinary operations on functions. 33.2 Proposition: l2 (A ) is a real linear space.
F UNCT I O N SPACES A N D
t.2 1 03
Proof: By the properties of addition and multiplication of real numbers, in terms of which f g and af are defined, we need only show that .C2 (A ) is closed under addition and multiplication by a real number.
+
For addition , we have (/ + g)2
=
f2
+ 2fg + g2 , so that
£ (f+ g)2 dm =£ �dm + 2L fgdm +£ g2dm. ��dm
� dm
Since f and g are assumed to be in .C2 (A ), are finite . and g2 it suffices to show fg E .C2 (A). This follows from the fact that we require f and g to be measurable and from the inequality Vg I < t (/" +g2), so that Vg < •. (See Exercise
Thus
� ldm
26.26.)
a E dl. f E .C2 (A ). Then JA (af)2dm = a2JA �dm < •.
For multiplication by real numbers, let
0
33.3 Corollary: If f and g E .C 2 (A ), then fg E .C2 (A ). 33.4 Corollary: lf/E .C2 (A ), then fE .C(A ). Proof: Exercise 34.26.
0
To prove that II lh is a norm, we first note that it arises from an inner product : f • g = fg ; that is, ll fl h = .JR. This is defmed and is finite for f,g E .C2 (A ) by Corollary
� dm
33.3.
33.5 Proposition : f • g == f.t fg Proof:
33.6 Corollary:
dm i s a n inner product o n .C2(A ). 0 Exercise 34.27. Remember that f = g means f = g a.e. on A .
I I l h is a norm on .C2 (A ).
For .C2(A ), the Cauchy inequality and the triangle inequality (also known as the Minkowsky inequality) have the following form. 33.7 Theorem (Cauchy Inequality) : For f,g E .C2(A ), If • g I < 11/lhllglh , or
JL fgdm J <Jf.. t'-dmJL g2dm.
1 04
L E B ES G U E I NTEGRAT I O N A N D FOU R I E R S E R I ES
For f,g E .C1 (A), .{.1/g ldm < llflhll g l h· Proof: Exercise 34.28.
33.8 Corollary:
33.9 Theorem
or
0
(Minkowsky inequality): For f,g E .C1 (A),
JL (f+g)1dm � +jjAg1dm.
We have shown that .C1 (A) is a real inner product space with f • g = f.t[gdm, and norm II flh = Vf.._ /1dm. We can therefore talk about orthogonal and orthonormal sets (see Example 32.17(3)), and about con vergence in the norm. We will delay a discussion of orthogonality until later; it leads in a natural way into the study of Fourier series. But let us explore .C1 convergence now. Let {/,.} be a sequence of functions in .C1 (A), and let /E .C1 (A) . Then, translating our general discussion of convergence in the last section into this context, we say that {/,.} converges to f in .C1 (or "in the norm II lh," or "in the mean") if and only if " 11 /-/,. l h = 0, ,.lim + ... i.e., lim f (f-h" )1 dm = 0. n +:...IA What is the relation between .C1 convergence and other notions of con vergence we know of for sequences of functions? Uniform convergence surely implies .C1 convergence (Exercise 34.29). Pointwise convergence does not imply .C1 convergence (Exercise 34.30), and .C1 convergence does not imply pointwise convergence (hence does not imply uniform con vergence). (See Example 27.2, where {!,.} converges in .C1 to1 0, but does not converge pointwise for any x. This example shows that .C convergence does not even imply pointwise convergence almost everywhere.) On the other hand, if {/,.} converges to f in .C1 , then a subsequence {f,.k } con verges to/pointwise a.e .. (See the proof of Theorem 33.1 1 .) Although .C1 convergence does not imply uniform convergence, it has one important property usually associated with uniform convergence
F U NCT I O N SPACES A N D 1.2 1 05
in the theory of the Riemann integral; it is possible to integrate an 1.1 convergent sequence term by term. 33.10 Theorem :
Proof:
1 Given fn• f E l1 (A), if nlim + fn =f in 1. , then
f d = �L n m L tdm.
By Cauchy's Inequality (Theorem 33.7),
I L tndm - Ltdm j = IL
= IL
0
The next question is one of completeness: is l 1 (A) a Hilbert space? The answer is yes. The following theorem is often called the Riesz-Fischer Theorem. 33.1 1 Theorem :
If A is bounded and measurable, then l1 (A) is complete.•
Let {fn } be a Cauchy sequence in l1 (A). We will fmd a sub sequence {/nk} of { fn } which converges pointwise a.e. to a function f. Then we will show that f E l1 (A) and that Unl converges to f in 1.1 . Proof:
For the subsequence, note that since {fn} is Cauchy, given e = t. there is an integer n 1 such that for n > n 1 , 11/n 1 -/n 111 n 1 such that for n > n 1 , ll fn 1 -fnll 1 <(t)'. In this way, we obtain a subsequence {fnk }, k = 1, 2, , such that for idl lc, . . .
Letting g be the constant function 1 in the Cauchy Inequality (or rather, in Corollary 33.3), we obtain
*The theorem works
even
if
we
allow inf'mite-valued functions (see note for Theorem
28.2). 1f/ e .t(A.), then / is f'mite a.e. on A .
1 06
L E B ES G U E I NT E G R AT I ON A N D FOUR I E R SE R I ES
Thus, k� l � link - fnk +l ldm < ../m(/1). By Corollary 28.4 to the Mono tone Convergence Theorem, we may interchange the summation and in tegration, to obtain
...
Hence k�- l link - Ink+ 1 1 E .C(A), hence is finite a.e. Since absolutely ... convergent series are convergent, f(x) = fn 1 (x) + k"i2 t (fnk + 1 (x) - fnk (x)) is well defined for almost all x E A . We can let f(x) = 0, say, on the ex ceptional set of measure 0 where the series for f(x) diverges. Note that for almost all x E A , m
f(x) = fn 1 (x) + � 'f.: 1 (fnk+ t (x) -fnk(x))
= fn 1 (x) + mlim (-fn 1 (x) + f,._ · ·m (x)) = lim fn k (x). +"" k +""
Now let us show that f E .C2 (A). Since f = (f -Ink) + Ink ' and Ink E .C 2 (A), it suffices to show that f - Ink E .C2!A). Now for i > k, llfn1 - fnkll 2 < (fJ, so that ilim llfn; -fnklh < {t) . Hence by Fatou's + ... Lemma (28.7),
k
£. (t-fnk)2dm < �L (fn1 -fnk)2dm (�) 2 . <
So t -Ink E .C2 (A).
Finally, Un } converges to/in .C2 since
k
The first term on the right side is < (t)2 , and the second term can be made arbitrarily small (for large enough n, k) since {fn } is a Cauchy sequence. 0
Let us now restrict ourselves to a closed interval (a, b] -certainly a bounded measurable set. We know that any continuous function is in .C2 ([a,b]) (why?). Also, iff: (a,b) -+ 6l is the .C2 limit of a sequence of continuous functions, then f E .C2 ([a,b]) (Exercise 34.33). As we shall prove, the converse is also true ; if/E .C2 ([a, b] ) , then /is the .C2 limit of a sequence of continuous functions. Put another way, we say that C( (a, b]) is dense in .C2 ([a,b]); that is, given f E .C 2 ([a,b] ), and e > 0, there is a function g E C([a,b]) such that llf -gll 2 < e. To prove this, we begin, as we often do, with open sets.
F UNCT I O N SPACES A N D £2 1 07
If G C [a,b] is open and e > 0, then there is a continuous function g such that II XG - gl h < e; that is, �a,b ) ( XG - g)2 dm < e2 •
33.12 Lemma :
Proof: Write G = V (a1,P1) (Theorem 7 .2), with the (a1,P1) disjoint. We define g(x) = 0 for x 'fE. G. For each i = 1 , 2, . . • , we attempt to make l(a i, b ;)(XG - g)2 dm < e 2 /21• Then countable additivity of the integral (I'heorem 24.3) will give the result.
• I I I
.,
I
Given (a1,b1) , therefore, let 8 be a positive number less than t (b1 -a1), otherwise undetermined as yet. Then defmeg(x) = 1 for a1 + 8 < x < b1 -8, and let g be linear between the points (a;,O) and (a1 + 8,1), and between (b1 - 8,1) and (b1,0). Then g is continuous, and since O < :xa -g < 1 on (a1,b1), we have
1a;,b;) (XG -g)2dm < 1a,,b;) (XG -g)dm = 2 (�8)
=
8.
(See shaded triangles in the figure.) Therefore, if we let 8 < e 2 /21 in the above construction, our result will follow. o Now we extend our result to measurable sets. Given A C [a,b] measurable, and given e > 0, there is a con tinuous function g such that II XA glh < e.
33.13 Lemma:
-
Proof: Find an open set G C [a,b] such that A C G and m(G) < m(A) + 6 , where 6 is a positive number to be determined. Then, by the previous lemma, there is a continuous function g such that II XG glh < e/2. Therefore,
-
1 08
L E B ES G U E I NTEGRAT I ON A N D F O U R I E R SE R I ES
II XA - g lb < II XA - xGib + II XG - gll 2 < {h a.b J (XG - XA Y dm)Ya + e/2 = (�a,b) (XG - XA)dmt + e/2 = (m(G) -m(A))'Ia + e/2 < 8'/a + e/2 (Why is ( XG - XA )2 = ( XG - x.t )?) Taking 8 = (e/2)2 gives the result.
0
Given f E .C2 ([a,b]), and given e > 0, there is a continuous [a,b] -+ 6l such that II f - gil 2 < e. Proof: Using the decomposition f = f+ + f_, we need only prove the result for f+ and f- separately; that is, we may assume f :> 0. But in this case, there is a monotone increasing sequence of non-negative simple functions {fn } which converges pointwise to f (Theorem 1 9 .4) . Therefore gn = (f - fn)2 defines a sequence, converging pointwise to 0, such that 0 < gn < [2 • By the Dominated Convergence Theorem (Theorem 28.9), nlim+oo�a,b) gndm = 0. Thus 11 1- fn lh = .JJ.1 b) gndm can be made arbitrarily
33.14 Theorem :
function g:
a,
small, for large enough n. Since we have approximated /arbitrarily closely by simple functions, wenneed only prove the result for simple functions, which are of the form r=� 1 c1x.t r. , the A; being measurable. Therefore, we need only find a continuous function g1, for each i = 1 , . . . ,n, such that llctXA 1 - gAb < e/n. This follows from Lemma 33 .1 3 . See Exercise 34.35 for details. o
34.
34.1
Exercises Prove the following for any vector space V. (a) oU 0 for all u E V. ( b) 0 is unique; in fact, if 0' E V had the property that 0' u some E V, then 0 01• (c) Additive inverses are unique ; in fact, if 0, then ( d) aO 0 for all a E 6t (e) ( -I )u for all v.
=
ii
=
= -ii
=
ii + v =
+ = ii for v = -ii.
ii e
34.2
Verify that the Examples 32.2 are all vector spaces.
34.3
Let M [ 0, 1 ) -+ 6l j { has a maximum and a minimum value}. Prove that M is not a vector space under the ordinary addition and multiplica tion by a real number.
34A
Verify that Examples 3 2.4 are all norms.
= {!:
FUNCT I O N SPACES A N D £2 1 09
34.5
Prove that if II II is a norm, then
lllilll - n v n l < n il --vn. 34.6
Verify that properties ( 1 ) - (4) of the metric d follow from the properties of ll n.
34.7
Prove the statements in Example
34.8
Prove that if /,. C( [ O, l ) ) and /,. -+ f in the lub norm ( Example 32.4( 2)), then /,. -+ / in the norm of Example 3 2.4(3) (called the .C 1 norm) .
34.9
Prove that any convergent sequence in a normed linear space is a Cauchy sequence.
3 2.6( 1 ) and ( 2).
E
34.1 0 Prove that 6!" is complete in its usual norm ( Example 32.4( I )). (You may use the fact that 6l is complete.)
34.1 1 Prove that 8(A ) is complete in the lub norm ( Example 3 2.4( 2)}. (Given
a Cauchy s,equence in this norm, its limit must be the pointwise limit of the sequence.)
34.1 2 Prove that C( [ 0, 1 ) ) is not complete with the .£1 norm ( Example
32.4(3)).
34.13 Prove that 6!( [ 0, 1 ] ) is not complete with the .£1 norm ( Example
32.4(3)).
(Hint : let /,.(x)
= x" .)
( Hint: what could
6!( [ 0, 1 ) ) ?)
34.14 Prove that if
{!,.} converge to pointwise which would not be in
u . v is an inner product, then u. (v + w) = ii . v + ii . w,
u • (aV) = a(u • V) , and 0 • v = 0 for all u,v, w E V, a E tR.
34.15 Verify that Examples 3 2. 1 1 are inner products. 34.16 Show that the
.£1
product by letting
norm (Example
32.4(3)) does not come from an inner
34.17 Prove that if u • v is an inner product, then ll u ll =
( 1 ), (2) and (3) of the norm (Definition 32.3).
34.18 If
..[ri7"U has properties
II ull = �. show that the Parallelogram Law holds:
110
L E B ES G U E I NTE G R AT I O N A N D F O U R I E R SE R I ES
p(x) = ax 2 + bx + c, if p(x) > 0 for show that the discriminant of p (namely b1 - 4ac) is < 0. (b) Give a proof of the Cauchy Inequality by considering
34.19 (a) Given a quadratic polynomial all
x,
p(x) = (u + xv) •. (u + xv), and using part (a). 34.20 Verify the statements of Example 32. 1 7. (Hint for (3):
sin(a + b) + sin(a - b) = 2 sinacosb cos( a + b) + cos(a - b) = 2 cos a cosb cos(a - b) - cos(a + b) = 2 sinasinb.) 34.21 (a) Prove that an orthogonal set in an inner product space is linearly
independent. (b) Prove that if {v. , . . . ,'v,.} is an orthonormal basis for .. - - v E V can be expressed as v = l: ( v • v;) v1•
34.22 Let
lo(x) =
\
2n + 1 < x < 2n for some integer n -1 if 2n < x < 2n + 1 for some iilteger n 0 otherwise. 1 if
{ 10 J. , . . . } is
g = J.0,1 1 1gdm. Given a point (xoJio) and a line ax + by = c in IR1 , use the ideas of section 32 to find the point on the line closest to (xo,y0), and the distance from (x0,y0) to the line. 3 I·
(b)
then any
1= 1
Let l,.(x) = lo(2"x) for n = 1 , 2 , . . . . Prove that the set orthonormal in /,( [ 0, 1 ] ) with respect to the inner product
34.23 (a)
V,
Repeat (a) for a point and a plane in IR
34.24 Find a non-measurable function I such that
•
11 is measurable.
34.25 Show that the relation "equal a. e. " is an equivalence relation. 34.26 Prove Corollary 33.4. 34.27 Prove Proposition 33.5.
F UNCTI ON SPACES A N D £.2 1 1 1
34.28 Prove Corollary 33.8. (Hint : let
h(x) =
l
l if f(x)g(x) ;> 0 -1 if f(x)g(x) < 0.
Then {gh = lfg l. Apply Theorem 33.7.) 34.29 Prove that if In , { E .C2 (A ) and In -+ { uniformly on A , then fn -+ f in .C2 • 34.30 Show that pointwise convergence does not imply .C2 convergence. (See
section 27.)
34.31 (a) Suppose that f is bounded and measurable on a bounded measurable
set A . Prove �1!!1.E .C2 (A). (b) Let {(x) = 1/v lx I for x + 0, {(0) = 0. Show
{ E l( [ - 1 , 1 ] ) \ .C2 ( [ -l , l ] ). 34.32 Let fn .f,g E .C2 (A ), and fn -+ f in .C2 • Prove that lim L fngdm = J,.. fgdm. n +• "' .
34.33 If In E .C 2 (A ) ·for
all
n,
and In -+ f in .C 2 , then show that { E .C2 (A ).
34.34 If fn E .C2(A ), fn -+ f in .C2 , and g is bounded and measurable, then
fng -+ fg in t?- .
34.35 Give all of the details of the proof of Theorem 33. 14. 34.36 Show that the unit sphere of .C2 ( [ 0, 1 ] ) is not compact. (The unit sphere is E .C2 ( [ 0,1 Dl If 11 1 = 1 }. Find a sequence Kn such that lf gn lf = 1 and
{!
l f gn
-Krdl ;> 1 for
n
+ k.)
34.37 For { E l(A ), define the .C 1 norm by If fli t = .ft lfldm . As for .C 2 (A), we
identify functions which are equal a.e. on A . (a) Show that I f I f 1 is a norm. (b) Prove that l(A ) is complete in the norm If II 1 • (c) Show that C(A ) is dense in .Q:A) in norm If 11 1 •
34.38 Generalize 34.3 7(c) to show that if V C .C2 (A ) is dense in .C2 (A ) in norm
lb . then V is dense in .C(A ) in norm If 11 1 • (Hint : show that .C2 (A ) is dense in l(A ) in norm If ft .)
II
34.39 Suppose that fg E .C( [a,b ] ) for all {E .C2( [a,b ] ). Show that g E .C2 ( [a,b ] ). ..
34.40 Let l: l f fn 11 2 n
=1
..
< •. Show that nl: 1 fn converges absolutely a.e., =
112
L E B ESGUE I NT E G R A T I O N A N D FO U R I E R S E R I ES
11 /l l x < ll fll 2v'mm for / E l2 (A ). Hence if /11 -+ f in £2 , then /11 -+ / in £ 1 • (b) Find a se�uence {/11} which converges to 0 in £1 but does not con verge in £ .
34.41 (a) Show that
34.42 If A is bounded and measurable and p a positive real number, define £P(A ) = {f:A -+ R [ / is measurable and fA l! IP dm < •}. Note that £2
and £1 are special cases. (a) For f E £P(A ), define II !lip = fl.ft lfJP. As in £2 we consider two functions equivalent if they are equal almost everywhere. Show that properties ( 1 ) (3) of definition 32.3 hold for £P(A). The remainder of this exercise will deal with proving property (4) of 3 2.3 for J!P(A ) when p > 1 . (b) Prove that i"b 1 - A < Nl + ( I �)b for 0 < � < 1 and a and b non negative real numbers. (Hint: Take the log of both sides of the inequality and use the concavity of the graph of log). (c) Prove Holder's Inequality : if p and q are real numbers greater than 1 such that 1 /p + 1 /q = 1 and if / E £P(A ) and g E J!'l(A), then fg E l1(A ) and fA lfg l < 11 /llplllllq· (Hint: let lfJP = a and Lf lq = b and � = 1 /p in part ( b) above). (d) Prove property (4) of 3 2.3 for £P(A ) with p > 1 . This is known as Mi)\towski's Inequality and states that II/ + g l l 11 < 11/llp + IIKIIp· (Hint: note that (/ + g)P = f(f + g)P - 1 + g(f + g)P - 1 , and apply Holder's Inequality to each term of the right hand side.) (e) Show that J!P(A ) is a real linear space. It turns out that these £P spaces are complete, but it is beyond our present scope to show this. -
-
CHAPTER
8
The JJ Theory of Fou rier Series
35.
Definition and EJCamples
The study of Fourier Series amounts to approximating certain func n tions by trigonometric polynomials of the form k1:= O (akcoskx + b"sinkx) or trigonometric series of the form k1:= O (ak coskx + bksinkx). -
Because the sine and cosine functions are useful in many physical applica tions involving waves (e.g. spectroscopy, propagation of sound waves, heat equations), the study of Fourier Analysis is of great importance to physical scientists. In the next Chapter we will discuss one of these applications. For now we restrict ourselves to 1.2 where the study of Fourier Series leads to particularly nice results. We shall also restrict our attention to the in terval [""''I', 1r] since trigonometric functions are periodic of period 211' (i.e., sinx = sin(x + 211') for all x). by
Recall from Chapter 7 that 1.2 [-11', 1r] has an inner product defined
and the resulting norm 1 13
1 14
L E B ESG UE I NTEG R AT I O N A ND F O U R I E R SE R I ES
Also recall that { u 1 , • • • , un, • • ·} is an orthonormal set in t.'· if and only if u; • ui = 0 whenever i � j, and II u�h = 1 for every i. There are many orthonormal sets in 1.2 [ -w, 1r] but in Fourier Analysis one uses the set
{
1
cosk.x c smk.x, Vc 'fr I
. �· . y2 V 'fr
1r
1
.
}•
k= l
(see example 32. 1 7 and exercise 34.20). Many of our results hold for any (countable) orthonormal set in 1.2 , and we will often use the more general notation {u;} : 1 • 35. 1 Lemma:
{
cosk.x c smk.x, Vc 'fr
1
1
. r.:;:: , V �'fr V 'fr
.
1
is an orthonormal set in 1. 2 (-1r, 1r] .
}-
��
Note: In doing Fourier Analysis, it is often helpful (and traditional) to
use the Riemann notation for the Lebesgue integral. That is, we will write �bf(x)dx instead of ���, b ldm, even though f may not be Riemann integrable. Proof:
First,
fw _l_ cosk.x dx = 1_ J wcos kxdx - sink.x l w = 0' -w .J2ir ../i 7r>/i -w k'lf
-
and similarly
J w sink.x dx y21r -w 1
--=
=
0.
Also, using the trigonometric identity
cosA cosB 21 cos(A + B) + 21 cos (A -B), =
f w ��dx lfw cos(m + n)x dx + lfw (cos(m - n )x dx 2 -w ...[; ...[; 2 -w 1r w 1r =
T H E .t2 T H E O R Y OF F O U R I E R S E R I ES
1 15
which = 0 if m :F n and = 1 if m = n (Verify!) Similarly
I
sin mx sin nx ,.. --
_,..
and
l O if m
-- dx =
.JW .J;
:F n
1 if m = n
f,.. sin""mx � dx = O "" _,..
for all m. n.
0
If you have any trouble (Riemann) integrating in the above proof, see Exercise 34.20 of Chapter 7. Given a finite orthonormal set { u 1 , ,u,. }, Theorem 32.1 8 says that the best approximation of [by a sum of the form k=:t 1 CtUt is the proiection of f on the subspace spanned by {u 1 , , u,.}. Therefore, for a countable orthonormal set { ut l;= 1 , we are motivated to try to approximate f by :t 1 CtU k where Ct = f • Ut· That is, we suspect that the best approxima k= tion for f in terms of the Ut (in the /.2 norm) is obtained by using co efficients of the form Ct = f • Ut· We give these a special name. •
•
•
II
�
•
•
•
ao
35.2 Definition :
Let {u"};= 1 be an orthonormal set in /.2 (-,r, 1r) and let
f E /.2 • Then the nth generalized Fourier coefficient off with respect to the set {ut l t= 1 is c,. = f • u ,. = f1 _ ,.. ,.. . r · u,.dm = f�,..f(x) • u ,.(x)dx. *
,
set is
The generalized Fourier series off with respect to the orthonormal
Throughout the remainder of the text we will be primarily interested in the particular orthonormal set
� 1 sin/a cos/a t • } Vii ' .;; ' "" � k= 1
*Note that this inner product is always deimed for f E £2
(see 33 .3).
118
LEBESG U E I NTE G RAT I O N A N D F O U R I E R SER I ES
In this case we drop the word "generalized," and the Fourier coefficients are ao = rr f(x) �dx, J_ w v 211' fw sinh' {jk = -fff(x) yt=11' dx,
-
k>
O.
The Fourier Series associated with f is I + � a cosh' + sinh' ) . ao .J21r "'" -k�=l k-.J; � It is conventional to write the Fourier Series for fin the fonn
-- ( ao 2
where
• Jff -fff(x)dx,
ao = 1r
35.3
R
+ k=��l (akcosh' + bk sinb)
• Jff 1r -ff f(x)cosh'dx,
a" = -
ff
b " = 1r-1 J-fff(x)sinh'dx
and we will adopt this practice. It is easy to see that the same series results. Note however that {t,cosh',sina};= • is not an orthononnal system so that this alternative method is merely a notational convenience. The most interesting questions about Fourier Series have to do with convergence. Given a function fE 1.2 [-,, 11'] , we have shown how to obtain its Fourier Series, but we do not know whether this series converges (unifonnly, /.2 , or at a point) and if it converges at x whether it converges to the value f(x). The remainder of the text will deal with these questions. For now let us look at some examples of Fourier Series. Example: Let 0 for -11' < x < o f (x) = { l for O < x < 11'. Then l sinb- l w = 0 1. f k =F O J -1 1 11' k o a" = - f(x)cosh'dx = - f cosh'dx = 11' - w 11') o 1 l w = 1 if k = 0. -x 11' 0 tr
tr
{
T H E £.2 TH E O R Y OF FOU R I E R SE R I ES
Similarly, b"
=
_ ! coskx
11'
k
"' =
Io
{
1 17
1 2 ;r 0 k for k odd
0 for k even. Thus the Fourier Series associated with/is
Notice that /(0) 1 , but at 0 the Fourier Series equals 4-· In fact, since we are using Lebesgue integration, any functions f and g equivalent in 1.2 (i.e. f(x) = g(x) for a.e. x) will have identical Fourier Series. This brief discussion should prepare the reader for some of the problems involving pointwise convergence of Fourier Series which we present in Chapter 9. =
Let f rationals. Then
35o4 Example:
=
XQ
on
[-11', 11'] ,
the characteristic function of the
-11' /(x)sinkxdJc 0 and -11'•J"'-f(x)coskxdJc = 0
If"'
=
-71'
for all k (why?). Thus the Fourier Series for /is identically 0. Notice that f(x) does not equal its Fourier Series at each rational number x. Let f(x) = lx I on [-11',11'] . Then each b1c is 0 because f is an even function (see Exercise 38.1). Also a0 11' (verify) and
35.5 Example:
=
J_,.. lx lcoskxdJc ,..
1 a" 11' =
-
Thus the Fourier Series for /is ! _i
2
11'
[cosx
-11'Jo
2 =
+
,..xcoskxdJc
-
=
cos3x + cosSx + 9
25
{ o
for k odd 04 for k even. fc'l
n
0
OJ .
More examples are to be found in the exercises at the end of this Chapter.
118
L E B ESGUE I NTEG R A T I O N A N D F O U R I E R SE R I ES
36. Elementary Properties
We first restate a result from Cfl;apter 7 which shows that the gen eralized Fourier coefficients of f yield the best approximation of f in 1.2 [-11', 1r] . Although we are primarily interested in the orthonormal set j 1 sinkx coskx } "" l .Jr,r ' ..;; ' "" k= 1 ' we will use the general notation for clarity. Let {u"};'=1 be an orthonormal set in n/.2 [�,1r] . Define, for /E 1.2 [-1r,1r] , ck = J.( - '11' '��' l fu kdm and sn(x) = kI:=Ockuk(x) [snn is called the nth generalized Fourier approximation of f.) . Let tn(x) = kI:=Odkuk(x) where the coefficients are arbitrary. Then sn(x) is a better /.2 approximation of /than tn(x). That is, �-'11','11' ) ( ! - sn i dm < �-'11','11' ) ({- tn)2dm. D Proof: This is just Theorem 32.18 of Chapter 7.
36.1 Theorem :
•
36.2 Corollary: Proof:
Proof:
With the above hypotheses/.1-'11','11' 1t2dm > kE= Ocl for every n. See Corollary 32.19.
D
This follows since Corollary 36.2 above holds for aU n.
D
(Riemann-Lebesgue Lemma) klim+ oo ck = 0. Proof: fr -'ll','ll' l2dm bounds the positive term series � c"2 by ko D Corollary 36.3, so the series converges and the result follows.
36.4 Corollary:
In the /. 2 sense, successive generalized Fourier approximations improve as n increases (i.e. fr -'11','11' ) (f-sn)2dm > fr-.'��' l (f - sn + t )2dm}
36.5 Corollary:
T H E .t2 TH E O R Y OF FOU R I E R S E R I ES
This follows from Theorem 36.1 since �-"'·"'I ({- s,+ 1 )2dm < fr-"'·"' 1(!- t, )2dm where t, = s, + Ou,+ 1 ·
1 19
Proof:
36.6 Corollary:
D
For the orthonormal set { _1_ sin/a cos/a } ..
..tiir ' ..;:; ' ..;; k= 1
{1)
where
Bessel's Inequality becomes
ak = 1
-
"'
f"' f(x)cosladx -71'
and
bk =
I J"' .f(x)sinladx,
-
"'
-71'
and (2) Riemann-Lebesgue becomes (" f(x)cosladx = klim+!...1("-11'f(x)sinladx = 0. .... )_11' klim Bessel's Inequality shows that an !.2 function f has Fourier coef· ficients which obey That is, iff E !.2 , then converges. The converse is also true: if there are numbers {a;};. 0 and { b;};. 1 sue� that converges, then these numbers are the Fourier coefficients of some func tion f E !.2 • The key to the proof is the completeness* of !. 2 (see Theorem *For this reason this result is often called the Riesz·Fischer Theorem.
1 20
L E B ESG UE I NTEG RATION A N D FOU R I E R SE R I ES
33.1 1} and the fact that £2 convergence permits integration term by term (Theorem 33.10). We prove the theorem for any (countable) orthonormal set which of course includes the special case of trigonometric series. (Riesz·Fischer Theorem) Let {u1}� 1 be an orthonormal set in £ 2 [-,r,w] and suppose r=� l cl is a convergent set of real numbers.
36.6 Theorem :
Then there exists a function / E £ 2 [-w,w] such that c1 = f • u1 and the partial sums s,. = c 1 u 1 + • • + c,.u,. converge to f in the £2 sense. ·
+ c,.2 by ortho Proof: For n > m, (lis,. -sml h )2 Cm l2 + normality of the u1 (calculate this!). Thus {s,.} is a Cauchy sequence in £ 2 so that by completeness of £2 (33.1 1) there is an fE £2 such that =
+
·
·
·
For , > i, and by Exercise 34.32 of Chapter 7. So c1 �-'" '" lu1dm. That is, the c;'s are indeed the generalized Fourier coefficients off.· 0 =
37.
!2 Convergence of Fourier Series
We have seen that Fourier Series provide a means of approximating a function / E £2 • In fact, given an orthonormal set {u1}i:- 1 in £2 [-w,w] , the ge �eralized Fourier coefficients c" J1_'" •'" lu�cdm yield partial sums s,. k=:E l c"u" such that 11/-s,.ll 2 < n t -t,.lh where t,. is any other series ,. of the form k:E= l d�cu k . It follows (Corollary 36.5) that successive partial Fourier sums get closer and closer to f in the £2 sense. We would hope in addition, that lim s,. =fin the sense of £ 2 • n +• =
=
In fact this is true. It takes a fair amount of work to reach this result, how ever, all of it interesting in its own right. We begin with a definition.
T H E £2 T H E O R Y OF FOU R I E R S E R I ES
121
A series of numbers 11:E= 1 a" is said to have (C,l ) sum (or Cesdro sum) s if lim oPI = s where n +oo PI s 1 + s2 + · • • + sPI 0PI = and s = a" . I PI n 11: = 1
37.1 Definition :
({oPI } is called the sequence of arithemetic means.)
Many divergent series converge (C,l). For example, 1 -1 + 1 -1 + · · · diverges, but converges to t (C,1 ) since oPI = t when n is even and oPI = PI2+Pit when n is odd (see Exercise 38.7).
37.2 Example:
If a series l:a" converges to the real number s, then it also converges (C,1 ) to s. Observe that 1 s -sPI I ls - oPI 1 = -1 11s - s 1 ) + · · · + 1s - sPI ) 1 < -1 ls -s 1 1 + · · · +-l
37.3 Example:
n
\i
_
\i
n
n
Given e > 0, there is an N such that for k > N, Is - s1r: I < e. Also there exists an M such that Is -s��: I <M for k N, Is - oPI 1 < N-M + n -Ne . n
--
n
Thus Is - PIlim +OO oPI I < e.
We will obtain the remarkable and useful result that for a continu ous function, the Fourier partial sums converge uniformly in the (C, l ) sense. (Note that there may be points at which the Fourier Series may not even converge in the usual sense-see the next chapter.) We now need two unexciting, but necessary trigonometric identities.
37.4 Lemma:
(1) For (J ::1= 2mr for any integer n,
1 22
L E B ES G U E I NT E G R AT I O N A N D F O U R I E R S E R I ES
(2) For 9 ::/= mr for any integer n,
.2 sin9 + sin39 + • • • + sin(2n - 1)9 = s�sm n89 .
Using the trigonometric identity sin(B +A) - sin(B -A) = 2sinA cosB, and letting B = k8 and A = � , we obtain Proof :
sin (k + i) 9 - sin ( k - i ) 9 = 2sin ( i 9 ) cosk9.
Letting k = 0,1 , • • • ,n and adding, we obtain
sin ( n + i ) 9 + sin (�) = 2sin (�) [1 + cos9 + cos29 + • • • + cosn8]
from which.part (1) follows easily. (2) Using the relation . cos(A -B) -cos(A + B) Si"nA smB = 2 we obtain sin9 [sin9 + sin39 + • • • + sin(2n - 1)9] = 1 - co; 2n9 .
0 Using the cosine double angle formula, this equals sin2 n9. We are now ready to prove some important results about trigono· metric series. The ultimate goal is to 2show that for f E £2 [ -w, 1r] , the partial sums s,. converge to f in the £ sense. On the way we will prove that for f continuous, the (C,l ) sums o,. of the Fourier Series converge uniformly to f. To achieve these results we need the following integral representations of s,.(x) and o,.(x).
Note: From now on we will find it necessary to use expressions such as f(x + t), where /E £2 (-w,1r) and x e [-fr,fr] . Since x + t need not be in (-w,1r) , /must be defined outside this interval. In this case, we assume f is extended periodically (with period 21r) to aU of R. That is, f(x + 21r) /{x) for all x E R. Of course, this may make it necessary to change the value at either _,.. or 1r, but this will not affect any integrals involving f. =
TH E .&2 T H E O R Y OF FO U R I E R SE R I ES
1 23
Let { E 1.2 [-.,.., 11'] . * Then the Fourier partial sums
37.5 Lemma :
s,. (x) = !
11'
where
[ [t( 0
x
+ t) + {(x - t)]D,.(t)dt,
0
if t = 2n11', n an integer
( 4)- t otherwise. 2�n ( � )
D,.(t) = �n n +
(D,.
is called the Dirichlet Kernel of order n . Note that D,.(-t) = D,.(t),.). Proof:
s,.(x) =
By defmition,
,. a° + I (at coskx + b tsin kx) 2 k= l
f,.,.. . f(t)dt + -11' tI,.= l ( coskxf,.. f(t)cosktdt + sinkxf,.. {(t)sin ktdt1 [ 1 ,. 1f f(t) -2 + I coskxcoskt + sinkx�nkt dt 11' J 1 211'
=-
1
_ ,..
_,..
=-
k= l
_ ,..
[ f(t) [ -21 + I cosk(x - t)] dt
=!
11'
_,..
37.4 1 = 1r
_,..
t
k= l
[ f(t) in ( n + 4x)<x; t- t ( ) _,..
J
2�
*Actually, / E .& ( -w,w] is enough here.
* *This integrand is not def"med for x only on a set of measure zero,
so
-
t
=
dt.**
2mw,m an integer.
that the integral exists.
0
However, this occurs
124
L E B ESG U E I NT E G R AT I O N A N D FOU R I E R SER I ES
[. ( j
By exercise 26.35 and exercise 38.14,
1 x+• •.(x) = _ [ f(x - u) 11'Jx - • 1
=
-
=
-
=
11'
1
SID
n+2 u 2sin !! 2
(
du
J." f(x -u)Dn(u)du = tJo f(x -u)Dn(u)du + -
11:
-·
-·
1 "
11't f"0f(x + t)Dn(-t)dt + 11'-f0 f(x - t)Dn(t)dt
t f" [f(x + t) + f(x - t)]Dn(t)dt.
-
11'
37.6 Corollary:
0
Let
for sn(x) defined as in the Lemma. Then On (X) =
where
11't f"0 [f(x + t) + f(x - t)] Kn(t)dt,
-
0
for t = 2m11', m an integer
sin 2 ( nt )
Kn (t) =
· h ·-
2n sm 2 2 (Kn
is called the Fejer Kernel of order n.) Proof: Clearly, n-1
-
so
i
1 " - [f(x + t) + {(x - t))Dt(t)dt n kI 11' =O 0
1 On(X) =
we need only show that
n-1
2
sin (nt/2) Dk (t) n tI 2n sin 2 (t/2)" =o
!
=
T H E 1.2 T H E O R Y OF FOU R I E R SE R I ES
But
1 25
n n-1 1 - l Dk(t) = I I sin k + 21 t 2n sin(t/2) k= =O O
(
n- k:I
-
)
by definition of Dk(t), and n-1 n n 2�n t/2) I sin k + - t = I sin j --1 t = I sin(2j - 1) ! = 8� I 2 k=o 2 2 sm(t/2) 1= 1 /= t
(
)
(
by 37.4.
)
·
0
We are nearly in a position to prove the desired result about (C,1) convergence. All we need are two easy results concerning the Fejer Kernel Kn(t). 37.7 Lemma:
( 1)
for n = 1 ,2 •
K 11" J(f 0,11' ) n(t)dt = 1
� • •
and
(2) if t satisfies 0 < 6 < It I < 11", then Kn(t) < Proof:
Sn(X) = I .
2nsin� (5/2)"
(1) Apply Corollary 37.6 to f(x)
=
1 , observing that all
(2) Use the definition
and note that on [5 ,11"] the minimum value of sin 2 (t/2) is obtained when t = 6. 0 Let /be a continuous function on [-., 11"] (thusfE .C2 [-11",11"] ). Then an converges to /uniformly on [-.,1r) , where
37.8 Theorem :
(i.e. the Fourier Series off converges uniformly (C,I) to f .)
1 26
L E B ESG U E I NTEGRATION A N D F O U R I E R S E R I ES
Proof: Since/ is continuous on [-11',11'] , it is bounded and uniformly continuous. Thus there exists a number M such that lf(x) I < M for all x E [-11', 11'] . By uniform continuity, given any e > 0, there exists a 6 > 0 such that lx - y I < 6 in [-11', 11'] implies lf(x) -f(y) I
Now
�fo''f(x + t) � f(x - t)K,.(t)dt -f(x) I 31:,7 1 �t" f(x + t) ; f(x - t) K,. (t)dt -f(x)� "K,. (t)dt I fo � s: + + f(x2- t) - 2/(x)K,. (t)dt I < � f(x t) � 1"f(x + t) + f(x - t) - 2/(x)K,. (t)dt I + � 2 < � � s: K,. (t)dt
la,. (x) -{(x) l 37�
+
� f" [ lf(x + t) l + lf(x - t) l+ 2 1f(x) l] dt 4n11'sm (6/2)Ja
< �2 +
.
I
[4Mdt.
2n11'sin 2 (6/2) a
The second term is clearly < 1" for n large enough, so the result follows.
0
We are finally in a position to prove the main theorem of this section.
Let fE l2 [ -n,11'] and s,. be the nth partial sum of the Fourier Series for f. Then s,. converges to fin the l2 norm. That is, given any e > 0, there exists an N such that n > N implies II s,. -.fll 2 < e .
37.9 Theorem :
Proof: Given e > 0, by Theorem 33.14 there exists a continuous function g such that II f - glh < f· By the previous theorem, a,. (of the Fourier series for g) converges to g uniformly on [--.r, 11'] , so by Exercise 34.29 there is an N such that II aN -glh < 1"·
TH E £2 T H EO R Y OF F O U R I E R SER I ES
1 27
By the triangle inequality, II/ - aN i h < e . Since aN is a trigono metric polynomial, Theorem 36.1 says that llf - sN i h < ll f - aN ih < e , and by Corollary 36.5 I f- sn 11 2 < e for > N. 0 n
37.10 Corollary
(Parseval's Formula): If/E .C2 ( [--w, 71'] ), then
Proof:
2 ao 1 2. I + -71' 1 f1 2 = 2 k = l (a ,? + bk )
Exercise 38.20.
We have now seen that the study of Fourier Series is rather nice when viewed in the .C2 setting. We have seen that .C2 is a complete inner product space in which one can integrate convergent series term by term. The continuous functions are dense in .C2 (in the .C2 norm), and the Fourier Series of each .C2 function converges to that function (in the .C2 norm). In most applications, however, one is interested in pointwise convergence. We have seen that .C2 convergence does not imply pointwise convergence. The study of pointwise convergence of Fourier Series does not yield such elegant results as those for .C2 • In fact, no nontrivial necessary and suf ficient conditions for convergence at a point are known. In the next Chapter we will produce several of the well-known sufficient conditions for pointwise convergence of Fourier Series. 38.
Exercises
38.1 (a)
(b)
Let / be an even function in .C 2 ( [ -71',71') )-that is, {(-x) = f(x) for
all all
x E [ --w,71'] . Show that in the Fourier series for /, b k = 0 for
k.
If / is odd-that is, f(-x) = -{(x)-show that a k = 0 for all k.
38.2 If /E .C( [ --w,71') ), why is /(x) cosnx E .C([ --w , 71' ) )? 38.3 Find the Fourier series for the following functions.
(a) (b) (c)
(d)
( e)
/(x) = x on [ --w, 71'] . f(x) = 3 cos 2x + 4 sin 5x + 2. f(x) = 3. 0 �f -71' < x < o fi(x) = X lf O < x < 71'. if -71' < x < o f(x) = sin x if O < x < 71'.
{ {o
38.4 On graph paper, carefully draw graphs of the first four or five terms of
the Fourier series for one or two of the functions in Exercise 3 8.3. On the same graph, draw the sum of the first two terms, of the first three terms, and so on.
1 28
L E B ESGUE I NT E G R A T I O N A N D F O U R I E R SE R I ES
38.5
Prove Corollary 36.6 in detail.
38.6
(a) Use Bessel's Inequality and the Fourier series of Example 3 S.3 to show that
� l /(2k - 1 ) 2 < 71'2 /8.
k= 1
(b) Use Example 3 S . S to get an upper bound on 38.7
(a) Prove that n� (-1 )" converges to -1 (C, l ). =O 2 (b) Does the series nE 1 n(-1 )" converge (C; l )? =
� l /(2 k 1 ) 4
k= 1
-
.
-
-
38.8
If 0 < a,. < b,. for all n and n E1 b,. converges (C, l ), show that nE a,. also =1 = converges (C, l ).
38.9
(a) If a,. > 0 for all n, show that o,. < o,. + 1 for all n. (b) If a,. > 0 for all n, and nE a,. = •, show that nlim o,. = •. This =1 +• problem shows that a series with non-negative terms converges to a real number if and only if it converges (C, l ) to that number.
38.10 Alter the proof of Lemma 3 7.4 to show that for 8 ::1: n71', n an integer,
,.
1 I1 sin k6 = 2-:-SID 8 ( I + cos8 - cosn8 - cos(n + 1 ) 8).
k=
-
38. 1 1 (a) Show that E coskx diverges.
k=O • (b) Use Lemma 3 7.4 to show that the partial sums of the series E cos lex k=O are bounded, for x ::1: 2n71', n an integer. (c) Use Lemma 3 7.4 to show that E coskx converges (C, l ) to 0. k=O (Hint: calculate o,. . See the proof of Corollary 3 7.6, or use Exercise 3 8 . 1 0 .)
-
38.1 2 Show that if E a k converges, then
k= 1
lim
(
N... k-1 -I N
N + • k= 1
)
a" = o.
38.13 Draw the periodic extensions of each of the following functions to all of fi. Is the resulting function continuous? (a) f(x) = x 2 on [ -n- , 71'). (b) f(x) = lx I on [ -71',71'). (c) f(x) = O �f � < x <. 0 X lf O < x < 71'
{
TH E £2 T H E O R Y OF FO U R I E R SE R I ES
1 29
38.14 If /E .C( [ -1T, 1T] ), extended periodically to R, show that
fx+
w
.X - 11'
J,..
f( t) dt = _ f( t) dt. ,..
(Hint: draw a picture.)
38.1 5 Prove Lemma 37.7 in detail. 38.1 6 Prove the following analogue to Lemma 3 7 . 7 : (a) D,. (t)dt = 1
! 1;
(b) For 0 < cS
< l t i < 1T, ID,. (t) l <
l /2 lsin(c5 /2) I.
38.17 Using Exercise 3 8. 1 6 and Lemma 3 7 . 5 in place of Lemma 37.7 and
Corollary 37.6, mimic the proof of Theorem 3 7 . 8 and attempt to prove that s,. converges uniformly to f for f continuous on [ -1T, 1T] . Does the proof work?
38.18 (a) Show that if I is continuous on [ �.1T] and its Fourier coefficients
are all zero, then t= 0. (Hint : Theorem 37 .8.) (b) Show that if I and g are continuous on [ -1T, 1T] and the Fourier coefficients of f and g are identical, then /(x) = g(x) for all x. (c) If f and g E .C2 ( [ -1T, 1T] ) have identical Fourier coefficients, what can you conclude? Prove your assertion.
38.1 9 Recall that the Taylor's series for cosx (or sinx), expanded about x = 0,
is a power series with infinite radius of convergence, and that the series converges uniformly to cosx (or to sinx) on [ �. 1T] . (a) Show that any trigonometric polynomial can be uniformly approxi mated by a polynomial on [ �, 1r ] ) ; that is, given a trigonometric polynomial t and any e > 0, there is a polynomial p such that l t(x) - p(x) l < e for all x E [ �, 1r ] . (b) Prove the celebrated Weierstrass Approximation Theorem : every continuous function on a bounded closed interval can be uniformly approximated by a_ polynomial. (Hint: first justify restriction to the closed int·�rval [ -1T,1T] . Then approximate with a trigonometric polynomial, and use part (a).) (c) Describe how you would construct a polynomial p such that l lx 1 - p (x) I < .0 1 for all x E [ -1T,1T] .
38.20 Prove Parseval's Formula: If f E .C2 ( [ �, 1T ] ) , then
!II l 1T
t
2 2
=
a o2 +
2
I (ak2
k= l
+ bk2 ).
Note the similarity to Bessel's Inequality. (Hint: Theorem 3 7 . 9 says that lim II s,. - /II� = 0. Compute II s,. -/II�.) 38.21 (a) Show, using Parseval's Formula, that
(Example 3 5.3.)
1: l /(2k
k= t
- 1 ) 2 = 1r2 /8.
1 30
L E B ESGUE I NTEGRAT I O N A N D F O U R I E R SE R I ES
(b) Find the sum of � l /(2k - 1 )4 . (Example 3 5 . 5) k= l (c) Find l: l /k2 • (Let {(x) = x on [-11",11"] .) k= l 38.22 If {, g E £2 ( [ -11",11"] ), with Fourier series
a o2 � a,. cosnx + /J,. sinnx, - + -' 2 n= l
a o2 � - + -' a,. cosnx + b ,. sinnx, 2 n= l respectively, show that
(Hint: Consider li t+ gll�. ll f - gll� and Parseval's Formula.) 38.23 Compute the Fourier series of f(x) = ex on [ --.,11"] and of g(x) = x on
[ -11", 11"] , and use Exercise 3 8.22 to obtain the sum of n� 1 /( 1 + n2) =O
.
f E £2( [-11",11"] ) and x E [-11",11"] , show that f�,/(t)dt is equal to the sum of the series obtained by integrating the Fourier series of { term by term. (Hint: let g(t) = 1 if -11" < t < x and g(t) = 0 if x < t. Use Exercise 3 8.22.)
38.24 If
38.25 Use Example 3 5.3 and Exercise 3 8.24 to show that
� cos(2n - l )x + 1 , for x E [ -11",11"] . l x I = 11" - ( 4/11") -' n= l (2n - I ) 2 38.26 (a) Use Example 35.5 and Exercise 3 8.24 to show that for 11" > x > 0,
� sin(2n - l )x .
1rx - x 2 = (8/11") -' n= l
(2n - 1 ) 3
(b) Let x = 11"/2 to obtain the sum of
i
(- 1 )
"
n = l (2n - 1 ) 3
.
38.27 If {is continuously differentiable on [--.,11"] , show that the Fourier series
of { 1 can be obtained from that of f by differentiating term by term. (Hint: integrate by parts.)
lim+J(x) and f(x0-) = lim_J(x) exist, for x +x o x +xo x0 E (-11",11") and { E .C( [ --.,11"] ), then nlim- a,. (x 0 ) = -1 [ f(x0+ ) + f(x o- )] . + 2
38.28 Show that if f(x0+ ) =
T H E £ 2 T H E O R Y O F FOUR I E R SE R I ES
131
-
38.29 Suppose that :E a n and n :E b n are absolutely convergent series. Show n= l :l
that
converges absolutely and uniformly on [ -w,w] , and its sum is continu ous. (Hint: Weierstrass M test.) 38.30 Show that if f is continuous on [a,b ) and the Fourier series of f con
verges at x, then it converges to the value f(x) at x. (Hint: Theorem 37.8.)
38.31 Conclude from the two previous exercises that the Fourier series for
lx I converges to lx I for all x E [ -,.., ,.. ] .
38.32 Prove that if
ao -+ � � (a cosnx + sinnx) 2 n= l n is the Fourier Series of a continuously differentiable function, then n:E= l nan and n:E= l n b n converge (See Exercise 38.27).
CHAPTER
9
Pointwise Converg en ce of Fou ierie r Series 39. An Application :
Vibrating Strings
In Chapter 8 we studied the rather abstract but elegant !. 2 theory of Fourier series and obtained a result about (C, l ) summability along the way. Of course, Fourier series did not arise as a neat way of expanding !. 2 functions, but rather as solutions to a certain class of physical problems. We will examine, rather informally, one such problem : vibrating strings. Suppose that a string is stretched under tension along the x-axis with fixed endpoints at x = 0 and x 11'. If the string is stretched out of its original shape and released, it will tend to vibrate; these vibrations will (under certain conditions) be transmitted to our ears via pressure waves in the air, and we will hear a musical tone. =
In order to analyze the situation mathematically, we introduce a function of two variables y = F(x, t), equal to the vertical displacement of the string at position x E [0,11'] and time t. Theoretical mechanics gives us a partial differential equation satisfied by this function under certain con ditions (for example, displacements must be small, there must be no fric tion, etc.). The equation is ( *)
for a constant c which depends on the physical properties of the string . Rather than attempt to solve this equation in general, we will examine some specific solutions under rather stringent assumptions. 1 33
1 34
L E B ES G U E I NTEG R AT I O N A N D FOU R I E R SE R I ES
Suppose for the moment that at time t into shape of a sine curveF(x,O) bsinx, ==
=
0 the string is deformed into
x E [0,11'].
Assume also that the string is at rest in this position at t 0 so that F1(x,O) 0 for all x E [0,11'] . Then if the string is released from this ==
==
y
"
position at t 0, it can be shown that it vibrates sinusoidally within the envelope y == ±b sinx. In particular, ==
F(x, t) = b sinxcosct for x E [0,11'] , t > 0. It is easily verified that this function satisfies the basic differential equation (*) as well as the conditions F(x,O) b sinx, ==
F1(x,O) 0. ==
Notice that since coset goes through an entire cycle (from + I to -I) whenever t changes by 211'/c, the frequency of the vibration (number of beats per unit time) will be c/211'. Now suppose instead that our original deformation was of the form F(x,O) = b " sinnx, a sine curve with n "humps," in [0,11'] , and suppose again that our initial velocity at each point is 0 so that F1(x,O) 0. Then a solution in this case is =
PO I NTWISE CON V E R G ENCE O F F O U R I E R SE R I ES
1 35
F(x, t) = b,.sinnxcosnct.
(Verify!) That is, we have a sinusoidal vibration at each x which is bounded by ±b,. sinnx. This vibration has n + I nodes (points at rest) and its frequency is nc/27r. Y l ••
"
_,,
The sequence of frequencies
c 2c 27r ' 27r '
generated above is called the harmonic or overtone series for the string. The first frequency, c/27r, is called the fundamental or first luzrmonic. The second, 2c/27r, is called the first overtone or second luzrmonic, and so forth. We have stated that if the string is initially deformed at zero velocity into the shape y =th b,. sinnx, then the string will vibrate purely with the frequency of the n harmonic. The question now is what happens when the string is initially de· formed, again with zero initial velocity, into an arbitrary shape F(x ,0) f(x). (Of course f cannot really be arbitrary; it must still obey the special assumptions such as small amplitude which were made in deriving the basic differential equation. Also f(O) and {(1r) should equal zero and the function must be continuous.) The solution could be obtained if we were able to write/in terms of a sine series: f(x) = n:E= l b,. sinnx. =
-
1 36
LEBESGUE I NTE G R AT I O N A N D F O U R I E R S E R I ES
Since we know all about the vibrations generated by b, sinnx, it would be reasonable to hope that we could superimpose the vibrations resulting from each b, sinnx; that is F(x, t) =
-
:E
PI = I
b, sinnxcosnct.
Indeed, formal differentiation term by term (assuming it is justified) shows this is a solution. The meaning of this series solution is that the string vibrates simultaneously with all of the frequencies nc/21f, for n = 1 ,2, • • • That is, all harmonics are present, each with different amplitude b, (some of which may be zero). The fundamental frequency usually determines the pitch we hear, and the amplitude of the overtone frequencies determines the quality or timbre of the note. .
Without discussion let us assert that similar results are obtained if we do not assume that the initial velocity of the string is zero. Our problem, then, has been reduced to the question of whether a given f can be represented by a sine series
In other applications (where f is not necessarily 0 at x = 0 and x = 1r), we must consider representations of {by trigonometric series f(x) = -o + a
2
�
k
k= l
(ak coskx
+ b k sinkx).
I f we assume that f has such a representation, and that term by term integration is legitimate (as for /.2 functions), then we find, as in the proof of Theorem 36.6, that a k and bk are the Fourier coefficients of f. In other words, if we are looking for a representation by trigonometric series, the only reasonable candidate is the Fourier series. Now, we know that every f E l [-,r, 1r] has a Fourier series. The question is, for which Lebesgue summable functions f and which points x does this Fourier series converge to f(x). In the following sections we will give some partial answers to this question. It is gratifying that most of the nice functions we meet in daily life do have Fourier series which converge to the function at least for most points.
PO I NTWISE CO N V E R GENCE OF FOU R I E R SE R I ES
1 37
40. Some Bad Examples and G ood Theorems
Since we can change a function on a set of measure zero without changing its Fourier series, it is obvious that Fourier series do not always converge to f(x) for each x . If we restrict ourselves to continuous func· tions, however, we cannot alter a function on a set of measure zero without destroying continuity. It is tempting to believe, therefore, that the Fourier series of a continuous function f converges to f(x) for every x. The following outline of an example (due to Fejer) shows that this is not the case. Given a positive integer n and x E [-11', 11'] , let cos(2n - t)x 1 L � + cos(n + l)x + + " 1 n n-1
40. 1 Example:
•
•
•
cos (2n + l)x _cos �(2_n_+....2)� ..;;;.. x 1 2
cos(3n)x n Then it can be shown (though not easily) that { s,. (x)} is uniformly bounded, i.e. there is a positive number M such that ls,. (x) I < M for all n,x. , let n " = 2� , and consider the series Now for k = 1 ,2, •
•
_ •
•
•
_
•
I -\s,. .. (x). k ..
k= l
Since
this series converges uniformly by the Weierstrass M-test to a continuous function {. Hence
/(x) =
(
)
cos3x _ cosSx cos6x 1 cos2x 1 2 2 + 1 16x + cos 17x + + cos31 1x _cos33x + !/cos 1 IS 4 \ 16 cos48x + ! cos2S 6x + cos34x 9 256 16 2 _
•
_ •
_
+
.
.
•
•
•
_
•
•
) (
•
•
•
)
1 38
LEBESG U E I NTEG RATI O N A N D F OU R I E R SE R I ES
Let g1 (x) be the ;th term of the series obtained by dropping the parentheses that is,
I g1 (x) t= 1
=
cos2x + cos3x _ cos Sx I I 2
_
cos6x + !cos I 6x + ! cos I 7x + 4 IS 4 I6 2
0
0
Then this cosine series is the Fourier series for /(e g. for p > 0, .
1
ap = 11'
-
tr ( I k1 2 '""(x)) cospx dx = -I I• I2 tr '"" (x)cospx dx .) J k= 1 '��' k= 1 k J •
_ ,.
-ff
However, the series •.l:= Og1 (0) diverges. We show it it not a Cauchy series by noting that given N, there is an N 1 > N such that for some k, -
I
1
gN 1 (0) = -2 o - ' k n rr
Therefore, if Sn is the nth partial sum of •.�= 1 g1(0),
(
I I I I = -2 - + -- + -- + o k n rr n rr - I nrr - 2 =
1 I 7r 2 -log nrr = -log2 k2 k2
=
o
o
I ) f"" -dt t
I + -I + I > 2 k2
1
I k2 log2 = log2. k2
- o
Therefore, the Fourier series of f fails to converge at 0, and thus does not converge to /(0). There are even worse examples involving continuous functions. If E is any set of measure zero in [-11',11'] then there exists a continuous func tion whose Fourier series diverges (unboundedly) on E (see Buzdalin, V V ; "Unboundedly Diverging Trigonometric Fourier Series of Continuous Functions," Math. Notes 7 (1970), pgs S-I2.) .
.
For many years mathematicians asked whether the Fourier series of an arbitrary continuous function had to converge at any point. Finally, in
PO I NTWISE CONVE R G E NC E OF F O U R I E R SER I ES
1 39
1966, Lenart Carleson showed that the Fourier series of every £ 2 function (hence of every continuous function) converges pointwise a.e. [see Carle son, Lenart; "On Convergence and Growth of Partial Sums of Fourier Series," Acta Mathematica 1 1 6 (1 966), pgs 135-1 57) . It should be noted that there exist summable (necessarily non-£2 ) functions whose Fourier series diverge everywhere.
Thus, theoretically, things are not too bad. However, to apply Fourier series to physical problems, one often needs to know whether a particular function has a convergent Fourier series at a particular point. We now present some sufficient conditions under which the Fourier series of f converges to f(x) at x. At first the hypotheses may seem ex ceedingly strong, but one must remember that many applied problems involve functions which have nice properties. The obvious property to try (since continuity failed) is differentiability.
/ E l2 ( , 11') and let / be differentiable at x0 E [-n',11'] . Then the Fourier series for f converges to f(x0) at x0•
40.2 Theorem : Le t Proof:
--,r
From Lemma 37.5 , the n th Fourier partial sum is "
Sn(X o) = -11'1 10 (f(x o + t) + f(x o - t)]Dn(t)dt.
(Recall that we assume f is extended periodically when necessary as in section 37 ) Also, for the particular case f(x) = 1 , we obtain .
I = � f" Dn(t)dt
,.. J o
for n
=
1 ,2,3, •
• • .
Thus for our general f, differentiable at x0 ,
Now
Sn(Xo ) -{(x o ) = ,..! Jro (/(X o + t) + {(X o - t) - 2/(Xo ))Dn(t)dt.
h (t) = f(x o + t) + f(x o -1 t) -2/(x o ) 2 sin 2 t can be defined so as to be continuous at t 0. (Compute . f(x0 + t) -{(xo ) + f(x o -t) -{(x o) t � t t+O 2 Sln 21 t . =
· -----
.
140
L E B ESGUE I NTE G R AT I O N A N D F O U R I E R S E R I ES
Note that this is where the differentiability of fat x0 is used. See Exercise 42.4.) Therefore, h E .C 2 [0,1r) (see Exercise 42.5). Thus,
== ![ = �J: h (t)sin (n + t) tdt.
-'n(x o ) -{(x o ) 11' 0 [f(x0 + t) + f(x0 -r) + 2f(x0)]Dn(t)dt
Thus by the Riemann-Lebesgue Lemma(Corollary36.4),an(x0) -f(x0) -+ 0 as n -+ Note that the Riemann-Lebesgue Lemma holds in this case by Exercise 42.6. 0 •.
We now present a condition on f(x) which will guarantee uniform convergence of the Fourier series for F. We first remind the reader that a function is said to be C2 on an interval [a,b] if its second derivative is continuous throughout [a,b]
.
Let f bt: C2 on [ -u 11'] • Then the Fourier series off converges uniformly to f on [ -u, 1f] .
40.3 Theorem :
,
Proof : The previous theorem guarantees pointwise convergence of the Fourier series to f(x) at each x. To show uniform convergence recall
an
=
-I f"' f(x)cosnxdx
11'
_,..
and
-I f"' f(x)sinnxdx
bn = 1f
_,..
which in this case can be taken to be Riemann integrals. Integrate by parts twice to show that there exists a positive number M such that lan I and Ibn I are both < M/n 2 • By the Weierstrass M-test, the Fourier series converges uniformly. (Where is the C2 hypothesis used?). 0
In the next section we will present a more general sufficient con dition for convergence of Fourier series. However, we again note that many useful functions are differentiable or C2 so that the rather easy results above provide all the necessary knowledge about convergence of Fourier series in these cases. 41 . More Convergence Theorems
In this section we shall examine the local (i.e. near a point) behavior of Fourier series. In order to proceed we need to know that the Riemann Lebesgue Lemma (Corollary 36.4) holds for f E .C [ -u 11'] • ,
PO I NTW I S E CO N V E R G E NCE OF F O U R I E R S E R I ES
14 1
{Riemann-Lebesgue): If /E .C(-11', 11'1 and {at};= o and { btl;= I are the Fourier coefficients of{, then tlim at = tlim bt = 0.
41.1 Lemma
+oo
+•
Proof: Given e > 0, there exists a simple function g on [ -11', 11'1 such that � 1/(t) - g(t) ldt < e. (This follows from the definition of the in tegral for fE ..C.)
Now g E ..C2 , so that if A t. Bt are the Fourier coefficients of g, then tlim A t = tlim Bt = 0 by the Riemann-Lebesgue Lemma for .C2 • But +00 +• for each k,
by the choice of g above. Thus
= tlimat -
0 and similarly
bt = tlim -
0. 0
We are now in a position to prove the Riemann Localization Theorem which says that the behavior of the Fourier series at x depends only on the values off arbitrarily close to x. This is somewhat surprising in view of the defmition of the Fourier coefficients of f as integrals over the entire in terval [ -11', 11'1 . If two functions in .C( """1,1' 11'1 agree in a neighborhood of x, then their Fourier series either both diverge at x or both converge to the same value at x, even though the functions may differ greatly outside the neighborhood. {Riemann Localization Theorem): Let f E .C[-11', 11'1 and x E [-11', 11'1 , and let 6 be any positive real number. Then
41.2 Theorem
lim
f'��'
n +oo Ja Proof:
[f(x + t) + f(x - t)1Dn(t)dt = 0.
Consider g(t)
==
�
0 on [0,6) f(x + t) + f(x - t) on [6,11'1 2sint/2
·
Clearly g E .C[-11',11'] since 2sint/2 > 2sin 6/2 > 0 on [6 ,11'1 . Thus
1 42
LE BESG U E I NT E G R AT I O N A N D FO U R I E R SE R I ES
lim
n +-
f'��'
J&
(f(x + t) + f(x - t)]D,(t)dt = 0
by the Riemann-Lebesgue Lemma and Exercise 42.6.
0
If f E .C( -'11', 'II'] is identically 0 on some open interval con taining x, then the Fourier series of f converges to 0 at x.
41 .3 Corollary:
Proof:
Suppose f = 0 on (x - 6 ,x + 6 ). Then the n th partial sum '11' s,(x) = 1 (f(x + t) + f(x - t)] D, (t)dt 'II' 0
1
-
=
1 'II' (f(x + t) + f(x - t))D, (t)dt. 'II' &
J
-
So by the Theorem, nlim s,. (x) 0. +-
0
=
and f g on an open interval containing x, then either both Fourier series converge to the same value at x or both diverge at x.
41 .4 Corollary: If f,g E ..C [-'11', '11' ]
=
Proof: Use f - g in Corollary 41 .3. Note that the Fourier series of f - g equals the Fourier series of f minus the Fourier series of g by the
definition of Fourier coefficients and linearity of the integral.
0
This local property of Fourier series is in marked contrast to power series results. Recall that if two power series have the same values over an interval, then they are equal on their entire interval of convergence. But Corollary 41 .4 says we can alter f drastically outside the interval (hence obtaining greatly different Fourier series) and yet retain the same values inside the interval. We will now obtain a generalization of a result in section 39 concern ing differentiability of f at x. Here we will require only that f have general ized left and right derivatives at x. We first need to define these carefully. Let f be a real valued function with right and left hand limits f(x + ) and f(x-) respectively at x. Then the generalized right and left hand derivatives offat x are respectively
41.5 Definition :
PO I NTWISE CO N VE R G E N CE O F FOUR I E R SER I ES
JjR '(x) = r
t +�
41 .6 Example:
Let
f(t) -{(x+ ) t -x
1 43
and
�
x 2 + 1 on [-11', 0) f(x) = 10 at 0 x - 2 on (0,11') . Then /(0+ ) = -2, /(0- ) = 1 , /R'(O) = 1 and fL'(O) = O (Verify!). Notice that f has no ordinary right and left hand derivatives at 0 since f is not continuous from either the right or left at 0. Let f E .C[-w, 1r) and fR'(x) and fL'(x) both exist. Then the Fourier series for {converges to
41 .7 Theorem :
Proof:
Exercise 42.9.
This is almost exactly like the proof of Theorem 40.2. See o
As an example of the above theorem, recall from Example 353 that the Fourier series for /(x) =
l O on [-w,O) 1 on [0,11']
has the value
Note that the above theorem truly generalizes Theorem 39.2 since if f is differentiable, it clearly has IR' and 1£' and thus its Fourier series con verges to
1 44
L E B ESG U E I NT E G R AT I O N A N D F O U R I E R SE R I ES
42.
Exercises
42. 1
In reference to Carleson's result (section 39), show that if the Fourier series of [ E .C 2 ( [ -11',11') ) converges pointwise a.e., then it converges point wise a.e. to f. (Hint : Show, as in the proof of Theorem 33 . 1 1 , that a sub sequence of the sequence of Fourier partial sums converges pointwise a.e. to f.)
42.2
Consider the Fourier series of Example 3 5.3 . (a) Show that 11'/4 = I - 1 /3 + 1 /5 - 1 /7 + . . . (let x = 11'/2) . (b) Let x = 11'/4 to obtain another series expression for 11'/4.
42.3
Using the Fourier series for x 2 on [ -11',11'] , find the sum of the series
See what other numerical series you can sum with this Fourier series. 42.4
Prove that h( t) =
f(x o + t) + f(x o - t) - 2/(xo)
I
,
2 sin 2 t
for f differentiable at tinuous at t = 0. 42.5
x0
E [ """11' ,11'] , can be defined so as to be con
If h(t) = f( t) /g(t) for t :1= t0 , t0 E [a,b J , g(t) :1= 0 except at t = t0 , f E .C 2 ( [a,b ) ) , g continuous, and lim f( t) /g{ t) = c E Q, then t + to h e .t2 ( [a, b ) ).
42.6
If h e .C 2 ( [ """11', 11') ), show that lim•fo'll' h( t) sin(n + t ) tdt = o. n+
42.7
Fill in the details of the proof of Theorem 40.3.
42.8
42.9
Given f E .C( [ """11' , 11') ) and e > 0, show that there is a simple function [ -11', 11'] such that f� 1/(t) -g(t)l dt < e. (Hint : apply the definition of the integral for r , t-." )
g on
Prove Theorem 4 1 .7.
42. 1 0 Look at the Fourier series of Example 35.3 and 3 5. 5 to verify Theorem
4 1 .7 at points of discontinuity.
42.1 1 (a) For x E [ 0, 11'] , find the Fourier series for the function X (o,x)·
(b) Obtain from (a) the expression, for x E [ 0,11'] ,
PO I NTWISE CONVE R G E N C E O F FO U R I E R SE R I ES
.!!_..=!
2
(c) Let x
=
1 45
..
.I sin nx n= l
n
.
= w/2 in (b) to obtain an interesting series.
42. 1 2 (a) Show that if a trigonometric series
� (a,. cosnx + b,. sin nx) -+ k 2 n= l ao
converges absolutely at c, then it converges absolutely at -c. (b) Show that if a trigonometric series converges absolutely at c 1 and c2 , then it converges absolutely at c 1 + c2 • (c) Show that if a trigonometric series converges absolutely on any open interval, then it converges absolutely everywhere.
Appendix
Logic and Sets
The connectives and quantifiers of logic, as commonly used in mathematics, are closely related to operations on sets. For example, the definition B = {x l x and x expresses the relationship be tween intersection and the logical connective "and." Of course B \B {x lx and x f:. This last set is lx or x B}, and A {x called the relative complement of B in
=
An EA E
EA
E B} AU = EA B}. A. Given a sequence of sets {A,.};=O one can form the union or the '
intersection of the sequence by making use of the logical connectives a ( "there exists'') and V ( "for all'') :
,90A, = {xl (an)(x EA,)} ,Q0A, = {xl (Vn)(x EA,)}. The reader can verify such identities as the following:
11\ (,QoA ,) = ,Qo (B\.4 , ) B\ (nn A ,) = nU (B\A ,) =O =O BU (nn=OA ,) = nn=O(B U A ,). 1 47
1 48
LE B ESGUE I NT E G R AT I O N A N D FO U R I E R SE R I ES
O pen and Closed Sets
A set A of real numbers is open (in IR) if for each x E A , there is an open interval (a,b) such that x E (a,b) a11d (a,b) A . A set B is closed if R\B is open.
C
Sometimes it is desirable to focus attention on a subset X of IR and to speak of subsets A c X as being open or closed relative to X. A set A C X is open in X (or "in the relative topology of X'') if there is a set U, open in IR, such that A = U n X. (Example : is open in the relative = ( -1 ,t) n [0,1 ) .) The reader can verify topology of [0,1 ] since that a set A C X is open in X if and only if for every x E A there is an open interval (a,b) such that x E (a,b) and (a,b) n X c A .
[o.t)
[O.t )
Given x E X, a neighborhood ofx in X is a set A containing x.
C X, open in X and
Bounded Sets of Reel N umben
u
A real number is an upper bound of a set of real numbers X if (Vx E X) > x] . A number is a lower bound if (Vx E X) < x] .
[u
2
[2
It is useful to allow the two symbols • and -• to be used as upper and lower bounds. Of course • is considered to be larger than any real number and - is considered to be smaller than any real number. Thus • is an upper bound for any set of real numbers and - is lower bound for any set of real numbers. Since the null set 0 has no elements, it has - as an upper bound and • as a lower bound. It is a fundamental property of the real numbers (The Completeness Axiom) that every set of real numbers has a least upper bound (an upper bound that is less than every other upper bound) and a greatest lower bound (a lower bound that is greater than every other lower bound). It is easy to see that no set can have more than one least upper bound nor more than one greatest lower bound. The least upper bound of a set X is denoted lub(X) or sup(X), the latter standing for "supremum." The greatest lower bound of X is denoted glb(X) or inf(X), the latter standing for "infimum." Note that for any non-empty set X of real numbers, glb(X) < lub(X) but that lub(0) = - and glb(@ = •.
APPE N D I X
1 49
Countable and Uncountable Sets Infinite sets are often encountered in mathematics, and it is some times useful to distinguish among "different sizes" of infinity. The smallest infinite sets are said to be countable or countably infinite. They can be put into a one-to-one correspondence with the natural numbers N { 0, 1 ,2, • • • }. Any infinite set which is not countable is said to be uncountable. For example, the integers z = { • • • -2, -1 ,0, 1 ,2, • • • } are countably infinite. A one-to-one correspondence with N which proves this can be given as follows : =
Z:
0
1
-1
2
-2
3
-3
4
*
*
t
t
t
t
*
*
N: o
2
3
4
5
6
7
This illustration shows the reason for the word "countable" in re ferring to such sets. The one-to-one correspondence amounts to writing out all the elements of the set in some systematic and non-redundant list, one after the other. They can then be "counted ." A reader who is not familiar with these ideas might suppose that every infinite set is countable. However, the set of all real numbers is un countable. Indeed any interval of real numbers is uncountable. We will illustrate with the open interval (0, 1 }. The method of proof is known as Cantor's Diagonal Argument. Suppose that (0, 1) were countable. Then its elements could all be listed, one after another. Let us write such a list as r . , r2 , r3, • • • Then each r1 can be represented by an infinite decimal of the form r1 = O.ana12a13 • • • , where each aq is an integer and 0 < aq < 9. (Some elements of (0, 1) have two decimal representations : for example, .2 = . 1 9999 • • • . We can avoid some difficulty by agreeing always to choose one of these representations-say the one with the infmite string of 9's.) •
Now we create an infinite decimal r = .d 1 d2 d3 • • • as follows : d1 is chosen to be different from a u (for example, d 1 = 5 if a u + 5 and d 1 = 6 if a 1 1 = 5). This guarantees that r + r 1 since the two numbers differ at the first decimal place. Now choose dl in the same fashion so that dl + all · Then necessarily r + rl In general, d1 is chosen so that d, ::/= ail , which guarantees r ::/= r1• We have created a number r E (0, 1) which is not equal to any number on the list! So our assumption that the list was complete is false, and (0, 1) must be uncountable.
O
.
1 50
APPE N D I X
The fact that between every two real numbers there is a rational number might lead one to suspect that the set of rational numbers Q is uncountable as well . However, Q is indeed countable, as we shall show. To do so, we will make use of the following results : if A 0 ,A 2 ,A 2 , • • • is a countable collection of countable sets, then their union U
t=o
A1 is count-
able. The proof of this result will be discussed in the next paragraph. To prove that Q is countable , let A 0 = [0, 1) n Q, A 1 = [ 1 , 2) n Q, A 2 = [-1 , 0) n Q, A 3 = [2,3} n Q, A 4 = [-2, -1) n Q, etc. Each A 1 is countable, as the following listing shows:
{
1
1
2'
3'
2 3'
4'
3 4'
4 5'
6'
1
[n n + 1 ) n Q = n n + - n + - n + - n + - n + -
'
'
}
n+- n+- n+- n+- n+- • • • .
2 5'
1
5'
3 5'
1
The list is complete, since every possible denominator appears with every meaningful numerator : Therefore U
i= O
A 1 = Q must be countable.
To demonstrate that a countable union of countable sets i s countable, we list the elements of each A 1: a10, a11 , a12 , • • • We can then list the elements of
a04, • • •
P A 1 as
r= O
•
follows : a00 , a0 t . a 1 0, a02,a u , a2o• a03 , a 1 2 • a 2 1 > a 3o •
(The rule being followed is to list the elements whose sub scripts add up to 0, then those whose subscripts add up to 1 , etc.) This seems straightforward once the basic idea is understood. Actually there are a couple of difficulties with the proof. First, there may be duplications in the final list if the A1 are not disjoint. This difficulty is easily overcome by simply eliminating duplications as they arise. •
A more fundamental problem arises from the fact that there are infmitely many ways to list each A1• In specifying a particular listing of the elements in A 1 we are in fact making one arbitrary choice (of a particular listing) for each i = 1 ,2,3, • • • . This process of making in fmitely many arbitrary choices at the same time cannot be justified on the basis of ordinary mathematical principles. An additional principle, called the Axiom of Choice, 1 must be invoked. Although the reader may fmd this principle trivial, that feeling is probably based on his experience with finite situations, where the Axiom of Choice is not needed. Use of 1
The Axiom of Choice states that one can form a set consisting of exactly one
element from each set in a given infinite collection of nonempty sets. For a more
complete treatment of the Axiom of Choice
K.S. Miller (Krieger,
1975).
see
Elementa of Modern A lgebra by
APPE N D ! X
151
the Axiom of Choice often leads to results which are obvious and com forting, but sometimes rather paradoxical and startling conclusions can arise. These results, coupled with a fundamental philosophical uneasiness with any principle which allows one to accomplish infinitely many steps all at once, have caused some mathematicians to reject the Axiom of Choice and the conclusions which follow from it. Even mathematicians who are happy with the Axiom of Choice are sufficiently sensitive to note specifically where it is being used in their proofs. It should be noted that the Axiom of Choice is not needed to prove that Q is countable. (However, the Axiom of Choice is needed in 7 .7 to show the existence of a non-measurable set.) The listing of [n, n + 1) n Q which we gave has the same form for every n and is specified in advance. Therefore no choices need be made in listing i=UOA 1• The reader should verify that the first few terms of the resulting listing, following the rule . 2 4 · - 21 • 2 • • • · g�ven eauer, are as fio11ows.· 0 '21 , 1 ' 31 '23 ·- 1 •3•3 1· • Real Functions
Let X and Y be two sets of real numbers and let f and g be functions which map X and Y respectively into �. (It is accepted practice to call X the domain off and Y the domain of g). One can define new functions as follows: t + g : x n y-+ � by (f + g)(x) = /(x) + g(x) fg: X n Y -+ � by (/g)(x) = /(x-.(x) ffg : X n {x E Y lg (x) :1= 0} -+ � by (//g)(x) = f (x)fg(x) f • g : { x E Y lg (x) E X} -+ � by (/ • g)(x) = f(g (x))
This last is called ''/composed with g." If f: X -+ �. and A C R, then {x E X l/(x) E A } is called the "pre-image of A under r' and is denoted / 1 (/t ) . This notation does not implicitly assume that /has an inverse (is one-to-one); the notation applies to any function. The reader should verify the following results: ,- 1 ( ? A t) = ? /1 (/t ,)
/ 1 (6M) = �\ f- 1 (A ).
1 52
APPE N D I X
If f: X -+ IR, and B C X, then f(B) = {f(x) lx E B} . Note that r 1 (f(B)) :J B, with equality holding if f is one-to-one. Similarly , ff.r 1 (A C A , with equality holding provided f is onto.
)
A function f: X -+ IR is bounded if f(X) c [m,M] for some m,M E IR. A function f: X -+ 1R is increasing on [a,b] C X if
(Yx,y E [a,b] )(x < y � f(x)
(Vx,y E [a,b] )(x f(y)). Also, f is monotone on (a,b] if f is either increasing on [a,b] or decreasing on [a,b] . A function is piecewise monotone on [a,b] if there are finitely many points a = x0 < x 1 < • • • < X n = b such that the function is monotone on [x1_ � ox1] for each i = 1 ,2, • • • , n. A function f: X -+ R is continuous at x E X if
(Ye > 0)(3:6 > O)(Vy E X)( lx -y I < 6 � 1/(x) -f(y) I < e). This is equivalent to saying that for every neighborhood V of f(x) in IR, there exists a neighborhood U of x in X such that f(U) is contained in V. If f is continuous at every point x E A C X, then f is said to be continuous on A. This is equivalent to saying 1 (V V open in R)(r (V) is open in A). A function f: [a,b] -+ 1R is piecewise continuous on [a,b] if there are finitely many points a = x0 < x 1 < • • • < Xn = b such that f is con tinuous on [x1_ 1 ,x1] for each i = 1 ,2, • • • , n . I t is a major result in the theory of real functions that if f is con tinuous on [a,b] , then f is uniformly continuous on [a,b] . This means that given e > 0, there is a 6 > 0 which works for every x E [a,b] ; in other words,
(Ve > 0)(96 > O)(Vx,y E [a,b] )( lx -y I < 6 � lf(x) -f(y) l < e).
APPE N D I X
1 53
Real Sequences A sequence {an } of real numbers has real limit L ( lim an n +oo
=
L) if
(Ve > O)(tiN) (Vn > N) ( I an -L l < e). Also,
an = • if (VM)(iJ.N)(Vn > N)(an > M), nlim +oo and
= - if (VM)(tiN)(Vn > N) (an < -M). nlim +oo an Not every sequence has a limit of course. A sequence {an } is increasing if (Vn)(Vm)(n < m � an < am ) , and {an} is decreasing if n < m implies an > am . A sequence is monotone if it is either increasing or decreasing. Every monotone sequence can be shown to have a limit which is either a real number or ± • . This limit is the lub of the set {an l n = 1 ,2, • • · } if the sequence is increasing, or the glb of the set {an l n = 1 ,2, • · · } if the sequence is cJecreasing. If {an } is any sequence of real numbers, define
for n = 1 ,2, • • • . Then it is easy to see that {b n } is a decreasing sequence , { en} is an increasing sequence, and bn > an > Cn for all n . Therefore the monotone sequences { bn }, { en} both have limits. The limit of { bn} is called nnm an or lim sup an , and the limit of { en } is called lim an or lirn n inf an .
+•
n + oo
The reader can verify that
n +oo
+•
(1) lim an < nllm an , +oo n +oo an (and, in that (2) {an } converges if and only if lim an = nurn �• n +oo case, nlim an = lim an = nHiii an), +• +• n +oo (3) if an < dn for all n, then lim an < lim dn and nnm dn , an < nnm +• +• 11 +• II + • lim = (4) n (-rzn) - nIlm an . +oo +oo Sequences of Functions A sequence { /n } of real-valued functions, all with the same domain D, converges pointwise to f: D -+ cR if nlim fn(x) = f(x) for all x E D; i.e . +•
1 54
APPE N D I X
(Yx E D)(Yc: > O)(ti N)(Yn > N)( lf(x) - fn(x) J < c:). The sequence {fn } converges uniformly to f: D -+ tR if the N can be chosen so that it will work for all x E D at the same time. That is, (Yc: > O)(tiN)(Yn > N) (Vx E D) ( If(x) - fn (x) I < c:). The sequence {fn } is said to be pointwise bounded if (Vx E D)(tiM)(Yn)( lfn(x) I < M) . The sequence is uniformly bounded if M can be chosen to work for all x E D at once : (tiM)(Yn) (Yx E D)( Ifn (x) I < M).
Biblio g raphy
Apostol, T ., Mathematical Analysis, Reading, Mass., Addison-Wesley, 1 957. Asplund, E., L. Bungart, A First Course in Integration, Holt, Rinehart, and Winston, New York, 1 966. Bartle, R., The Elements of Integration, John Wiley and Sons, New York, 1 966. Burkill , J., The Lebesgue Integral, Cambridge Tracts No. 40, Cambridge University Press, New York, 1961 . Goldberg, R., Methods ofReal Analysis, Waltham, Mass., Blaisdell, 1 964. Halmos, P., Measure Thoery, Princeton, D. Van Nostrand, 1950. Hewitt, E., K. Stromberg, Real and Abstract Analysis, New York, Springer Verlag, 1 965 . Lebesgue , H., Measure and the Integral, San Francisco, Holden-Day, 1 966. Munroe, M., Introduction to Measure and Integration, Reading, Mass., AddisonWesley, 1953. Royden , H., Real Analysis (2nd Edition), New York, Macmillan, 1968. Rudin, W., Principles of Mathematical Analysis, New York, McGraw-Hill, 1953 . Scanlon, J ., Advanced Calculus, Boston, Heath, 1967. Sprecher, D., Elements of Real Analysis, New York, Academic Press, 1 970. Temple, G., The Structure of Lebesgue Integration Theory, Oxford, Clarendon Press, 1 97 1 . Williamson, J., Lebesgue Integration, New York, Holt, Rinehart and Winston ,
1962.
1 55
INDEX
a-ary expressions 42 absolute convergence 1 1 1 , 145 additivity of Riemann integral 6 of Lebesgue integral 70 a.e. 50, 57 almost everywhere 50, 57, 80 axiom of choice 1 8-21 , 1 50 B(A) 94-96
Banach space 97 Bessel's Inequality 1 1 8 binary expressions 42 Borel 1 3 Borel sets 33, 57 bounded functions 1 52 bounded sets 38 Buzdalin, V. V. 138
C(A) 94-99, 1 09 (C , 1) 1 2 1 c2 140 Cantor 1 3 , 36 Cantor set 33-35 , 43 , 56 Caratheodory 36
arithmetic 35 Carleson, Lenart 1 39, 144 Cauchy 2 Cauchy inequality 97, 103 , 1 OS, 1 10 Cauchy sequence 96 Cesaro sums 1 21 characteristic functions 7 , 1 17 closed sets 3 1 , 148 compact 24, 45 complements 22, 26 completeness axiom 148 completeness of normed linear space 96 component of x in G, 1 8 composition of functions continuous functions 6, SO, 1 52 continuous a.e. 84 convergence absolute 1 1 1 in the mean 1 04 in the norm I I 11 2 1 04 t} 104, 1 20 of a sequence of functions 96, 1 04, 1 53 of a sequence of vectors 96
cardinal
1 67
1 58
I ndex
convergence continued uniform 96, 104, 1 54 countable 9, 149 countably additive 20, 29, 3 1 , 40, 67,
73
countably additive class of sets 33 decreasing functions 1 52 decreasing sequences 1 53 dense 1 1 1 derivatives, generalized right and left hand 142 Dirichlet kernel 1 23 distance 94 Dominated Convergence Theorem 78,
81 , 86, 1 08
dot product 97
E 18 Egoroff's Theorem 85-87 equal a.e. 1 02. 1 1 0 equivalence 1 02, 1 1 0 even function 127
f+ 70 f- 70
Fatou's Lemma 8 1 Fejer kernel 1 24 Fourier coefficients 1 1 5-1 1 8 Fourier partial sums 1 23 Fourier series 1 1 3-1 1 8 function spaces 93 Fundamental Theorem of Calculus 1 ,
inne r measure 22 inne r product 97 integrals, upper and lower 10 Jordan content 1 7 , 27
£(A) 78, 94, 1 03 , 1 1 1 .1! 1 109, 1 1 2 .1!2 101 , 1 1 1 .t2 (A) 102, 1 1 1 J!P(A) 1 1 2 least upper bound 148 Lebesgue 13 Lebesgue integral 59 Lebesgue measure 38, 40 Lebesgue sum 14, 60 l.eibnitz 1 length of vector 94 limit 3 lim 53 , 1 53 n + oo
lim 53 , 1 53
n+•
linearity of Lebesgue integral 65, 71 o f Riemann integral 6 linear space 93 Lusin's Theorem 85-87
m 22, 33 m . 22 m* 19, 33
harmonics 135 Hilbert space 99, lOS Holder's Inequality 1 1 2
measurable function 47-49 measurable set 13, 2 1 , 25 , 29 measure 1 5 , 22 for bounded sets 38 for unbounded sets 40 measure zero 25, 35, 64, 73 metric 95 Minkowsky inequality 104, 1 1 2 Monotone Convergence Theorem 72,
improper integral 88 increasing functions 1 52 increasing sequences 1 53
monotone function 6, 1 52 monotonicity of the Lebesgue intetral 65, 71
7
greatest lower bound 148
78
I ndex
monotonicity continued of m., m* 22 of the Riemann integral 6 neighborhood 148 Newton 1 non-measurable set 3 1 norm 2, 95, 109 norm 11 2 1 03 normed linear space 95 nowhere dense 43
II
odd function 1 27 open sets 1 8, 3 1 , 148 orthogonal set of vectors 99, 1 1 0 orthonormal set of vectors 99, 1 10,
1 14
outer measure 1 7 , 19 overtones 135
P(A) 94 Pn(A) 94-98
Parallelogram Law 1 09 Parseval's formula 1 27 partition 2 perfect set 43 periodic extension 1 22, 1 28 piecewise continuous functions 7, 1 52 piecewise monotone functions 7, 1 52 pointwise bounded 1 54 pointwise convergence 54, 104, 1 53 preimage of A under f 1 5 1 projection of a vector onto a subspace
1 59
Riemann integrability 82-84 Riemann integral 1 -3 Riemann -Lebesgue Lemma 1 1 8, 141 Riemann Localization Theorem 141 Riemann sums 2, 3 Riesz-Fischer Theorem 105, 120 set function 1 5 se quence Cauchy 96 of arithmetic means 121 o f functions 96, 104 of numbers 1 53 of vectors 96 a-algebra 33 simple function 53 , 60-62 step function 4 subadditivity 23 subsequence 1 04 summ able 70 symmetric difference 44 trigonometric polynomials 1 1 3 trigonometric series .1 13 unbounded functions 68 unbounded sets 40 uncountable 149 uniform convergence 55, 96, 104, 1 54 uniformly bounded 1 54 uniformly continuous 1 52 unit sphere in .C2 ( [0, 1 ] ) I l l
1 00 Q, the rational numbers 8, 1 50
vector space 93
tR, the real numbers 1 8
�A) 94, 97, 109
tR" 94-101 relative complement 147 representation of simple function 54 Riemann 2
Weierstrass Approximation Theorem
1 29
Weierstrass M test 1 3 1 , 137, 140 Z, the integers
149