PURE AND APPLIED MATHEMATICS A Wiley-Interscience Series of Texts, Monographs, and Tracts Founded by RICHARD COURANT Editor Emeritus: PETER HILTON Editors: MYRON B. ALLEN III, DAVID A. COX, HARRY HOCHSTADT, PETER LAX, JOHN TOLAND
A complete list of the titles in this series appears at the end of this volume
Lebesgue Meas ure and Integration An Introduction Frank Burk
A WILEY-INTERSCIENCE PUBLICATION
John Wiley & Sons, Inc. NEW YORK I CHICHESTER I WEINHEIM I BRISBANE I SINGAPORE I TORONTO
This text is printed on acid-free paper. @) Copyright© 1998 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4744. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, E-Mail:
[email protected]. Library of Congress Cataloging in Publication Data: Burk, Frank. Lebesgue measure and integration: an introduction 1Frank Burk. p. em. -- (Wiley Interscience series of pure and applied mathematics) "A Wiley-lnterscience publication." Included bibliographical references (p. ) and index. ISBN 0-471-17978-7 (cloth: acid-free paper) I. Measure theory. 2. Lebesgue integral. I. Title. II. Series: Pure and applied mathematics (John Wiley & Sons : Unnumbered) QA312.B84 1998 515'.42 -- dc21 97-6510 CIP
Printed in the United States of America 10 9 8 7 6 5 4 3 2 I
For Janet
Contents
xi
Preface Chapter 1. Historical Highlights
1
Rearrangements I 2 Eudoxus (408-355 B.C.E.) and the Method of Exhaustion I 3 l.3 The Lune of Hippocrates (430 B.C.E.) I 5 l.4 Archimedes (287-212 B.C.E.) I 7 1.5 Pierre Fermat (1601-1665): xr/qdx = bp/q+lj(pjq + 1) I 10 1.6 Gottfried Leihnitz (1646-1716), Issac Newton (1642-1723) 1 12 l.7 Augustin-Louis Cauchy (1789-1857) I 15 1.8 Bernhard Riemann (1826-1866) I 17 1.9 Emile Borel (1871-1956), Camille Jordan (1838-1922), Giuseppe Peano (1858-1932) I 20 1.10 Henri Lebesgue (1875-1941), William Young (1863-1942) 1 22 1.11 Historical Summary j 25 1.12 Why Lebesgue I 26 l.1 1.2
Jt
vii
viii
CONTENTS
32
Chapter 2. Preliminaries Sets I 32 2.1 Sequences of Sets I 34 2.2 Functions I 35 2.3 Real Numbers I 42 2.4 Extended Real Numbers I 49 2.5 Sequences of Real Numbers I 51 2.6 Topological Concepts of R I 62 2.7 Continuous Functions I 66 2.8 Differentiable Functions I 73 2.9 2.10 Sequences of Functions I 7 5
87
Chapter 3. Lebesgue Measure 3.1 3.2 3.3 3.4 3.5 3.6
Length of Intervals I 90 Lebesgue Outer Measure I 93 Lebesgue Measurable Sets I 100 Borel Sets I 112 "Measuring" I 115 Structure of Lebesgue Measurable Sets
I 120
Chapter 4. Lebesgue Measurable Functions 4.1 4.2 4.3 4.4
Measurable Functions I 126 Sequences of Measurable Functions I 135 Approximating Measurable Functions I 137 Almost Uniform Convergence I 141
Chapter 5. Lebesgue Integration 5.1 5.2 5.3 5.4 5.5
126
147
The Riemann Integral I 147 The Lebesgue Integral for Bounded Functions on Sets of Finite Measure I 173 The Lebesgue Integral for Nonnegative Measurable Functions I 194 The Lebesgue Integral and Lebesgue Integrability I 224 Convergence Theorems I 237
CONTENTS
ix
Appendix A. Cantor's Set
252
Appendix B. A Lebesgue Nonmeasurable Set
266
Appendix C. Lebesgue, Not Borel
273
Appendix D. A Space-Filling Curve
276
Appendix E. An Everywhere Continuous, Nowhere Differentiable, Function
279
References
285
Index
288
Preface
This book is intended for individuals seeking an understanding of Lebesgue measure and integration. As a consequence, it is not an encyclopedic reference, or a compendium, of the latest developments in this area of mathematics. Only the most fundamental concepts are presented: Lebesgue measure for R, Lebesgue integration for extended real-valued functions on R. No apologies are made for this approach, after all, it is the proper foundation for any general treatment of measure and integration. In fact, no claim to originality is made for any of the mathematics in this book, but we do accept full responsibility for any mistakes or blunders in its presentation. It is old mathematics after all (standard graduate fare for the last forty or fifty years), but particularly beautiful. It deserves a wider audience. Lebesgue measure and integration, presented properly, reveals mathematical creation in its highest form. Motivation has been the dominant concern, and understanding will be the final measure. Where to begin? As a concession to understanding the subtleties of measure, and the effort required for such, I have taken the least upper bound axiom as a starting point. (Besides, it would be difficult, if not impossible, to improve on Landau's (1960) book, Foundations of Analysis.) The formal prerequisites are a basic calculus course and a course emphasizing what constitutes a proof, standard methods of proof, and the like. In reality, a curiosity for things mathematical and the "need to understand such," is both necessary and sufficient. The arrangement of topics is standard. The historical struggle to give a xi
xii
Preface
rigorous definition of "area" and "area under a curve," resulting in Lebesgue measure and integration, is the subject of Chapter 1. (Tribute is paid to our mathematical ancestors by understanding and studying their results.) Mastery of this material is not necessary for subsequent chapters. After all, it is an "overview," written with the benefit of hindsight. The reader may return from time to time as she understands "measurable", "Borel", "Lebesgue Dominated Convergence," and so on. Mathematical concepts (undergraduate analysis) that are useful for the understanding of measure, measurable functions, and integration, are developed in Chapter 2. Chapter 3, measure theory, is the essence of this book. Here an elementary, but rigorous, treatment of Lebesgue measure, as a natural extension of "length of an interval" and as a subject of interest in and of itself, is presented. Set measurability is via Caratheodory's Condition. Measurable functions, motivated by the necessity of "measuring" inverse images of intervals as discussed by Lebesgue [Ma], are defined and developed in Chapter 4. The last chapter, Chapter 5, begins with the Riemann integral, developed from step functions. Replacing "step" with "simple" results in the Lebesgue integral for bounded functions on sets of finite measure. Some incisive observations and we have the celebrated convergence theorems that permit the interchange of "limit" and "integral", and justifies "Lebesgue" for those with such a need. (By the way, if at any time you are confused or lack a sense of direction, I apologize; for a solution, reread the master [Ma].) Finally, appendices A-E present other topics of beauty and inspiration to mathematicians, testament to the wonderful creativity of the human mind. This book may be used in many ways: especially as a text for an undergraduate analysis course, first-year graduate students in statistics or probability, and other applied areas; a self-study guide to elementary analysis or as a refresher for comprehensive examinations; a supplement to the traditional real analysis course taken by beginning graduate students in mathematics. I want to thank my good friend and colleague, Gene Meyer, for his countless hours of discussions and suggestions as to topics, and what would or would not be appropriate for a book of this nature. Accolades to Debora Naber. She had the arduous task of translating my handwriting into the final manuscript. I thank my parents, Glen and Helen Burk, whose constant encouragement has been a source of strength throughout my life. Finally, I thank my wonderful wife Janet, who somehow finds the time to encourage my dreams while rearing our five beautiful children(Eric, Angela, Michael, Brandon, and Bryan.).
Even now there is a very wavering grasp of the true position of mathematics as an element in the history of thought. I will not go so far as to say that to construct a history of thought without profound study of the mathematical ideas of successive epochs is like omitting Hamlet from the play which is named after him. That would be claiming too much. But it is certainly analogous to cutting out the part of Ophelia. This simile is singularly exact. For Ophelia is quite essential to the play, she is very charming-and a little mad. Let us grant that the pursuit of mathematics is a divine madness of the human spirit, a refuge from the goading urgency of contingent happenings. -Alfred North Whitehead Mathematics, rightly viewed, possesses not only truth, but supreme beauty--a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show. The true spirit of delight, the exaltation, the sense of being more than man, which is the touchstone of the highest excellence, is to be found in mathematics as surely as in poetry. What is best in mathematics deserves not merely to be learned as a task, but to be assimilated as a part of daily thought, and brought again and again before the mind with ever-renewed encouragement. Real life is, to most men, a long second-best, a perpetual compromise between the real and the possible; but the world of pure reason knows no compromise, no practical limitations, no barrier to the creative activity embodying in splendid edifices the passionate aspiration after the perfect from which all great work springs. Remote from human passions, remote even from the pitiful facts of nature, the generations have gradually created an ordered cosmos, where pure thought can dwell as in its natural home, and where one, at least, of our nobler impulses can escape from the dreary exile of the natural world. -Bertrand Russell
Lebesgue Measure and Integration
Historical Highlights
Some of the major discoveries in quadratures that culminated with the Lebesgue-Young integral are presented in this chapter. Our purpose is twofold: 1. We want to acknowledge our appreciation and gratitude to the thinkers of the past. It is hoped that the student will be motivated to continue these threads that distinguish civilization from barbarism. Neglect of mathematics works injury to all knowledge, since he who is ignorant of it cannot know the other sciences or the things of this world. And what is worse, men who are thus ignorant are unable to perceive their own ignorance and so do not seek a remedy. ~Roger Bacon
2. The student will see the process of mathematical creation and generalization as it applies to the development of the Lebesgue integral. Reason with a capital R =Sweet Reason, the newest and rarest thing in human life, the most delicate child of human history. ~Edward Abbey
If this material is too difficult on the first reading, relax. It will make sense after Chapter 5.
1
2
1.1
HISTORICAL HIGHLIGHTS
REARRANGEMENTS
The figures below demonstrate the general idea of "rearranging "; in the first example, a circle rearranged into a parallelogram. This method has been known for hundreds of years.
2nr ;1
// r ;/
2nr 2nr /Y-----"-.~/--
----,/--- -------,~------- "/,--------~-/-----~v--- -----v ,..-------~
I'
r
2rcr """(-- - - - - - - - - - - - -
~
--------~--3>-
M~--------- ~----- ----J, ~ (a)
(b)
(2nr)r
=nr 2
1.2
EUDOXUS (408-355 B.C.E.) AND THE METHOD OF EXHAUSTION
I~ (c)
(d)
b
(e) "Scaling" (c)
1.2 EUDOXUS (408-355 B.C.E.) AND THE METHOD OF EXHAUSTION "Willingly would I burn to death like Phaeton, were this the price for reaching the sun and learning its shape, its size, and its substance." -Eudoxus
3
4
HISTORICAL HIGHLIGHTS
Eudoxus was responsible for the notion of approximating curved regions with polygonal regions: "truth" for polygonal regions implies "truth" for curved regions. This notion will be used to show that the areas of circles are to each other as the squares of their diameters, an obvious result for regular polygons. "Truth" was to be based on Eudoxus' Axiom: Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continuously, there will be left some magnitude which will be less than the lesser magnitude set out.
In modern terminology, let M and f > 0 be given with 0 < E < M. Then form: M, M- rM = (1 - r)M, (1- r)M- r(1 - r)M = (1 - r) 2 M, ... , where 1/2 < r < 1. The axiom tells us that for n sufficiently large, say N, (1 M < t:, a consequence of the set of natural numbers not being bounded above. Back to what we are trying to show: Let c, C be circles with areas a, A and diameters d,D, respectively. We want to show ajA = d 2 jD 2 , given that the result is true for polygons and given the Axiom of Eudoxus. Assume aj A > d 2 j D 2 . Then we have a* < a so that 0 < a - a* and a* /A= d 2 jD 2 . Lett:< a- a*. Inscriberegularpolygonsofareaspn, Pn in circles c, C and consider the areas a- Pn, A - Pn:
rt
a- Pn
Now, double the number of sides. What is the relationship between a- Pn and a- p 2n?
a- Pn
a- P2n
1.3
THE LUNE OF HIPPOCRATES (430 B.C.E.)
5
Certainly a- p 2n < l/2(a- Pn)· We are subtracting more than half at each stage of doubling the number of sides. From the Axiom of Eudoxus, we may determine N so that 0 = 2 sin2 1>
2
and
that is,
14
HISTORIC AL HIGHLIGHTS
2
2 x = 2z /(1 + z
).
Leibni tz knew xz X
= J zdx + J xdz:
1---- ------ ---- ----fzdx fxdz
z
Hence
t1
= }o 2 zdx
l
il
2i 2 dz - -1 [ 1+z 1 o -2 =
4 2 2 }ot z (1-z +z 2
1
1 1 1 1 =2-3 +5-7 +···
···)dz
(long division)
(integr ating termwise),
and by adding I /2 to both sides,
WOW!! It would be difficult to name a man more remark able for the greatne ss and universality of his intellectual powers than Leibnitz. -John Stuart Mill
L7
AUGUSTIN-LOUIS CAUCHY (1789-1857)
15
Note:
1
1
1
1
S7r = 1 X 3 + 5 X 7 + 9 X
}}
+ ...
and l
gln 4 = 2
1 X
I
4+ 6
X
I
8 + 10
X
12 + ...
Joy in looking and comprehending is nature's most beautiful gift. -Albert Einstein
1r 1 I 1 1 I Newton: - = 1 +-----+-+-- .... 4J2 3 5 7 9 11
Since Newton routinely integrated series termwise, and (1 4 2) ( 1-x4 +x 8 _.,, ) = 1 +x 2 -x4 -x6 + .. ·, (1+x)=(I+x
+ x 2 )/
{I 1 + x2 dx = 1 + ~ - ~ - ~+~+_I - .... }o 1 + x4 3 5 7 9 1I The reader may complete the argument by observing 2
1+x -______,. 1 + x4
-
I [
2
1
I - J2x + x
2
+
I
1 + J2x + x
] 2
and
evaluate the appropriate integrals with the substitution
J2 1 x+- =-tanB. 2 J2 [regarding Newton] Nature to him was an open book, whose letters he could read without effort. -Albert Einstein
1.7
AUGUSTIN-LOUIS CAUCHY (1789-1857)
Cauchy must be considered the founder of integration theory: "Calculation" was henceforth to be replaced by "existence".
16
HISTORICAL HIGHLIGHTS
In 1823 Cauchy formulated a constructive definition of the integral, reminiscent of Cavalieri's and Fermat's approaches, but with a fundamental difference: In place of specific functions (x 2 , x 113, ... ) he started with a general function/ on [a, b] and formed the sum
where a= x 0 < x 1 < x 2 < · · · < Xn-i < Xn = b is a partition of[a, b]. The integral of Cauchy was to be the limit of such sums as the maximum of the xi - xi-!, II D.x II, approaches zero:
l
b
n
f(x)dx =
lim
ll~xll--->0
Lf(xi_J)(xi-xi-1)· i= 1
The question: what properties off would guarantee the existence of this limit? Cauchy argued that iff is continuous on [a, b], then this limit would exist. We also mention that Cauchy proved the Fundamental Theorem of Calculus, and thus settled integration for continuous functions on closed, bounded intervals. Cauchy's Construction:
a= xo
XJ
Xn-1
Xn =
The sole aim of science is the honor of the human mind, and from this point of view a question about numbers is as important as a question about the system of the world. -C.G.J. Jacobi
b
1.8
1.8
17
BERNHARD RIEMANN (1826-1866)
BERNHARD RIEMANN (1826-1866)
While investigating Fourier series and the attendant questions of convergence, Riemann (1854) asked the question: "What is one to understand by J:f(x)dx?". His answer (which is the most widely used definition of the integral), the so-called Riemann integral:
{b f(x) dx }a where
lim llb.xll->0
=
t
j(xJ(x;- Xi_J),
i=l
x; is an arbitrary point of [x;_ 1, xd and f
is bounded on [a, b].
Riemann's Construction:
a=
XQ
Xj
Xj
-
Xi-1 X;
X;
Xn
= b
Having given a constructive procedure (much like Cauchy's) Riemann then says: "Let us determine the extent of the validity of this concept," and asks: "In what cases is a function integrable and in what cases is it not?" His subsequent investigation resulted in a weaker requirement than Cauchy's requirement of continuity. He showed that the limllb.xlt_,o 'E-7=d(x;)(xi- xi_ I) exists iff f is bounded on [a, b] and for each pair of positive numbers t: and 8, there exists an ry > 0 such that whenever Pis a partition of [a, b] with II ~x II < f!, then the sum of the lengths of those subintervals [x;_ 1 , xi] with sup [X 1- I l X·] I
f- inf f > 6 [Xi- I ,x;]
is less than E. In other words, a bounded function/ is Riemann integrable on [a, b] iff large oscillations off are restricted to "small" sets. Riemann then gives an ingenious example of a function/ that is discontinuous on a
18
HISTORICAL HIGHLIGHTS
dense set of points of R, but is nevertheless Riemann integrable. Riemann's Function:
f(x)
=
f n=l
(n~)
,
n
where (x) denotes the difference between x and the nearest integer if xis not of the form k + l/2, k an integer; otherwise (k + 1/2) = 0. We carefully discuss this function. Let ¢n(x) = (nx), n = I, 2, ....
n = 1.¢ 1 is discontinuous at x = (2k + 1)/2: ¢((2k + 1)/2 -) = +1/2, ¢((2k+ 1)/2) = 0, ¢((2k+ 1)/2 +) = -1/2.
1 2
0
1 2
n = 2. ¢ 2 is discontinuous at x = (2k+ 1)/4: ¢((2k+ 1)/4 -) = +1/2, ¢((2k+l)/4)=0, ¢((2k+1)/4 +)=-1/2.
¢n is discontinuous at x = (2k+ 1)/2n: ¢((2k+ 1)/2n -) = +1/2, ¢((2k+l)/2n)=0, ¢((2k+l)/2n +)=-1/2.
?
2n
and so on.
° 2n
1.8
BERNHARD RIEMANN (1826-1866)
19
Where isf(x) = "£f(nx)jn 2 continuous? By the Weierstrass M-Test 2 (Theorem 2.11), the sequencefm(x) = "£~ 1 (nx)jn converges uniformly to! on R. So if x0 is not of the form (2k + 1)/2n; n, k = I, 2, ... , then ¢n is continuous for all n, and since uniform limits of continuous functions are continuous (Proposition 2.12), f is continuous at x 0 . It remains to is discontinuous on the dense set of points show f {(2k + I)j2n; k, n = 1, 2, ... }. Denote g(x-) - g(x+) by ~g(x). n=l.Letx=(2k+l)/2; ~¢ 1 =1, ~¢ 2 =0, ~¢ 3 =1,andingeneral, ~¢ 21 = 0, ~¢21-1 =I:~' 1,0, 1,0,.... Thus ~f((2k + 1)/2) = 2 I/1 2 + I/3 2 + I/5 2 + ... = Jr /8. Let x = (2k + 1)/4; ~¢ 1 = 0, ~¢ 2 = 1, ~dJ 3 = 0, ~¢ 4 = 0, ~¢ 5 = 0, ~¢ 6 =I, ~¢ 7 = 0, etc., repeating in a pattern of !o, I,O,O!, 0, 1,0,0, .... Thus
n = 2.
~f (2k + 1) = __!__2 + _12 + __ 12 + ... 4
(n)
X=
= ··· =
(2k
+ l)/2n;
~¢2n
= 0, · · ·,
~dJI
2
6
10
~¢2 = · · · = ~rPn-1 = 0, ~¢n = 1, ~¢n+1 repeating m the pattern 0, 0, · · ·, 0, 1 ,
=
0,0, · · · ,0,0, 0, ···.Hence n
n
~f (2k + 1) _ 1 . 2n . - n2
i-
I . I (3n) 2 + (5n) 2 + ...
" .Note: In the calculation of~~ we have used uniform convergence:
hm L: = 2: lim", to show that f is discontinuous on the dense set of points { (2k + 1)j2n; n, k = 1, 2, ... }, where the "jump" at (2k + 1) j2n is
20
HISTORICAL HIGHLIGHTS
D..f( (2k + I) /2n) = 1/ n2 x 1r 2 j8. Otherwise, f is continuous. Since 2 2 1r j8n > 8 for only a finite number of values of n, we can "cover" with a finite number of small intervals, and thus f is Riemann integrable. By the way, what is Jd "£(nx)/n 2 dx? Observe that Jd ¢n(x) dx = Jd (nx) dx = 0. Can we conclude 1
la 2::-
(nx) dx = n2
2:: la
1
1 2n o (nx)dx = 0?
Yes, because of uniform convergence we may interchange""£" and "J" (Theorem 5.3). This completes our discussion of Riemann's Function. During the latter part of the nineteenth century we had several reformulations of the Riemann integral by Peano, Jordan, Volterra, Darboux, and others that we "lump" together under "Darboux sums". -b b Darboux (1875) introduced upper Ua f(x) dx) and lower (J _a l(x) dx) Riemann integrals for a bounded function on [a, b]: n
b
J
-
f(x) dx = scp
a
p
L
·~J
l~
inf
[x;_ 1 , x;]
I·
(xi- xi~!),
over all partitions P of [a, b]. I is said to be Riemann integrable iff l(x) dx. This definition is equivalent to Riemann's but is _a f(x) dx = generally easier to apply.
t
J:
1.9 EMILE BOREL (1871-1956), CAMILLE JORDAN (1838-1922), GIUSEPPE PEANO (1858-1932)
Peano tried to connect integrability of a nonnegative function with the "area" of the setS= {(x,y) I a< x < b, 0 < y N. By definition, lim Xn = supS, and this is what we wanted to show. The other parts of the conclusion are left to the reader. •
= 1 + 1/1! + 1/2! + · · · + 1/n!. Obviously Xn < Xn+l and since xn < 1 + 1 + 1/2 1 + · · · + 1/2n < 3, we have convergence. Of course we recognize the limit as e. Example 7:
0, then lim sup Xn < a. 4. If Xn > a - E for every E > 0, then lim inf Xn > a. 5. If lim xn = x IS a real number, oo, or -oo, then lim inf Xn = lim Xn = lim sup Xn- (If x is a real number, x- E < Xn < x + E for all n > N(E) (Apply 3 and 4). The other cases follow easily.)
2.6.6
Problem
For the sequences (xn), calculate lim infxn.lim supxn: 1. 2.
Xn
3.
Xn
4
·
5.
1,2,3, 1,2,3, 1,2,3,· .. . 1, 1' 2, 1' 2, 3, 1, 2, 3, 4, ... .
Xn
1
1 1 ' '2'
1
1 1 1
1,2'3' '2'3'4' ...
1 1 3 1 7 1 15 2' 4' 4' S' 8 ']6' 16'
Xn :
Xn
1 1
1
1
1
-'
~n
6n2
=-, - 1 + -
,
n = 1, 2, ....
n = 1, 2, ....
2.6. 7
Problem
(xn) is a bounded sequence of real numbers. 1. If lim sup Xn = M, then 1. for every E > 0, Xn < M of n.
+ E for all but a finite number of values
2.6
SEQUENCES OF REAL NUMBERS
55
Hint:
M+e ii. xn
>M -
Hint:
E
for infinitel y many values of n.
If finite, ....
2. If lim inf Xn = m, then i. for every t: > 0, Xn > m - E for all but a finite number of n. ii. xn < m + t: for infinitel y many values of n.
2.6.8
Problem
If (xn) is any sequenc e of positive real number s, then liminfx n+i/ Xn < liminfx nl/n < limsupx nl/n < limsupx n+J/xnIf lim sup Xn+d Xn = oo, right hand inequal ity is obvious . So suppose limsupx n+ 1 /xn = M. Then Xn+dxn < M + E '\In> N, I.e., Xn+k < (M + t:) Xn for all k > 0.
Hint:
2.6.9
Problem
1. If (xn), (Yn) are sequenc es of real number s with Xn < Yn for all n, then
i. lim sup Xn < lim sup Yn ii. lim inf Xn < lim inf YnHint:
Xn
= sup{xn,Xn+l' .. .} < sup{yn,Yn+i' .. .} = Yn-
that lim supxn N(E), k = 1' 2, .... Example 12: Xo = 0, Xr = 1, xn = (xn-l + Xn-2)/2, n > 2. Then I Xn+l- Xn I= (1/2t. But Xn+k lies between Xn+l and Xn for all k > 1. Thus I Xn+k - Xn I < 1/2n for n > 0 and k > 0. Example 13: Cauchy.
xn = 1 + 1/2 + · · · + Ijn. Then X2n- Xn > 1/2 and not
14: Example kj2Jn +k.
Xn
= ..fii.
Then
I Xn+k -
Xn
I= Jn + k- Vn >
J; sin (t)jtdt. Form> n, Xm- Xn = J: sin (t)/tdt = Xn 2 -cos(t)/tl~ -J;cos(t)j t . Thus lxm-xnln. This sequence is Cauchy, and converges (lim xn = 1rj2).
Example 15:
2.3 For a sequence (xn) of real numbers the following conditions are equivalent:
THEOREM
I. (xn) converges to x: lim Xn = x, x a real number. 2. Cauchy condition: for every E > 0 there exists a natural number N(E) so that -E < Xn- Xm < Efor all n, m > N(E). 3. -oo < liminfxn = limsupxn < oo.
2.6
SEQUENCES OF REAL NUMBERS
57
The arguments are sketched. I Xm- Xn 1 N(E). Thus Xn < Xm + E and lim supxn =lim Xn < Xm for any m > N(E), i.e., lim sup xn is a lower bound for the set {Xm, Xm+ 1, ... } , that is, lim sup Xn < bn < lim inf Xn. 3 :=:::::::}- 1 Proof"
liminf
2.6.11
Xn
= limsup Xn
Comment
If we have a convergent sequence of real numbers (xn), xn~x (real), then equivalently I Xn- xI ~o. If we let Yn =I Xn- xI, and form(~), (Yn), we have ~ = 0 for all n and Yn decreases monotonically to zero. Thus 0 < Yn < Yn and Yn+l < Yn· The usefulness of this result may appear as: "If we are given that a sequence converges, then we may suppose the sequence is nonnegative and monotonically decreasing to zero." Sequences are frequently given in the form a 1 + a2 + · · · +an + · · · . The usual terminology is to say we have a series. 2.6.12
Definition
A series is a sequence of numbers denoted by L an or L ak. A series L ak is said to be convergent to the real numbers if the sequence of partial sums (sn); s1 = a!, s2 = a 1 + a2, sn = a 1 + · · · +an, converges to s, and we write
Thus lim (s- sn) = 0 is sometimes written oc
lim Lak n+!
=
0.
58
PRELIMINARIES
If (sn) does not converge, the series is said to be divergent. Of course a series 2.::: ak with nonnegative terms ak > 0 either converges or has limit oo since the sequence (sn) of partial sums is a nondecreasing sequence. In other words, a series 2.::: ak of nonnegative terms converges iff its partial sums are bounded.
2.6.13
Comment
Will convergence of a series be affected if we rearrange (change the order of) the terms? And, if we have convergence for the rearranged series, will it be to the same number? For example, the series 1 - 1/2+ 1/31/4 + · · · = ln2. If we "rearrange,"
(If Sn = 1- 1/2 + 1/3- · · · + ( -1)"+ 1 1/n, then sj 11
=(1-~)-~+(~-~)-~+···+( 1 _ 1 )--1 2 4 3 6 8 2n-1 4n-2 4n 1111 1 I 1 =--- +--- + ... + - - = - s2 2 4 6 8 2(2n - 1) 4n 2 n
and, consequently, lim s3n =lim 1/2 s2n = 1/2 ln 2. Since s3n - s3n+l-o and s3n - s3n+2 -o, s~ -1/2 ln 2.) Fortunately, our applications will be restricted to series with nonnegative terms, and in this situation, rearrangements do not affect convergence: Suppose 2.::: ak = Sand 2.::: a'k = S*, with every term of the original series (ak > 0) occuring exactly once in the new "rearranged" series 2.::: aic. We show S = S*. Let a < S. Then a < sN for some N because the sequence of partial sums (sn) is nondecreasing. Then a < s~ where M is chosen so large so that every term a 1, a 2, ... , aN occurs in s~. Thus S < S*. An analogous argument shows S* < S. In conclusion, the terms of a nonnegative series may be arranged in any order without affecting the sum, finite or infinite. In our development of Lebesgue measure we will on occasion deal with double series, Lk(Ln akn), and again questions of convergence and rearrangements will be important. For example, are there conditions so
2.6
SEQUENCES OF REAL NUMBERS
59
that
+
a21
+ al2 + ... + aln + .. . + a22 + ... + azn + .. .
+
akl
+ ak2 + · · · + akn + · · ·
all
"Rows then Columns"
+··· + a12 + a22 + · · · + ak2 + · · · a11 +a21 +···+akl
+
aln
"Columns then Rows"
+ azn + ... + akn + ... ?
We certainly want each term of the original series akn to appear exactly once in the "rearranged" series. This suggests we need a 1 - 1 map of IN x lN onto lN. The reader might refer to Proposition 2.3. Possibilities:
:>; (:>;ak,) ~ (a11 +a"+···+
a 1,
+ .. ·)
+ (a21 + a22 + · · · + a2n + · · ·)
+ (akl + ak2 + · · · + akn + · · ·)
1 ?
· a 11
+ a21 + a 12 + a 31 + a22 + a 13 + · · · ,
that is, correspond the (k + n- 1)(k + n)/2- (k- 1) term of the right hand side with akn· This map is 1 - 1 from lN x lN onto lN. If we used the map (k,n)+----'>2k- 1(2n- 1), then does Lk(Lnakn) = a11 + a 21 + a 12 + a 31 + a 13 + a 22 + ···?As before, additional restrictions
60
PRELIMINARIES
must be imposed, as the next example shows. 1
-1
0
0
0
~
0
0
1
-1
0
0
~
0
0
0
1 -1
0
~
0
l
l
l
l
1
0
0
0
"Rows then Columns"
l 1
0
"Columns then Rows" The rearrangement theorem we are looking for is given in Stromberg. We include the argument for completeness. Suppose that akn is nonnegative and that ¢ is any 1 - 1 mapping of IN onto IN x IN. Then PROPOSITION
2.4
=
L aq,(i). i
These sums may be finite or infinite. Proof Let o: be any real number so that o: < 2:; aq,(i). We will show o: < l:k(l:n akn) and conclude L: aq,(i) < l:k(l:n akn)· Since akn are nonnegative, the partial sums of 2:; aq,(i) are nondecreasing. Thus we have I so that o: < 2:{= 1 aq,(i)· Select K, N so large that
Hence 0:
(i) < L:~1 ac/>(i)· Thus L:k(L:n akn) • < L:i ac/>(i). The argument is complete.
62
PRELIMINARIES
2.7 2.7.1
TOPOLOGICAL CONCEPTS OF R Definition
The open interval (a, b) is the set of real numbers between a and b. The intervals ( -oc, a), (b, oo) and ( -oo, oo) are to be considered open. A set G of real numbers is said to be open if, for each p E G we have a positive real number 8, so that (p- 8,p + 8) c G. In other words, G is open if every point of G is in the center of an open interval wholly contained in G. R is open and the empty set 0 is open. As far as set theoretic operations, "finite" intersections and "all" unions of open sets are open. 2.5 1. If G1 , G2 are open subsets of R, then G1 n G2 is open. 2. If {GoJ is any collection ofopen subsets of R, then UGa is open.
PROPOSITION
Proof 1. Suppose X E Gl n G2. min{ 81, 82}.
Then
(x- 8;, X+ 8;) c G;.
Let
8=
2. If x E UGm then x E Gao for some a 0 . Since Gao is open, we have (x- 8, x + 8) C Gao for some 8 > 0. Thus (x- 8, x + 8) C Gao C u~.
Note:
• We cannot strengthen 1:
n (-1/n, 1/n) ~~
=
{0}.
The next theorem reveals the structure of open sets in R. Every nonempty open set of real numbers is the union of a countable collection of mutually disjoint open intervals.
THEOREM 2.4
Proof Let G be a nonempty open set of real numbers and p any point in G. Because G is an open set, we have 8 > 0 so that (p- 8,p + 8) c G.
2.7
63
TOPOLOGICAL CONCEPTS OF R
Define a= inf{y I (y,p) c G} and b = sup{z I (p,z) c G}. Then a, bE Re. If a is a real number, then a tf. G since the openness of G would require (a- 81 , a) c G for some 81 > 0. This contradicts GLB. Similarly for b. Regardless, we have an open interval /(p) associated with every point p E G. If I(p 1), I(p 2 ) are associated with distinct points p 1,p2 E G, then I(pJ),I(p 2 ) are identical or disjoint since a 1 < a2 < b 1 would imply a 1 = a2 and b 2 = b 1, when / 1 = (a 1 , bJ) and h = (a 2 , b2 ). The reader may argue the other cases. Thus G is the union of a collection of disjoint open intervals. Select a single rational number from each interval. This rational number will correspond to exactly one interval because the intervals in our collection are disjoint. Thus the collection is countable and we are done. •
2.7.2
Definitions
A set will be called closed if it is the complement of an open set. The closed interval [a, b] is the set of real numbers between and including a and b, and is closed (complement of ( -oo, a) U (b, oo )). The sets ¢ and R are both open and closed, [0, 1) is neither open nor closed, and the reader may show (using De Morgan's Laws) that "all" intersections and "finite" unions of closed sets are closed. "Finite" unions cannot be strenghtened: U2[1/n, 1- 1/n] = (0, 1).
2. 7.3
Definition
A point p is called a limit point of a set A if for every 8 > 0, there is an x E A, x f. p, such that -8 < x- p < 8. In other words, pis a limit point of the set A if every open interval centered at p contains a point of A different from p.
2.6 A nonempty set F of real numbers zs closed contains all its limit points.
PROPOSITION
iff
it
Proof Let F be a closed set and p a limit point of F. We will show P E F. We argue by contradiction. Assume pis a limit point of the closed set F and p tf. F. Then pis a member of the open set Fe. Thus we have 8 > 0 so that (p- 8,p + 8) c Fe, that is, (p- b,p +b) n F = 0. This contradicts the definition of p being a limit point of the set F. Now we
64
PRELIMINARIES
suppose F contains all its limit points. We show F must be closed, or equivalently, Fe must be open. Let p E Fe. Since F contains all its limit points, and p E Fe, p is not a limit point ofF, that is, we have a t5 > 0 so that (p- t5,p + 8) n F = 0. In other words, given a point in the complement of F we have an open interval centered at that point m the • complement ofF, and hence the complement ofF is open.
2.7.4
Definition
A collection {Ga} of open sets covers a set A if A c UGa. In this case, the collection {Ga} is called an open cover of A.
2. 7.5
Definition
A set of real numbers is compact if every open cover of the set contains a finite subcover: If A C UGa, then A C u7= 1Ga,, A compact, Ga, open.
2.7 (LINDELOF) contains a countable subcover. PROPOSITION
Any open cover of a set of real numbers
2.7
TOPOLOGICAL CONCEPTS OF R
65
Let A be any set of real numbers with open cover {Ga} : A c UGa. We want to show Proof
A
c
U
Ga.
countable
Pick pEA. Then p E Gn for some a and since Gn is open, (p- 8,p + 8) c Ga for some 8 > 0. Pick rational numbers r 1 , r 2 so that p - 8 < r 1 < p < r2 < p + 8. Thus we associate with each point p of A an open interval I(p) = (r 1 , r2 ) with rational endpoints. Since the set of all possible intervals with rational endpoints is countable, the collection B = {I(p) I pEA} is countable. But each member of B may be associated with exactly one member of {G 0 ,}, say, G~. Thus the collection {G~} is countable and covers A. • We now prove one of the most important theorems of analysis. THEOREM
2.5
(HEINE-BOREL)
A set of real numbers is compact
iff it
is
closed and bounded.
Suppose K is a nonempty compact subset of R. We show K is closed and bounded. Since K c U (-n, n), a finite subcollection covers K, that is, K c (-N, N) for some Nand hence K is bounded. We still must show K is closed. Equivalently (Proposition 2.6), will show that any point not in K is not a limit point of K. Pick p r;j. K and form closed sets Fn = [p- 1/n,p + 1/n], n = 1, 2, .... Then p = n[P- 1/n,p + 1/n] Fnt = U F,;. But then K is covered by the and since p r;j. K, K c collection of open sets {F,;}. Since K is compact, we have K c F{UF{U· · ·UFJ. for some N, or K c FJ, since F{ c F{ c · · · c FJ,. No point of K is in [p- 1/ N,p + 1/N]. Thus pis not a limit point of K. In R, compact sets are bounded and closed. Now suppose K is closed and bounded. To show K is compact, we must show every open cover of K reduces to a finite subcover. Let {Ga} be an open cover of K. By Proposition 2. 7 (Lindelof), we have a countable subcover: K C U Ga,· Define sets Gn and Fn as follows: Gn = Ga 1 U Ga 2 U · · · U Ga" and Fn = K G~. Gn is open, Fn is closed, Gn c Gn+! and Fn+l c Fn for all n. If Fn is empty for some n, we have a finite subcover. So assume Fn i= 0 for all n and form the closed set Fn. Each set Fn is a closed, bounded (Fn C K), and nonempty subset of K. If we define Xn = supFn, then Xn E Fn since Fn is closed, the sequence is nonincreasing and bounded below (Fn+l C Fn C K). Let x 0 = inf{x 1,x 2 ••. }. Claim: x 0 EnFncK. Otherwise, (x 0 -8,x0 +8)nFN=0 for some 8>0 Proof
n
(n
n
n
66
PRELIMINARIES
and some N. Since Fn+I c F,, x 0 + 8 is a lower bound for {xN, xN+l, ... } and hence {x1, x2, ... }. But x 0 is the greatest lower bound. So Xo E Fn c K, which implies Xo E K, Xo (j_ Gn for all n. But the collection {Gn} covers K. Thus Fn must be empty for some n and the proof is complete. •
n
2.6 BOLZANO-WEIERSTRASS real numbers has a limit point.
THEOREM
Every bounded infinite set of
Proof Let E be a bounded infinite set of real numbers. We show E has a limit point. Since E is bounded, we have a closed bounded interval [a, b] that contains E. Suppose for every x E [a, b] we could find an open interval I(x) centered at x that contains only a finite number of points of E. The collection of such open intervals would be an open cover of the compact set [a, b]. Thus a finite number of such intervals would cover [a, b], and hence E. By assumption each of these intervals contains a finite number of points of E. So E would be finite. But E is infinite. Thus we have at least one point of [a, b], say x 0 , so that every open interval I(x 0 ) centered at x 0 contains an infinite number of points of E. But then x 0 is a • limit point of E, and this is the conclusion we wanted.
2.8
CONTINUOUS FUNCTIONS
2.8.1
Definition
A real-valued function f defined on a set E of real numbers is said to be continuous at c E E if for every t > 0, we have a 8(c, t) > 0 such that X E (c- 8, c + 8) E implies f(x) E (.f(c)- E, f(c) + t). The 8 usually depends on both c and E.
n
f f(x) X
c.:8
• •c
c+-8
f(c) - E
f is continuous at c
{(c)
{(c)
+E
2.8
CONTINUOUS FUNCTIONS
67
I
f(c) + E
~-
f(c)
- - -
- - -- - -
· - -- -----
! {(c) - E ··· -
c-8 c c+8
f is continuous at c Iff is not continuous at c E £, then we have E > 0 so that for every b(c, E) > 0, there is an x E (c- b, c +b) n E for which f(x) > f(c) + Eor f(x) O x 0. Then (/(c)- E,/(c) +E) is an open subset of R. By assumption, f- 1((/(c)- E,j(c) +E)) is open relative to E and contains the number c. Select b > 0 so that (c- 8, c +b) n Eisa subset of f- 1((/(c)- E,j(c) +E)), that is,
J ((c - 8, c + ti) n E) c (f (c) - E,/ (c) + E) . Thus f is continuous at c. This completes the argument. Example 23:
f(x) = 1/x, 0
<x
n. But by Bolzano-Weierstrass, the sequence must contain a convergent subsequence (xnk) converging to some c E [a, b]. Since f is continuous at c, f(xnk)---tf(c). But this contradicts lf(xnk) I> nk. A continuous function on a closed bounded interval must be bounded. We will now show f assumes a maximum. Let M =sup {f(x) I x E [a, b]}. M is finite by the argument above. Since M- 1/n is not an upper bound, we have Zn so that M- 1/n f(b). • We now show continuous functions preserve compactness. 2.10 Iff is a real-valued continuous function on the compact set K, then f(K) is compact. PROPOSITION
Sketch: Let {Ga} be an open covering of f(K) : f(K) C UGa. Because f is continuous, f- 1(Ga) is open, and K cj- 1 (UGa) = uj- 1 ( Ga). In other words, the collection {f- 1 ( G"')} is an open cover of the compact set K. By Reine-Borel, a finite number, say f- 1 ( Ga 1 ), ••• , f- 1 ( GaJ suffices to cover K: K C U ~J- 1 ( Ga ). But then, f(K) C N " Ui=l Ga;· • Proof
I
2.8.2
Definition
f be a real-valued continuous function with domain
E. Then f is uniformly continuous on E if for every E > 0 there exists 6( E) so that -E 0. We must determine 8(E) > 0 so that -E 0 so that -E/2 < f(x) - f(c) < E/2 whenever X E (c- 8(c), c + 8(c)) n K. The open sets { (c- 8(c)/2, c + 8(c)/2)} form an open cover of the compact set K. By Heine-Bore!, a finite subcollection of these sets will cover K: K c U [: 1 (ci- 8(ci)/2, ci + 8(ci)/2). Define 8 as the minimum of the set: {8( cJ) /2, 8( c2 ) /2, ... , 8( eN) /2}. Pick any points x, yin K satisfying -8 < x- y < 8. We will show -E 0. So
Proof
limf(x)- f(c) > 0. X- c X-+C Then we have 8 > 0 so that a< c- 8 < c + 8 O.
Let x = c + 8/2. Then f(c + 8/2) > f(c). We have a contradiction to • /(c) being the maximum. The reader may complete the argument. 2.9 Let f be a real-valued continuous function on [a, b], differentiable on (a, b), and satisfy f(a) = f(b). Then there exists c, a< c < b, with f'(c) = 0.
ROLLE'S THEOREM
Proof
The figure indicates why such a result might be true.
a
c
b
74
PRELIMINARIES
Since f is continuous on [a, b], [a, b] (Proposition 2.9),
f assumes a minimum and maximum on
a< xM, Xm 0.
. (. )) and hm hm nx n
. sm . (nx Calculate lim hm
SEQUENCES OF FUNCTIONS
+
h-->0
hh- ln(x))
and
"Limit of the derivative" is not necessarily the "derivative of the limit". 2
n
Example 31:
Let ln(x)
=
L k=O
(1
x
2
k"
+X )
Calculate lim (limln(x)) and lim (limln(x)). x-->0
n
n
x-->0
"Limit of the sum" is not necessarily the "sum of the limits".
Example 32:
Let ln(x)
= n2x(I - x 2 t, 0 < x < I.
Calculate lim( lim tln(xi) b.xi) and n
116-xll-->0
i= 1
"Limit of the integral is not necessarily the integral of the limit".
2.1 0.3
Definition
A sequence of real-valued functions Un) is uniformly convergent to a function I on a set E iffor each E > 0, we can determine a natural number
78
PRELIMINARIES
N(t.) so that f(x)- E <J~(x) N(t.) and all x E E. We write limfn = f(unif). The essential difference between pointwise convergence and uniform convergence is the difference between "local" and "global": the N(x, t.) in pointwise convergence generally depends on a particular point x ("an N for each x"); theN( E) of uniform convergence "works" for all points of E ("one N for all x"). It is nonsense to talk about uniform convergence at a point. Uniform convergence deals with behavior on a set of points. The figure below illustrates the definition .
/
,. .,.,. ....
------- ....
............
............
/
...... ........
/
......... .............
___ .,.,..,
...
...""
/
f+E
f (-E
fn
E
The reader should show that an equivalent definition for uniform convergence may be stated as follows: limfn =.f(unif)
-¢=*
lim sup n
E
l.f(x)
-fn(x)
I= 0.
Generally speaking, uniform convergence preserves good behavior:
Example 33:
"Bad"
{,,(x)
->
~ { ~:
"Good"
x irrational x rational
O<x 0. Then Iimfn(x) = f(x) l+nx
= 0.
( l.J_) n n
0
limfn = f(unif) since lim (sup E
Example 37:
fn(x) =
2
n~
If~- f I) =lim~= 0.
I+nx 2
(
n
;
n
x > 0. Then limfn(x) = f(x)
1 ,1) n
0 I
Convergence is not uniform: lim ( s~p I fn - f I ) =
li~ I =
I.
= 0.
81
82
PRELIMINARIES
Example 38:
fn(x)
=
2n 2 x 2 2 1 +nx
;
x > 0. Then limfn(x)
= J(x)
- 0.
1 n
(- ,n)
0
Convergence is not uniform: lim( sup n
2.1 0.4
I fn
- f
I) = limn = oo. n
Problem
1. fn(x)
=
0 <X< 1.
Xn, xn
2. fn(x)
= (I + xn)'
0 <X< 1.
Show (fn) converges uniformly on [0, I - E], 0 < converge uniformly on [0, I]?
f
< I. Does Un)
n2x 3. fn(x) = ( I+ n3x~")' 0 <X< 1. Does (fn) converge uniformly on [0, I]? How about[E, I], 0 4. fn(x)
< f < 1?
= nxe-nX, x > 0.
Does Un) converge uniformly on [0, oo )? How about [E, oo ),
f
> 0?
A useful test for determining uniform convergence for a series of functions was developed by Weierstrass. 2.11 (Weierstrass M-Test) A series ofreal-valuedfunctions Lfk(x), fk defined on a set E, converges uniformly onE if I fk I < Mk onE, k = I, 2, ... and 2:_ Mk converges. THEOREM
•
2.10
Example 39:
Let Mk = Example40:
2x
I: 1 k3 2' -oo < x k + X
SEQUENCES OF FUNCTIONS
83
< oo;
k;12 . We have uniform convergence on R. 2:(-x)\ -1 < x < 1; k
"..... __ ..,.
~
''
''
I
/
I
I
I 1 -X+ x 2
/'
1 . 1 +X
''
--"'-~,--
''
''
''
1
Weierstrass isn't helpful here. Define gn(x) = I - x
--X
-1-
x 2 · · · + (-xt,
84
PRELIMINARIES
- 1 < x < 1. Then
gn(x)= =
1- ( -xr+l
1 +x 1 1 +X
, -1 < x < 1, and lim ( sup
-l<x 0.
~ li:n[~ (I -
1;
kx)
l
Note that fn(O) = 0. Does limfn(O) = f(O)? We have completed the review necessary for the succeeding chapters . . . ., there is no study in the world which brings into more harmonious action all the faculties of the mind than the one [mathematics] ... -J. J. Sylvester Many arts there are which beautify the mind of man; of all other none do more garnish and beautify it than those arts which are called mathematical. -H. Billingsley
~ sin (ka) 2
~ -oc
k2
-1 -
00
-oc
2
sin (xa) X
2
dx, 0 0, I= f- 1 ((1/4, 9/16]). 2. f(x) =sin (x), 0 < x < 21r and I= f- 1(( v'3/2, oo)). 3. f(x) =cos (x), 0 < x < 21r and I= f- 1 (( -oo, -1]). x2
0 < x < 1 and S =f- 1 ((9/16, 5/4)). 1 <x< 2 3-x ' What "length" would be reasonable for S?
4. f(x)
=
{
'
5. f(x) =x 3 andi=f- 1 ((-oo,-8)). 6. f(x) =I x I and I= f- 1 ( ( -oo, 0]).
3.1.3
Problem
If I 1 and I 2 are two intervals with I 1 c h, show l(I1 ) < l(h) (monotonicity).
3.1.4
Problem
If I, I 1 , h, · · · , In are bounded open intervals with n
I
c
UIkl
then !(I)
0. The intervals (a - E, a + f.), ( b - E, b + f.), I 1 , h, ... , Ikl ... form an open cover of the compact set [a, b]. By Heine-Bore! (Theorem 2.5), a finite subcollection, (a- E, a+ E), (b - E, b + f.), In 1 ,!112 , ••• ,l11k will cover [a, b], and thus (a, b).
Proof
Application of Problems 3.1.3 and 3.1.4 yields k
!(!) < 2E + 2f. +
I:>unJ i= I
Since this holds for any f. > 0, we conclude /(I)
0, it follows (GLB) that the set of all such numbers possesses a nonnegative greatest lower bound. 4. We could require the lengths of the intervals h to all be less than a given positive number 8 since each Ik can be subdivided into disjoint subintervals each of length less than 8 without affecting the sum of their lengths. 5. f.L* is a set function, whose domain is all subsets of R, 2R, and whose range is [0, oo], the nonnegative extended reals:
\
0
!
0
R
,u'(A)
00
6. If (h) is any countable cover of A by open intervals, then since the infimum is a lower bound,
> 0, there is a countable collection (h) of open intervals such that A c U h and
7. If f.L*(A) <ex:, then corresponding to each
E
:)0
1
JL*(A)
0 and by what we just showed and monotonicity, b- a< f-l*([a,b])
< b- a+ 2E.
Thus f-l*([a,b]) = b- a. The cases when I= (a,b], or [a, b) may be completed by the reader. We have then the Lebesgue outer measure of any bounded interval is just its length. Finally, let I be an unbounded interval, for example, (a, oo). For any positive number C >a, (a, C) is contained in (a,oo). Consequently,
C- a= !((a, C))= J-l*((a, C))< f-l*(I) since (a, C) is a bounded interval. Thus f-l*(I)
= 00 = l(I).
The Lebesgue outer measure of any interval is its length. Translation invariance, property 7, is based on the fact that length,/, is translation invariant: If I= (a, b), then c +I= (a+ c, b +c) and 1(1) = l(c+l). If I is (b,oo), (-oo,a), or (-oo,+oo), then c+l is (b+c,oo), (-oo,a+c), or (-oo,+oo), respectively, and again !(I)= l(c +I). If A is an arbitrary subset of R with A c Uh, then
c+AcU(c+h),
3.2
LEBESGUE OUTER MEASURE
97
and
Jl*(c+A) < Ll(c+h) = Ll(h). This tells us that Jl* (c +A) is a lower bound for the "lengths" of covers of A, and because J.L*(A) is the greatest lower bound of such numbers,
By starting with a cover (Jd of c +A, we have A c U(h- c). The reader may finish the argument. Countable subadditivity remains. We must show
for any sequence (Ak) of sets of real numbers. Of course if the series 2:::: Jl*(Ak) diverges the argument is immediate, so assume 2:::: Jl*(Ak) < oo and let E > 0. For each nonempty Ak choose an open cover (hn) so that
and 00
Jl*(Ak)
~· (X n [ VE•l) + M' (X- [ E, l)
=
~>'(Xn£,) + 1:
independent of n.
(
x- [QE•l)
(monotonicity)
3.3
LEBESGUE MEASURABLE SETS
109
Therefore
1t(x) >~>'(X n £ >
1)
+ p' ( x-
[
QE,l)
t!(xn [ l) +p'(x- [ l) QE1
QE1
(subadditivity).
The reverse inequality follows from subadditivity (Theorem 3.1). We have completed the argument for showing M is a O"-algebra of subsets of R. 2. We are left the task of showing countable additivity of J..l* when restricted to members of M. Let (Ek) be a sequence of mutually disjoint sets from M. We must show
But in part 1 we showed
for any X
c
R. Replacing X with R,
Finite additivity holds. Then,
N and lim J-L(Ek) = oo. Since EN C UEb J-L(EN) < J-L(UEk) and thus J-L(UEk) = J-L(lim Ek) = oo. So we may suppose J-L(Ek) < oo for all k. Since
and the sets Ek+l - Ek are mutually disjoint, we have oc
p.(UEk) = J-L(Et)
+L
p.(Ek+l - Ek)
(5)
k=l
= J-L(EJ) + L[J-L(Ek+t)- J-L(Ek)]
(7)
k n
= J-L(EJ) + li~ L[J-L(Ek+J)- J-L(Ek)] k=!
9. Proceed as in part 8: Write a disjoint sequence of sets and use part 5:
Thus
J-L(EI)- J-L(nEk)
= J-L(£1 - (nEk)) = LJ-L(Ek- Ek+J) 1
= L[J-L(Ek)- J-L(Ek+J)] I
Since J-L(£1 )
< oo, we may subtract and the conclusion follows.
3.5
"MEASURING''
10. We essentially use part 9. Recall limsupEk
=
n (U
m?.l
Ek) ·
k?.m
Then
UEk ~ UEk ~ .. · I
and Em
c
U Ek .
k?.m
2
Thus
and, hence,
11. Again liminfEk =
n cn Ek
I
Thus J.L(
Ek
2
n Ek)· Then m?.l k?.m U(
c · · · and
n
Ek
I
n Ek) < J.L(Em). Hence
k?.m
c E1,
n
Ek C E2, ...
2
117
118
LEBESGUE MEASURE
But
n
= f.L(u
m?:1 k?:m
Ek)
= f.L(lim inf Ek)
by part 8.
12.
• Appropriately, it is time to "calculate"; we take specific sets and compute their measure. Enjoy! 3.5.1
Problem
The reader should discuss the Lebesgue measurability of the sets that appear below and calculate their Lebesgue measure when appropriate.
1. Ek=
ck~l)'l} {X
I sin(x)
0,
1 , a
(xk-1 +_a_) with Xo =a> 1; nEk. xk-1
8. Ek = ( ka* - 1, kbt - 1), 1 < a < b; lim inf Ekl lim sup Ekllim Ek.
9. Ek = (akl bk), ak =
1 ( -1 )k- 1 1 - + ··· + k , bk 2
1
=
1-
1
3 + 5 + · · ·+
1
(-l/+ ; 1·1mm . f£kl 1·1msup Ek• 1·Im E k· 2k- 1
10. 1-.i, ~ ck!~l/k' (1 + !t} nE,. 11. Rational numbers; any countable set. 12. Irrational numbers in (0, 1): Positive measure, but does not contain an interval.
13. Cantor set: uncountable set of measure zero. 14. Numbers in (0, 1) without 7 in their decimal expansion.
15. Any set of numbers with Lebesgue outer measure zero (any subset of such a set). 16. The symmetric difference of two measurable sets £ 1 and £ 2 . (£11'::!.£2 = (£ 1 - £ 2 ) U (£2 - £ 1)) f.L(£ 1f::..E2 ) = 0 iff f.L(£ 1
17. If E Lebesgue measurable, f.L*(A- E)= f.L*(A)- f.L(E). 18. f.L*(A U B)+ f.L*(A
-
£ 2 ) = f.L(E2
f.L(E)
-
£ 1) = 0.
< oo, and E c A, then
n B) < f.L*(A) + f.L*(B).
120
3.6
LEBESGUE MEASURE
STRUCTURE OF LEBESGUE MEASURABLE SETS
We are comfortable with the familiar; closed, open, nowhere dense, countable, etc. Where does Lebesgue measurability enter the picture? Are there relationships between topological properties (open, closed, etc.) and Lebesgue measurability? The next theorem shows that Lebesgue measurable subsets of Rare "almost open", "almost closed", and so forth. 3.5 equivalent: THEOREM
For an arbitrary subset E of R, the following statements are
1. E is Lebesgue measurable in the sense of Caratheodory; 2. Given E > 0 we can determine an open set G C R with E C G and J-L*(G- E)< t: ("exterior" approximation by open sets); 3. Given E > 0 we can determine a closed set F c R with F c E and J-l* (E - F) < t: ("interior" approximation by closed sets); 4. Thereis(hset B1 withE c B 1 andJ-L*(B 1 - E)= 0 (B 1 is a countable intersection of open sets; if we relax "open", we can obtain a very good approximation by Borel sets); 5. There is an Fa set B 2 with B2 c E and J-l* (E- B2 ) = 0 ( B2 is a countable union of closed sets; if we relax "closed", we have very good approximations/rom the "inside" by Borel sets).
Proof 1.=::}2. Assume E is a Lebesgue measurable subset of R with J-L(E) < oo. From the definition of Lebesgue outer measure, we have an open cover G = Ulk so that E c G and J-L*(G) < J-L(E) +E. Since G = (G - E) U E and G is Lebesgue measurable (Theorem 3.3), J-L( G) = J-L( G - E) + J-L( E). Because J-L( E) < oo, we may subtract and obtain J-L(G- E) < t:. If J-L(E) = oo, let Ek =En [-k, k]. Ek is Lebesgue measurable, J-L(Ek) < oo, and by what we just showed we have an open set Gk so that Ek c Gk and f-t(Gk- Ek) < t:/2k. Since E = UEk and E c UGk = G, it follows that J-L*(G- E)= J-L(G- E)
3. 5.===>1. We must show J.L*(X n E)+ J.L*(X- E)= J.L*(X) for every subset X of R. Let E be an arbitrary subset of R and B 2 the Fa set guaranteed by part 5: B 2 c E, ~-t*(E- B2) = 0. Since B2 is Lebesgue ~-t*(X n E)= ~-t*((X n E) n B2)+ 3.3), (Theorem measurable J.L*((X n E)- B2). Because (X n E)- B2 c E- B2, J.L*((X n E)B2) = 0, and we have ~-t*(X n E)= ~-t*(X n B2). On the other hand, J.L*(X- E)= J.L*( (X- E) = ~-t* (0)
n B2) + ~-t*( (X- E)- B2)
+ ~-t* ((X- E)
- B2)
< ~-t*(X- B2). Thus, ~-t* (X
n E) + J.L* (X- E) < J.L* (X n B2) + J.L* (X- B2) = J.L* (X)
because B 2 is Lebesgue measurable. This yields Caratheodory's Condi• tion onE since ~-t*(X) < ~-t*(X n E)+ ~-t*(X- E) by subadditivity. 3.6.1
Problem
1. E = {rational numbers in (0, 1) }. What is F of part 3 of Theorem 3.5?
122
LEBESGUE MEASURE
2. E = {irrational numbers in (0, 1) }. How about F of part 3 of Theorem 3.5? 3.6.2
Problems
1. Show that E c R is Lebesgue measurable iff we have Borel sets B 1 , B 2 satisfying B 2 c E c B 1 and f-l*(B 1 - B 2 ) = 0. Hint:
B1 - B2 = (B 1 -E) U (£- B 2 ).
2. Let E be a Lebesgue measurable set with f.-L(£) < oo. Show that we have a decreasing sequence of open sets ( Gk) so that limf.-l(Gk) = f.-l(E). Hint:
E
c Ok, f.-l(Ok- E) < ljk, and form 0 1,01 n 0 2 , 0 1n
02 n 03, ... 3. Let E be a Lebesgue measurable set with f.-L(E) < oo. Then we have an increasing sequence of closed sets (Fk) so that lim f-L(Fk) = f-L(E). Borel sets, being countable unions and intersections, with complementation, of open and closed sets are familiar. The next result relates Borel sets and Lebesgue measurable sets. It is a corollary of Theorem 3.5, but it so succinctly describes Lebesgue measurable sets that we choose to call it a theorem. THEOREM 3.6
Every Lebesgue measurable set of real numbers is the union of a Borel set and a set of Lebesgue measure zero (Lebesgue measure is the completion of Borel measure).
Let E be a Lebesgue measurable set of real numbers. We then have a Borel set (:Frr) B so that B c E and f-L(E- B) = 0 (Theorem 3.5). But E = B U (E - B); B is our desired Borel set and E - B is the Lebesgue measurable set with Lebesgue measure zero. • Proof
The last theorem of this chapter tells us that sets of finite Lebesgue measure are "almost" finite unions of intervals. Suppose E is any subset of R with f-L*(E) < oo. Then Eisa Lebesgue measurable set of real numbers iff we have a finite union of open intervals U so that
THEOREM
3.7
f-l*(E- U) for any
E
> 0.
+ f-l*(U- E)< E
l
3.6
STRUCTURE OF LEBESGUE MEASURABLE SETS
123
Proof We first assume E is Lebesgue measurable and construct a finite union of open intervals U so that 11 * ( E - U)
+ 11 * ( U -
E)
c} n {x E E if(x) < c}. 5. Let G be any open set in R. Then f- 1(G) is a measurable subset of E. Hint: 1 1 G = Uh, h disjoint open intervals and f- (G) = f- (Uh) = uf- 1(h). Again, continuous functions deal with topological properties of inverse images whereas measurable functions require measurability of inverse images. Different measures could presumably yield different measurable functions. We give two examples (neither necessary for later material) to illustrate that, in general, we must be very careful about the specific measure being discussed. Example 1:
Define an outer measure a* on subsets A of R by
a*(A) =
number of elements in A if A is finite . . fi . { oo, I"fA IS m mite.
Using Caratheodory's Measurability Criteria (3.3.1) the reader may show that every subset of R is measurable. Then, for any extended realvalued function f, the set {x I f(x) > c} is a subset of R and thus measurable. That is, every extended real-valued function is measurable, relative to this specific measure. Example 2:
Define an outer measure {3* on any set B of real numbers by:
{J*(B) = { 0, 1' The reader may show that 0, R are the only measurable subsets of R. Let f(x) = x, x E R. Then {x I f(x) > 0} = (0, oo) is not 0 orR, that is, the identity function is not measurable. In fact, the reader may show that only constant functions are measurable relative to this measure. Hopefully Lebesgue measure will give us something between these two extremes, that is, enough interesting functions to build a "useful" theory of integration. Caution: In what follows a "measurable set" means a "Lebesgue measurable set of real numbers", a "measurable function" means a
130
LEBESGUE MEASURABLE FUNCTIONS
"Lebesgue measurable function". For any function, the domain will always be a subset of Rand the range will be a subset of R or Re (real-valued or extended real-valued). Using Definition 4.1.1 to show a function is measurable is illustrated by the next example and problem.
Example 3: 1. f is constant on a measurable set E, say, f(x) = k, x E E. Then
0,
{x I f(x) > c} =
c>k
{ E, c < k.
The sets 0 and E are measurable. Thus f is a measurable function on E. 2 X
2. f(x) =
2
x
'
1.
c < 0 {x I f(x) > c} = (-oo,2- c); open intervals are measurable, and J-l( ( -oo, 2- c))= oo.
u. O c} = (-oo, -y'C) U {1}. IV. 2 < c {xI f(x) > c} = (-oo,-y'C). f is a measurable function.
4.1.3
Problem
Please show the following functions are measurable on their respective domains and calculate J-l(j- 1( ( c, oo] ) ) . 1. f(x)
I
= -,
O<x c}isameasurable subset of E. If A = 0, the proof is immediate since the empty set is measurable. Otherwise, let x EA. Then f(x) > c, and because f is continuous at x, we have 6(x) > 0 so that for z E (x- 6(x), x + 6(x)) n E, f(z) > c. Roughly speaking, given any point in A, we have an interval centered at that point also in A. That is,
A=
U((x- 6(x), x + 6(x)) n E) xEA
=(U(x-b(x),x+6(x))) nE. xEA
In other words, A is the intersection of an open set (measurable) and the measurable set E. The set A is measurable and the argument is complete. • We can weaken continuity to continuity except on a set of measure zero, commonly referred to as "continuous almost everywhere." Sets of measure zero do not affect measurability of a function.
132
4.1.4
LEBESGUE MEASURABLE FUNCTIONS
Definition
A property is said to hold almost everywhere on a measurable set if the set of points where it fails to hold has measure zero. In particular, two functions f and g are said to be equal almost everywhere if they have the same domain and p, ( {x lf(x) -1- g(x)}) = 0. We sometimes write f = g a.e. on E.
4.3 Suppose f and g are extended real-valued functions defined on a measurable set E. Iff is a measurable function onE and if g = f except on a set of measure zero, then g is a measurable function on E. PROPOSITION
Proof Let c be any real number. We must show { x E E I g(x) > c} is a measurable subset of E. Define A = {x E E I f -1- g}. By assumption, A is measurable with measure zero. Then g = f on the measurable set E- A, and {xEE I g(x)>c}={xEE-A I g(x)>c}U{xEA I g(x)>c} ={xEE-A lf(x)>c}U{xEA I g(x)>c} =({xEE lf(x)>c}n(E-A))
U{xEA I g(x)>c}. The set {x E A I g(x) > c} is measurable since it is a subset of a set of measure zero (Problem 3.5.1 ). Because f is a measurable function on E, {x E E I f(x) > c} is a measurable subset of E, as is E-A. • 4.1.5
Problem
Show that iff is a measurable function on a measurable set E and if A is any measurable subset of E, then f is a measurable function on A. Hint:
{xEA I f(x)>c}={xEE I f(x)>c}nA.
Every Riemann integrable function defined on [a, b] is a measurable function on [a, b]. PROPOSITION 4.4
Proof The reader may recall (Theorem 5.1) that a bounded function f on [a, b] is Riemann integrable iff the set D of its discontinuities has measure zero. Then f is continuous on [a, b] - D, hence measurable on [a, b] -D. Define g to be f on [a, b] - D and zero on D. Then g is a
4.1
measurable function on [a, b] ( {x D I f(x) > c} U {xED I g(x) > c} c > 0 or D if c < 0 since g = 0 on measure zero. Thus f is measurable
E [a, b]
MEASURABLE FUNCTIONS
133
I g(x) > c} = {x E [a, b]{xED I g(x) > c} = 0 if
and D ) , and f = g except on a set of • by the previous proposition.
It is not true that measurable functions are Riemann integrable (Problem 4.1.3). Not all functions are measurable functions, but, since nonmeasurable sets of real numbers are "few and far between," we would expect the same for nonmeasurable functions.
4.1.6
Problem
Let N be a nonmeasurable subset of [0, 1] and define f(x) =
1' { -1
xEN
xE[O,l]-N.
'
Show f is a nonmeasurable function on [0, 1]. Note: If I = 1 on [0, 1], If I is a measurable function, f is not. Most operations performed with measurable functions preserve measurability. PROPOSITION
4.5 Suppose f and g are real-valued measurable functions,
defined on a measurable set E, and k is any real number. Then the following functions are measurable functions on E: f+k, kf,
Proof 1.
1/1,/2 , ~(g#O g
on E), f+g, f·g, f(g#O on E). g
We sketch the arguments:
{xEE I f(x)+k>c}={x EE I f(x)>c-k}.
n. k
=
0, then kf
= 0 and {x E E I kf(x) > c}
If k > 0, {x E E I kf(x) > c} reasoning if k < 0.
= {x
E E [ f(x)
=
0, c > 0
{ E, c < 0.
> cjk}. Use similar
134
LEBESGUE MEASURABLE FUNCTIONS
111. {x E £1
lf(x)
I> c} =
c} U {x E E lf(x)
E
{~I
2
IV.
c
E,
{x E Elf (x) > c} = {
c y'c}, c > O.
{xEEig(x)>O},
v.
VI.
{x E £11/g(x) > c} =
c=O
n {x E Elg(x) < 1/c},
c>0
{x E Elg(x) > 0} U {x E Elg(x) < 1/c},
c < 0.
{x E Elg(x) > 0}
U( {x E Elf(x)
{xlf(x) < g(x)} =
< -c}, c > 0.
< rk} n {x E Efrk < g(x)} ),
'k
rk rational. The denseness and countability of the rationals is convenient isn't it. Then {x E E I f(x) + g(x) > c} = {x E E I c- g(x) c} = {x E E If (x) > c} U { x E E I g(x) > c} and {x E E I min (f(x),g(x)) > c} = {x E E lf(x) > c} n{xEEig(x) >c}. {x
2. {xE£11/(x)l>c}=
{ E,
E
Elf(x) > c}
c 1.
3. f(x) = sin(x). If we replace measurability of
f and
g in Proposition 4.5 with
continuity, we still have a valid proposition. The operations performed with measurable functions in Proposition 4.5 in no way distinguish measurable functions from continuous functions. It is in limiting operations on sequences of measurable functions that fundamental differences begin to appear.
4.2
SEQUENCES OF MEASURABLE FUNCTIONS
Pointwise limits of sequences of continuous functions need not be continuous (fk(x) = xk, 0 < x c}
U {x E
=
E
I fn(x) > c}.
n=k
The sequence (Jk) is a nonincreasing sequence of measurable functions and since lim supfk = lim}k = inf{.fi Ji, ... }, measurability follows from part 1. An analogous argument yields measurability for lim inffk· 3. liminffk = limfk = limsupfk· 4. Now suppose a function f onE is the almost everywhere limit of (Jk) and letA = {x E E I limfk(x) not defined or limfk(x) I f(x) }. The set A has measure zero. Define a new sequence offunctions (gk) on Eby
X
E A,
and let g be given by g(x) = {f(x), 0,
xrtA XE
A.
Since each function gk equals a meusurable function, fkl almost everywhere on E, gk is measurable (Proposition 4.3). If x E A, limgk(x) = 0 = g(x). If x rt A, limgk(x) = limfk(x) = f(x) = g(x), that is, limgk =gonE. By part 3 g is measurable onE and then by applying Proposition 4.3 again, f is measurable onE since it is equal tog almost everywhere on E. •
4.2.1
Problem
Calculatefk, ]k, limsupfkl liminfjb and limfk where appropriate, for the given sequences of functions:
O<x< 1. 2. fzk(x) =
k , (1 + 2x)zk
hk-I (x)
=
(1 - k x_ 2
)zk~l
1
,
0 <X
Kn on E'Kn· Each of the sets E kI ,Ek2 , ... , E'k, ... is "almost" E-A. We will n show that
is "almost" E- A and that we have uniform convergence on Ef.
~t( (E- A)- nE'Kn)
=
~t(U( (E- A)- E'KJ)
0 be given. We show 0 < gk < 8 on Recall 1/N < 8. that so N Choose sufficiently large. E'iN = {x E E- A I gKN < 1/N}. But E'iN :J nEKn and gk < gKN for all k > KN, that is, 0 < gk < 8 for all k > KN and all X
E
nEKn· n
•
The proof is complete.
By the way, where in the above argument did we use the assumption J..L(E) < oo? The example, fk = X[k,k+IJ with f = 0 shows J..L(E) < oo is necessary. 4.4.1
Problem
Using Example 4, the reader should repeat the argument of Egoroff's Theorem with fk(x) = x\ 0 < x < 1 and f(x) = 0, 0 < x < 1: gk(x) = xk, 0 < x < 1 and limgk = 0 on [0, 1). Show E k = [0, 1) for all k and E kr = E l. ii. Show E~ = [0, 1/2(!/k)) and K 2 > 277 if, ior example,£= .01. 111. Show E'k = [0, 1/n(l/k)) and (1 - Ej2n)K" < ljn.
1.
(1' 1)
1
n
E~
0
I
~----~~~~1
less than _e ·
2n
X
4.4
ALMOST UNIFORM CONVERGENCE
145
In 1912 Lusin showed that measurable functions are "almost" continuous. We use Egoroff's Theorem to establish this result.
4.3 (Lusin, 1912) Iff is a real-valued measurable function, defined on a set E offinite measure, then we may construct a closed subset Ec of E so that !-L(E- Ec) 0, we have step functions¢ and '1/J, ¢ < f < 'ljJ on [a, b]. so that PROPOSITION
0 < 1b '1/J(x) dx -1b ¢(x) dx
= 1b['I/J(x)- ¢(x)]dx <E .
.___ ___. a= X0
Xj
-
-
-
--
-~
-- -
~-
.______
Xn-
1
x, = b
5.1
THE RIEMANN INTEGRAL
153
1b[ 1/;(x)- ¢(x) ]dx ='--i_
Proof Assume the bounded function f is Riemann integrable on [a, b] and let E > 0 be given. From 5.1.5 and the definitions of le~st upp~r bound and greatest lower bound we have step functions ¢ and 1/J, J < f < {/;, so that bf(x) dx--E = Jb f(x) dx- E < 1b J(x) dx < Jb f(x) dx 2 2 _a a _a 1
rb
E
. ' 2 b' b' b' Thus 0 < fa 1/;(x) dx- fa c/J(x) dx = fa [1/;(x) - ¢(x)] dx < E. Now let E > 0 be given and assume we have step functions ¢and 1/J, ¢ 0 is given. We have 8 > 0 so that for X E (xo- 8, Xo + 8) n [a, b], -E 0. So w1 (x 0 ) = 0. Conversely, suppose w1 (x 0 ) = 0 for some x 0 E [a, b] and let E > 0. Then inf{ w1 (I) I I any open interval containing x 0 } = 0 implies we have an open interval I* containing x 0 so that 0 < w1 (l*) < E. Since I* is open, choose 8 > 0 so that (x 0 - 8, x 0 + 8) c I*. Then X E (xo- 8,xo + 8) n [a,b]===?- E 0. This implies that the set of discontinuities off, say, D, may be written as D = U{x E [a,b]l w1 (x) > 1/n}. n
What do we know about the sets Dn = {x E [a,b]l w1 (x) > 1/n}, n= 1,2, ... ? We claim Dn is a closed set in [a,b], or, equivalently, D~ = {x E [a, b]l w1 (x) < 1/n} is an open set in [a, b]. Let xED~. We will determine a 8 > 0 so that (x- 8, x + 8) n [a, b] c D~. The argument is very close to what we did above. Since w1 (x) = inf{w1 (I) I I any open interval containing x} < 1/n, we have an open interval I* containing x so that wr(I*) < 1jn. Because I* is open, there exists a 8 > 0 so that (x- 8, x + 8) n [a, b] c I* n [a, b]. We are done if we can show (x- 8, X+ 8) n [a, b] c D~. Let z E (x- 8, X+ 8) n [a, b]. Then w1 (z) <w1 ((x-8,x+8)) <w1 (l*) < 1/n;zED~. The preliminaries have been lengthy, but the proof of the main theorem is now fairly easily constructed if we keep this idea in mind: We can cover the discontinuities off with a set of small measure. On this "small" set, f
5.1
THE RIEMANN INTEGRAL
161
may be badly behaved, but f is bounded. On the rest of the interval, a "large" set, f is well-behaved (continuous). Integrability is determined by the behavior off on the "large" set. It's time to begin in earnest. We first suppose f is continuous a.e. on [a, b], that is, the set of discontinuities off, D, has measure zero, and let If I < Bon [a, b]. We will show f is Riemann integrable on [a, b]. Because Dn = {x E [a,bJI w1 (x) > 1/n} is a subset of D, Dn has measure zero. Thus we have Dn c Uh, h open intervals, L:: /(h) < Ej4B. But Dn is a closed and bounded subset of [a, b], hence compact. By Heine-Bore!, we have a finite subcover of Dn by the open intervals, h, that is, m
Dn C
Uik,h ' ' i=l
open intervals. Then the set [a, b] - (hi U h 2 U · · · U hJ is a .finite union of closed intervals 1 1, 1 2 , ••• , h· That is,
Recall that all points of Dn are in hi U h 2 U · · · U Ikm· Thus wf(x) < 1/n on J 1 U J 2 U · · · U h. Can we now define step functions ¢, 1/J in the obvious way: ¢(x) = inf/ on the intervals J1, ... , h, hi, ... , lkm,
1/J(x) =sup/ on the intervals J1, ... , h, hi, .. ·, hm? The problem here is that we need some kind of estimate for supf(x) - infj(x), J;
1;
having only the information that w1 (x) < 1/n on each h Does "pointwise" small, w1 (x) < 1/n, imply "global" small, w1 (Ji) < 1/n, or something like this? Again, we call on Heine-Borel. Since w1 ( x) < 1/ n for each x E h we have an open interval containing lx so that w1 (Ix) < ljn. But then the collection {lx} is an open cover of the closed, bounded, and hence compact set, h A finite subcover must cover h We have a partition of Ji, and the "sup of/"-"inf off" over any of the subintervals of this partition is less than 1jn. This is exactly what we need! (Clearly this does
162
LEBESGUE INTEGRATION
not imply w1 (JJ < 1/n!) Do this for each J,., i = 1,2, ... , L. We have a finite collection of subintervals whose union is J 1 U h u · · · U J L' and on any one of these subintervals, say J*, w1 (J*) = sup{f(x)
I
x E J* n [a, b]} - inf{j(x)
Now define step functions
I
x E J* n [a, b]}
¢, {/; in the obvious way (finite number of
subintervals is critical to being a step function): A
¢ = infj on the subintervals of J1 ,Jz, ... , h and hi, ... ,lkm, 1/J =supf on the subintervals of J1, ]z, ... , h and hi, ... , hm. Then
1
b [1/J(x) A
- <j>(x)]dx 1/n}, if we can show J-L(Dn) = 0, we would be done, since a countable union of sets of measure zero is a set of measure zero. Fix ann, say, N, and consider the set DN
= {x
E
[a, b]l w1 (x) > 1/ N},
with
E
> 0.
Because f is Riemann integrable on [a, b] by assumption, we have step b functions ¢
A
A
A
5.1
THE RIEMANN INTEGRAL
every point of D N is in 'D 1 or a member of the set {a, x 1 , x2, So
... , Xn-1,
163
b}
For I:v I , some point of DN is in each subinterval, i.e., each subinterval ' ' of'D 1 containsapointxwithw1 (x) > 1/N.Butthenw-¢> 1/Nonthis subinterval. Conseq uen tl y,
~>
L
>
~L
(lengths of subintervals that contain points of DN ),
D1
so I:v 1 (xk- xk_ 1) <E. The intervals of 'Dh along with the finite set {a, x 1 , ... , Xn_ 1 , b} contain all points of DN. So J..L(DN) < E, and the proof of this theorem is complete. Congratulations! • We have characterized Riemann integrable functions. We now show that the integral properties that hold for step functions (Proposition 5.1) remain valid for Riemann integrable functions. This theme, step functions first (easiest), and then more general functions as combinations or limits of easier functions, is repeated throughout this chapter. 5.2 If bounded functions f and g are Riemann integrable on [a, b], and k is any real number, then
THEOREM
1. (kf) is Riemann integrable on [a, b], and
1b
(kf)(x) dx =k
1b
f(x) dx
(homogeneous);
2. (f +g) is Riemann integrable on [a, b], and
(additive);
3.
1b
f(x) dx
n = lim
h
Jm .
The argument is as follows: Pick 0 < ¢n ¢n on E. We will show
Because ¢n is nonnegative and simple, we have
where E = U~=l (En Ek) and the nonempty En Ek are mutually disjoint measurable subsets of E. So we must show
5.3
THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS
205
But the integral is additive on the domain (Proposition 5.5). Thus
Our claim will be justified provided we can show lim { m
}EnEk
Jm > ckp,(E n Ek),
If ck = 0, immediate. Suppose ck > 0. Let a be any number between zero and one. (This idea of 0 < a < 1 has been attributed to W. Rudin.) We construct a sequence of sets (Bm) as follows:
Bm = {x
E EnEk I
¢m(x) > ack}.
Bm is measurable, Bm C Bm+l since Jm < ¢m+l> and En Ek = UBm (If p E E n Eb ¢n (p) = ck and consequently limm ¢m (p) > ck > acb i.e., ¢m(p) > ack form sufficiently large, in other words, p E Bm for m sufficiently large.). So the sequence Bm is an expanding sequence of measurable sets that "fills" En Ek. Then the sequence (p,(Bm)) is monotone increasing with (Theorem 3.4). But the sequence UEnEk ¢m) is also monotone increasing with
Therefore,
But this holds for any a between 0 and 1. Thus
and the argument is complete. The reverse inequality is obtained by interchanging ¢m and ¢n. 5.6 The Definitions A and B of the Lebesgue integral of a nonnegative measurable function are equivalent. PROPOSITION
206
LEBESGUE INTEGRATION
By the Approximation Theorem 4.2, we have a monotone sequence of nonnegative simple functions, 0 < ¢n < ¢n+I onE, with Proof
lim(/Jn =f onE, and
{
}E
Jn < }E{ ¢n+I·
We show
Suppose 0 < ¢*
otherwise
2. It's natural to consider monotone decreasing sequences. Can we draw any useful conclusions in this setting. Suppose It > / 2 > · · · . Then -/1 < -h < · · ·, or 0 1.
'
f+ =
0<x 0, IE (kft =IE kf+ = k IEf+ < x and IE(kf)~ = k IE f~ < oo because kf+, kf~ are nonnegative measurable functions (Proposition 5. 7). By definition, (kf) is integrable on E.
Furthermore, IE(kf) = Idkft - IE(kf)~ = k IEf+- k IEf~ = k IE f, where the last equality is the definition off being integrable on E. If k < 0, (kft = (-k)f~, (kft = (-k)f+, IE (kft = -k IEf~ < oo, and IE (kfr = -k IEf+ < oo, that is, (kf) is integrable on E. Again,
2. Since f, g are integrable on E, (Proposition 5.9). Because IE
If I , I g I are integrable on E If I= JE If I+< oo, fE I g I=
236
LEBESGUE INTEGRATION
fEigl+ 0, and J(o,oo) x/(ex- 1) = Jo,oo) 2::: 1 xe-kx, where
x(ex- 1)- 1
Hint:
Jro,oo) xj(ex- 1) = the first equality follows from, "sets of measure zero do not affect the Lebesgue integral".
m. Show
tv. Show
x2
lc
[O,oo) ex - 1
h
1 . L 3 2 k 1
= -
1
x"'
--=((a+ 1)r(a+ 1),
[O,oo) ex - 1
a> 0.
(Zeta-Gamma)
Recall tan- 1 (x) =
lo
x
1
1+
t.2
dt
-1 0 and fk(t) =
cl+l /k {
t
O< t 0, we have
f
lim
fk
=
}[o.oc)
f
Iimfk
}[o,k]
v.
(1
[O,oc)
e-t
2)2 = (lim =lim( -
4
VI.
1 )( 1 ) fk
lim
[O,oo)
r
}[o,oo)
gk
[O,oc)
fk·
r
}[o,oo)
gk)
Appendix A
Cantor's Set
When you have eliminated the impossible, whatever remains, however improbable, must be the truth. -Sir Arthur Conan Doyle
A.1
CANTOR'S SET
The German mathematician Georg Cantor (1845-1918) is regarded as the creator of set theory. Earlier (Chapter 2) we used the "Cantor diagonalization argument" to show the set of real numbers is uncountable. In this appendix, the beautiful Cantor set, a source of so many examples and counterexamples in mathematics, is developed. We begin with a constructive (geometrical) process applied to the interval [0, I]:
Step 1.
252
Divide [0, 1] into three equal parts and remove the "middle
A.1
CANTOR'S SET
253
third", the open interval (1/3, 2/3): 2 3
l
0
3
2 3
1
3
1
1
0
2 3
3
1
The closed set F 1 = [0, 1/3] U [2/3, I] is left. Step 2. Divide the two closed subintervals of F 1 into three equal parts and remove the "middle thirds," the open intervals (1/9, 2/9), and (7 /9, 8/9): 0
1
~
7
8
9
9
9
9
1
•
0
2
1
2
7
8
9
3
3
9
9
•
1 9
•
2 9
•1
•
2 3
3
~
1
•
I
9
•
§ 9·
F2
The closed set F 2
=
[0, 1/9] u [2/9, 1/3] u [2/3, 7/9] U [8/9, 1] is left.
Continuing in this way, after n- I steps we have deleted 1 + 2 1 + 22 + ... + 2n- 2 = 2n-i - I disjoint open intervals, and have left the closed set Fn-i consisting of 2n-i closed intervals, each of length I/3(n-l).
•1
254
APPENDIX A
Step n. Divide the remaining 2n-l closed intervals into three equal parts and remove the "middle thirds." The closed set that is left, Fn, consists of 2n closed subintervals, each of length 1j3n.
A The Cantor set, C, is what's left in [0, 1] after deleting "middle thirds," namely, DEFINITION
The constructive approach is complete. We now give an analytic definition of C. DEFINITION
B
The Cantor set, C, is the points of [0, 1] which have a base three (ternary) expansion without the digit 1. That is, for x E C, ' an= 0 or 2
0
Why two definitions? The idea here is that depending on the property to be discussed, Definition A will be preferable to Definition B and conversely. It's like the statement; Prove 0 = 1r- 1r3 /3! + 1r5 /5! - · · · . Who believes it, but when we recognize this is sin( 1r), with the "unit circle" interpretation of sine and the "power series" interpretation of sine-no problem. Our task at hand: reconcile Definition A and Definition B, the constructive versus descriptive interpretations of the Cantor set. Some discussion of ternary expansions is appropriate. We will show that every number in the interval (0, 1) has at least one, and at most two, ternary expansions. Recall x = .a 1a2 a 3 • • ·an · · · means an = 0, 1, or 2 and 3 X= I: an/3n. The integers 0, I' 2 are called the ternary digits and x = .a 1a2 ···an··· a ternary expansion of x.
r
3
Claim:
Every number in (0, I) has a ternary expansion.
Step 1. Since 0 < x < I, x will be in one of the "thirds": (0, I/3) or [1/3,2/3) or [2/3, 1). So 0 < 3x < 1 or I< 3x < 2 or 2 < 3x < 3. Let a 1 = [3x], where [ ] is the greatest integer function. Thus a 1 = 0 or I or 2
A.1
CANTOR'S SET
255
and a 1 < 3x < a 1 + 1, that is, a1
al
1
3 -
3
3
-~<x 0. What if A is any set of positive Lebesgue measure: JL(A) > 0? (Every subset of a set of Lebesgue measure zero is Lebesgue measurable and has Lebesgue measure zero.) Since 00
A c U(-n,n), JL(A n (-N,N)) > 0 n=i
for some natural number N, and thus An ( -N, N) must be uncountable since any countable set is Lebesgue measurable with measure zero. Select any countable subset of A n (- N, N). This countable set will play the role of the "rationals" in the proceeding construction. Mimic the previous arguments to establish that every set of real numbers with positive Lebesgue measure contains a Lebesgue nonmeasurable subset.
272
APPENDIX B
Over the years attempts having been made to enlarge the class of Lebesgue measurable sets [SS]. The "idea" would be to find a measure >. so that 1. >.(E) defined for all E c R (all subsets of Rare "measurable"); 2. >.([0, 1]) = 1; 3.
>-CQ En) =~>.(En), En mutually disjoint (countable additivity);
4. >.(E +a) =>.(E) if E measurable (translation invariant). As has been shown, Lebesgue measure satisfies parts 2,3, and 4, with requirements 3 and 4 forcing abandonment of 1. How about replacing 3 with
finite additivity, in an attempt to "measure" all subsets of R? We have the following results [Ba]. There exists a set function >.defined for all subsets of R such that 1'. 2'. 3'. 4'.
0 .(E) < oo for all E c R >.([0, 1]) = 1 >. is finitely additive >. is translation invariant.
Then Banach, Hausdorff showed the above result holds in Rn iff n = 1, 2. In 1933 Kolmogorov [Ko] developed probability theory from an axiomatic standpoint with countable additivity being of fundamental importance: We want countable additivity, even at the expense of not being able to "measure" all sets of real numbers.
Appendix C
Lebesgue, Not Borel
The unexamined life is not worth living. -Socrates
C.1
LEBESGUE, NOT BOREL
We show that there exist Lebesgue measurable sets that are not Borel sets: the class of Borel sets, B, is properly contained in the class of Lebesgue measurable sets, M. We will use the PROPOSITION
C.l Iff is a continuous, strictly increasing mapping of R
onto R, then f maps Borel sets to Borel sets. Proof The argument is indirect. Let C be the collection of all subsets of R whose image under f is a Borel set. That is, if A E C, then f(A) is a Borel set. We will show that this collection C is a sigma algebra that contains the closed intervals, i.e., B. We claim Cis a sigma algebra of subsets of R.
Suppose A E C. We claim AcE C. Since f is strictly increasing, f is 1 - 1, and consequently f(Ac) nj(A) = 0. Thus R = f(R) = f(A U A c)= f(A) Uf(Ac), that 273
?74
APPENDIX C
is, R- f(A) = f(Ac). Because A E C, f(A) is a Borel set. Thus R- f(A) = f(Ac) is a Borel set, that is, Ac E C. Next, suppose (Ak) is a sequence of sets in C. Since f(UAk) = Uj(Ak), and each f(Ak) is a Borel set (since by assumption, Ak E C), the countable union of Borel sets is a Borel set, that is, f(UAk) is a Borel set. In other words, UAk E C. Cis a sigma algebra. We now show closed intervals are in C. Suppose I is a closed interval. Then f(I) is an closed interval: f([a, b]) = [f(a),f(b)J (Intermediate Value Theorem, increasing). Thus C is a sigma algebra containing all closed intervals. Because 8 is the smallest such sigma algebra, 8 c C. Borel sets are mapped to Borel sets. We are ready to prove the
C.l There is a Lebesgue measurable set of real numbers that is not a Borel set.
THEOREM
Proof Let
(x) = 0, x < 0 and 1>(x) = 1, x > 1. Form a new function f by defining f(x) = x + 1>(x), x E R. Then because 1> is continuous and nondecreasing on [0, I], f is a continuous, strictly increasing mapping of R onto R: the function f maps Borel sets to Borel sets by the proposition we just proved. Let C denote the Cantor set. We show f( C) 1s a Borel set and p(f( C))= 1. 1. f( C) is a Borel set.
Cis a closed set, Cis a Borel set, and since f maps Borel sets to Borel sets, f( C) is a Borel set. In particular, f( C) is a Lebesgue measurable set and [0, I] - f( C) is a Lebesgue measurable set. What is the Lebesgue measure off( C)? 2. p(f(C)) = 1.
p( /([0, I] - C))
=
p( f(Uh))
=
LP(f(h)) = LP( (ak
=L
=
p( Uj(Ik))
+ 3 a b - ab - 1 · 7r = a b
(23 - ab 7r-
) 1 '
and since ab > 1 + 37r/2, 2/3- 1rj(ab- 1) > 0. We have then that as K---+oo,hr--.Oand[ (f(x+hK) -f(x))/hK [increaseswithoutbound. The function f is not differentiable at x. The last example we present is due to Patrick Billingsley (American Mathematical Monthly, 1982). His argument is developed around the following observation: Iff is differentiable at x, and if aK < x < {JK, (f(fJK)- f(aK))/ then fJK- aK---+0, 0 < fJK- aK, , where ({JK- aK)---+f (x). He then proceeds to exhibit sequences (aK ), ({JK) such that (f({JK)- f(aK))/(fJK- aK) does not have a limit.
Note:
The observation follows from the equation:
Now, suppose fo is defined by fo (x) is the distance from x to the nearest
E.l
AN EVERYWHERE CONTINUOUS, NOWHERE DIFFERENTIABLE, FUNCTION
283
integer: 1 2·
i
------~~---~
-1
1
2
o!
1
---~~
1
2
This function, f 0 , is periodic on [0,1) with period one and vanishes at the integers. Most importantly, fo is linear on the intervals [m/i, (m + 1)/i], [m/2 2 , (m + 1)/22 ], ... , [m/2n, (m + l)/2n], ... where m = 0, ±1, ±2, ... and n = 1, 2, .... We say fo is linear on any interval whose endpoints are successive dyadic rationals. Billingsley's Function:
f(x)
=
f=!o(~:x) . n=O
This function is continuous by Weierstrass M-Test. Fix x. Since we have an integer N so that N < x < N + 1, 0 < x - N < 1, we may as well assume x E [0, 1). Considering dyadic expansions, we have: a1
=0
or 1.
ai
So, let
= 0 or 1.
284
APPENDIX E
Now form the difference quotient
Consider the first sum: 0 < n < K - 1.
and
But fo is linear on [m/2K-n, (m 0 < n < K- 1. Thus
+ 1)/2K-n] as long asK- n >
I:J(2n,8K)- fo(2na.K) 2n(,8K- a.K) n=O
=
1, that is,
I:±l. n=O
The second sum vanishes because 2n,8K = (m + 1)/2K-n and 2na.K = (m/2K-n) are integers forK < n, i.e., n > K. So, if differentiable at x, we would have the nonsense that f (x) = I; 0 ± 1 ! This concludes our treatment of continuous, nowhere differentiable, functions. I
References
[AJ] [A 2 ] [AB] [Ba] [Be] [Bel] [Bo] [Boy] [Bu] [BK] [Ch] [CJ] [CR] [DH] [Dud]
Apostol, T.M. (1966). Calculus. Waltham: Blaisdell Publishing. Apostol, T.M. (1974). Mathematical Analysis. Reading, MA: AddisonWesley. Asplund, E., & Bungart, L. (1966). A First Course in Integration. New York: Holt, Rinehart, and Winston. Banach, S. (1923). Sur le probleme de Ia mesure, Fund. Math. 4. Beckmann, P. (1971). A History of7r. New York: St. Martin's Press. Bell, E.T. (1937). Men of Mathematics, Simon Schuster, 1937. Boas, R.P. Jr. (1960). Primer of Real Functions. Mathematical Association of America. Boyer, C.B. (1985). A History of Mathematics, Princeton University Press, 1985. Burkill, J.C. (1953). The Lebesgue Integral. Cambridge: Cambridge University Press. Burrill, C.W. & Knudsen, J.R. (1969). Real Variables. New York: Holt, Rinehart, and Winston. Chae, S.B. (1980). Lebesgue Integration. New York: Marcel Dekker. Courant, R. & John, F. (1965). Introduction to Calculus and Analysis, Vol. I, New York: Wiley Interscience. Courant, R. & Robbins, H. (1980). What is Mathematics?, Oxford: Oxford University Press. Davis P.J., & Hersh, R. (1981). The Mathematical Experience, Birkhauser. Dudley, R.M. ( 1989). Real Analysis and Probability, Belmont: Wadsworth. 285
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REFERENCES
Dunham, W. (1990). Journey Through Genius, New York: John Wiley & Sons. (Ed g) Edgar, G.A. (1990). Measure, Topology, and Fractal Geometry, New York: Springer-Verlag. [cEdw] Edwards, C.H. Jr. ( 1979). The Historical Development of the Calculus, New York: Springer-Verlag. [hEdw] Edwards, H.M. (1969). Advanced Calculus, New York: Houghton Mifflin. [EG] Evans, L.C., & Gariepy, R.F. (1992). Measure Theory and Fine Properties of Functions, Boca Raton: CRC Press. Folland, G.B. (1984). Real Analysis, New York: John Wiley & Sons. [F] [Fl] Fleron, J.F. (1994). A Note on the History of the Cantor Set and Cantor Function, Magazine, April. French, A.P. ed., (1980). Einstein: A Centenary Volume, Harvard [Fr] University Press, 1980. Gelbaum, B.R., & Olmsted, J.H. (1964). Counterexamples in Analysis, [GO] San Francisco: Holden-Day. Goldberg, R.R. (1963). Methods of Real Analysis, Waltham: Xerox. [Go] Halmos, P.R. (1962). Measure Theory, Princeton: Van Nostrand. [Hal] [Har] Hardy, G.H. (1967). A Course of Pure Mathematics, Cambridge: Cambridge University Press. Hobson, E.W. (1959). The Theory ofFunctions ofa Real Variable, Vol. 1 [Ho] and 2, New York: Dover. Klambauer, G. (1979). Problems and Propositions in Analysis, New [KJ] York: Marcel Dekker. Klambauer, G. (1973). Real Analysis, New York: American Elsevier. Kline, M. Mathematical Thought from Ancient to Modern Times, Oxford University Press, 1972. Kolmogoroff, A.N. (1956). Foundations of the Theory of Probability, [Ko] 2nd ed., ed. Nathan Morrison, Chelsea. Korner, T.W. (1989). Fourier Analysis, New York: Cambridge Uni[KoJ versity Press. Kramer, E.E. (1982). The Nature and Growth of Modern Mathematics, [Kr] Princeton, NJ: Princeton University Press. [Lan] Landau, E. (1960). Foundations of Analysis, New York: Chelsea Publishing. Larson, L.C. (1983). Problem-Solving Through Problems, New York: [Lar] Springer-Verlag. May, K.O. (1966). Henri Lebesgue, Measure and the Integral, San [Ma] Francisco: Holden-Day. McLeod, R.M. (1980). The Generalized Riemann Integral, Mathema[Me] tical Association of America.
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Index
A
almost everywhere 132 almost uniform convergence 141 Archimedes 7, 25 axiom of choice 266 B
Banach 100, 272 Bernstein 266 Billingsley 282 Billingsley's function 283 Bolzano 66, 281 Bolzano-Weierstrass theorem 66 Borel20, 21, 23, 26, 91, 112, 113 Borel measure 21 Borel set 112,113,114 Borel sigma algebra (Borel a-algebra) 112 bounded set 42
c Cantor 39, 48, 49, 280, Appendix A Cantor function Appendix A Cantor set Appendix A Caratheodory's measurability criteria or condition 100 288
Caratheodory's theorem 105 Cauchy 15, 16, 25, 56 Cauchy integral 16 Cauchy sequence of real numbers 56 Cavalieri 10, 25 Cellerier 281 characteristic function 13 7 closed interval 63 closed set 63 collection of subsets of a set 34 compact set 64, 65, 70, 71 complement of a set 33 "condensation of singularities" 280 continuous function 66, 67, 68, 69, 70, 71,72,73,74,79, 126,156,261 almost everywhere 131, 159 convergence theorems 237 convergent sequence of functions 75 of real numbers 51, 52 convergent series 57 countable additivity 88, 89, 272 countable set 37 countable subadditivity 95 countable union of sets 40
INDEX
D
Darboux 20, 26 Darboux sums 20 De Morgan's laws 34 difference of sets 33 differentiable function 73, 74 Dirichlet function 27 discontinuous function 67, 68 disjoint sets 33 divergent sequence of real numbers 51 divergent series 58 domain of a function 36 E
Egoroff 141 Egoroffs theorem 141 empty set (0) 32 equivalent sets 37 Eudoxus 3, 25 Eudoxus' axiom 4 everywhere continuous nowhere differentiable function Appendix E extended real numbers 49 F
Fatou's theorem (lemma) 222, 237, 240 Fermat 10,25 finite set 37 function 35, 36 Billingsley's 283 Cantor Appendix A characteristic 13 7 continuous 66, 67, 68, 69, 70, 71, 72, 73, 74, 79,126,156,261 continuous almost everywhere 131, 159 differentiable 73, 74 Dirichlet 27 discontinuous 67, 68 domain of36 everywhere continuous nowhere differentiable Appendix E inverse image of 36
289
Lebesgue integrable 181, 182, 183, 224,225,233 Lebesgue measurable 126, 127, 132, 133, 134, 135, 138, 141, 145, 196 monotone 138, 158, 260, 261 nonnegative simple 198 one-to-one 36 onto 36 range of 36 Riemann integrable 152, 153, 154, 156, 158, 159, 163, 168, 170 Riemann's 18 simple 137, 138 step 148, 149 uniformly continuous 71, 72 Weierstrass' 281 fundamental idea of Lebesgue 23, 184 Fundamental Theorem of Calculus (FTC) 12, 170, 172 G
GLB43 greatest lower bound of a set 42 greatest lower bound property (GLB) 43
H Hankel280 Harnack 93 Hausdorff 100, 272 Heine-Bore! theorem 65, 91 Heine's theorem 72 Hippocrates 5 I
infimum of a set 42, 43, 50 infinite set 37 integral Cauchy 16, 25 Darboux 20, 26 Jordan 21, 22, 26 Lebesgue 22, 23, 26, 147, 173, 175, 181, 193, 194, 198, 224 lower and upper 22, 179, 180 properties of 189, 206, 235
290
INDEX
Newton-Leibnitz 12, 25, 27 Riemann 17, 26, 147, 148, 152, 192 lower and upper 20, 150 properties of 163 Young 22, 23 intermediate-value theorem 70 intersection of sets 33, 34 interval closed 63 length of 90 open 62 inverse image of a function 36 J
Jordan 20, 21, 26 Jordan integral21, 22, 26 lower and upper 21 Jordan measure 21 K
Kolmogorov 272 L
LDCT 29,239 least upper bound of a set 42, 50 least upper bound property (LUB) 43 Lebesgue 22, 23, 26, 159 Lebesgue dominated convergence theorem (LDCT) 29, 239 Lebesgue, fundamental idea 23, 184 Lebesgue integrable function 181, 182, 183, 224, 225, 233 Lebesgueintegral22,23,26, 147,173, 175, 181, 193, 194, 198,224 lower and upper 22, 179, 180 properties of 189, 206, 235 Lebesgue integration 147 Lebesgue measurable function 126, 127, 132, 133, 134, 135, 138, 141, 145, 196 Lebesgue measurable set 100, 101 structure of 120, 122 Lebesgue measure 23, 87 Lebesgue monotone convergence theorem (LMCT) 211, 237 Lebesgue nonmeasurable set 101,
Appendix B Lebesgue, not Borel, set 23, 114, Appendix C Lebesgue outer measure 93, 95, 99, 105 Lebesgue sums 23, 24 Leibnitz 12, 13, 14, 25 length of an interval 90 Levi 211 limit of a sequence of functions 75 of a sequence of numbers 51 of a sequence of sets 35 limit inferior of a sequence of functions 135, 222 of a sequence of numbers 53 of a sequence of sets 35 limit point of a set 63, 66 limit superior of a sequence of functions 135 of a sequence of numbers 54 of a sequence of sets 35 Lindelofs theorem 64 LMCT 211,237 lower Lebesgue integral 179 lower Riemann integral 20, 150 LUB43 June 5, 7 Lusin's theorem 145 M mean value theorem 74 measure 87, 89, 221 Borel2l Jordan and Peano 20, 21 Lebesgue 23, 87 "measuring sets" 115 method of exhaustion 3 monotone function 138, 158, 259, 261 monotonically decreasing sequence of sets 34 increasing sequence of sets 34 mutually disjoint sets 34 N Newton 12, 15, 25
INDEX
nonnegative simple function 198 0 one-to-one (I -1) function 36 onto function 36 open cover of a set 64 open interval 62 open set 62 p Peano 20, 26 Peano measure 20 R
range of a function 36 rearrange double series 60 series 58 rearrangements 2 reflection of a set 43 Riemann 17, 26, 192 Riemann integrable function 152, 153, 154, 156, 158, 159, 163, 168, 170 Riemann integral!?, 26, 147, 148, 152, 192 lower and upper 20, 150 properties of 163 Riemann's function 18 Rolle's theorem 73 Rudin 205, 212
s "scaling" 3 Schoenberg 276 sequence of functions almost uniform convergence 141 convergent 75 limit inferior 135, 222 limit of75 limit superior 135 measurable 135 Riemann integrable 168 uniform convergence of 77, 79, 82, 168 of numbers
291
Cauchy 56 convergent 51, 52 divergent 51 limit inferior 53 limit of 51 limit superior 54 of sets limit inferior 35 limit of 35 limit superior 35 senes convergent 57 divergent 58 set 32 all subsets of 34 Borel 112, 113, 114 Borel sigma algebra (Borel aalgebra) 112 bounded 42 bounded above 42 Cantor Appendix A closed 63 compact 64, 65, 70, 71 complement of 33 countable 37 De Morgan's laws 34 difference 33 disjoint 33 empty 32 equivalent 37 finite 37 GLB43 greatest lower bound of 42 greatest lower bound property (GLB) 43 infimum 42, 43, 50 infinite 37 intersection 33, 34 least upper bound 42, 50 Lebesgue measurable 100, 101 Lebesgue nonmeasurable Appendix B Lebesgue, not Borel 23, 114, Appendix C Lebesgue outer measure of 93, 95, 99
292
INDEX
limit point of 63, 66 LUB43 member of32 mutually disjoint 34 nonempty 32 open 62 open cover of 64 reflection of 43 subset 32 supremum 43, 50 uncountable 37 union of 33 upper bound of 42 sigma algebra (O"-algebra) of sets 103 simple function 137, 138 Solovay 266 space-filling curve Appendix D step function 148, 149 supremum of a set 42, 50 Suslin 23, I 14, Appendix C T ternary expansion 254 theorems approximation of measurable functions 138 Bolzano-Weierstrass 66 Borel sets are Lebesgue measurable 113 Cantor 48 Caratheodory I 05 characterization of Lebesgue integrability 187 characterization of Riemann integrability I 59 convergence of sequences of real numbers 56 Egoroff 141 Fatou 222, 237 Fundamental Theorem of Calculus 170, 172 Heine 72 Heine-Borel 65 Intermediate-value 70 Lebesgue characterization of Riemann integrability I 59
Lebesgue dominated convergence 239 Lebesgue monotone convergence 211, 237 Lebesgue, not Borel Appendix C Lebesgue outer measure 95, 105 Lindeloff 64 Lusin 145 Mean value 74 "measuring sets" 115 Riemann integrability implies Lebesgue integrability 181 Rolle's 73 structure of Lebesgue measurable sets 120, 122 structure of open sets in R 62 Weierstrass M- Test 82 translation invariance 87, 95
u uncountable set 37 uniformly continuous function 71, 72 uniformly convergent sequence of functions 77, 79, 82, 168 union of sets 33 upper bound of a set 42 upper Lebesgue integral 22, 179, 180 upper Riemann integral 20, 150
v Van Vleck 266 Vitali 23, 99, 105, Appendix B Volterra 20
w Weierstrass 66 continuous nowhere differentiable function 281 Weierstrass M-Test 19, 82 y
Young 22,23
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