Measure Theory (Including Banach Space, Hilbert Space, Spectral Theory)

Mathematical Analysis- m

'MEASURE THEORY Space, Hilbert Space, Spectral Theory) [For Honours and Post-Graduate Students]

(Including Banach

By

KESHA WA PRASAD GUPTA B. Sc. HO;IS., M.Sc. Department of IIfathematic!>, P.P.N. College, Kanpur.

KRISHN A Prakashan Mandir 11.

Shivaji Road.

Meerut-250001 (U. P.) INDIA.

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MEASURE THEORY Thirteenth E dition 1994-95 All rights reserved by the author.

Price Rs. 55'00 only. Published by KRISHNA Prakashan Mandir, 11, Shivaji Road, Meerut-l (U.P.) India and Printed at Fine Art Press, Meerut.

PREFACE

How to define the measure of a set in a pure Mathematics is a modern concept and it is needless to state that there is hardly a book by any Indian author on this topic. of writing such a book under the

Hence the necessity

caption,

'The Theory of

Lebesgue Measure and Integration' was taken because of the huge and unbearable cost of foreign books on the subject. The subject being very modern in

nature, all needed defini­

tions have been given at the beginning of the chapters, and though apparently the volume looks

very thin but it

covers the entire

course as laid down in various Indian Universities. It is needless to emphasise that the style adopted in the book is lucid, cleat, easy and clearly understandable to

the students. A

good many solved and unsolved examples have been given in every chapter so that students may have enough practice in the subject. I am grateful to Dr. (Mrs.) P. Srivastava, matics,

B. H. U.,

Dr. S. N. Lal

and

Reader in Mathe­

Sri Ram Bechan Ram

Lecturers in Maths., B. H. U., for their valuable guidance. also thankful to my friends and well

I am

wishers Dr. M. P. Jaiswal,

Prof. R. G. Gupta, S. P. S. Bhadauria for their valuable sugges­ tions.

Thanks are also due to the printers and publishers

who

took pains in bringing out this work. This book may have some defects and

I shall thankfully

receive constructive suggestions for its improvement from all those interested in studying the subject. Kanpur

K. P. GUPTA PREFACE TO THE THIRTEENTH EDITION

The present edition has been thoroughly revised and enlarged. This edition solves the purpose of Measure Theory and Functional Analysis (Modern Analysis) both. All the chapters have also gone

Substantial changes.

An attempt has been made to remove any

type of mistakes of previous edition. K. P. GUPTA

'l'f''f'!ff'flm'l fircrivr~ I fet!' CltTT"~ ~~ ~~~ II f~ f~~U( fiffet~~ fifU~ I f~q"~mn~m ~s~ II

We adure you the guardian of the south-east quarter and Ruler of the whole uni~'erse, eternal bliss personified, the omnipresent and all pervading Brahma manifest in the form of Vedas. We worship Lord Shiva shining in His own glory, devoid of material attributes, undifferentiated, desire less, all pervading consciousness, having nothing to wrap about hifl1se/f except ether.

- Authors and Publishers

Contents Chapters

1. 2. 3. 4. 5. 6.

Basic concepts of set Basic set operations Functions and sequences Bounded. derived, open and closed sets on the real line Countability of sets Measure and outer measure Ring of sets, a-ring of sets Algebra of sets a-algebra of sets. Finitely additive Translation invaJ iant Postulates for all ideal measure function Measure space Completeness of measure function Caratheodory's postulate for outer measun: Measurable set Problems related to measure functions Problems related to ring of sets, a-algebra

7.

Lebesgue measure of a set Measure of open and closed intervals Measure of rectangle and a parallelopiped Exterior and Interior measure Some Theorems and solved problems Equivalence of Lebesgue-exterior measure and Caratheodory outer measure Fundamental Theorems Limiting sets Covering in the sense of Vitali Vitali's covering Theorem Sets of the Type Fa and Gil Borel measurable set Unsolved problems

8.

Measurable functions Measurable functions. Almo,l cverywerc Equivalent functions Characteristic function Simple function. Limit supelior and limit inferior Some Theorems and solved problems Lebesgue measurable functions Borel mensurability of functions Little Wood's three principles

Pages

1 6 11 16 21 46 46

47 47 49 49 50 50 50 51 51 63

73 73 74 74 75 95 99 100 106 107 110 110 III

115 lI5 115 116 117 118 136

136

( ,i ) Chapters 9. The Lebesgue integral of a function Lebesgue integral of a function Lebesgue integral of an unbounded function First mean value Theorem Countable additivity property of the integral

10. Theorem on convergence of sequences of measurable Functions Convergence in measure F. Reisz's Theorem D. F. Egor's Theorem Lebesgue bounded convergence theorem Lebesgue's dominated convergence theorem Beppo-Levis's Theorem Faton's Lemma

11. Absolute continuous functions. Indefinite integral and Differentiation Continuous function, Absolute continuous function Indefinite integral. Dift'erentiable Increasing and decreasing functions Functions of bounded variation Lebesgue point Problems related to functions of bounded variation Problems related to absolute continuous functions Problems related to indefinite integral Fundamental theorem of integral calculus Problcms related to Lebesguc point of a function

12.

V-Space Conjugate number, LP-space Norm of an elemcnt of LP-space Convergence sequence. Cauchy sequence Completeness of Ll'-space, Banach space Some theorems Holder's inequality Minkowski's inequality Schwarz's inequality LP-space is a normcd linear space Reisz Fisher theorem Some theorems

13.

Further theorems on Lebesgue Integration Integration by parts Second mean value theorem; Relation between Lebesgue integral and Stieltjes integral Cumulative distribution function Lebesgue-St ieitjes integral

Pages

140 140 144 147 150

176 176 178 179 181 183 185 186

194 194 195 195 195 196 197 205 214 219 220

225 225 225 226 226 226 229 231 232 237 238 241

246 246 248 250 254 255

(

vii

)

Chapters 14. The Weierstrass approximation theorem and semicontinuous function Bernstein polynomial S. N. Bernstein's theorem Weierstrass approximation theorem Semi-continuous function Theorem on semi-continuous functions

15. Signed measure

Pages

256 256 256 260 261 262

265

Signed measure. Positive and negative sets 266 Theorems on positive and negative sets 267 Hahn decomposition theorem 272 Singular measures 273 Jordon decomposition. Absolutely continuous measure function 274 Radon Nikodym theorem 274 Lebesgue decomposition theorem 277

16. Product measure

280

17. Fourier series

284

Periodic function. Trigonometric series 284 Finite discontinuity 284 Fourier series and Fourier Coefficients. Even and odd functions 285 Parsevel's identity for Fourier series " 292 Riemann--Lebesgue Theorem 294 Dirichlet's integral 295 Summation of series by arithmetic means 297 Fejer's integral 297 Summability of Fourier series. Fejer Theorem 298

18. Banach space

301

19. Hilbert Space

352

20. Finite Dimensional Spectral Theory

416

SYMBOLS AND THEIR MEANING Symbol 3 =>

iff & ¥ E

rf; :::>

c o N R R+ Q Def.

Q+ s.t. U

n

e.g. w.r.t. an~a

!no (S)

me (8) mi (8) In

(8)

L-integrable R-integrable E (f(x) > a)

t

f(x) dx

a.e.

Meaning there exist or there exists implies is implied by implies and is implied by if and only if and for every belongs to does not belong to is a superset of ·is a subset of null set Set of natural numbers Set of real numbers set of positive real numbers set of rational numbers Definition set of positive rational numbers such that Union intersection is cardinally equivalent to sequence consisting of points a" a2. a3, a4, ... sequence consisting of points at, a2, 03, ... for example with respect to the sequence converges in limit to a outer measure of a set S Exterior measure of a set S Interior measure of a set measure of a set S Integrablelin the sense of Lebesgue Integrable in the sense of Riemann {x E E : f(x) > a}

Lebesgue integral orr(x) over E

AI

almost everywhere complement of A

Tx

T(x)

1 Basic Concepts of Set The basic idea of a set is the origin point in the study of Modern Algebra. 1'0. Der. A set is a collection of well distinct objects. The objects of a set are called the elements or members of that set and their membership is defined by certain conditions. The elements of a set can be anything: mango, pen, sun, moon, river etc. Examples. (i) The set of interior points of a circle of unit radius with centre at the origin. (ii) The set of pen, ihkpot, pencil. (iii) The set of Ram, Mohan, Shyam, Bhole Shanker. (iv) The students of Banaras Hindu University. (v) Collection of letters a, b, c, d, e forms a set. Remark. (i) Set and aggregate both have the same meaning. (ii) The elements of a set must be distinguished from one another. 1'1. Notation. S~ts are usually denoted by capital letters A, D, C, D, E, P, Q, X, ... and their elements are denoted by corresponding small letters

a, b, c, d, e, p, q, x, ... It is not necessary that the elements of a set A are denoted by a. We can also write the symbol b for the general element of the set A. If a is an element of a set A, then this fact is denoted by the symbol a E A. The symbol E is used for anyone of the following phrases: (i) belong to. (ii) belongs to. (Iii) is in. (iv) are in. "belongs to", determined from the context "a E A" is read as "a belongs to A". If a is not a member of A, then we shall write a rt. A

2

BASIC CONCEPfS OF SET

It is customary to put a vertical line "I" or. an oblique "I" througt. a symbol to indicate the opposite meaning of the symbol. If we write "Let x E X", then the meaning of this is "Let x be an element of X". Remark. (i) It should be noted that order is not preserved in case of set, where as the Older is necessarily preserved in case of a sequence. That is to say, each of the sets {I, 2, 3}, {3, 2, I}, {t, 3, 2} denotes the same set. (ii) The repetition of an element does not change the nature of a set, i.e., each of the sets {I, 2, 2, 3}, {I, 3, }, 2}, {I, 2, 2,3, 3, 3} {t, 2, 3} denotes the same set. 1'2. Set of sets. It may happen that the elements of a set are sets themselves. A set consisting of a number of sets is called set of sets. Set of set,; is also expr.:ss·ed by saying "family of sets" or "class of sets". In order to make a distinction between a set and a family of sets, we use bold letters. A, D, C, P, ...... in case or family of sets. For example {A, B, C, D, E} is a family of sets. 1'3. Subset. If every element of A is also an element of a set B, then A is called a subset of B. This relationship between A and B is denoled by the sFmbo I A C R and is read as "A is a subset of B". or "A is contained in R". "Let A C !l" is read as .. Let A be a suhset of B".

Symbolically A is called a sub,et of B iff any a E A ~ a E B. Examples. (i) {a, {b}, {c}, {b, c} are subsets of the set la, b, c}. (ii) Evidently any a E A ~ a E A. Hence, by def. of subset, A C A for every set A. 1'4. Super set. A C B is also expressed by writing R::> A and is read as "B contains A" or "B is a superset of A". 1'5. Equality of sets. Two sets A and B are said to be equal if every elemeut of A belongs to B and every element of R :l1ongs to A. This relationship between A and B is denoted by the symbol A= B.

B.4.SIC CONCEPTS OF SE r

3

Sy,mbolically A=B iff any x E A ~ x E B. Alternately A C B, B C A {:> A=B. If A=B is not true, then we write A::;eB. 1'6. Proper subset. A is called a proper subset of B if A c B and A::;eB and this relationship is denoted by the symbol ..4 ~ B and is read as "A is a proper subset of B". From this definition it follows that every element of If is an element of Band B contains at least an element which does not belong to A. Examples. (i) {a, b} is a proper subset of la, b, c}. (ii) Since A C A and A =A and therefore A is not a nroper subset of A, i.e., A k A is not true for every set A 1'7. Finite set. A set consisting of a finite number of elements is called a finite set. Example. {a, b}, {I, i, 3}, {a1' a2, aa, ...... , a.} are aU finite sets. 1'8. Infinite set. A set consisting of an infinite number of elements is called an infinite set e.g., N, Q, R, Z, ... are all infinite sets. 1-9. Null set. A set containing no element is called a null set and is denoted by symbol.p. It is also called empty set or void set. Examples. (i) {x: x:;t:x1=~. For x::;ex is not possible for any element x. (ii) {x E R: X2 < O}=.fo. For the square of any real quantity ;;iii 0 and so x· < 0 is impossible for any x E R. Remark. There will be only one empty set and so we shall always speak of the empty set instead of an empty set. 1'10. Power set. The family consisting of all subsets of a set is called the power set ot'that set and is denoted by P, e.g. {B: B C A} is the power set of A and so we write P={B: B C A}.

Evidently the value of P varies with the value of A, i.e., P depends upon A and so we denore the power set of A by the symbol P (A). Finally, P (A)={B : B C AJ.

4

BASIC CONCEPTS OF SET

e.g. it A={l, 2, 3}, then

P (A)={ {I}. {2}, {3}, {I, 2}, {2, 3}, {3, I}, {I, 2, :}, q, }

Ex. 1. Prove that, C A for every set A. Solution. Since, does not contain any element and therefore the condition any x E ' " x E A is fulfilled and hence the result follows. Ex. 2. If a set A contains n elements, then P (A) contains 2" elements. Solution.

The total number of elements of P (A)

=nCO+nC1 ~-"C2+···+nc" =(1 + I)", by Binomial expansion

=2". 1'11. Universal set. If all the sets under consideration are subsets of a fixed set, then this fixed set is called universal set or universe of discourse and is denoted by the symbol X or by U. Singleton set. A setjconsisting of only one element is called a singleton set e.g. each 6f the following sets is a singleton set {a}, {b}, {I}, {2}. 1'12. Indexed set and Index set. Let At be a non-empty set for each r in a set 6. In this ca~e the sets AI' A 2 , 04 3 , ...... , A. are called indexed sets and the sets 6.={l, 2,3, ...... , n} is called irdex set. Here the suffix r E II of Ar is called all index. Such a family of sel." is denoted by {Ar : r E 6.} or {Ar} r E 6. Example. Let A 1 ={a, b, c}, A 2 =:c, d, /, m}, A3={m, n, q}, Ac={p, q, r, sl, A.;=(a, I, d, f}, 6={1, 2, 3, 4, 5}.

Here we find that 3 a non-emp y set At "f r E 6.. Hence 6 is called an index set and the sets AI' A!, A3 , AI' A; are c:.llied indexed sets. 1'13. Heriditary property. A nOll-empty family A={Ar} of sets is said to be heriditar y if Ar C A., A, E A => Ar E A. 1'14. Definition. A family {A.} of sets i5 sa id to be p.·rwise disjoint if

Ar n

..4.=,

¥

r, sEll S.t. r=l=s.

BASIC CONCEP rs OF ~ ET

Problems Set 1. Prove the following: (i) A C t/J => A=t/J (ii) A c B, B C C ,. A C C. 2. Are the sets ~, to} and {t/J} different? 3. If A={l, a}, then find P (A). 4. State whether each of the following statements is correct or incorrect. Here S is any set and S::;6~. (i) S E P (S), (ii) S C P (S), (iii) {S} E P (S). Answer.

3. 4.

{t/J, A, {I}, (i) correct.

{a} }

(ii) incorrect.

(Iii) incorrect.

2 Bas ic Set 0 perations Introduction. From arithmetic we know that(0 a+b is defined as the sum of the members a and b. (ii) a-b is defined as the difference of a and b. (iii) ab is defined as the product of a and b. (iv) ~ is defined as division of a and b, provided b#O in the last-case. In a similar fashion we will define the operations of union, intersection, difference, product in case of sets. Union. The union of two sets A and B, defined by A U B, is defined as the set of those elements which either belong to A or to B. Symbolically, A U B={X : x € A or x € B} A U B is usually read as 'A union B'. The following have the same meaning: union, sum, logical sum, join. The union of a finite number of sets A10 AI.· .. · .. , A,. is denoted by fl

Al U A2 U ............... U A,.

or by U Ar

,=1

CD

Similarly Al U A. U .......... U A .. U ...... =

U A, •

•

Also U Ar={x: x E Ar for some value of r, I ~ r ~ n}. '=1

In this diagram the shaded portions represents A U B. Such type of diagram is called Vend diagram. Examples. (i) P the set of points which lie on the line AB and Q that of the line CD, then P U, Q is the set of those points which lie either on AB or CD.

~IC SET OPlRATION3

If A={l, 2}, B={a, b, c, I}, then A U B={l, 2, a, b, c, I}, {2, a, b, c, I}={ I, 2, a, b, c}. Remarks. (i) Some writers write A+B in place of A U B. (ii) The meaning of "for some element a E A" or "for at least one element a in A" is the same. Intersection. The intersection of two sets A and B, denoted by A n B, is defined as the set containing those elements which " belong to A and B both. Symbolically, A n B={x : x E A and x E B} A n Breads" A intersection B". The intersection of a finite number of sets AI' A2 ,· ........ A.. is denoted by (ii)

,. or by n Ar • r-l

n

By def.

n Ar ={x : x E Ar

r=l

"f r S.t.

1

~ r ~ n}.

Examples. (i) If A={a, b, c}, B={l, ?, a}, then A n B={a}. (ii) If A={l, 2, 3}, B ={c, d, e}, then An B=t/I. In the diagram the shaded portion represent A n B. Remarks. (i) Some writers write A.B for A n B. (ii) The following words have the same meaning: (iii) The phrases ''for every r" and ''for all values of r" both have the same meaning. (iv) It is ea-;y to prove that ,A n B=B n A AU B=B U A (A U B) U C=A u (B U Cj (A n B) n C=A n (B n C) (v) From the definition it is quite obvious that and 'X E A n B ~ x E A and any x E A n B x E B so that A nBC A, A nBC B.

8

BASIC SET OPERA TJONS

(vi)

Evidently any a E A => a E A U B. and any a E B ~ a E A u B. so that A C A U B, B C A u B. Observe the difference between (v) and (vi). Disjoint sets. Any two sets A and B are said to be disjoint sets if their intersection is ,p. More precisely A and B are disjoint sets if A n B=t/I, e.g. {l, 2}, {a, b} are disjoint sets. Difference or sets. The difference of a set A w.r.t. a set Bis a set which contains only those elements of A which do not belong to B and is denoted by the symbol A-B. Symbolically A-B={x: x E A and x f1. B}. Similarly we define B-A={x : x E B and x ft. A}. Example. (i) If A={l, 2, 3, a, b}, B={a, b, c, d}, then A-B={l, 2, 3}, B-A={c, d}. Complement of a set. The complement of a set A w.r.t. the universal set X is defined as the set X -A and is denoted by X'. Symbolically, A'=X-A={X : x E X and x It is also clear that any x E A' ~ X ft A. Remarks. (i) X'=rjl For X--X=rP (ii) ,p'=X. For ",'=X-,p=X (iii) A U A'=X, A n A'=c!> (iv) A-B=A n B', B-A=B n A' For A-B={x: x E A, x ft B} ={x : x E A, x E B'} =A n B' Similarly B-A=B n A' (v)

ft

A}

(A,),=A

=> x E A or x E B x rt. A U B => x rt. A and x (/. E, Symmetric of difference sets. The symmetric difference of two sets A and E, denoted by A 6 B, is defined as A A R=(A--B) U (B-A)

(vi)

x E A U B

9

8ASIC SET OPERATIONS

Example.

. If A={l, 2, 3, 4}, B={3, 4, 5, 6} then A-B={I, 2}, B-A={5, 6}, A 6 B=(A.-B) U (B-A)={1,2} U {5,6}

={l, 2, 5, 6}. Distributive law. The union of sets is distributive w.r.t. the intersection of sets and conversely, I.e. (A U B) n C=(A n C) u (11 n C) (A n B) u C=(A u C) n (11 u C). De-Morgan's law. Let AIX be a non-empty set for each IX in an index set 6.. Then (i)

(

u AIJ. )' = ( () ~IX' ) (I.E 6 «ELl

( () ArJ. )' = U Aac' (I.E 6 IXE6 For proof refer to author' book on "Set theory and Related Topics." Ordered pair. An element of the form (a, b) is called an ordered pair. The elements a, b in the ordered pair (a, b) are called the first element and the second element respectively. Equality of ordered pairs. Let (a, b) and (c, d) bc any two ordered pairs. We define (a, b)=(c, d) iff a=c, b=d i.e. (a, b)=(c, d) a=c, b=d. Product of sets. Let A and B be any two given sets. The product of A, B is defined as the set consisting of all ordered pairs (a, b) where a E A, b E B and is denoted by the symbol A x B. Symbolically, AxB={(a, b): a E A, b E B} A X 11 is read as "A cross .B" . The product set A x B is sometimes called Direct product or Cartesian product. The reason for the name Cartesian product is that the set A X B can be plotted in a Cartesian plane. Similarly BXA={(b, a) : b E B. a E A} (ii)

rr

J'

_ cJ ___ ~{(L

bl

_____.fb n ..,.. .----- A

-

------::::

)(

10

BASIC SET OPERATIONS

Evidently A x B::j:.B xA. If either B or A equal to ~, then AxB=t/J. Example. If A={I, 2}, B={a, b, c}, then AxB={(l, a), (I, b), (I, c), (2, a), (2, b), (2, c)} AxA={I, 2}x{l, 2} ={( I, J), (I, 2), (2, 1), (2, 2)}. From this example it is clear that if A contains 2 elements and B contains 3 elements, then A x B contains 2 X 3 = 6 elements. Generalising this result, we find that if A contains m elements and B contains n elements, then A X B contains m x n elements. This fact implies that A x B is an infinite set if either A or B or both are infinite sets. Product sets in general. The product set of the sets, A, B, C is denoted by A X B x C and is defined as A·xBxC={(a, b, c) ; a E A, b E B, c E C}. Generalising this result to a finite number of sets, we have A1 xA2 x .. x An={(al' a2 , ••• , an) : ar E Ar ¥ r, s. t. l~r~n} Examples. (i) The product R x R [or simply R2J is defined as Euclidean plane (of two dimensions). (ii) The product set R x R x R [or simply R3J is defined as Euclidean space of three dimensions.

3 Functions and Sequences 3'0. Function. Suppose X' and Yare two given sets. By some given rule, if each element x E X corresponds to a unique element Y E Y, then that rule is called a map of X into Y and is denoted by f. We write f: X... Y. This is read as "f maps X into Y" or . 'I is a mapping of X into Y". Mapping sometimes called operation or map or correspondence or function.

Let an element y E Y be corresponded by an element x EX, then Y is called the image of x and is denoted by f (x). Thus y=/(x). "/(x) is read a'S 1 of x" Here x is defined as the pre-image of y. The set X is defined as the domain of the ftinction/ and Y is called co-domain of j, I(X) is cabed the range or the image set off. It is also expressed as /(X)={/(x) : x E X} Evidently /(X) C Y. Remark. From the definition of a map it is clea.r that a map ./: X_ Y is said to be well-defined if (i) any element x EX=> f(x) E Y (ii) any element x EX=> a unique element/(x) € Y. In other words 'no element of X can have more than one I-image, i.e., given any x E X, we can not find two or more than two elements Yl' Y2 E Y such that Yl-/(X), Y2 f(x). (iii) Two or more than two elements of X may have the same/ -image in Y. i.e., if Xl. x 2 • Xa, , .. E X, then, Examples. gram:

/(x1)=/(X z) =/(Xa) = ... (i) Let a map /: A-B be defined by the dia-

12

" FUNCtIoNS' AN[j SEQUENCES

f

8

Here a E A ~ 3 1,2 E B such that 1 =!(a)=2. This ~ ! is not well-defined. (iii) The map ! A~B defined by the following diagram is well defined.

Aj~~f _"; f. ------..-= 10--b

•

I

domain of f {a, b, c} range of 1={2, I} Co-domain of/=(2, 1, -I). 3'1. Onto and into mappings Letf: X---,>-Ybe a map. I is called an 'into' map if 3 at least one element in Y which does not have a pre-image in X. i.e.,fis an 'into' map if/(X) is a proper subset of Y. i.e., if!(X) C Y. / is said to be an 'oIllO' map if !(X)= Y, i.e., if evcry element in Y has a pre-image in X. 3-2. One-one and many-one mappings. Let: X --;. Y be a map. / is called a one-olle map if Here

/(X1)=(X 2) ! Xl> X 2 01'

E X ~ XI=X~,

equivalently

Xl :;CX2 : Xl> X 2 E X ~ f(xd:;tf(x 2} if different elements In X have different images in Y. . / is called a many-ol/e l1lap if dilfercnt clement s in X hav~ the same i-image in Y i.e., if

i.e.

j(x})=f(x") : Xl, x 2 E X ~ X'*X 2'

13

pUNCflONS AND SEQUENCES

Examples. gram:

(i)

Let f: A ~B be a map defined by the dia-

e j

Here

f(A)={f(x) : x E A} ={x, y, z}=B.

Also /is one-one. Finally, f is one-one onto map. (i) The map f: A~ B defined by the follow ing diagram is many-one into map. fj

A

3·3 Real niued map. A map is said to be real valued map if a set is mapped into the set R by it. That is to say, any map f: A ~R is called a rea' valued map. 3·4. Set Function. A function is called a set Junction if its domain consists of a family of sets e.g., the map f: {A. : i E N}-+N defined by the formula

is a set fu nctio n. 3·S. Real Valued Set Functi.n A function is called real valued set function if its domain is a family of sc!s and its co· domain is the set of real numbers R. e.g.,

a map f:

{Ar

t

r E L;..

-?

R

is real valued set function and this map is said to be finitely additive if

14

FUNCTIONS AND SEQUENCES

I (

U Ar

,=1

)= UI ( ,=1

Ar )."

The map I is said to be couatable additive if

I (

~1

Ar

)= r~ I (

Ar )-

3'6. Extended real valued set function. An extended real valued set function is a real valued set function which also takes either of the value -{-

00, -- 00.

_[-00,00] such that f(Ar}=+oo or for some value of r. It may be noted that 00 and -00 both are not real numbers. i.e.,

a map

I: {

Ar}r E 6.

-00

Step function. We divide the interval [a, b] by means of points Xl' X 2, ...... , X" s.t. a=xo < Xl < X 2 is said to be bounded if 3 a number M > such that I a" I < M ¥ n I::. N. It is easy to prove that Convergence => boundedlles~. 3'10. Metric spa.:e. Let X7:-tP be any given space. Let X, y, z E X be arbitrary

°

Fl.JNt::TIONS AND

~EQUENCfS

15

A function P: XxX .... R having (he properties listed below (i) p (x, y) ;;;. 0 (ii) P (x, y)=O iff X= y (iii) p (x, y)=P (y, x) (iv) P (x, y)..Lp (y, z) ~ P C is said to be monotonic increasing or monotonic non-clccrcasing if an ,,;;; an+! ¥ 11 E N. A s~quence < an > is said to be monoton'c decreasing or monotonic non-increasing if an+! ~ an ¥ 11 E N.

A sequence < an > is said to be monotonic or monotone if it is either monotonic increasing or monotonic decreasing. Notice that a sequence will be either monotonic increasing or monotonic decreasing. 3'12. Axiom of Choice. Cartesian product of a family of non-empty sets is non-empty. Axiom of choice may also be stat·~d as "there exists a choice fu nction for any non-empty fami Iy of nOll-empty sets".

4 Bounded Sets, Derived Sets, Open Sets and Closed Sets on the Real Line Rea) Jine. The elements of the set R can be represented by means of points on a straight line. The origin of this straight line is taken to represent the number o. The real numbers x > 0 are represented on the right of lero, where as the real numbers x< 0 are represented on the left of o. Each point of this straight line represents a unique real number and each real number determines a unique point on the line. For this reason the set R is referred to as the real line or real axis. This is why, we use the words point and number conversely. Remarks. Throughout this chapter we shall take R as the universal set provided nothing is contradictory stated. O.,en and closed intervals. Let a, b E R be arbitrary. We write [s, b]={x E R : a x E D(A) or x fI. D(A). e.g. the limit point of the set {J, j, 1, t, ... } is O. Condensation point. A point x is said to be a condensation point of a set A if every neighbourhood containing x contains an infinite number of points of A. In accordance with this definition a finite set has no condensation point. Closed set. A set A is said to bl! closed if every limiting point of A belongs to the set A itself. Symbolically, a set A is said fo be closed if D(A) C A.

Examples.

(I)

{O,

1,1, I, 1, ...... }=A is a

closed set.

For

D(AJ={O} C A. (2) A closed interval is a closed set.

For D«(a, bJ)=(a, b] C la, bJ. (3) Every real number is a limit point so that ration'll limit points belongs to the set Q. This => D(R)=R C R D (Q)=R B---A.

fA

(x) .... x is one-one

onto

For

A-B => 3 a one-one map f: A ~--+ B.

=> f- 1 : B :$

onto ---+

A is also one-one.

B-A.

(iii) transitive: A-B, B-C:o> A--C. onto

For A-B, B-C .. 3 one-one mapsf: A ---+ B onto

and g : B - - .. C. onto

=- gf: A - - - C is also one-one, .. A--C. given relation satisfies all the conditions of an equivalence relation. Hence the result !"t. Cardinal numbers. [Kanpur 71, 70; Utkal 5] The relation A-B in the family of sets is an equivalence relation. Hence the rela.tion decomposes the family of sets into di!;joint equivalence classes. Every equivalence class defines a unique cardinal number. In other words the cardinal number of a class of equivalence sets is any representative of the class. The cardinal number of a set is sometimes called its power or potenc~'. If P is any finite set consisting of p elements, then its cardinal number is defined as p. The cardinal number of ~ is defined as O. A cardinal number corresponding to a finite set is called a finite cardinal number. A cardinal number corresponding to an

rt means that the

COUN'fABILIrY OF

~ETS

23

infinite set is called a transfinite cardinal number. It follows from these definitions that all the transfinite cardinal numbers a"C 'greater than any fi!1ite cardinal number. Notation. The cardinal numbers of the sets P, Q, R, S, ........ . arc denoted by the corresponding small letters P. q. r, s, ........ . Also the cardinal number of a set A is denoted by I A I or card A. Then, by definition / P 1= card P=p, I Q /=card Q=q. The cardinal numbers of the set N and the set of all real numbers [0, l~ are denoted by a and c respectively. Then . / N 1=8, I [0, l] /=c. The symbol Xo (read as aleph nUll) is also used to denote I N /. This symbol was originally used by Cantor. If P""Q then by definition of cardinal number, 1 P/·=IQI· 5'2. Sum of cardinal numbers. Let P and Q be any two sets with cardinality p and q S.t. their intersection is empty i e., I PI=p.1 Q I=q, P () Q=; Then the sum of p and q is defined as p+q=1 P u QI· More generally, let P". be a cardinal number corresponding to the set p~ ¥ CIt in an index set 1::,.. Further suppo'se that. . Prx. ("\ P~=,p >tf Of, ~ € 6 s.t. 1X:;t=~. We define the sum of cardinal numbers as E p",= U P", J -EDCltEI::,. • Examples. (I) 1/ A={l, 2}, B={5, 6, 7), then 1A 1=2, I B 1=3 and A () B=~ so that 1A U B 1==1 A 1+1 B 1- 2+3-5. Also A u B={I, 2, 5, 6, 7} and hence 1 A U .B 1-=5. This is a verification of out definition 5'2.

(ii)

Then

...

Let

A={a, b, c}, B:oo{a, p, q, r, s}.

A U B=(a, b, c} U {a, p, q, r, s}. ={a, b, c, p, q, r, s}. 1A

u

B

1=7.

24

COUNfABILITY OF SETS

Evidently 1A 1=3, 1B 1=5 and A n B::;t:~. :. By definition 1A UBI =I=- I A 1+ I B I i.e., 7 ;::3+ 5. This example also verifies our definition of sum of cardinal numbers and it also shows that An B=; is a necessary condition for defining the rule 1AuB 1=1 A 1+1 B I. S·3. Prodlict of cardinal numbers Let P and Q be any sets s.t. I PI =p, 1 Q I=q. We define the product of cardinal numbers p and q as pxq= I PxQ I. More generally, let Pa. be a non-empty set for each Gt in an index set A and let pa.=1 Pa.. Then we define the product of cardinal numbers as npa.=j x Pa.I. where ilpa. stands for the Pl.P2' Ua· .... and xPrJ. stands for the cartesian product. Example. Let P={a, b}. Q={a. J, 2, 3}. Then I P 1=2, I Q 1=4, pnQ,ptP. By definition, 1 P x Q I =2 x 4=8. Now I Px Q 1=1 {(a, a), (a, I), (a, 2), (a, 3), (b, a), (b, 1), (b. 2), (b, 3)} I

=8. This verifies our definition of product of cardinal numbers. !§·4. Definition. A set A is called an infinite set if A is equivalent to a proper subset of itself. In the language of notation, a set A is called an infinite set if A -.I to a proper subset of A e.g. N, Q, R, C etc. 5'5. Definition. A set A is called a finite set if it is equivalent to {I, 2, 3, ... ,n} for some value of n E N. S·6. A set A is called a denumerable set if A ..... N. e.g. tal' aa, a8,·· ... } is denumerable, since this set -- N under the map an .... n. Moreover this is a sequence and therefore a sequence is a/ways denumJrable.

A denumerable set is sometimes called enumerable or countably infinite .

•: A.....,N ~ 1A 1=1 N I => I A 1=8. Also A.-..J N ~ A is denumerable. Hence a denumerable set can also be defined as : A set A is denumerable if I A 1=8.

25

COUNTABILITY OF SETS

We shall always use this very definition in future 5'7. Definition. Countable sets. A set is called (:ount4ble or atmost countable if it is finite or denumerable. Example. (i) A={I, 2, 3, 4}. Then A is finite so that, by definition, A is countable. (ii) Let A={a1 • a2 • aa,·"} so that A is denumerable. Hence by definition A is countable. Here observe tbe difference between coulitable and countably infinite. 5 8. Definition. A set A is ca.lled an uncountable set if A is an infinite set and A is not cardinally equivalent to N. e.g. Rand C are uncountable sets. The following have the same meaning: Non countable, non-denumerable, non-enumerable, uncountable. 5'9. Definition. A set A is said to have the power of contil1um if A r- [0, 1]. Also we say that the cardinality of A is (:. Theorem 2. To prove that

I AxB 1=1 B 1+1

B 1+1 B I.. ·to Proof. Let A and B be any two sets. AxB={(x, y): x E A, Y E B} U

=

{x, y) : )' E B}

xEA Then AxB

I

I A I terms.

1=\

u {(x, y) : y E B} x E A

... (1)

For a fixed x E A. consider the map .f: 8~{(x. y) : )' E B} given by .f (y)= (x, y)

~ )'

E B.

Evidently.f is one-one onto. Then B '" {(x: y) : y E B} This ~ 1 B 1 =1 {(x, y) : y E B} I. In this event (') shows that I AxB 1=1 B 1+1 B 1+· .. to

1A 1terms.

Q.E.D. Verification of the theorem 2. Let A={l, 2}, 8-(1,2, 3}. Then

1 A 1=2, I B 1=3 I AxB 1=2.3=6

A > 1 A 1 -«. Solution. Let« be any cardinal number. Let II E;;: 1 A 1

I B I .s;; I A I B- to a subset of A or B-A IAI

E;

I A I -I B I.

which

... (1)

IBI

=> A- to a subset of B or A-B. ...(2) The statements (I) and (2) taken together imply the required result. Ex. 2. If () and {J are cardinal numbers such that fI. ~ f3 and {J ~ ar, prove that at = 11. [Banaras 1969] Solution. Let A and B be two sets s.t. 1 A 1 =a, 1 B 1 =-~. Also let at ~ {J. (J ~ ac. To prove that Cl""'~. we have to prove that: A I - 1 B I. GlC;;B=> IAI ~ IBI lO A -- to a subset of B or A-B ... (1) ~~Cl"IBI~IA: ~ B- to a subset of A or B-A. . .. (2) The statements (I) and (2) imply the required result. Theorem 3. A set A is countably infinite iff A can be put In the form of the sequ[!nce tOI' O 2 , tl3 • .• 1 of tlte distinc: elements. Proof. A set A is countably infinite iff 3 a one-one onto map f: N ~ A i.e. iff (i) m.;f:n and m, n E N ~ f (m)-:;cf (n). (ii) Given any element b E A, 3 an e1em'!nt a C N s.t. f(a) b i.e., iff f[N]=A and f is one-one. 111

27

COUNT ABILI j Y OF SETS

or iff given any element n € N. 3 Of! € A lI.t.j(n):::a n andf is one-one. Above means A={al. a2 • aa •... }' where a10 02' 0a.· .. are distinct elements. Theorem 3ft. Every subset of a cvuntable set is coufttable. [Meerut 1987, 88] Proof. Let A be a countable set. Then A is either finite or enumerable. (I) When A is finite. Then every subset of A is also finite and so is countable. (ii) When A is eoumerable. Then A can be written as ... (1) A={x1, XI''''} Let B be a subset of A. (i) If B=;. then evidently B is countable. (ii) If BtF"', then B is expressible as B={x. l , x n2 ,· .. }, where XIII E A. Evidently B is enumerable, by virtue of (l). Theorem 4. The set of all real numbers iI' the closed illferval [0. 1] is not denumerable. [Banaras 1971 ; Kallpur 68] Proof. Let A denote the set of an real numb.:rs in tne eto.ed interval lO, I J. SymbolicallY we write

A={x E R : 0 ~ X ~ l}=[O, 1). To prove that A is uncollntable. Suppose not. Then A is couhtable. This implies that every element of A must appear in the sequence x"' .. .of distinct elements. i.e., B={xlo x 2 , xa, .. ·} ... {I) Write decimal expansion of those xl's as follows: x 1=0'xu Xl 2 Xu xu .... · X 1m . .. .. X10 X 2 , Xa,· ......

X~=O.X21 X 22 X 23 X~, ...... x:.,. ..... .

X.,,=0·Xm1 XtII~

X"'a X""

..•..•

XIII"",

where XII E {O, 1. 2, 3, 4, S, 6, 7, 8, 9} ¥ i and j and each decimal contains an infinite number of non-zero elements. We can write 1 and i in two ways as given below: J 1=1'0000... ... ... . 1 1-0'5000 ....... ..

28

COUNTABILIrY OF SETS

J t =0'99999 ...... =0'9 } 1 1=0'49999 ...... =0'49

and

... (2) But in the present case we shall write the decimal representation of the elements A in the form (2). Construct a real number ~=O, ~l> ~a ... ~". ... s.t. Xu = 5, write ~l =4 if and if Xu :f.:5, write fl = 5 in general if x",,,,=5, write ,,,,=4 and if x",,,,;t:5, write f",=5. In either case, it is clear that x".".;t:'''' ¥ m. . .. (3) Since ~1It=4 or 5 ¥ m E [0, 1], i.e., ~ E A ... (4)

ea.

:. e

x",=~ ~ O.X"'1 X.I xma ...... =O'~l ~l! ~a""" => xmm=~", T m. Contrary to (3). :. x",:;;c~ ¥ m.

e

In this event (I) shows that ~ A. Contrary to (4). Hence the required result follows. Q.E.D. Theorem S. To prove that the set R is uncountable. [Ufkal 6S ; Puajab 62 ; Vikram 61] Proof. Firstly we shall show that I R/=c. We know that cardinality of the set of real numbers in the open interval (0. b) is c, so that I (a, b) /=c ¥ 0, b E R s.t. a < b. This .. \ ~ )i=c. ...(1) The set R is the set of all real numbers lying in the open interval (-00, (0), i.e., R=(-O, co).

(-r'

Define a mapj:

(-~, ~ )~

(-00,00) by the formula

/(x)=tan x. Evidently j is well defined. Also it is easy to verify that! is one-one onto. :.

This

i.e. This

(-~, ~)-

I(

(-00, (0).

~ -~, ~ )=1 (-co,

00)

c=/ R I.

with (I).

~

in

a~ordance

R is uncountable,

I· Q.E.D.

29

COUNTABILITY OF SETS

Ex. 3. To ~ow that lor every real number x, the real numbers in lhe semi-open interval [x, x+ I) lorm an uncountable set. Solution. Let x E R be arbitrary. Define a map I: [x, x+I)_[O, 1] by the formula.!(X)=X-x, Evidently l(x)=x-x=O l(x+l)=x+l-x=1 :. I is well defined. f(XJ=/(X2 ) ; Xl> X 2 E rx, x+l) => X1-X=XI-x This => I is one-one. I is a continuous map => I is an 'onto' map. :. [x, X+I) '" [0, 1) This => I [x, x+l) 1=1 [0, I) ,. But the set of all real numbers in [0, 1) form an uncountable set and hence the set of all real numbers [x, x+ I). Proved. Theorem 6. (i) II Ai is countable infinite set, then fI

U A. is countable infinite and hence deduce that n.a =a. 1=1

-

(ii)

If Ai

is countably infinite set lor i = 1,2, 3, ... ...... then

U Ai is counfably infinite and hence deduce that

8+8+8+ ...... +to 8 terms=a. [Meerut 1972; Kurukshetra 69] Or Union 01 countable collection 01 cOll'Jtable sets is countable. [Meerut 1987, 86] Proof. (i) Let A. be a countably infinite set. Let

A=U" A • • =1

To prove that A is countably infinite. Let the elements of A; be displayed as Al : au a12 Q I8 ...... 0 1....... . ,42 :

au

A8 : 0al

at~

a 21 .... .. 0 2" ..... .

aaJ 0 83 , ..... as" ......

.

Write B={au , 012'''' ... a 1" .. • ; 0 21 , a22' ...... ; 0"1' a2,,, .. ·}· The S(~t B considered as the set of distinct elements is coun:. , B I =8. ...( 1) tably infinite.

30

COl!NTABILlTY OF SElS

n A}= r/> for i;f=j, then evidently I A I = I B, If Ai n A, t:.d> for i :;t:j, th'.!n A is equipollent If A,

=a.

with some subThis:? I A I ~ ! B I -'-~a or I A I E;;;. a ... (2) Al C A and 1 Al I =8 ... (3) :. a= I Al I ~ I A I or a ~ I A I Combin:ng (2) and (3), we get II ..; I A I ~ a, which:? I A I = a. In either ca"" 0 1,,; .. }. Clearl.t' A is expressible in the form of a sequence of distinct elements and hence denumerable so that ! 11 =a. Secondly we assume that A, n AFi=t/J for i .,cj. In this case A is equipollent with some subset of N' so that ! A I ~ I N I or I A I ~ a. . .. (4) But Al C A and I Al I ~ 2.

31

COUNT ABILITY OF SETS

so that

a= I Al IE;;/ A ' or a ~ I A / This => / A / =a, by (4) and (5).

... (5)

This proves the required result. Deduction. Further assume that B. n A,=~ for i.:l=j. Then from what ha'i been done it follows that

I A 1 -a::>·

IUA.I=a => ; I 1

'~1

=.

/ A. 1

=> a+a+ ... +... to a terms=a This completes the proof. . Theorem 7. To prove that N X N is countable, or to prove that a2 =a.a=a. Proof. To prove a.a=a is the same as to prove I NxN 1= 1 N ,. The elements of N x N may be displayed a'i follows: (1.1)

(1,2)

(2, I)

(2. 2)

(1,3) ...... (I,m) ... (2, 3) ...... (2. m) ...

(n, I)

(n, 2)

(n. 3) ...... (n, m) ...

If we write then

(i)

.

At={(i. 1). (I. 2). (i,3) ....... } •

NxN= u A.

\ ,-I

(ii) A. nA,=q. for i#j (iii) A, is a countably infinite "V i .

..

These facts lead to the conclu,ion that U

Ai is

countabl~

infinite i.e., N )( N is countably infinite and hence NxN ...... N . This => / NxN /=/ N I Ex. 4. Prove I P (A) 1=2 / A / ;f A is any finite set. Solution. Let A be any finite set with cardinality n. P (A) is the family of all subsets of A. / P (A) / =·co+ ..C'+ ..C3+ ........('" =(1 + 1)"=2"=2 I A / :. I peA) / =2 I A I . 1 heorem 8. (I) If A, is non-enumerable set ¥ .. e: :01.

32

COUNTABILIlY OF SETS

-

then U AI is non-enumerable and hence deduce that ~ /

1=1

(U)

c+c+c+ ...... to a terms=c. If Ai is non-enumerable set for 1 B,..., N, => N '" B. B U C ,.., N, N '" B ~ B U C '" B. Also, P '" P (by reflexivity). Now, B U C r- Band P '" P. Therefore B U CUP'" B U P which => I B lJ CUP 1= 1B U P 1

=> 1B U CUP 1=1 A I => 1A U C 1=1 A 1 => cx+n=« => n+GI=«.

:

A-B=P

on using (I)

Hence the proposition. Theorem 11. Prove a+«=GI. CIC being any transfinite cardinal number. Or, If an enumerable set is added to an infinite set, the power of the infinite set is unaffected. Proof. Let A be any infinite set with cardinality" so that I A 1=«. Let A n N=.p. We have to prove that I A U N 1=1 A I. If we show that «+a=2, ,the result will follow. A is an infinite set => 3 B C A s.t. 1 B 1=a .. We have A U N=(A-B) U B U N=(A-B) U (B U N) or A U N=(A-B) U (B UN). . .. (1) BUN, being a finite union of countably infinite sets, is countably infinite. :. BUN'""" N. But B -- N. .: B is enumerable. The relation M ,..., N. where M and N are any two sets, is an equivalence relation in the family of sets.

36

COUNT ABILIfV OF SETS

:. B --' N => N -." B BUN ,.., N, N I"W B => BUN - B. Now BUN - Band A-B - A-B. Combi ning these tw:> we have, (A--B) U (B U N) ,.... (A-B) U B Using (1), we get A UN,.. , A. This => I A U N 1==1 A ,.

(by symme try) (by transiti vity) (by reflexivity)

=> lZ+a=~. Q.E.D. Ex. 5. If we subtract an enumerable set from a non-enumer-

able so:t, then the remaining set is non-enumerable. [Agra 19661 Proof. Let A be a non-en umerab le set. Let B be enumerab1e set. Let A-- B=P. We claim P is non-en umerab le. Suppos e the contrar y. Then P is an enume rable set. Moreo ver A =:B U P. A, b~ing a finite union of counta ble infinite sets, is an cnum~rable set_ A co:ttra.diction. For A is nonenum~rable. Hence the require d result follows . Ex. 6. If an enumerable set is subtracted from all enumerable set, the rema;n;ng set .will be eltllmerab Ie. Proof. Let A a'1d B b:>th be enume rable sets. To prove that A - B is enum;! rable. Let A -B=P . We have to prove that Pis counta bly infinite. Supp0'ie the contrar y. Then P i'i non-en umerab le. A=B U P. A is non-enumerable, being a finite union ,of a non-en umerab le and an ~numerable set. A contrad iction. For A is enumerable. Hence the require d result follows. Ex. 7. If (I. ;s any transfinite card;n~l nu-nber, then a Proof. Let A h~ any infillite set with cardina lity It, i.e., 1 A I ~7a:. To prove thrtt a
or

Q+ is cardinally equivalent to Q- under the map T!_ -+~~ q

q

:. I Q+ I = I Q- I =8 i Q I = I Q+ U Q- u to} I =8+n+l =(a+a)+1 :. associative law holds. =a+l=a :. I Q I =8, which:::> Q is enumerable and hence countable.

Ix. 9.

Prove that the et 01 integers is countable. Proof. To prove that Z is countable, it is enough to prove that I Z I =8.

Write

Z+={l, 2, 3, ...... } Z-={-I,-2,-·3, ... }. Then Z= Z+ U Z- U {O}. Also Z+, Z-, to} arc disjoint sets :. I Z I = I Z+ i + I Z- I + : to} I = I Z+ I + I Z- I + I to} I Z+ N under the mapping n -+ n Z- ,.... N under the mapping -n -- n so that I Z+ I = I N I , I Z- I = I N I I Z+ I = I Z- I = I N I =8. or Substituting these values in (I), we get I Z i =8+8+1=(a+a)+I=a+:=8.

... (l)

"oJ

:. : Z I =a. Ex. to. Prove thai a < c. Solution. To prove that 8 < c. We know I N 1 - a, R i =c, NCR. NCR;} ! N I ~ R i :) 8

. .. (2)

Q.E.D.

I

~ C.

38

COUNTABt-LlTY OF SETS

N is not cardinally equivalent to R under any mappi·ng. :. a:;t:c. Now a:;t:c, a ~ c ~ a < c Theorem 12. (Cantor's theorem) : To prove that I A I A,.., peA) so that 3 one-one map onto

I:

A

---l-

P (A)

Take B=={x E A : x ff. f (x)} x E A .. I (x) E P (A), by (3) .

.. f

... (3) ... (4)

(x) C A, since P (A) is the family of all subsets of A.

Also B C A, according to (4). Thus, Be A,/(x) C A for any x E A. This . 3 CIt E A. s.t.f(II.)=B. CIt e: A. => 11. e: B or tx ft. B. Consider the case in which 11. e: B. . .. (5) Now, /It e: A. ell e: B => ell ft. I (a.), according to (4) . • ell ft. B, Contrary to (5). .: I (tx)=B. :. The possibility Ie e: B is ruled out. Consider the second possibility in which tx fF. B. .. .(6) Now, • e: A. => ell ft. B ... II ¢ I (tx) .: I (II)-B. . . II e: B according to (4). Contrary to (6). :. The possibility CIt (/:. B is also ruled out.

COUIIITABILITY OF SETS

39

It amounts to saying ~ E A does not imply at E B or at B. Again we get a contradiction. It means that our initial assumption is wrong. This means that (2) is true. Remark. If I A I =n, (finite cardinal number) then the last theorem implies that n < 2n.

rt.

[Kanpur 1910; Banaras 69] ":

I

P (A)

I

=2 1 A '=2 n •

Ex. 11. Let X be any set. Let A (X) be the family of all the cizaracteristic fUllctions of X. 7 hell prove that the power set of X is equil'alent to A (X). . Solution. Let A (X) denotc the family of all the characteristic functions of a sct X. To prove that A (X) f"oJ P (X). Let B C X be arbitrary. Let X8 denote the characteristic function of B relative to X. Then X B: X ... {I, O} s.t. X B (x)=

JI if x E B

to if x

(/:. B

Evidently X B is an arbitrary element of A (X) and B is an arbitrary element of P (X). Defined a mapf: A (X) - P (X) by requiring thatf(XB)=B. This shows that f maps the characteristic func~ion of B relative to X into the set B. Evidently f is onc-one onto. Ex. 12.

A (X) '" P (X). Q.E.D. Prove that the set of algebraic numbers is countable. [Punjab 65; Kanpur 72; Bhagalpur 68]

Proof.

First we shall show that the set of polynomials P (x)=aO+a1x+a2 x 2 + '" +amxm with integral coefficients is enumerable. Let m be a fixed positive integer. Let Pmn qenote a polynomial of degree m in which I ao I + I a1 I a2 I + I + ... + I am I =n, ... (1) where n is any positive integer. Then Pm2 denotes a polynomial of degree m with coefficients s.t. I 00 I + I a1 I I a2 I +... + I am I = 2.

+,

40

COUNTABILITY OF SETS

Write Pm= U {Pm,. : n EN}. Then P. is a countable union of polynomials Pm. of fixed degree m in which the coefficients are subjected to the condition (1). Since N is enumerable and therefore Pm is enumerable. Write

P=u

{Pm: mEN}

P is enumerable, being a countable union of enumerable sets. Evidently P is the set of aU polynomials with integral coefficients. Secondly we shall show that the sct of algebraic numbers is countably infinite. Write

00

P'= U {p" (x) : p,. (x)=O} II-I

'

where P.. (x) stands for AJlQiynomial of degree n with in,tegral coefficients. From what has been done it follows that P' is enumerable. An algebraic number is defined as the root a polynomial P (x)=O with integral coefficients. Take A,,={x : x is a, solution of p,. (x)=O}. The A" is finite. SilWe a polynomial of degree n has at most n roots. A= U {Aa : EN}. Define Evidently A is the set of algebraic numbers. A being a countable union of countable sets, is countable. Since A is not finite and hence A is enumerable. This completes the proof Ex. 13. To prove that the set 0.( a/l irrationals in any interval is non~enumerable. [Raj. 1965; Agra M.Sc. Statistics 65J ,Proof. We know that the set of aU real numbers (i.e. the set of all rationals and irrationals) in any interval is uncountable. We also know that the set of all rationals in .any interval is enumerable. Then the complement of the set of rationals relative to the set of real numbers in any interval is uncountable, i.e., the set of irrationals in that interval is uncountable. For complement of an enumerable set w.r.t. a non-enumerable set is non-enumerable. Ex, 14. Show that the set of a/l transcendental numbl?rs ill alH' interval is non-enumera~/e. . (Punjab 1965; Agra \i.Sc, Statistics 65J

41

COUNTABlLln OF SETS

Proof. We know that the set of all real numbers (i.e. the set of all algebraic numbers and transcendental numbers) in any in· terval is non·enumerable. We also know that the set of all algebraic numbers in any interval is enumerable. Then the complement of the set of all algebraic numbers in any interval relative to the sct of aU real numbers in that interval is uncountable, i.e., the set of alI transccndentals ·in that interval is uncountable. For the complement of enumerable set reia'jve to a non-enumerable set is non-enumerable. Ex. 15. Two enumerable sets are equivalent. Proof. Let A and B ~e both enumerable sets.

To prove that By hypothesis,

A -- B. A r-- N, B "-' N.

Since the relation A """"' B is an equivalence relation in the family of sets. (by symmetry) B -- N ~ N ""'"' B. By transitivity A,... N, N '""" B => A -- B. Theorem 11. Prove that C.c=c. fBanaras 1967] Proof. Let A be the set of all real numbers in the closed interval [0, 1]. In symbol A=[O, IJ. Then I A I =c. To prove that c.c=c, it is enoguh to show that

IAXA/=IAI· Write B={(O, x) : x E A}. Define a map I: A---'>- B by the requirement

1 (x) = (0,

..,. x E A

x)

Evidently 1 is one-one onto so that A -- B. Evidently

:.

I A i '-= / B I B c A X A. IB I ~ IAXA I I A I ~ I AxA I

.

... (1)

... (2), on using (I> Let x, yEA be arbitrary. Then x and y can be uniquelY written as x=0·X1X S X3 ••• y=O·yly2Y.··.··.··. in the form of infinite decimals which do not end wit:. zer~. That is to say, this infillite decimal contains non-7ero digit'>. We write i=0·4999, .... =049 instead of t=0'500

or

42

COUNTABILITY OF SETS·

Define a map /1

: A x A~A by

the formula

It (x, Y)=O·X 1Y\X2Y2XaYs .. ·· .. Evidently /1 is one-One. This => lAxAI~IAI

Combining this with (2), we get IA:~[AxAI~IA[

or

IAxA!=IAj.

Find the power 0/ an aggregate

Ex. 16. M

2m

'

0/ numbers

M and m being positive and integral.

Solutio~.

Let

A={~

Q.E.D. gil'en by

[Agra 1964J

: M, mEN}

To determined the power of A.

AM={~

Write

:mE

N}

yo

ME

N

Elements of AM can be displayed as follows:

Al : ~

1

}a ... }........

2 A2 : 2i

2

2

3 A3 : 2i

.2i

22

23

3

3

28

... 22ft ...... ."

n

23

Obviously (i) AT is enumerable (ii) Ar n Ar'=r/J

...

¥ ¥

3

2ft ...... n

...

2ft ....

rE N. r, r' E N s.t. r-:pr',

(iii) A= U A r • r--l

Being an enumerable union of enumerable sets, A is enumerable and hence its cardinal number is a i.e., the power of the given set is a. Ans. Ex. 17. Show that the set 0/ points in the closed interval [2, 4] and in the open-interval (J, 2) are cardinali} equivalent. [Banaras 1970J Solution. Consider the map

---r

/: [2, 41- [I, 2] s.t. f(x) = ( ,:-2 ) + 1.

f is continuous => / is onto.

43

COUNTABJLlTY OF SETS

I'.() f( Xl) =J~X2

::>

~12-~+I-_X22-·2+1

=> X 1 =X 2 •

This says that! is one-one. Thus f is one-one onto. Hence [2,41 '" [I, 2J. This . .. (1 ) => I r2, 4J I = I [l, 2J I . ... (2) But I [1, 2J I = I (1,2) I . Combining (I)' and (2), we get the required result. Theorem 12. (Dedekind~peirce). El'er)" infinite set is equivalent to its proper subset. • or Every infinite set can be put into a one-one correspondence wilh a proper subset of itself. Proof. Let S be an infinite set. Then S contains a denumerable subset, say, Write S*=S-{al, a z, ... } and Sl=S-{a1} Then SI is a proper subset of S. Also Sand S1 both are infinite sets. Define a map f S-+SI S.t. f (a,.)=a'+ 1 for n= I, 2, 3, ... and /(s)=s ¥ s E S*. Evidently / is one-one onto map Hence S is equivalent to SI' It follows that S is equivalent to its proper subset. Note. This fact was used by Dedekind to define infinite sets as follows. A set is iJ~rinite iff it call be put hlto a one-one correspondence with a proper subset of itself. Theorem 13. (GaliJeo's Paradox). Any denumerable set can be put into a one-olle correspondence with a proper subset of itself.

Proof. Let A be a denumerable set. Then A is expressible as A={at. a3 , aa,· .. }. Write B=A-{al}={a2 , aa, ... } Define a map f : A-+B s.t. f(a,.)=o"+1 forn=l, 2, 3, ... Evidently f is one-one and onto map. Also B is a proper subset of A. Hence the theorem.

44

COUNTABILl1Y OF SETS

Theorem 14. Every nOll-empty open set 011 real line is the countable union of nan-empty disjoint open intervals. [Meerut 1988,86] Or A bounded set of lion-overlapping intervals is countable. Proof. (i) Let G C R be an open set and x E G be arbitrary. Then 3 an open interval I", with centre x such that x E I., C G. Let C. denote union of all such open intervals. Then C.,= u tI",CG}. Evidently COl is the largest open interval containing x such that x E C., C G. Moreover, each x EGis contained in /'" and each / .. is contained in G. The suggests that G= U C"'. We have either CIII n C,=t/> or C., n CY=l=~, ¥ x, y E G. Consider the case CIII n Cy::j::.cp. Then C", U Cy is an open interval containing both x and y such that C.. C (CIII U Cy) C G, C, C (C", u C,) C G. By definition of Ce; and C y , we have (Co: u Cy) c C.. ' (CIII U Cy) C Cy. Hence C",=C~ U Cy and Cy=CIII U ell, i.e. C.. =Cy Thus, '" x, y E G, we have either Cm=C, or CIII n C,=r/I, Also G=U Cm• This declares tl-at G is the union of pair-wise disjoint nonempty intervals. It remains to prove that the intervals C", form a countable collection. (m Let G={G,.} be a collection of open subsets of R such that (i) G,.:/ t/J. 'V n, Oi) G.. n Gm=t/> for Il ,em. We know that each G •• contains an infinite number of rational and irrational numbers. Let K=lK,,}, where K" is the set of all rational numb;!rs in Gn. D::fine a map f: K - G such thatf(k,,)=G n • . .since for each k .. E K, there is a unique open set G.. E G, md conversely. :. f is one-one onto. Therefore K -- G. But K considered as the set of rational numbers U KA , is .::ountable. Since K C Q. Hence G is countable. Theorem 15. ~r {G,.} is all)' cdlectioll ~ non-empty disjoinl open sets ofR, then prOl'e that (G,,} is c(Juntable. {Kanpur 1979, 771 Proof. For proof, sec (ii) part (Of '[ h.:orcm 14 Problem 18. SIIolI' ,IIat a cOllllfahle set is a Boret set.

COUNTABILITY OF SETS

45

Solution. Let A be a countable set so that A is expressible as A={a1 , a2 , a~, ... } {ar}={x: x=ar }

= U ~ x : ar ~ x < an + ~}

"=1 ( n =Countable inters!?Clion of closed sets. {or} Also A= U r EN These facts prove that A can be obtained by the formation of countable union and intersection of closed sets and open sets and hence A is a Borel set. 5'10. Continum Hypothesis. It states that there exists no set having its cardinal number between a and c. But this hypothesis does not assert that there exists no set with cardinal numb.;r greater than c. In this connection Cantor had proved that the pow.:!r set peA) of any set A has cardinal number greater than the cardinality of A. Consequently the cardinality of P (R) is greater than c.

6 Measure And Outer Measure 6'0. Def. Boolean ring. (or ring of sets). [Gujarat 70] A non-empty family A of subsets of X is called a ring 'of sets or B-ring if any A, B E A ~ A-B E A, A U BE A. That is to say, ring of sets is closed under the formation of unions and differences. Remark. (1) A n B=.A.-(A-B), A t:. B=(A--B) U (B - A), where A t:. B stands for the symmetric difference of A and B. These equations dec/are that A n B E A, A t:. B E A if A is a ring of sets. (2) ", E A. A' E A where A'=X-A (2) A, B E A .. A U B E A. 6'3. Def. a-algebra of sets (or a-field). A non-empty family A of subsets of X is called a-algebra of sets or a-Boolean Algebra of Borel Field if (i) A E A => A' E A GO

(ii) Ai E A =>

U At E A.

Remark. (1). Each algebra (a-algebra) A is a ring (a-ring) because if A and B are in A, then A-BEA, since A--B=(A'u B)'. (2) A ring A of subsets of X is an algebra in X iff X E A. For if A is an algebra, then X = A' U A where A E A so that X E A. Conversely if X E A, then A' E A ¥ A E A. For A'=X -A so that A is an algebra.

6'4. Der. Semi-ring. A family A of subsets of X is called a semi-ring if (1) tP E A and (2) A, B E A=> A-B E A (3) AI E A S.t A, n A1=f# for i-:f::j .. 3 A E A GO

6·S. Det. Monotone class. A non-empty class of sets A is called a mO:lOtone class if for every monotonic sequence < AI > of sets in A, we have lim An E A. Examples If A.,. C A.,. pi, then is an increasing sequence. 00

For increasing sequellce, we write . If A= u

n-l

we denote it as An t A. If An+-! C An, then




An, then

is decreasing

GO

sequence and we write and if n An=A, then we write ft=J

An.j, A If M is a monotone class, then AE M for every sequence

in M. Any a-ring is a monoton~ class. A monotone class is a a-ring iff it is a ring. , 6'6. DefinItion. Complete lattice. The set R.=R U {-co, ~}=[-oo, co]

48

MEASURE AND OUTER MEASURE

is called Extended set of real numbers. Briefly R. is called complete lattice. 6'7. Definition. Set Function. Refer to D!finition (3'4), Chapter 3, on p~ge 13. 6·S. Definition. Extended real valued set function. Refer to Definition (3'6), Chapter 3, on page 14. 6'9. Definition A fU!1ction (set function) is said to be finite function (finite set function) if all its functional values are finite. Def. Additive Class. [Kanpur Statistics 75] Let Q be a space and A be a cla~s of subsets of ~:I. A is said to be finitely additive class of sets if (i) cf, E A (ii) A, B E: A ~ A-B E: A, A U BE: A. This definition is parallel to the definition of ring of sets. Hence we can say that every ring of sets is finitely ajditive class and vice-versa. Def. Completely Additive Cla..s. (Kmpur Statistics 77, 75) Let Q be a space. The class A is said to be a complete ly additive class of sets if (i) r/J E A (ii) A E A '* A' E: A (iii) If is any sequence in A, then U An E: A. All these condition => m (A U B)=m (A)+m (B). m is called finitely additive if A. E A for i=l, 2, ... , n S.t. A. n AI=1> for i=l=j =>

(3)

m( (4)

UA,

1=1

)= i:

1=1

m (A,)

m is called Countably additive if

A, E A ¥ i E N s.t. A,nAI=1> for i:;f;j (5)

~ m( UA,)=;m (A,) i=l

i=l

m is monotone if ¥ A, B E A s.t. A C B => m (A) EO;; m (B)

m is subtractive if A, BEA s.t. A :J B, m(A) is finite => m(A-B)=m(A)-m(B) (7) In is Countably subadditive if

(6)

A, E A ViE N =>

In (

'~1 A.) ~ '~l m (Ai)

m is translation invariant if A E A, x EX=> m (A+x)=m (A), where A+x={y+x : yEA} (9) m is regular if AEA, E > 0 => 3 G, FE A S.t. m (G)=E~m(A) ~ m(F)+" 6'11. Postulates for an ideal measure function. A measure is an extended real valued set function m defined on g·ring A of subsets of X s.t. (I) meA) ~ 0 ¥ A E A. In particular m (1))=0. (2) A, B E A s.t. A C B ~ m (A) ~ m (B) (3) A, B E A s.t. A n B=1> => meA U B)=m(A)+m(B). (4) For an intervall, m(l)=l (/)=length of the intervall. (8)

so

MEASURE

(5) If

OT,!!.ER MEASURB

is a sequence of pai rwi se d i sjoi n t sets in A, then

m

(6) m

AND

(u ) i=1

A, =

E '=1

m (A,)

is translation invariant, i.e.,

where

-

m (A+x)=m (A) A+x={y+x: yEA}.

Definition.

6'12.

Measurable space.

y.

x€X,

[Kanpur 1969]

subsets of X and m is m eas ure on A, then the tripl e (X, A, m) is called a measurable sp ace. If A is

a

algebra of

If m assumes only the values 0 and 00, then the m ea sure space (X, A, m) is called degenerate. Any set A E A is called a measurable set ( relative to the

m).

measure

Definition. Let

6'13.

m be a measure function on a a-ring A

of subsets of X. (1) Completeness

of me�sure function. Th e measure function is said to be complete if B e A s.t. m(B)=O, A C B =- A � A i.e., if all the subsets of a set of me asu re zero are measurable. In this case (X, A, m) is call'Jd a comple te measure space. m

(2)

An y set A€ A i9 said to have finite measure if m(A)< IX..,

(3)

The measure of any set A e A is said to be C7-finite if n EN> s.t.

3 a sequence < A" E R: 00

,

(ii)

m (A,,)

(i)

A C U A..

(4)

The measure fu nct io n

(5)

The measure function

11=1

< 00 \j n e N.

m is cal led finite or a-finite accor­

ding as the measure of every set is finite or a-finite. a-finite)

(i) (ii)

m

is called totaUy finite (totally

if X e A i.e. A is algebra of sets. m(X) is finit,e, (�r a-finite).

6''(4. Herid �ta.ry froperty. A n:1n-empty class S={Si} of sets is said to. have. heriditary if SI C S,. S� e S => SI E S. 6'15.

Caratbeedory's postulate for outer measure. [Kolbapur 70; Banaras 66,65] o

Let an extended real valued set fu nction mo defined on 'U

51

MERSURE AND OU fER MEASURE

heriditarya-ring A={A.} have the following properties: (i) mo(A,) ~ 0 and mo(Ar)=O if A,=p (ii) Ar C A. mo(A,) ~ mo(A.) (iii) mo (u Ar) 1:mo(Ar) r

I'(A)=I'(11) if Xo E A or A. Again if x. e: B, then Xo e: A or Xo ,. A. Hence if Xu e: B, then ",(A)= 1 or O. This ... ",(A) ~ I'(E). For 0 < 1. Finally A C B ~ I'(A) ~ p.(E). (ii) To prove ",(A U 8)=I'(B)+I'(B) if A n B=.p. We have Xu ¢ A U R or Xo E:: A U B x. ,. A U B => Xu ~ A and x. ~ B. Conscquenrly I'(A U B)=O => ",(A)=O=",(.8). :. ,..(A U 8)=I>'(A)+I'(B). In the second case, Xu e A U .B => Xo E A and Xo ¢ B or Xu e: B and Xu ¢. A,for A n B=rf.. :. I'(A U B)=1 => ,.(A)= 1,1'(8)=0 or I'(B) = I, 1'(04)=0• . This => I'(A u B)=,...{A)+I'(B). Hence in both cases, (iii) is proved {iv) To prove p.( U A .. ) C;;; E 1'(14,,), where A" C X.

Xo

e

x,,.

x.,.

"

"

We have x. ¢. U A. or Xo E U A.. If Xo ~ U A", then x. ~ An ¥ n :. If 1'( U A.. )=O, then ,,(...4,,)=0 ¥ II so that

..

i: ~(A,,)=O+O+O+ ... +0=0 :.

p,(U A .. )=1: ",(A,,). "

II

In the second case if Xo E U A", then x. one value of If or Xo E A" V-

ft.

e

A

n.

for at least

MEASURE AND OUTER MEASURE

i.e.,

p{A

no

)= 1

or p{An)= 1 ¥ n.

Hence if p( U An)= I, then

.

E p{An )=O+1+0+ ... +0=1 n 00

l.: (J.(An) = 1 +l-f-} + ...

or

+ 1=00.

"""I

This or

~ p

(U An)=E P{An) p (U An)<E p.(An) ~ p(U An)~Ep(An)'

From (i), (ii), (iii) and (iv), it follows that p is an outer measure function. Theorem 3. An outer Ineasure is monotonic and a - sub additive. [Kanpur M.Sc. Statistics 1976; Banaras 69] Proof. Let A be a a --ring of sets which are subsets of X. Let (X. A, mol be a (outer) measure space. (i) To prove that mo is monotonic. For this we have to show that lno(A) IE;; lno(B) ~ mo(C) ::;:;; me(D) ~ ...... if A C Be C cD ..... . Consider the case in which A C B. Notice that B=(B -A) U A, (B-A) n A-t/l. This::> mo (B)=lno (B-A)+mo (A) ~ mo (A). For Ina (B-A) ;;;;;, 0 ::> mo (B) ~ mo (A) ~ Ino (A) ~ Ino (B). Similarly B C C, C C D . ::> Ino (B) ~ mo (C). mo (C) ~ Ino (D).

Combining these results mo (A) ~ 111" (8) ~ mo (C) if

E; Ino (D)

(1) U E,= U At 1=1

i=l

n BI=~ for i~j.

(2) E, (3) B.

(2)

=> mo

(,~

=> mo

(U A,)== 'f

c

Et)= ,~ mo (D,) 1110

(E,) by (I)

~}; 1110

(A,) by (3)

i_I

t=l 00

,=1

mo

A.. ¥ n.

(u AI) ~ rr: • --1

1=1

mo (A,) .

Theorem 4.

If T be a subset of the set of real numbers s.l. and if p.(A)=m. (A n T) 'V A C R, then show that p. is an outer measure (Caratheodory's outer measure).

0< ma (T) < Proof.

00

Let A, B, C, T C R be arbitrary, but Tis fixed and

o< Also we define

Ii

mo (T)


m(B)

But

.s;;; 0

~ 0 ¥ B. m(B) ~ 0, !nCB) ~ 0 => m{B)=I.J.

m( 8)

Since any set of measure Zero is measurable and hence B is measurable. Proved.

Problem~.

Let A be a mo measurable subset of B,

< is a monotonic

lJcfiue

From this it hence Lim t .... exists. Also Lim E.. =E. For allY set TC X.

... (1)

sequence

T=TnG-t-TnG'.

...(ft.)

By Caratheodory' postulates for outer measure. i.e.

ma(T)=m\l(TnG-t-1'nG') ~ mo(TnG)+mo(TnG') moU') '- mo( . nG)+mo(TnG')

Tn

Sct Then

... (2)

G=e

T=,EU(TnG'):JE.. U(TnG'), by (1) (110(T) ~ 1110 [Ly, U (TnG')] Evidently d (En, TnG') > O. :.

This follows frum our construction (1). Then 1110 lE.. U (TnG')]=m o (En)+m o (TnG') Making n ..... oo.

... (3)

61

MEASURE AND OllTER MEASURE

Lim rn [E" U (T

n

G')]= Lim rno (E.. )+mo (T

n-+oo

n G')

=mo (E)+rno (T n G') =lno (T n G)+mo (T n G')

... (4)

Making ll-oo in (3) and using (4), mo (T) ;;;;, rno (T n G)+rno (T n G'). Combining this with (2), we get rno (T)'=rn o (T n G)+mo (T n G'). This => G is outer measurable. Proved. Theorem 9. If < E,. > is monotonic non-increasing sp.quence of measurable sets (outer measuralyle sets), then the limit set ('~

E=

n Ek is a measurable (outer measllrable) set and for every T

k~l

ojfinite outer measure, rno (T Proof.

(i)

n

E)=

Lim

n-co

rno (T

n

En).

To "how that Lim rno (T n-+oo

Lim rno (T

n-+ 00

n

En)=m o (T

n

n

E.. ) exists and

E).

L':!t T be a set of finite measure. Since El ::J E2 ::J £3 ::J ...... It follows that T n £1 ::J Tn E2 :J T n £3 ~ ..... . Therefore 1110 (T n EI ) ~ Ino (T n E2 ) ~ Ino (T n E3)";;;.··. Consequently < 1110 (TE,,): /lEN> is monotonic non-increasing sequence. In brief we shall denote Al n A2 n Aa n ... , by AIA2Aa ..... Similarly the set A U B U C is denoted by A+B+C. Also every member of < rno (TEn) : n EN> is non-negative. Hence this sequence has a limit so that Lim

n--.oo

1110

(TEn) exists_

00

~n~C~=>EC~=>TnEcTn&

"=1

=> TE

C TEn

Lim

=> rno(TE)

=> n--".>oo rno (TE)

n_---

~

< n-+oo rno (TEn)

Lim · T ak mg rno (TEn)=},., we get rno (TE) 'IV

rno (TE,,)

Lim

< .\

••

·.(1)

62

MEASURE AND OUTER MEASURE

The set T can be broken as T=TE+TE1' + E1TE2 ' +E2 TEa'+·· .. ··+&.-1 TE..' ...... (*) Making use of Caratheodory postulates for outer measure, Ino (T) ~ mo (TE)+m o (TE1')+m O (EITE2')-i-mo (F2TEs') ...... +mo (£"-1 TE,,') ... (2) Since each En is outer measurable, mo (T)=nto (TE.. )+mo (TEn') Ino (T)=rn o (TE1)+mO (TEl')

... (3) Taking E'II-l T for the set T and applying the condition of meac;urability to En, we get mo (EII - 1 T)=rn O (E.._1TE,,)+mo (En- 1 TE.. ·) or rno (E.- 1 T)=rn O (EnT)+rn o (En_lTEn') [For En-I::> En => En n En_1 =EnJ. or rno (E"-lTE,/)=rno (TEn- 1)-rn o (TEn). ...(4) Writing (2) with the help of (3) :md (4) rno (T) ~ 1110 (TE)+[rn o (T)-mo (TE 1)J+[rnO (TEl)-rno(rE~)] +[nto (TE2)--mO (TEa)]+· .... · =[mo (TEn-I) -rno (TEn)] =m o (TE)+rno (T)-mo (TEn). Taking limit as n-'t-oo, mo (T) ~ mo (TE)+m o (T)- ,\ ... (5) or ,\ ~ rno (TE).

Combining (I) and (5), we get A=rno (TE) Lim i.e. lila (TE,,)=mo (TE). n-'?oo (ii)

To show that E

co=

n

,,=1

En is measurabl:::.

For any set T C .Y, we have T=TE+TE' ... (6) From which we get rno (T) ~ mo (TE)+rn o (TE'). ...(7) From (6), TE'=T-TE using this in (*), we get TE' =T' E1 ' + El TE2 ' +~ TEa' + ..... + E..~1 TEn' This gives mo (TE):::;;; mo (TE1')+mo (E1TE/)+m o (E2 TE3')+ ...... + mo (t~'-lT£,.. )

63

MEASU!,E AND OUTER MEASURE

Writing this with the help of (3) and (4), ~ [mo (T)-mo (TE 1 )]+[mO(TEl)-(mO (TE2)]

mo (TE')

+[mo (TE2)-rno (TEs)]+···+ [rno (TE"-I)--m O (TE,.)l =mo (T)- rno (TEn)

Making

n--l>-rrJ,

mo (TE') or

rno (T)

m (T) -- Lim rno (TEn)

n-oo

rn (TE') + Lim rno (TEn).

n-oo

Lim mo (TE,,)=m o erE)

But or

~

~

Ino

(T). ~ rno (TE'H-1I10 (TE)

Combining (7) and (8), we get rno (T)=mo (TE')+mo (TE) ~ E is outer measurable. Problems related to ring of sets, a--algebra. Problem 8. A Boolean ring B containing X (whole space) ;s an algebra Solution. Let B be a Boolean ring s.t. X E B. To prove that B is an algebra we have to prove that ¥ A eX=> A' E B A, X E B => X --A=A' E B, by def. of ring. Problem 9. Intersection of two B-rings is a B-ring. [Gujarat 1970]

This

Solution. Let 8 1 and B2 be two Brings. To prove that 8 1 n B~ is a B-ring, we have to prove that A, B E Bl n B~ => A-B, A U BE Bl n B2

n B2 A, B E Bl and A, B E 8 2 => A -- B, A U B E 8 1 and A -B, A U B E B2 ~ A-B, A U BE 8 1 n B 2 • A, B E Bl ~

Problem ]0. EI'ery algebra is a ring. Solution. Let A be an algebra. To prove A is a ring, we have to prove A, B E A=> (1) A -B E A. (2) A U Be A. (2) follows from the definition of algebra. A, B E A ~ A', B'. BE A, by def. of algebra. ~ A' U B E A ~ (A U B), E A ~ AN n B'=A n B'=A-B e A

64

MEASURE AND OUTER MEASURE

Theorem

to.

A mOll%ne ring j..s a a-ring.

Proof. Let A be a monotone ring. Let Ai E A ¥ i. By assumption A is closed under the formation of finite unions and n

so Al , Al U A 2 •

••••.• ,

U Ai etc. belong to A.

1=1

Write B,,-=.· U" Ai' Then Bn E A. Also

>


be a sequence of sets in A. To prove that 3 a sequence < Bn > of sets in A. s.t.

B,.

n Bm=,p for m:;t;n and

00

00

U B.= U Ai.

>=1

i=1

By assumption, Ai E A for i= 1, 2, 3, 4, ...... 11-1

Write

Bn=A,. -

U Ai' 1=1

Then . 0 0 0 0

This shows that (I) U Bi= U Ai i=1

i=l

(2) B .. n Bm=,p for m=l=n. (3) Bn C All "I n. => Al U A2 E A ~ Al U A2 U Aa E A.

(i)

"-1

Generalising this, we see that U Ai E A. i=1

n-l

U Ai E A, An E A

~

"-1

An -- U Ai E A, by (iii)

Bn E A => < B" > is a sequence of sets in A. Also we have seen that < B" > has the properties (I) and (2).

Hence Proved. Problem 12. If I' is completely a-additive set function on a class E and if < En t > is any sequence of sets from the class E, thell lim (I' £11)=1' (lim E,,). [Kanpur Statistics 1975. 76J Solution. < Eit > is non-decreasing sequence and its limit 00

will be U Eft.

This increasing sequence is converted into sequ-

ence of disjoint sets as : Write Eo=rb, Bn =EII -En _ 1 for n=l, 2, 3, ... 00

00

"=1

II-I

Then we have U B.. = U E,,=lim En

70

MEASURE AND OUTER MEASURE

... ...

go

lim &=

u

n=1

00

En= U (E.. -Bn-l) 10=1

Problem 13. Let", be a measure on an algebra A. (i) If An C A"+I' An e A. 1 EO; n < <X1 00 and ~ U An E A, thell p.

ai)

(

If A" :::> A"+l' A" '" (A 1)

'" (


o. 7·2. Measure of a closed interval. Let La, bJ be the smallest closed interval containing a closed set F. Then we define m(F)=b-a-m(F'). F' being the complement of F w.r.t. the interval. To prove that the measure of a closed interval is its length. Let A=[a, b]. Then A is a closed interval and hence a closed set. [0, b] is the st:Jlallest closl1d interval containing A.

74

LEBESGUE MEASURE OF A SEr

Complement of A relative to [a, b]=A'=cp. For A=[a. b]. By definition, m(A)=b-a-m(A')=b-a-m(q,)=b-a-O =b-a i.e. m([a. b])=h-a. 7'3. Measure of rectangle. The area of an open rectangle R (a < x < b, c < y -< d) i.e. (b-a) (d-c) is defined as the measure of R. Thus m(R)=(b-a) (d-c). The area of a closed rectangle R (a ~ x ~ h, c ~ y ~ d), i.e. (b -a) (d -c) is defined as the measure of R. Thus m(R)=(b-a) (d-c). 7'4. Measure of a parallelopiped. The volume of an open parallelopied V (a < x < b, c < y < d, 1< z < m) is defined as the measure of V. Thus m(V)=(b-a) (d-c) (m -I). Similarly the measure of a closed parallelopiped V' (a ~ x ~ b, c ~ y ~ d, I ~ z ~ m)

is defined as its volume. m{V')=(b--a) Cd-c) (m-/). Thus 7'5. Exterior and interior measure. (Lebesgue measurable set) [Kan~ur 75. 73. 68] The Lebesgue exterior measure (or simply exterior measure) of any set A, denoted by m,(A), is denned as follows: m,(A)=inf. {meG) : G :J A, G is open} ..• (1) From this it follows that meG) > me(A). Hence given. > 0, 3 an open set G :J A s.t. m(G) < me{A)+,. . .. (2) If the set A is contained in an interval (a. b), then

o ~ m,,(A) ~ b-a. Also and

Al C A2 =:> m.(A 1 ) m (0) ~ m [(a+., b-e)]=b-e-(a+e) => m (G) ~ b-a-2E. . .. (4) By (1), m (Ol)+m (0 2) < m. (A)+111. (A')+2e. Using (3), m (G) ~ m (Gl)+m (G 2 ) < m. (A)+m, (A')+2e or m (0) < m. (A)+m_ (A')+2 •. Using (4), we get or b-a-2f. < m, (A)+m, (A')+26 or b-a m, (A') < m. (,.,)+4& or m, (A)=b- a-m. (A{) < m. (A)+4. or m. (A) < m, (A>+4e. Since e is arbitrary and hence making E~O, we get the required result m. (A) ~ m. (A). Theorem 2. A sei .A is measurable (i.e. Lebesgue measurable) iff its complement A' is measurable. Proof. Let a set A be contained in an interval (a, b). Then we know that ... (1) m. (A)=b-a-m. (A').

76

LEBESGUE MEASURE OF

~

SET

Let A be measurable so that mi (A)=m, (A)=m (A). . .. (2) To prove that A' is measurable, we have to show that m, (A') =mi (A'). . .. (3) Writing (1) with the help of (2), m (A)=b-a-m, (A') or m. (A')=b-a-m (A). ...(4) (1) is true ¥ A. Replacing A by A' in (1), we obtain mi (A')=b-a-m, (A). For (A,),=A. Using (2), we get mt (A')=b-a-m (A). ...(5) Since, R.H.S. of (4)=R.H.S. of (5) Hence, L.H.S. of (4)=L.H.S. of (5) i.e. m, (A')=tni (A'). This => A' is measurable. (ii) Let A' be measurable so that ...(6) me (A')=mi (A/)=m (A'). To prove that A is measurable By virtue of (I), we have tnt (A)=b-a-m, (A') and mi (A')=b-a-m, (A). In view of (6), these equations become mi (A)=b-a-m (A') ... (7) m (A')=b-a-m. (A). ...(8) Equating the two values of m (A') b-a-m. (A)=b--a-m, (A). mi (A)=m. (A). or This => A is measurable. Proved. Note. Let a measurable set A be contained in [a, b). Then In; (A)=b-a-m. (A') m (A)=b-a-m (A') or m (A)+m (A')=b-a. or For A is measurable => A' is measurable. Theorem 2. (a) Evay bounded open set and bounded closed set are measurable. lMeerut 1987, 86] Proof. Let G be a bounded open set. Form G by means of a finite number of disjoint closed intervals. Then the sum of the lengths of these intervals is equal to the length (measure) of G. Hence m, (G)=m (G) ... (1) (i)

77

LEBESGUE MEASURE OF A SET

Thus G can be obtained in an open set consisting of G itself, and contains a closed set s.t. difference of measures of closed set and open set is arbitrarily small. Consequently m. (G)=m (G)=m. (G)

This ~ G is measurable. Since complement of a measurable set is measurable. Also complement of open set is closed set. This ~ Gf is measurable and G is closed. Theorem 3. . Let A and B be any two sets such that A e B. Then (i) mi (A) ~ m. (B) (U) m, (A) ~ m. (B). Proof. Let A e B. (i) To prove that mi (A) ~ m; (B). By definition of interior measure, Ini (A)=sup {m (F) : F is closed, Fe A} lni (B)=sup {m (F) : F is closed, FeB}. Let S be the set of numbers consistjng of measures of all closed subsets of A. Similarly we define the set T for B. Then m; (A)=sup S, mi (B)=sup T. . .. (1) F is any closed set s.t. F e A => F CAe B. For A C B. f

~

Fe B

:;. F is closed subset of B.

This shows that any m (F) E S ~ m (F) E T. Hence SeT. Consequently sup S ~ sup T. For least upper bound, i.e. supremum of a subset of any set can not exceed the least upper bound of the set itself. sup S ~ sup T =? mi (A) O. EI: is a meastJrable 3 a closed set F1: and an open set G" S.t. F; C cI: C G"o m (Gd-m (Fd

< T"

Write F=F1 U F2 , G='G 1 U G2 • Then F is a closed set and G is an open set so that F and G are measurable sets. F1, C Ek C GI •

:::>

2

2

2

/:~'l

1:=1

le·,1

U Fl. CUE/;. C U

G,:.

=>FCEcG => FeE, E c G => mi (F) m. (E)-mt (E) ~ m (G)-- m(F). ...(1) F'=A finite intersection of open sets

G -F=G n =An open set=A measurable set m (G-F)=m (G)-·m (F).

...(2)

Similarly m (G/ -x > - y. Proved. Theorem 7. If El and £2 are measurable subsets of [a, b] then prove that E2)' m (E1 )+ m (E2)=m (El U Ez)+m (El [Meerut 19S8, 87; Kanpur 79, 88; Kolhapur 70; Banaras 69] Solution. Let El and E2 be measurable subsets of[a, b) so that m.(E1)=m,(E1)=mr E1 ) } m.(E2 )=mi(Ez)=m(E2 ) ... (*) Firstly we shall prove that (i) m"(£1)+m,,(E2)~m.(EI U E 2 )+m.(E'1 n £2) (ii) mi(E1) .. mi(B2)e.l11i(F."t U E2)+mi(E1 n E2)· Prove this as in Theorem 6. Wr:ting (i) and (ii) with the help of (*) (i)' II1(E 1 )+m(E2 ) ~ m(£1 lJ E2 )+m (E 1 n £2) (ii)' IJl (E1 )+I11(E2 ) ~ m(EI U EzH-m(E l E2 ).

n

n

81

LEBESGUE MEASURE OF A SET

For finite union and intersection of measurable sets are meac;urable so that El U E2 and El n E2 both as measurable and so m. (El U Ea)=m, (El U Ea)=m (El U E2 ) me (El n E 2)=m, (El E2)=m (E1 nEs). Combining (i)' and (ii)' we get the required result. Problem. If El and E9 are measurable subsets 0/[0, b], prove that El U E 2, El n E2 and E1 -& are measurable and show also that (i) m(h.;)+m(E2)=m(El n E2 )+m (El U E2 ) (ii) m(E1 -E2 )=m (E1 )-m(E2) if E z C El [Kanpur M.Sc. 1979] Solution. (i) Prove as in theorem 7. (ii) Since difference of measurable sets is measurable :. E1 -E2 measurable set

n

(E1 -E2 ) n E2 =rp m [(E1 - E2 ) U E2J= m (E1 -E2 )+m(E2) or m(E1 )=m (E1 - E2)+m(E2) or m(El) - m(E2 )=m (E1 -E2) Problem 2. If A, B, 'C, are p.-measurable sets, show that p.(A)+p.(B)+ p.(C)=p. (A U B U C)+ p. (B n C) +p. (C n A)+p. (A n B)-p. (A n B n C). [Banarsas III, 1970] Solation. Here we shall use the following lemma: Lemma. If E1 , E2 are measurable subsets of [a, b], then m(E1 )+m(E2 )=m (El U E2 )+m (E1 n E2 )· Now we come to the proof of the problem. Let A, B, C be measurable subsets of [a, b]. Since finite union and intersection of measurable sets are measurable and hence A n B n C, A U B U C, A U B, A n C etc. are all measurable. Taking E]=A U 8, E2 =C in the lemma, we get m (A U B)+m (C)=m (A U B U C)+m[(A U B) n C]

:.

... (1)

Again, by the lemma. m (A)+ m(B)=m (A U B)+m (A n B) >r m (A)+m (B)-m(A n B)=m (A U B). Hence m (A U B) n C] =m[(A n C) u (B n C)]

... (2)

82

LEBESGUE MEASURE OF A SET

==m (A

n C)+m (B n C)-m (A n C n B n C).

[This follows from (2)] =m (A n C)+m (B n C)-m (A n B n C) or m[(A U B) n C]=m (A n C)+m (B nC)-m (A n B n C) . ... (3) Writing (1) with the help of (2) and (3) m (A)+m (B)-m (A n B)+m (C) =m (A U B U C)+m (A n C)+m (8 n C)-m(AnBn 0, 3 an open set Gn S.t. E. C G,., m (G n ) < m. (E;.)

E

+ Tn

LEBESGUE MEASURE OF A SET

83

An arbitrary union of open sets is open '1nd therefore -"

]; G, is an open set s. t.

'~1

... (1)

This follows from the fact that En C G", ¥ n. (1) =>

m.( UE, )-E;;; m, (U G, .)=m(u G r~1

r=1

'=1

00

r )'"

1: m(G,)

r_l

~ m. (,! E. ) ~ !,m (G,) .-- n~-2 Stage 1st 2nd 3rd

Number of intervals dropped 1 2

22

nth 2"-1 2,,-1 I" The sum of the lengths of the intervals dropped upto nth stage =/1 +2/2 +22 / 3 +", +2"-1 I" =.\- n~2' by assumption.

The limiting sum of the dropped intervals

LEBESGUE MEASURE OF A SE r

= nl:oo

l

A-

87

n~21=~-0=A'

Let A denote the set of end points of the interval {III} with their limiting points, then A is a non-dense perfect set. Also m(A)=m [0, I]-length of dropped intervals=l-A. If we take A=i, then m (A) = l-i=i. The set A formed as above is a non-dense perfect set with measure 1. Theorem 9. 'Any set A is measurable iff an open set G containing A and a closed set H contained in A can be so determined that I G I - I H I < 6, where I G I stands for length of open intervals in G. ' Proof. Definition of exterior measure => given. > 0, we can find an open set G :J A s.t. I G I < m, (A)+e/2. ...(1) Similarly we can find a closed set H S.t. m. (A) < I HI +e/2 or - I HI < (E/2)-m, (A) ... (2) (1)+(2) gives I G I - I HI < e+m. (A)-m. (A). ...(3) I. Let the set A be measurable so that m.(A)=mi (A). To prove 1G 1- 1if I < e. ...(4) Now (3) => I G 1- I HI < e+O. Hence the result. II. Let (4) be true. To prove A a'l measurable. The definitions of exterior and interior measure show that m. (A) ~ 1 Gland mi (A) ~ I HI. SO that m. (A)-m; (A) < 1(;1 - I HI < e, by (4). But E is arbitrary and so making e~O, m. (A)-m; (A) 0 or m. (A) < m, (A). But m. (A) ~ n1i (A) is true ¥ A. Hence m. (A)=m; (A) so that A is measurable. Problem 9. A subset E of [a, b) is measurablt> ~ffgiven e>O, :I open sets G1 , G2 S.t. G1 :J E, O2 => E', m (G1 n G2 ) < e. Solution. Let E C [a, b]. The definition of exterior measure sllg~ests that 3 op(!n sets G1JE, OJ :J E' s.t. m(G1 ) <me (E)+6/2, m (G 2) < m~ {£,)+./2. This => m (G1)+m (G~) < me (E)+m. (E )+e. But m (OI)+m {G 2 )=m (G l U G2 )+m (01 n O2 ), Hence m (G 1 U G2 )+m (G 1 n G2 ) < m. (E)+m, (£')+e .. (I)


E' ~ G1 UG 2 ::> EUE'=[a, b] => G1 UG 2 ::> [a, b] => G1 UG 2 =[a, b] as G1 U G. c [a, b] => m (G 1 U G2 )=b-a.

Now (1) => b-a+m (G 1 n G2) < m. (E)+m, (£')+_ ... (2) I. Let E be measurable so that mil (E)=mi (E). . .. (3) To prove

m (G 1

n

G2 )


E => m. (E) ~ m (Gl ). Similarly m. (E') ~ m (G 2 ) Hence mll(E)+m.(E');:;,:;; m(G1 ) +m(G2)=m(G1 U G2)+m(G1 nG2) or mil (E)+mll (E') ~ b-a+6. This ~ m. (E) ~ b-a-m. (E'H'E=m. (E)+ •. Making £_0, we get m. (E) ~ m, (E). But m. (E) ;;;;, m, (E). This => m. (E)=mi (E) s> E is measurable. Theorem 11). If A ill a countable set then prove that m.(A)=O. Proof. Let A be an enumerable (countable) set. To prove m, (A)=O. By assumption, A is expressible as A={xft : n EN}. or

Let each point x" be enclosed by an open interval of length in such a manner that these intervals form a family of disjoint open intervals. The measure of this family of sets is

_/2

ft

-

E

(J)

..1!.1 _/2"= I -(t) =f. This => A can be contained in an open interval of length _. This Q m, (A) EO; _. But II! is arbitrary and hence making 5 .... 0, we get mil (A) But m. (A) ;;iii 0 ¥ A. Hence m. (A)=O.

~

O.

Corollary l. Since the set Q of rational numbers is countable and so m. (Q)=O. Also m. (N)=O=m. (I). Corollary 2. Since the length of the interval (0, 1) or [0, 1] is 1 so that m. «0, ]»=m ([0, l])=l:;t:O.

89

LEBESGUE MEASURE OF A SET

Hence [0, 1] is not countable. Corollary 3. The converse of Theorem lO is not always true. Example. Outer measure of Cantor's set is zero but it is uncountable set. Also Cantor's set is itself measurable. Hence any set with outer measure different from zero is uncountable. Theorem 11. lfme (A)=O, then A is measurable. Or, to prove that a linear set 0" Lebesgue exterior measure zero is Lebesgue measurable. [Banaras 1966] Proof. Let m. (A)=O. To prove A is measurable. We know that m. (A) ;:;at mi (A) ~ 0 Y A. This => 0 ;;;, m, (A) ;:;at 0 => rni (A)=O. Thus me (A)=O=m, (A) or m. (A)=m. (A), this => A is measurable and its measure is zero. Ex. An enumerable set is measurable and its measure is zero. [Kanpur 1976 ; Nagpur 64 ; Banaras 69] Solution. Prove as Theorem lO that me (A)=O. This => A is measurable and its measure is zero. Problem 7. The difference of two measurable sets is measur[Meerut 1986J able. Solution, Let El and E2 be measurable sets. To prove that EI-E2 is a measurable set EI-E2=El

n

E 2 '·

E2 is measurable => Ez' is measurable ~ El n E 2 ' is measurable ~ EI-E2 is measurable. Problem 8. If a measurable set G1 contains another measurable set G2 , then show that G1 -G2 is measurable alld its measure is 1I1(G1 ) -m(G2 )· Solution G~ is measurable => G2 ' is measurable => G1 n G2 ' is measurable => G1 -G2 =G 1 n G?' is measurable Remains to prove that m (G 1 - G 2 )=m (G 1 )-m (G 2 ). Since (G 1 -G2 ) n G2 =rp. Hence 111 [(G 1 -G2 )+G1 J=m (G 1 -G 2 )+m (G 2 ) or In (G1)=m (G 1 -GZ)+m (G 2 ) or In (G 1 }-m (G 2 )=m (G 1 -G2 ). Problem 9. A singleton set is a meas!lrable set and its measure is zero.

90

LEBESGUE MEASURE OF A SET

Proof. Let A={a} be a singleton set. To prove that A is a measurable set. Since every singleton set is a closed set and a closed set is a measurable set. Hence A is measurable. Furthermore A can be regarded as closed interval [a, a] which contains only the point a. By definition of measure m ([a, a])=a -a=O. m (A)=O. ' Problem 10. Sh~w that the Lebesgue measure of the following set is zero {x E R : 0 x",¢~ ¥ m =>~ it E ... (3) according to (I). ": f",=O or 2 ¥ m. This => t e: E ... (4), according to (2). Evidently (4) contradicts (3). Hence the required result follows. Tbeorem 12. Show that the Lebesgue exterior measure ;s a [Banaras 1967] Caratheodory outer measure. Proor. To show that Lebesgue exterior measure is Caratheodory outer measure, it is enough to show that

and if

(i)

m. (E) ;;. O.

(ii) El C E z => m, (E1 ) U F" C U Ek C U G" "=1 k~l "=1

=> Fe E C G. => FeE, E c G. => m, (F) ~ mt (E), m. (E) ~ m. (G) (Refer Theorem 3) => m(F)=mi(F) ~ mi (E), m. (E) ~ m. (G)=m(G) [For Closed sets and open sets both ore measurable sets] => m(F) ~ m, (E) ~ m. (E) ~ m (G) => m. (E)-m, (E) ~ m (G) -11'1 (F). . .. (2)'. G-F=G n F'. For G n F'=Grt(X-F)=GnX--GnF=G-F. =A finite intersection of open sets =An open set=A measurable set G={G-F) U F and (G -F) n F=r/>. :. m (G)=m (G-F)+m (F). From which, m (G-F)=m (G)-m (F) Similarly, III (G,. -F, )=m (G,) In (Ft ). Notice that G

n

Feu (G,. -FJ. This => k-l

III

"

(G -f) ~ E k=l

III

(G k - Fd

98

LEBESGUE MF.ASURE OF A SET

~

m (G)-m (F) .;;; E• [m (Gt)-m (Ft )] 11=1

.. m (G)-m (F) m. (E') ~ m (S,/) [For S. is measurable~ by theorem 13 and so S,,' is measurable]. => m. (E') ~ m (S,,')=b-a-m (Sn) =b-a- "E m (E,,) r=1

~

m. (E')

~

" m (Er) b-a- E r=1

n

=> E m (Er) r=1

~

b-a-m. (E')=m, (E),

100

LEBESGUL MEASUF.E OF A !'oET

:0

• m (E) m, (E) ;;. E 7'=1

Making n-+oo; we obtain m. (E)

~

00

E m (Er) 7'=1

=> m, (E) ;;;;.

E m (Er) ~ m. (E) ... (3), On using (I)

7'=1

.. m, (E) ;;. m. (E). m, (E) ~ m. (E) is always true. :. m. (E)=m. (E). This ., E is measurable and But

.. m (E)

~

00

E m (Ek ) ~ m (E), by (3). ~_1

This concludes the problem. Theorem 16. Second Fundamental Theorem.

If E 1 , E 2 ,···are

00

meosuroh/e sets, then the set U Er is measurable. 7'=1

Proof. E=

00

U E~ is a measurable set. k_l

Let

By De-Morgan's law,

F'=[ nE,; ]= UE t'

~=l

k-l

Ek is measurable

::>

its complement Ek ' is measurable. 00

=> 1: E,r' is measurable 1:=1

.. F' is measurable => F is measurable.

Lialtinl Sets. Theorem 17.

Suppose that


be a monotonic increasing sequence of measurable sets, so that 00

El C E2 C £3 C···· .. Also let E= U £1" lII=1

An enumerable union of measurable sets is measurable. Also difference of two measurable sets is a measurable set. Hence E and E,+l-E,. are measurable sets. Notice that

£=E1 +(E2 - E1 )+( E3 - £2)+'" +(E,-E,-I)+ ..... . 00

£=E1

i.e.

+ ,=1 ~ (E"+l-E,,)

is a disjoint union of measurable sets. Therefore 00

m (E)=m (El) + ,,_1 ~ m (Er+l-Er)

... (J)

" (m (Er+l)-m (E,)] m (E)=m (El )+ 1:

or

,=1

or

... (2)" .

For

00

2-' [m (Er+J--m (Er)]

,.=1

-=Lim [{m (E?)-1J1 (E1)}+{m (Ea)-m (E2 )} n-+oo

+ ... +{m (En)-m (Ell-1m

102

LEBESGUE

MEASU~E

OF A. SET

Lim m (B)= n_oom (En).

From (2),

Ex. Show that if < Eft > is sequence of measurable sets all contained in a set offinite measure and Lim En exists, then Lim Eft is measurable and m (Lim E,,)=Lim m (Eft). [Banaras 1967J Soilltion. Let < En > be a sequence of measurable sets. Then this sequence will be either monotonic increasing or monotonic decreasing. To be particular at this stage, we further suppose that the sequence is monotonic increasing so that E1 C E2 C E8

c ..... .

00

Let E= U Ek • Now prove as in the abOYe theorem 17 that .10-1

m

,lim En )=lim \n_oo n_oo

In

(E). /I

Theorem 18. If E 1 , Ez, ...... are measurable sets and 00

E= r_1 n E~ and if El ::> E2 :) ... , then Lim m (Eft), z.e. . m (Lim Eta ) = Lim m (En). m (E)= n~;rJ n_ 00 n_oo Or Let p. be a countably additive non-negative measure defined on a a-algebra of subsets of a set X such that I' (X) < 00. ~lrow that if {An: n E N} is a contracting sequence of I'-measurable subsets of X, then I'

[

n ] Lim n E N An == nE N

J.I. (An).

[Banaras 1964J Proof. Let < Eft > be a contracting (monotonic decreasing) sequence of measurable sets so that ~I ::> E2 ::> Ea ::> ...... Also let

00

E= n En.

,,-I

Tken E=

Lim

"_00 E.

103

LEBESGUE MEASURE OE A SET

i.e.

... (1)

is a disjoint union of measurable sets. Since an enumerable union and intersection of measurable sets are measurable. Also difference of two measurable sets is measurable. Now (1) is expressible as 00

m (E1)=m (E)+ l: m (Er-Er+1 ).

...(2)

r=1 00

m (E1)=m (E)+ E [m (Er)-m (Er +J )]

or

r=l

"-1

or

E [m (Er)-m (Er+l)]

r=l

or

... (3)

= n-+oo Lim [m (El)-m

)-riz (E.)1 + ... T[m (E"-l)-m (E

(Eq)]+[m (E2 ~

rI ) ]

= Lim

(3)

[m (E

) -

m (En)]

1 n-.oo can also be written as

Lim m (E1)=m (E)+m (E1)- n'+oo m (En)

or

m (E)=

Lim n~OO

III

(En).

Theorem 19. The set of all measurable (i.e. Lebesgue measurable) sets is a--algebra (or a-field). Proof Let A be the set of all measurable sets. Let A, B E A be arbitrary so that A and B are measurable sets.

To prove that A is a -a-algebra, we have to show that (i)

A' E A

104

LEBESGUE MEASURE OF A SET

(f'i) AU.:; .2 A (iii) if < En > is a sequence of measurable sets in A 00

u

then

E, E A.

~=1

Let all the sets belongiilg to A be contained in an interval (a, b). (i) Let A be measurable so that m. (A)=m, (A)=m (A) .... (1) To prove that the complement A' of A is measurable. ...(2) We know that m, (A)=b-a-m. (A'). This is true for every A. Replacing A by A', m, (A/)=b-a-m. (A) ... (3) writing (2) and (:S) with the help of (I). m (A)=b-a-m. (A') mi (A')=b-a-m (A) m. (A/)=b-a-m (A)=m, (A') From which, or m. (A')=m, (A'). This => A' is measurable. (ii) Let E 1 , E2 E A so that El and £2 are measurable sets and E=E1 U E 2• let To prove that E E A, we have to show that E is measurable. Consider an arbitrary number c > O. Ek is a measurable set => :I a closed set F" and an open set Gle e s.t. F.,. C E" C Gk , m (G,,)-m (FTe)

E

< T'

Write Then F is a closed set and G is an open set so that F and G are measurable sets, 2

FTe C Ei C G" => U Fit .=1

~

o

I

I

e i=1 u Ek e II_I u

G7r,

Fe Ee G.

=> FeE, E e G• .. m, (F) ~ m, (E), m, (E) m, (E) ~ m. (6) ~ m (G)

~ In

(G)

m (F) ~ [For m, (F)=m (F), m. (G)=m (G) and m, (A) m. (E)-m. (E) ~ m (G)-m (F). •.. (4)

LEEr~Gt'E

lOS

MEASURE OF A SEt

This

2

~ m

(G-- F) E;', l: m (G,,-Fre ) k=l

=> m (G) -m (F)

~

k:l 2

[11'1 (Gre)- 11'1 (Fk ) ]

e

< 2.2=.

=> m (G)- m (F) < , => m. (E)-m. (E) ~ m (G)-m (F)

=> m, (E)-m. (E)
m. (E)=m. (E) => E is measurable.




be a sequence of measurable sets and let 00

E=

l- En .

.. =1

To prove that £ is a measurable set. For the sake of convenience we suppose Ek nE/:'=0 for k:l=k'. Here the operation E stands for u. We know that

or or -

m. (E) Let

00

~ ~ .=1

m (E,,).

. .. (5)

•

S.. = 1.' Ek • 1c~1

El and E. are measurable sets. El U E. is measurable by (ii), => S.. is measurable

106

LEBI:SGUe MEASURB OF A SET ~

its complement S ..' is also measurable.

m (S..)=m (

.

i.e.

1:

Jt~l

EII)=

i:

k=1

m (Ek )

m (S..)= 1: m (Et ).

. .. (6)

11=1

S.. = 1:" EkCE

i.e.

Sn C E so that E' C S,.'.

1=1

E' C S ..'

~ m~ ~ ~

==-

(E')

~

(E)

~

me (S,.')=m (S,/) m~ (E') .s;; m (S,,')=b-a-m (S.. ) m. (E') ~ b-a-m (Sn) m (5..) .s;; b-a-m. (E')=m. (E) m (S.. ) ~ mi (E) II

~ nt.

m (S..)= }; m (EA:) .=1

~ 111.

(E)

"

~ E 11=1

m (Ek ).

.Making n-'?oo, .

m. (E)

.

~ .

00

L: m (Ek )

1:=1

;;;;.

me (E), on using (5).

mi (E) ~ m~ (E). But m. (E) ~ m. (E) is true for any set. Consequently m. (E)=m. (E). This ~ E is measurable. This concludes the problem. 7'8. Definition. Convering in the sense of Vitali. [Banaras IV 70] Let E ·be a set and M be family of closed intervals such that none of them is a singleton set. The set E is said to be covered by the family M in the .sense of Vitali if I.rf x E E and ¥ ~ > 0 ~ 3 a closed interval I E M s.t.XE/,

or

In

(I)


of subset of R, prove that m"'( U An) ~ I. 11/* (A .. ). [Banaras Ill, 1971J ,

113

LeBESGUE MEASURE OF A SET

11.

If A, B, Care I'-measurable sets, show that peA) +I'( B) +1-'(C) =,..( A U B U e) +1'(Bn e) + l'(en.4) (a)

+1'(.4 n B)-I-'(AnBnC).

(b) zero:

Show that the Lebesgue measure of the following set is

{x E R : 0


al ...(1) is measurable for every extended real number a. The set {x E E: / (x) > a} is denoted by the symbol E(/> a). Remark (1) The condition (1) of the above definition can be replaced by any of the three conditions of theorem 4, page 118. Remark (2) The measure of the set E (/ > a) may be finite or infinite. Remark (3) If E=R, then the set E (/ > a) becomes an open set. 8'1. Definition. Almost everywhere. A relation, which holds except ID a set of ,measure zero, is said to hold almost everywhere. That is to say, a property P is said to hold almost everywhere on the set E if 3 Eo C E S.t. (i) the property P holds 'V x e E-Eo (ii) m'(E.,'=O. The following have the same meaning (i) almost everywhere on the set E (ii) for almost all points of E. The phrase almost everywhere, is denoted by the symbol a. e. or p. p. S·2. Definition. Equivalent functions. Two functions/and g defined on the same set E, are said to be eqUivalent if m [E (f:,cg)]=O. That is to say, functions / and g defined on the same set E, are said to be equivalent if 3 A, BeE S.t. E=A U B,J=g on A,J:;t:g on B, m (B)=O.

116

MEASURABLE FUNCTIONS

.·3.

De6aition. Characteristic function. [Meerut 1990] Let A be a subset of a set E. The characteristic function KA of A is defined as : I if x E A KA(X)= { 0 if x E E. . A

The function Kix) is measurable iff A is measurable. Hence existence of a non-measurable set implies the existence of a nonmeasurable function. The function K,.(x) is also called indicator flUlCtJo. of A . ..... Deftnition Simple Function. [Kanpur Statistics 1977, 75] A function is said to be simple function jf the range of the function is finite. For example, characterist;c function. , A real valued function ~ is called simple if it is measurable and assumes only a finite number of values. If ~ is simple and has finite number of values Ill> 112, ..... «ta, then", is expressible a'l

.

~(x)=

E _, KAi '=1

where A.={x : ~(X)="I}' The sum, product alld difference of two simple functions are simple. 8'5.

Limit Superior aad Limit Inferior Let < Ar : r EN> be a sequence of sets.

The supremum

00

of this sequence, denoted by set U A, and the limit superior of r~l

__

__

,,",,00

this sequence, denoted by Lim A" is defined as lim A.= n U A... n=1 tn._II

The iufimum of the sequence, is defined as the set

00

n

A.. and

..=1

lilllit ioferior of the sequence is denoted by Lim A•• and is defined as

Lim A.. =

00

00

a-I

_::::IB

unA...

If Lim A.. = Lim A". then we sar that the limit of the sequeace exists and the common value is dC!noted by lim A... Thus Lim A.=Lim A,,=Lim All if Lim A .. exists.

117

MEASURABLE FUNCTIONS

Monotone Sequence. A sequence of sets is said to be non-decreasing (expanding) if A,. C A,,+! V n and is denoted by < A. t >. A sequence of sets is said to be non-increasing (contracting) if A.. +! C A,. ¥ n and is denoted by . In either case the sequence is said to be a monotonic sequence.

Theorem. Every monotone sequence of sets is convergent. Proof. Let be a non-decreasing sequence, then

By def. lim inf A.. = lim sup A.. =

n01

U

n=1

[n Am]= UA,. Ift=..

[In~" A.

... (1)

11=1

J

... (2) 00

But we see that for any non-decreasing sequence, U Aft is in,,=11

dependent of k, so we may take k= I. Lim sup A,,=

lienee, by (2),

n [u J= U R=l

m-.l

A",

m~l

A".=

U A...

n-l

00

Hence, lim sup An= U A.. =lim inf A.. .. =1 00

or lim sup A.. =lim inf A .. = lim A .. =

U An. n=1

Hence monotonic non-decreasin~ sequence is convergent. Similarly we can also prove that monotonic non-increasing sequence is convergent. Theorem 1. Let f be a measurable function defined over a 00

measurable set E,: ¥ kEN and E = U E/... 1 0), (iii) E (f < 0), (iv) E (f ~ a), where 0 is any real number and E (f > a) stands for the set (x E E :f(x)

>

o}.

Similarly we define the other three sets. To prove that these four sta.tements are equivalent. (i) ~ (ii).

,For

E(f~ a)= ~1

[(

Ef> a-

~)]

an arbitrary intersection of measura.ble sets is measurable. (ii)

~

(i).

For £

(I >

0)=

~1 E( f ~ a+ ~ [

)]

and an arbitrary union of measurable sets is measurable. (ii) (iii).

For E (f < a)=[E (I ~ a)J' and a set is measurable iff its complement is. Here complement of every set is taken w.r.t. the set E. (iii) ~ (iv). For

£(/~ a)=

n£(f< ~+~) n

n=1

and an arbi~rary intersection of measurable sets is measurable. (iv) ~ (iii). For

E(f < a)= ~1 [E ( f ~ a -

! )]

and an arbitrary uni"n of measurable sets is measurable. This completes the proof. 00

,

Remark. 1.1. {XEE :f(x);;tGt}= U {x E E :f(x) > «-(lin)} n_1

2.

E(/ :) if e > 0

E(

I < :)

E (el> a)= {

if e

< O.

Both the sets on R.H.S. are measurable. Hence E (ef > a) is measurable and so is el. Therefore el is measurable VeE R. (ii) To prove that I f I is measurable. [Indore 78J E ( III < a)=[E (I < a) U [E (j > -a)] if a ~ O. Also a finite union of two measurable sets is a measurable set. E is given to be measurable. Therefore the measurability of E ( II I < a) follows. [Kaopur 83J Consequently III is measurable over E. For I x I < a => -a < x < a. (iii) To prove that c+1 is a measurable function. E(c+l> a)=E(/> a--e) =E (I> b), on setting b=a-c:, = measu rable set. Hence the result (iii) follows. (iv) To prove that 12 is a measurable function. E (f2 < a)=E ( III < ya) if a ~ 0 But E( If I < va)=[£(/< va)] u [E(f> -va)J. : . .E(f2 < a)=[E(/< v'a)J U [EU> -ya)] if a ~ O. Hence the measurability o~~~onseqUentlY (2 is measurable over F. . -_·;i ,'- '0 . (v) Let / vanish no\\ here on E so that f (x):;i:O V x E E. . H ence fIeXists. E ( / > 0) ifa=O if a;;;.. 0 fE(/>O)]n[E(/ a --= [[E(f< O)J n lE(/< I/a]) U [E(/>O)] ifa a ) is measurable and so is ;

.

By the theorem, just proved cl is measurable function VeE R :;. ~I is measurable function, on taking c= - I.

.r is measurable =>

122

MEA SURA BLE FUNCTIONS

Theorem 6. Let f and g be measurable functions defined over a measurable set E. Show that f+g, f - g, fg ore measurable fune. tions over E. [Meerut 1986 ; Banaras 69, 67 ; Kanpur 79, 77, 71] Proof. Let f and g be measurable functions defined over a measurable set E. To prove that f+g, f-g, fg are measurable over E. Firstly we shall establish the following lemma: Lemma. If f and g are measurable functions defined over a measurable set E, then the set E (f > g) is measurable. [Kanpur 76, 7S, 72] The proof of the lemma starts. Let f> g, then by the corollary to the axiom of Archimedes there exists a rational number r S.t. f> r > g. Therefore E (f > g)= U [[E (f > r)] n [E (g < r)]]

reQ =An enumerable union of measurable sets =measurable set Hence the lemma. For (i) Q is an enumerable set (ii) fand g are measurable => E (f > r) and E (g a)=E( f> a-g) = measurable set, by the lemma just proved. For f and a -g both are measurable functions over E. Thus E (f+g > a) is a measurable set and hence f+g is a measurable function. II. To prove that f-g is measurable. f and g are measurable over E => f and - g are measurable over E q f-g is measurable over E . •: f+ (-g) =f-g.

123

MEASURABLE FUNCTIONS

III. To prove that fg is measurable over E. f and g are measurable functions over E => f +g, f - g are measurable functions over E => (f+g)2, (f-g)! are measurable functions over E . ~ (ftg)2_(f _g)2 is a measurable function over E =>

!

[(f - g)2-(f-g)2] is a measurable function over E

=> fg is a measurable function over E. Alternate method. To prove that f+g is measurable. Since for any real number «, 3 a rational number r s.t. f(x) < r < t1.-g (x) whenever f(x)+g (x) < t1. {x E E: f(x)+g (x) < «} =U ,. [(x E E: f(x) < r} n {x E E: g (xi < «-r}l. This union is taken over a set of rational numbers. It follows that f+g is measurable. Similarly we can prove that f - g is also measurable. Theorem 7. If f and g are measurable functions defined over a measurable set E and if g vanishes nowhere on the set E, tlren the quotient function

.~

is measurable Ol'er E.

[Kanpur 1979, 77J

Proof. Let f and g be measurable functions defined over a measurable set E. Also let g vanish nowhere on E so that g (x):;eO '" x E E. Then

I.

~ g

exists.

To prove that fg is measurable. lGujarat 1970J f and g are measurable functions => f+g. f--g both are measurable functions ~ (f +g)2 and (f - g)2 both are measurable functions => (f+g)L·(f g)2 is a measurable function => 1 [(f+g)Z-(f-g)2] is a measurable function fg is a measurable function.

=-

II. To prove that..! is measurable over E. g

Let a be any real number.

124

MEASURABLE FUNCTIONS

E(g>O) (E (g > 0)] () (E (g

>

[E (g

O)]U[(E (g

ifa=O

< I(a)] if a>O < O)]()[E Ig < I/o)]]

ifa

Hence E ( ;

a ) is measurable in every case.

Therefore) is measurable over E. g

III.

To prove that f

Evidently

g

is measurable over 1:;.

Lg =f. J.-. , ~g g

is measurable function, by II

= product of two measurable functions. =measurahle function, by J.

Hence the result (Ill). Pr0l1e that the space of measurable functions is closed under the usual operations of Arithmetic. [Bbagalpur 1968J Theorem 8.

Proof. The fundamental operations of Arithmetic are: addition. subtraction, multiplication, division. Let f, g be arbitrary members of the space of measurable functions. If we show thatf+g, f-g, fg and f

g

(if g vanishes nowhere

on E) are mea~urahle functions, the result wili follow. Of course the functions f and g are measurable over a measurable set E. Here write the proof of Theorems 6 and 7. Ex. 1. !If and g are real valued measurable functions on a measurable space X, prr l'e that the f U g, f () g are measurable [Gujarat 1970J functions. Solution. Let c he any real constant. Letl and g be real

va.lued measurable functions defined on a measurable set X so that X ( f > c) and X (g > c) are measurable sets. (i) To prove that f U g is measurable. X (f U g> c)=[X (f > c)] U [X (g > c)] =finite union of measurable sets =measurable set.

125

MEASURARLE FUNenONS

This proves the result (i). (ii) To show that f n g is measurable X (

rn

g

>

c)=[X (f

>

c)] () [X (g

>

e))

=finite intersection of mea!>urable sets =measurable set. Thi~

=> the result (ii).

Ex 1. Show that the set of all measurable functions form a vector space over R. [KaDpur 1914]

Solutio.. Let E be a m~asurable set. Let V be the set of all measuruble functions defined over E. Then Y={f:f: E+R is a measurable function}. To prove that Y (R) is a vector space. Firstly we shall prove the following lemmas. I.emma 1. Iff is measurable, then af is measurable, a being any real number. Also let c be any real number.

if a

>

0

if a


c) is measurable so that af is measurable. Lemma 2. Iff and g are measurable, then f+g is also measurable E(f>g)= U [E (! > r > g)], Q=set or rational numbers reQ U

[E (f

>

r)]

n

[E (g


ag is measurable => -g is measurable => c - g is measurable. Now E {f+g > a)=E ( r < a -g)=measutable set For E ( f> g) is measurable. Now we come to the main problem. Let!. g, hEY be arbitrary. Then a, b, c E R, R being the set of real numbers. (i) (Y, +) is abelian group l. !+g E Y, by Lemma 2. 2. 3 0 E V s.t.f+O=O+f=/.

126

MEASUI,f x. It means that 0 is a special case of a constant function and so it is measurable. Also (/+0) (x)-f(x) + 0 (x) =1 (x) +0=1 (x), this. 1+0-1. 3. -f E V s.t. -1+1=0, by Lemma 1. 4. f+g=g+/as (R, +) is an abelian group. 5. 1+ (g+h)=(f+g)+h as (R, +) is an abelian group. (ii) al E V, by Lemma 1. 6. (a+b)/=a/+b f 7. a (/¢-g)=a/+ag. 8. J.I=f. (6), (7). (8) follow from the fact that (R, +, .) is a field. From what has been shown it follows that V (R) is a vector space. Theorem 9. Let < /" > be a sequence 01 measurable /unctions defined over a measurable set E. Then sup {f1'/2'/3 ... } and inl { J;. h. la ....... } are measurable over E. [Kanpur 1986, 76] Proof. Let < In > be a sequence of measurable functions defined over a measurable set E so that E ( I .. > a) is a measurable set ¥ n, a being any real number. Write g(x)=sup {/,,(x) : n=l, 2, 3, ...... } h (x) =inf f/n(x): n=l, 2, 3, ...... } hex) can be defined in terms of supremum as follows: h(x)=-sup {-f,,(x): n=l, 2, 3, ...... }. To prove that g and h are measurable over E, it is enough to show that g is measurable over E. E (g

>

c

a)= U [E (fn ,.z:;l

=

""U

.=1

> a»)

{x E E :f,. (x)

>

a}

= An enumerable union of measurable sets =measurable set Theorem 10. Let < III > be a sequence of mea.mrabe functions defined over a measurable set E. Show that

sup {J;'/2""'f,.},

in/{fh!2"'" !.},

Lim!", Lim!n, are measurable Gver E.

Henc~ show that Lim /t. is

measurablover E if Lim!n exists. [Kanpur 1986, 76, 72, 70; Banaras 197., 68, 65]

127

MEASURABLE FUNCTIONS

Proof. Let:< In > be a sequence of measurable functions defined over a measurable set E. Step I. To prove that sup {Ir : 1 EO; r ~ n} and inf { J.. : ] :s:;;; r < n} are measurable over E. Define M (x)=sup {/r (x) : 1 EO; r < n} =

U" r-l

IT (x)

<


>

a) is a

a) is a measurable set V n.

Evidently E[M> a)= U" E[(fr

>

a)]

=A finite union of measurable sets =A measurable set. This proves that E (M > a) is a measurable set.

Step II.

To prove that

(Tm

In and Lim fn are measurable

functions over the set E. D.!fine Mk (x)=sup { f" (x)},

rBanaras 1965]

n;Pk 111,. (X)=

inf {I" (x)}. n;pk

Then Lim fn (x)=

inf {Mk (x)} k~l

Lim I .. (x)= sup {mk (x)}. -k ~ 1 By case (i), M,. (x) and mk(x) both are measurable functions ~ dcli.ned over E V kEN and ag lin, by case (i),

128

ME,\SUR,\BLE FUNCTIONS

inf {Mk(X)}, k;;;tl

SUp {mk(x)} k:?1

are measurable over E, i.e. lim/.. , Lim

f. both ate measurable

over E. Step. III.

Let Lim / .. exist: [Meerut 1971; Kolhapur 70]

To show that Lim /. is measurable over E. By case (ii), Llm"j., Lim/.. both are measurable over E. By hypothesis, Limf.=Lim /.=Limf,•. From what has been done, it follows that Lim /. is measur· able over E. Theorem 11. A continuous/unction defined over a. measurable set E is measurable. [Kanpur 76; Meerut 71.; Kolbapur 69] Or, Show that a continuous function defined in a closed interval is measurable. [Kanpar 1987J Proof. Let / be a continuous function defined over a mea· surable set E. To prove that / is a measurable function over E. Let a· be any real number. We claim E (f;;;' a) is a closed set. A=E (/ ~ a). . .. (1) Let To prove that A is closed, it is enough to prove that D (A) C A. . .. (2) Let Xo E D(A) be arbitrary, then Xo is a limit point of A so that 3 a sequence <x,.> whose elements x" E A lim s.t. n-+oo x,.=xo· / is continuous at xo, X,.-X o ~ /(x") ~ /(x o). This follows from Heine's definition of continuity

. .. (3)

By (t), x" E A => ft x,.) ~ a => lim j (X,.) ~ a n-+oo => /(x.) ~ a, according to (3), => Xd E A, by (I).

129

MIlASURABLE FUNCTIONS

Finally, any This

Xo

~ Xo E A. D(A) c A => A is closed ~ A is measurabl. => E( I ~ a) is measurable

E D(A)

~

=> I is a measurable function over the set E. Theorem II (a). Let I be a function delined on a measurable set E. Then f is measurable iff lor any open set G C R, 1-1 (G) is a measurable set. Proof. Suppose a function f is defined on a measurable set E. Let G C R be an open set, where R=set of real numbers. I. Suppose I is measurable on E. Aim. /-1 (G) is a measurable set. Since any open subset G of R is expressible as a countable union of disjoint open intervals. Hence G is expressible as 00

G= U In, where 1.. = (a,,, bit) n=1

/-1 (G)= U {x E E: I(x)

.

E

, ft }

But/(x) E I,,=(a n , b.. ) ~ /(x)E (a .. , b.) ~ an Xo E A. => D(A) c A => A is closed => A is measurabl. => E( f ~ a) is measurable => / is a measurable function over the set E. Theorem II (a). Let f be a function defined on a measurable set E. Then f is measurable iff for any open set G C R, f- 1 (G) is a measurable set. Proof. Suppose a function f is defined on a measurable set E. Let G C R be an open set, where R=set of real numbers. I. Suppose / is measurable on E. Aim. /-1 (G) is a measurable set. Since any open subset G of R is expressible as a countable union of disjoint open intervals. Hence G is expressible as Finally, any This

Xo

00

G= U I", where 1,,= (a,,, b,.) ,.=1

..

f- 1 (G)= U {x E E: f(x) E ,ft } Butf(x) E I,.=(a", h,,) => f(x)E (a,., btl)

=> an < lex) < b..

:. f- 1

(G)= U [E (/ > a.. )

"

f is measurable => E ( f

n

E (f


a)

or [-1 (G)=E (/ > a) Also f- 1 (G) is measurable. Hence E ( / > a) is measurable. Consequently / is measurable on E.

130

MIlO\SURABLIL FUNCTIONS

Remark. The converse of Theorem 11 is not ture. That is to say, a measurable function ne.ed not be continuous. Example. Consider a function f: R ... {O, I} defined by { lifO ~ x is a sequence of measurable functions [Meerut 1986] then show that Limf. is measurable. 0, To prove thqt the limit 01 a convergent sequenr.e 01 meas.urable lunctions is measurable. [K8n.,ur M. Se. P. 1986) Proof. Let < Ita' be a sequence of measurable functions. To prove that Lim f. is measurable. t. < I. > is a monotonic sequence and hence it is either monotonic increa'!ling or monotonic decreasing. If < /.1 > is a monotonic increasing sequence, then Lim f .. = sup {f.}. n ~ 1 If < I. > is a monotonic decreasing sequence, then ·Limffl= inf {lit}. n~1

Hence if we prove that sup {/1,f•• ...... } and inf{/l,h, ... · .. } are measurable. then the problem will be p-roved. Let g(x)=sup {fIJ., ...... } ... inf {h. fs •...... }. 00

Then £(g > a)= U [£(f.

>

a)]=An enumerable union of

measurable sets. Hence E(g > a) is mea'iurable and so g is measurable. hex) is expressible as h(x)=--sup {-I.(x} : n=l, 2, ...... }. This .. h(x) is also mea'!lurable as g is measurable. This completes the proof. Ex. Let < I. > be a sequ:mce 01 real valued measurable "=1

functions. If lim

n-+-IX)

Solution.

1.=1 eXists,

then show that I is also measurable.

[Kanpur 1975] Here write the proof of above Theorem 12.

MEASURABLF. FUNCTIONS

131

Tbeor,em 13. III is a meafurable function, defined on a measurable set E and if g and f are equivalent functions, then g is a measurable function on E. An Alternate Statement. Iff is a measurable function and if f=g almost every where, then g i~ measurable. Proof. Let E be a measurab'.e set and let a be any real number. Also let fbe a measurable function on E so that E (f > a)

is a measuPabte set. Further suppose that f and g are equivalent functions on E so that f=g a.e.· on E ... (1) To prove that g is measurable over E, we have to show that E (g > a) is a measurable set. (I) c> 3 ACE S.t. m (A)=O, f(x)=g (x) V- x E E-A=B (say). ~ f:#g on A. f=g on B, m (A)=O, A n B=0, E=A U B. m (A) =0 ~ A is measurable. E and A are measurable sets => B=E -A is measurable. f is measurable on E, BeE is measurable ~ f is measurable over B. Refer theorem 3, page 118. => A (f > a) is a measurable set => B (g > a) is a measurable set. For f=g on B c> B (f > a)=B (g > a) A (g > a)={x E A: g{x) > a} C A. Also m(A)=O. This c> A (g > a) is measurable. For every subset of a set of measure zero is measurable. E (g > a)=[A (g > a)] U [B (g > a)]. For E=AuB. = union of two measurable sets. =measurable set. :. E (g > a) is a measurable set. Ex. Answer the following questions in < Yes' or 'No'. Is a function equal to a measurable function almost everywhere itself measurable? [Kaapur 1974] Solution. Yes. (Refer Theorem 13). Theorem 14. To show thar the characteristic function of a set E is measurable iff E is a measurable set. [1\1eerut 1990, 88; Kolbapur 69; Banaras 69; Gujarat 10]

132

MEASURABLE FUNCTIONS

Or The set E and its characteristic function both are measurable or both non-measurable. Proof. Let E be a subset of a measurable set A. The characteristic function KII of E is defined as {I

KE(x)= 0

if x E E if· x E A-E

We shall show that (i) if KE is measurable, then E is measurable. (ii) if E is measurable, then KE ~s measurable. (i) Let KE be measurable on A so that A (K F. > 0) is a measurable set. . But E=A (KI> > 0). Hence E is measurable. (ii) Let E be measurable and a be any real number. Then 0 if a ~ 1 A (KE > a)= I E If 0 ~ a < I lAir a < 0 Every set on the R.H.S. is measurable.

r

Hence A(KE>a) is measurable and so KE is measurable on A Theorem IS. Iff and g be measurable real valued functions defined on X and F tf(x), g(x)]=h(x), x E X be real and continuous function on the Euclidean plane RI, show that h is measurable. LKanpilr M.Sc. 1984, 81; Kanpur 69J Proof. Let a be any real number. Let , G={(u, v) : F (u, 11) > a}. Then G is an open subset of R2 !oo that G is expressible as 00

G= U In, R~l

where is a sequence of open interval,> s t. /,,=:(11, }.) : an < II < bn , en < I' a,,}n{x : g(x) < ..

ot

0 ..


{x : g(x) E G} is measurable, by (1) => fog is measurable, by (2). Theorem 16. (E. Botel). Suppose a measurable function f defined on a closed interval [a, b], is finite almost everywhere on the sume inrerval. For all numbers. a • • > 0, there exists a continuousfunction", [a, b] such that m [(E If-; I ~ a)] < E. [Paojab 1969) Proof. Letfbe a measurable function defined on [a, b] S.t.

f is finite almost everywhere ~ [a,

b].

Case I. Letfbe bounded. Then 3 a number k > 0 s.t. If(x} I < k ... (1) ¥ X E [a. b]. Let a, • > 0 be arbitrary numbers. Fixing C1, 6 by ~hoosir,~ a positive integer m so large that (kIm) m [E ( If-cjll ;;;;. 0)] ~ m [E (f:Fg») +m [E( I g-.p E

I~

0)]

E

is a sequence of measurable functions, (Banaras 68J then -Lim/n , Lim I,. are measurable. Prove that if the functions I and g are finite measurable, then Show that if

15. 16. 17.

18.

19.

20.

21.

l+g,1 -g andfg are mea'iurable functions. [Banaras 67J If < fn > is a sequeGce of measurable functions show that Lim sup In, Lim inff" are measurable. [Banaras 67] (a) Define a measurable function on a set X, and show that the characteristic function of a measuraple set is 3t. measurable function. (b) If I and g are real valued measurable functions on a measurable space X, prove thatf+g,fg,f U g and! n g are measurable. [Gujrat 70J (a) Iff and g be measurable functions and c be a constant, prove that functions cf and f+ g arc measurable. (b) A real valued function wluch is continuous in an open interval is measurable. lMeerut 1972] When is a function said to be measurable? Prove that the limit of a convergent sequence of measurable functions is measurable. [Meerut 71] Prove that the following condition is necessary and sufiiclent that the function be measurable: tire set {x: I(x) ~ r} is measurable for every rational number r. [Kolbapur 69J A functionf is measurable ifr the set {x :f(x) < r} is measurable for every rational number. [Kolhapur 70J

9 The Lebesgue Integral of a Function Introduction. Lebesgue's definition of an integral is more general in comparison to Riemann's. It enables us to integrate those functions for which Riemann's method fails. That is to say, if a function is integrable in the sense of Riemann in an integral (0, b), then it is also ~ebesgue integrable over (a, b) but not conversely, i.e. if a function is Lebesgue integrable over an interval (a, b), then it is not necessarily Riemann integrable over (a, b). The phrase "Integrable in the sense of Lebesgue" is written as "L-integrable." From now onwards, we adopt the following conventions:

r f(x) dx=Lebesgue integral of/ex) over E.

,8

(R)

J E

lex) dx=Riemann integral of/ex) over E.

9'0.

To define the Lebesgue Integral of a function. [Meerut 1987; Kanpur 86,72, 70J First Method. Sets in two dimensional world are ref.:rred to as plane sets. In two dimensional world, an interval is taken to be a rectangle (in particular a space) and the area of the rect:lOgle represent'> the measure of the rectangle. Let lex) be an txtended real valued function defined over a measurable set E. We define Q U: E)=-, {(x, y) : x E E, 0 ~ y ~ Ax)}. Do (/, E)={(x, y) : x E I::, 0 ~ y < /lx)}. The function/(x) is said to be Lebesgue \ Integrable over E if {J (/, E) is measurable and the Lebesgue integral of lex) over E is dd'ined as the plane measure of D (/, E), i.e.

l,. /(x) dx=m [.0 (/, E)].

141

THE LEBESGUE INTEGRAL OF A FU:-ICTION

Second Method. Case (I). When f(x) is bounded over E. [Kanpur 1983,32, 77, 76,70 ; Indore 78] Let I(x) be a bounded measurable function defined over a measurable set E. We can take 0:

is

k~l

Write

u=sup {.flo}, V=inf {Sk} k~l k ~ 1 Then U ~ St, V eo;; SI. We claim U V. Suppose not. Then U> V.




0, we have U

I! - "2


By assumption, U> V so that we can take U-V> 2,. Then (4) reduces to 8 1 -S1 > • Making • -+ 0, A contradiction. Hence our initial assumption is wrong. This ~ U~ V. Thus S1

and we write

f/

f

(x) dx= Lim [f(x)]", dx. . .. (1) m-..o,.oo E If the second member of (I) is finite, then we say that function/ (x) is L-infegrable or summable over E. Thus we see that every non-negetIve unbounded measurable function is L-integrable but only functions baving finite integrals are called summable. E

In order to define the integral of any unbounded measurable function we proceed as follows: /(x) ~ 0 for x E El j(x) < 0 for x eEl' Let E=...oEI U E2 , then El n El=0. By countable additivity property of the integral, i.e. Theorem 3.

and

f

I>

l (x)]", dx~oJ

~

r/(x)]", dX+[ ~ [J~x)]", dx.

...(2)

Making m-+oo, we get

I/ 6

If

J

£1

(x) dX=J

El

lex)

/(x) dx=oo,

J

E,

dX+[ / (x) dx. £2

/(x) dx=

-00,

then

f f(x) dx=(l-oo)+(-oo). whi~h is meaningless. The integral of a function has a meaning if and only if one of the integrals on the R. H.S. of (2) is finite. Definition. Let f be a measurable function on [a, bJ. If both fiand/- are L-integrable on (a, b), then we say that/is L-integrable. In this case we write / E L (a, b). Thcorme I. ~r tile ordinate set Q (/, E) is measurable then / (x) is measllrable over the measurable sel, E. Proof. Let D E) be Lebesgue-measurable, E being a measurable set. To prove that (x) is measurable over E, is is enough to prove that E (j(x) >0) is measurable, a being any real positive number. mEN be arbitrary. Let Deline t.. . -tx E E: r," a}=E (f.> a)

«(.

r

146

THE LEBESGUE INTEGRAL OF A FUNCTION

Ea+l/m={X E E :f(x) > a

+1}=E ( f > a+k)

1=( a, a+k)

and Evidently and therefore

Ea X I ::J EOH/.,. X I. Also EJ X I ~ 0 n (R" X /) ::> E"+llm X I ... (1) where Rn=Rx Rx ...... (n factors). o is known to be measurable. R" is always measurable. An interval is always measurable and hene e I is measurable. From -this it follows that Q n (Rn x I) is measurable so that m. [.on (R"x/)]=m e [f.'n (R"xln..=m LOn (Rnxl)) ... (2) From (1) m, (Ea X /) ;;;. m; [SJ n (R" x I)] and m.[Dn (R"x J)] ~ m. (Eo+l/rn X I). Making use of (2), we get m. (Eo x I) ~ m [Un (R" x I)] and m [D n (RI> x l)] ~ m, (E,,+1/m X I). Combining the last two inequalities, we have ~

m, (EI> x I) ~ m [D n (R"E I)) ;;;. m. (E"tl,,,, x /) From which

m, (Fax I)

i.e. i.e. or

~

m. (E"aH;mx/) (I) ? ·m. (EO+1Im) X m. (I) m, (Eo) X m (I) ~ me (EaH,m) X m (I) m, (Eo) ~ m. (EaH/m)·

tn.. (Eo) Xm,

Making m-+-oo and observing that

Lim",-+o:> m, (Ea-tl/"')= m. (Lim,..-+- 00 E O +1I"') we get, m. (Eo) ~ me (Eo). But me (Eo) ~ m, (E.) is true lor any set E". Combining (3) and (4), we get m; (Ed)=m e (Eo). This proves that Eo is measurable, i.e., Hence proved. E (f > a) is measurable. Ex.1. The Lebesguf! integral of a bounded measurable tionf(x) defined over a measarable set E is the common limit upper and lower Lebesgue sums s alld S. Sol. To prove that

...(3} . .. (4)

Juncof the

THE LEBE("GUE INTEGRAL OF A FUNCTION

IE f

147

[in the limit]

(x) dx=s=S.

We know that s

Lf

~

(x) dx

~

S. ~

Taking limit as max. (YH1- y;.) s

tf ff

~

~

(x) dx

S=s

[in limit]

(x) dx=S=s

E

If a bounded measurable

First Mean Value Theorem 2. tion f satisfies the inequality II

then


,f(x)

t;;

lit-!

J(x) dx

f(x) dx+JR

eJ:

. .. (2)

f(x) dx. fltl

Since f(x) is bounded on the set E and therefore we can take Cl ~ lex) ~ (J on the set R II +1' On using the first mean value theorem, we get

~

•• m (R .. t l )

JR

f(x) dx ' ek(l) n e,p)=cp. m (e/:)=m (e1P»+m(e k (2».

155

THE LEBESGUE INTEGRAL OF A FUNCTION

Making max. (YTe+l - Yk)--+O, we have

f

j(x) dx=!

11

E~

dx+J

f(x)

E2

f(x) dx.

Theorem 4. Let f and g be any two bounded mcasnrable functions definerJ on a measurable ~et E. SholV that (i) (ii)

(iii)

J" t I

~

f(x) dx

L

L

[f(x)+g(x)] dx-=

CJ(x) dx=c

I>

if f{x) ~

g (x) dx

f{x) dx+

g.(X)

L, g(x) dx (Bhagalpur 66)

I

f(x) dx, c being any constant

E

Proof. Let I and g be any two bounded measurable functions, defined over a measurable set E. We can take. on the set E. at is a sequence of bounded measurable functions defined over E. I Kanpur M.Sc. P. 1985] Step (iii).

Let c he any constant.

To prove that

L

c(.\) dx=c

L

/(x) dx.

lIndorc 1978]

158

THE LEBESGUE INTEGRAL OF A FUNCTION

Let s be lower Lebesgue sum relative to [(x) with the scale Yo, YI' Y2' ... , Yn. Let c > O. Let s' be the lower Lebesgue sum relative to cf (x) with the scale CYo, ('YI' CY2' ... , cy". Evi af+bg is also bounded measurable. Now

Ie [a/(x)-1-hg(x)] d~=In alex) dx+Lbg(x) dx, by (i) =0

IE

f{x) dx+b

IE

g(x) dx, by (ii).

Proved.

159

THE LEBESGUE INTEGRAL OF A FUNCTION

Ex. Iff and g are integrable (boundC!d or unbounded) functions over a measurable set E (of finite or infinite measure), then prove that f+g is also integrabla over E and that

L

f+g=t

f+fE

g.

[Kanpur M.Sc. 7.3, M.Sc. P. 75] Solution. Theorem 4.

Here prove the equation

(5)

of the

above

Theorem 5. If two bounded measurable functions f and g are equivalent on a measurable set E, i.e., if f=g a.e. Of! E,

then they have the same integral, i.e.,

t

f(x) dx=

Is g(x) dx.

or, The bounded measurable functions which are equal almost everywhere have the same integral. Also prove that the converse of this theorem is not true. Proof. Letf(x) and g(x) be bound.:::d measurable functions defined over a measurable set E. Let f(x)=g(x) a.e. on the set E. To prove that

J fix) dx=J ,.

E

g(X) dx.

Our assumptiol1 implies that f(x)=g(x) ¥ x E E except in a set of measure zero. This fact is expressed by saying that f(x)=g(x) ¥ x E £1 C £ s t. m (£ -E1 )=0. Clearly E=(8-El) U E1 , (E-E1) () El=0. By countable additivity property of the integral,

/,. (f-g) dX=!(E_E1 ) (f-g) dx+JEl (f-g) dx. f(x)=g(x)

¥

. .. (1)

x E El => IEl (f-g) dx=JEl :f-I) dx.

... (2)

160

THE LEBFSGUE INTEGRAL OF A FUNe flON

m(E- £1)=0

~ [(£-£1)

C[-g) dx=O.

. .. (3)

Writing (I) with the help of (2) and (3), we get

or

t t f/

or

IE

«(-g) dx=

Let p

0, q> 0, then show that

J:;:,-:q dX=~ "-/+q +p~fq --p-l-)q +, ..

(i)

and deduce that (it)

LOg2~-'-1--}+~-}.+ _,

(iii)

4 =1-- 3 +5---=r+ ,.,

~

J

1

I

Solution. We have x P- 1( 1+XO)-l = x P -) (1 _.. xtl.!- x'q _x3Q +,.\4Q -

.,)

00

= XP-l

L' (x2I1Q ._ X(2,,+1)Q)

'.=0

=

, 00

1:

[X2"~l-p-l_X(2"+1)G-t-P-ll

,,=0

Integration vie Ids the .result .. /

Jo ~V-l 1T x

dx= ~ [

1

q

.. =0

X2 ..

q~::, _ _ _~~~~~q+p

11

(2 + I) q +p ~(j=O

'J.llq r P

1 +____ - __

J l

III = ____ P p',-q 1'+2:.1 XoP-l

o 1 +Xfl

1 ! I dt=-·- .--- -- + .-~p

p+q

p +i..q

p+3q

-.-1. -+.,. .v +3'{

.L I

'"

, .. (l)

Hence the result (i). Put p=q = I and observe that

J1 1' . dx= [ o I

+x

log (! +x) 11 I

log 2= I·· i

we get

=

Jo

+ t -- t +'"

Jog 2, Hence the rc,ult (ii).

Taking p= I, 1j=2 in (I) a.nd obscuing that

}11 I Q

+X2

dx= [ tan- J x

J1

r.: n ='0j',

165

THE LEBESGUE INTEGRAL OF A FUNCTION

we get Henc~

the result (iii).

sin x . b . .~. P rove that th e fiunctlOll IS /lot Le esgue integr-

Ex., 6

x

able over [0, 00[.

[Banarlls 1971]

T 0 Sh ow t hat th e f unctIon . sin-x.IS not L -mtegra ' bIe. · So1Ut 100.

x

-x

If' we show t hat sin.x.IS not summabl e over [0,00[, we can conclude the result.

For this we must show that

J

I

OO

o

sin x , 00.

!-I- - - !

X

Consider the integral

I"'"

sin ~ dx=

o

X

Jr.

£ r- 1

I~}..!l

X

J dx

X

(r-ll"

{y+

IJIE fll ~ Jill. E This completes the proof. Theorem 8, Let I be a bnunded measurable lunction deji'ned

If

I'

Thl'.I1 , E f \ ~ E I fl. [Meerut 1972, 71; Kolhapur 70; Banaras 71j Proof. Hert' write III Part of the about Theorem 7. Theorem 9a. Let f and g be unhounded measurable fUllctiuns defined over measurable set E. Show that

over a measurable set £.

(I) JEfdX=

1!1 fEkldx

0

E=I~l E .. Sot,

where

n £/=0 for k::j=k'. [Meerut 1975, 72, 71J Proof. Let f and g be unbounded measurable fUllctions defined over a Tll('asurable set £. Suppost', without losing generality, that/(x) ~ 0, g (X) ~ o. r or if the result is applicable to positive functions, it is equally applicable to negative ';unctions therefore it is so by addit;on in general case. Let mEN be arbitrary. Define [!tx)]m and [g (x)] ... as follows: Ek

when/ex) ~ when/ex) >

[!(X)]m={f(X)

and then we define

J E

m (f)

dx= Lim

Similarly define

f

E

g (x) dx= Lim

m·,.oo

m~oo

f

E

I

1:-

[f]

0

(g (xli", dx.

In

dx.

til

111

170

THE LEBESGUE INTEGRAL of A "UNCTION

Step

(i)

00

Let E= E E. s.t. Ek n It=l

To prove that

J

jJ

f{x) dx=

E

1:=1

Ek'={lJ for k�k'.

IE f(x)

dx.

"

[f (x)]", is bounded and measurable over fore, the integral of[f(x)]", is additive, i.e .• Since

I Making In-+«>

E

E

rf(x)]", dx

.=1

IE

E f(x) dx ;a

and there­

[f(x)]", dx

[f{x)�111 dx

II i: � 1.

J

J.:o�k

E

we get

'£1 IEk f(x)

. .. (1)

dx.

Again

IE [f(x)]", dx= ,,! lEk [f(x)]111

dx �

; 17;< f(x) dx Lk

.... 1

[f(x)].. (A)

...(2)

=

L

f dl", then 1> is countably additive.

Prove as above that

fEI dl'== E, I 1

Suppose f is

Ele

/"1"

(Kanpur 1986]

...(4)

THE LEBESGUb INTEGRAL OF A FUNCTION

But

:.

L

~(A)= I if> (E)=

dll, E=

I~l r/>(E

k)

or

~l E

1~ Ek

k=

rp (

171

k~ Ek )=~! ¢(E

k)

This proves that cP is countably additive. Theorem 9b. Let I, g be non-negatIve valued functions on la, b]. If I, gEL [a, b], thenl+g E L [a, b] and

Ie (/+g)=JE I+/ E g. [Kan{}ur 1972, 73; M. Sc. P. 75] Proof.

To prove that

IE (/+g) dx=fE Idx+JE g d.". Let 1\1 (x)=/(x)+g (x). Now we have to show that

IE

q, (x)

dX=/Ef(X) dx+fE g(x)dx.

Case I. When I(x) ? 0, g (x) ~ 0 so that Firstly we shall show that

[,p (x)]", Let

1 (x)

~ l/(x)1",+[g (x)]", ~

EO;; m, g (x)

4 (x)

~ O.

[1\1 (x)hlll.

< m ; then

. .. (3)

[J(x)]",=/(x), [g (x)]",=g (x) [1\1 (x)]m EO;; '" (x) =f(x)+g (x) =l f (x)]",+[g (x)] ..

i.e.

[1/1 (x)]",

< [f(x)] ... +lg (x)]"..

If either! (x) or g (x) is greater than m, then

4 (x) >

...(4) m so

that

r

[1/' (x).)",=m < f(x)]",+[g (x)lm [I), (x)]", < [f(x ]",+ [g (x)]"' From (4) and (5), it follows that [1jI (x)]", [hx)]",+[g (x)]", is always true. Exactly in a similar way, we can show that [!(X)]m+[g (x)]", ~ [tjI (X)]2"" Combining (6) and (7), we get the result (3). Integrating (3).and observing that

i.e.


0 and

tldX=O then/(x) is a e. zero. [Bhagalpur 1968] 14. D.!fine the Lebesgue integral of an unbounded function. on [a, bJ. Show that if

f(x)=~ (0
be a sequence of measurable function defined over a measurable set E. Let/be a measurable functicn defined over E S.t. (i) /(x) < 00 a.e. on the set E. (ii)

Lim

n-+oo

m[E(I/n-fl

~E)]=O

¥

E

>

O.

Then the sequence < f" > is said to converge in measure to the function! 1O· 2. Definition. Pointwise convergence. Let < fn > be a sequence of measurable function defined over a measurable set E. If there exists a measurable functionf over E S.t. Lim (,,(x) = lex) 'V x E E, n--+ CI,

then we say that the sequence




converges

pointwj',~ to

fon E. 1Q. 3. Definition. Convergence almost c\'erywhere. [Indore 1Q78! Let ,/ In / be a sequence of measurable functions defined

177

THEOREMS ON CONVERGENCE OF SEQUENCES

over a measurable set E. If 3 a measurable function f over E and (i) m (A)=O

a lIet A S.t.

lim I.n (X)=J\x ,rr) \f X E E-A n-+oo (i.e., converges pointwise to f on E-A) then we say that the sequence < f .. > converges to I almost everywhere on E. 10'4. Definition. Uniform Convergence. A sequencl:: < I .. > of measurable functions defined over a measurable set E is said to converge uniformly almost everywhere to a functionf if :I ACE s.t. m(A)=O and given E > 0, 3 no E N s.t. ¥ n ;;;::: no and V x E E--A => If..(x)-/(x) 1< e. Theorem 1. Let < fn > be a sequence of i1;ltegrable functions which converge zn mean to a lunction f, then f ..-I in measare. [Indore ,1970] Proof. Let < f,. > be a sequence of integrable functions defined over a measurable set E. Let En=E (1/.. -1 I ~ 8) where 8 > O. Then 1/,,(x)-/(x) I ~ 8 y X E En.

(1'\')

This => fEn I/n(x)-j(X) 1 dx Since/R~{in

lim n-+oo

... (1)

m(En).8

mean on E and hence

I

Ifn(x)-(x)

E. C E => fEn (2) and (3)

~

I dx=O

IIn-I I ~

... (2)

IE

11..- II

... (3)

~ IE" I/n-/I=O as n-+OO.

In this event (I) But m(En)

~

0,

=> m(En }.& ~

V Ell'

0

::;>

Hence

m(E,,) EO;; 0 as n-H>O.

lim n--jooo

m(En) =0.

This => m [E (/I. -II ~ 8)]=Oas 11_00 Y a> O. => 1,,-1 is measure. Theorem 2. Let < f,. > be a sequence of measurable lunctions

lim a measurable set E an d I,'f' n_oo thell show that I is measurable 011 E. Oil

Proof.

I.• (x) =J,I') (X a.e.

on E .

Let A={x E E: lim In (x):;i: I(x)}. Then b} def. n-+oo of almost everywhere, m(A)=O. Define, g..(x)=/n(x) if E A and g(x)=O if x E A.

178

THEOREMS ON CONVERGENCE OF SEQUENCES

This => lim gn =lim/n if x ,. A ~ g(x)=/(x) if x ,. A and g(x)=O if x E A. The above facts prove that < g,. > converges pointwise to g on E. Also gn is mea'iurable V n. Hence limit function, i.e., g is mea'iurable. Consequently I i'i measurable. Theorem 3. (F. Riesz). Let be a sequence ollunctions which converges in m'!asure to the lunction Ion a measurable jet E. Then there exists a subsequence which also converges to the functionl almost everywhere. [Meerut 1986j Proof. Let is a monotonic decreasing sequence of m:!a~urable sets. This declares that lim m(A,,)=m(B) .. (3). (Refer Theorem 18. Chapter 7, page 102). 00

It follows from (I) and (2) that m(A .. ) \

.

< 2;,,=nn/c.

Consequently

lim m(An) =0 .. (4) as En" is convergent. This => m(B)=O, by (3). Let y E E -B b~ arbitrary. Then y f!. B y ,. B => y t Ano for some natural number no => y If. E( lin:, II ~ Ok) 'V- k ~ nO

=> IIn~(") -/lY) ! < lim => k_oo

Ok

1,.··(y)=/(Y)

,?L

k ~ no

V Y

E E-B.

179

THEOREMS ON CONVERGIlNCE OF SEQUENCES

Also

m(B)=O.

lim

/ ..t=/

Hence k -+00

a.e.

on E.

Theorem 4. (D.F. Egorff). If a sequence 0/ measurable/unctions is convergent almost everywhere on a measurable set E, 0/ finite measure, then/or every 6>0,3 a measnrable set EoeE s.t. (i) m(Eo)< 8, (ii) the sequence is uniformly convergent on E---Eo· [Meerut 1988,90] Proof. Let < f. > be a sequence of measurable functions defined on a measurable set E s.t. < fra > converges almost everywhere to a finite measurable function / on E. Let lim/..{x)=/(x) V x E A. Then, by def. of a.e., m (E-..4)=O. Since every function is measurable on a set of measure zero and

E::' =

Ii [E (Ift{X) f,.-+/ on E

:. s.t. m

~

-f(x)

I 0, 3 positive (E-E~.. ) < e/2m Given«

(11m) )] and E"'=

]1 E:'.

(E - E"')=O. ..

integer no depending on .. and m

v

Define Eo=

U (E --E:

m=l

•••

)

g

Being countable union of measurable sets, Also Eo C E m (Eo)=m [

(1)

!?u is

measurable.

U (E-E"0m )] ~ E m (E-E"') < 110

'"= 1

E,

by (1).

m=1

This ¢ m(Eo)< 6. This completes the first part of the promblem E-Eo=EThus

l.,f

U (OE -Ernno )=E n

"'=1

I

(u

E"') ... =1 no

n ~ rio and x E E -Eo ~ x E E~

=> If.. (x) - f(x) I < 11m. Hence < f" > converges uniformly t%n £-Eo· Theorem 5. (D.F. Egorff). If a sequence of measurable functions converges to a/inite limit almost ef'erywhere ill a measurable

180

THEOREMS ON CONVERGENCE OF SEQUENCES

set E, then, given 8, we can lind a set 01 measure greater than m (E) -/l in which the sequence converges uniformly. Proof. Let < In > be a sequence of measurable functions defined over a measurable set E. Let this sequence converge to a finite measurable function! alntost everywhere on E. Let 8 > O. To prove that :i ~ measurable set ACE S.t. (i) meA) > m (E)-8 Oi) < In > converges uniformly on A. Consider a convergent series ~7)k of positive terms and a sequence < an > of positive terms converging to zero s.t. 171 > 172 > 173 > ........ . 00

An (17)= E E (1/k--f1 :? 0),0> O.

Write

k='TI

Then m (An (17)1=0 as n- oo • ...(1) For In-f. By virtue of (J). corresponding to every positive integer k, we can find another positive inte~er 7)k s.t. m [(ARk) (0)] < Tlk. Now we cho.))e a positive integer

ko

00

E

S.t.

7).


m (E) -

8. For m (B) < 8 ~ -m (8) > -8. :. m (A) > m (£)-8. Rem'lins to ShOW that Itt (x)-"/{x) uniformly on A. Let. > 0 be arbitrary. Then we can find an integer k S.t. k ;;;;t ko, 0" < c ailY x E A ~ y ~ B. For A=E--B. ~ x

f'I.

Ani

(a,)

~ .Y f'I. E ( Ilk ~ /It(x) fix) ~ /ft{x) - /Ix)

I I ~ ai) I < 0, < ~ !< E ¥ k

for

k;;;-:",

k ~ ni ~ n" x E A

~ In "'/uniformly on A. Let g be a function s.t./and g are equivalent on £ so that

m

[E ( 1--1= g)]=O.

. .. (2)

181

THEOREMS ON CONVERGENCE OF SEQUENCES

To prove that the sequence < In the function g. For this we have to show that

..

I.e.

>

converges in measure to

I/,.-g I ~ 6)J=0. . .. (3) Iln-g I ~ E) C E(f~g)+E( lin-II;;'.)

[E (

In

Evidently E( This => m [E ( I I .. -g I ';J 6)J ~ m [E (l;i=g)]+m [E ( Using (1) and (2), we get m [E ( 1 /..-g I ~ 6)J ~ 0+0=0 m [E (

I/n-g I ~

Il,.-I I ~ .)].

6)J ~ O.

,... (4)

But measure is a non-negative quantity so that m [E( Iln-g I;;" 6)] ~ O. . .. (5) Combining (4) and (5), we get the required equation (3). Proved. Theorem 6. Lebesgue bounded convergence theorem. Let < In > be a sequence 01 bounded measurable lunctions defined over a measurable set E s.t. I/.. (x) I < M >.f n E Nand>.; x E E. Let < /., > converge ill measure to a measurable junction I on the set E, M being a positive cOllstant. Then

Lim

n~oo

f

E

j,,(x) dx=J

E

j{x) dx.

[Meerut 1986; Kolhapur 76 ; Kanpur 85, 70, 73) . Proof. Let < /.. > be a ~equence of bounded measurable functions defined over a measurable set E S.t. Iln(x) 1 < M>,f n E Nand >,f x E E. ...(1) Then In is integrable over E >.; n E N, M being a positive constant. Let < In > converge in measure to a measurable function Ion E so that Lim m[E( If"

IH'OO

II ~E)]=O,

... (2)

for every. > O. To prove that Lim

1l~ 00

f

E

In (x)

dx=J !(x) dx. E

From (I) and (2), I/(x) I < M. Suppose 8 > 0 is arbitrary and set A.=E( lin-II ;;" 0), Bn=E( 1/.. -/1 < 0). Then E=A" U B.. and An n B,,=(lj.

"

.(3)

. .. (4)

182

THEOREMS ON GONVERGENCE OF SEQUENCES

The definition of convergence in measure ensures that Lim m (An)-O.

n-oo

. .. (5)

This follows from (2). By countable additivity property of the integral

t

lin-II dX=JAn 1/11-/1 dX+JBn lin-II dx.

We have lin-II o 00 [Meerut 1988, 87; Banaras 69, IV 70; Kanpur 1'14,82,69] Proof. Let < in > be a non-decreasing sequence of integrable functions defined over a measurable set E.

L et

Lim f," b e 'mtegra bl e over E.

11-+00

186

THEOREMS ON CONVERGENCE OF SEQUENCES

J

f

Lim I .. (x) dx. To prove that Lim In(x) dx= n->oo B B n_oo Since .< In > is a non-decreasing sequence and hence

11 ~/2

~/3 ~

..... .

This ~ 11 ~ In ¥ n ~ 1.. - 11 ~ 0 ~ I\In ~ 0 ¥ n, on taking ~ .. =/n-fl' Moreover < In > is a sequence of integrable functions implies that is a sequence of integrable functions. Finally is a sequence of non-negative integrable functions. Applying this to the Lebesgue's bounded convergence theorem, lim

f

n .... oo B or

lim

n ..... oo

or

lim

11_00

dX=

I

lim

Bn-oo

J (/. -fJ dx=I B

B

.1. 'f'n

dx-

E

E

dx=

I ..(x) dx=.

E

B

dX

lim (f.. -/l) dx

n-oo

f In J /1 f n-+oo r I n_oo' n_OO,E Lim

or

.f. '!In

lim In dx-

J 11 d'( E

lim (,,(x) dx.

Hence the result. Remark. The theorem 8 is also known as "Lebesgue's monotone convergence theorem". Ex. State and prove Lebesgue monotone convergence theorem. [Meerut 1988; Kanpur M.Sc. P. 81] 9. Fatou's lemma. Let < f" > be a sequence of non-negative integrable lunctions delined over a measurable set E S.t. Lim.Iniff,,,= ( a.e. on F .

(l')

n-+cc

I

Lim. n_oo ml I,,(x) dx
is an increasing sequence of non-negative integrable functions and hence by Beppo-Levi's Theorem 8,

Lim

n_cx>

i.e.

I

Ii

gn(X) dx=

=J

I

,r(X) dx

E

F.

Lim gn(x) dx

n-+ 00

=1 n-oo Lim inf

Lim gn(X) dx= n-+co Ii Using this in (1), we get

J

I

~

£

f(x) dx

J f(x) dx. E

Lim inf

n-oo

fn(x) dx

I

fn(x) dx.

£

Hence the result. Ex. State and prOl'e Fatou's lemmaIor integration. [8bagaJpur 1969 ; 8anaras 64] Problem 1. State the l.ebesgue's dominated convergence theorem and use if to e~'aluate the following integral: Lim f,,(x) dx n-'>- 00

J1

0

n3 / 2 x

' where fn(x) = 1 +n2x~' 0 ~ x ~ 1, n= 1,2, 3,... ...

,

[Kanpur 1976, 74, 72] Solution. For the statement, see Theorem 7. 1 n3 / 2 X2 1 ---:,. -=·I.(x) say 2x J• ~ x ,fn(x)=-. x l+n '1'. ,

Then fn(X)

~ Ij.(x).

Also 4(;\) is integrable in [0, 11·

188

THEOREMS ON CONVERGENCE OF SEQUENCES

Hence, by Lebesgue dominated convergence theorem, lim

Jlo fn(x) dx= =

-=

r

lim fn(x) dx

... (1)

·0

I: 11:: (r+~~2- ) 1: n~~ (In)( .~ :X

dx

)

2

112

=Jlo O. (·~2) dx=fl 0 dx=O. U+x

ADS.

0

Problem 2.

> 0, prove that

If rI.

11-.lim+00 J"0(1- nx- )"

X'- 1

dx=

10 00

e-rxCl -

I

dx

where the integrals are taken in the Lebesgue sense. Let fn(x) = ( 1-:

Solution. Then fn(x)

.p

~

0

0.

XCI-I.

0

lim 11_00

f,.(x) dx

J" (1--::')" x"- dx= Joo en

lim

l

Il_ex>

0

0

W

X~-1 dx.

Show that the theorem of bounded cotlv.!rgence

Problem 3.

applies to .(,.(x) = 1 Solution.

>

Hence, by Lebesgue dominated

J" f,.(x) dx= JOO

lIm

IX

X)" =e-X.

n r} (x) is Lebesgue integrable. convergence theorem, 11-+ ex>

Il-oo

x a- 1 •

(x) V n where Ij!=e-ID

Notice that lim (1-

i.e.

r

[Kanpur 1973]

::2X2for °< x < J.

lKaDpur 1972]

nx

fn(x) = 1 +n 2x 2

1

=-1--

'

I

'~---1---------2

nx +I1X [\/(nx) -V'(IIX)] Take M=i. Then I f,,(x) I ~ M. Hence, bv Lebesgue convergence theorem, lim lim f,,(x) dx. f,,(x) dx=

Jl

11

Il-+ ex>

L.H.S. of (J)=lim

'-.

0

0 11-+00

JIo 1+11\ : IlX

1 log (l +n2x2), dx=lim -2 1/

~

t

+2

..• (1)

(form~) 00

189

THEOREM is a sequence of non-negative measurable functions such that lim inff,.(x)=/(x), x E E, prow that

n-+XJ

_

f

E

Idp.

~

lim infI I" dr"

n-+ co

E

[Kanpur M.Sc. P. 1981]

3. State and prove Lebesgue's dominated convergence [Kanpur 1983] theorem, and use it to evaluate the following integral: lim 1 J j~{x) dx, where

n_ co

0

n3 / 2 x In(x)= l+n t \-2' 0 ~ x ~ t, n=l, 2, 3, [Kanpur 1976] 4. State the Lebesgue's dominated convergence theolem and usc it to evaluate the following integral: lim n3 '2 x (,,(x) dx where In (X) = 1 + 2 -2' 0 ~ X ~ 1.

fl

11-+ co

0-

n x

[Kanpur 1974] 5. (a) State and prove the theorem of bounded convergence for Lebesgue integrals. Show that this theorem is not true for Riemann integrals.

192

THEOREMS ON CONVERGENCE OF SEQUENCES

(b)

If ~ > 0, prove that

lim

t n(l_~)n xa-1 dX=J"'"

e-1JIj X,,-l dx, n 0 where the integrals are taken in the Lebesgue sense. [Kanpur 1973] 6. (a) Let < ,f,. > be a sequence of Lebesgue integrable functions in a set E such that 11-+00

1.

fn(x) ~ 0 for x E E and

almost everywhere in E.

lim n-;..oo

Prove that

J E

f

fn(x)=f(x)

f

~ n-+oo lim fn. E > for which strict

Give an example of a sequence < fn inequality holds in the above result. Hint. See Theorem 9, Page 186. (b) State and prove Lebesgue's theorerp of dominated convergence. Show that the theorem of bounded convergence f, (x)

applies to

nx =1+n20 and the theorem of dominated cqnvergence to g,. (X)=11 3 / 2 x/(1 +112X2) for 0 E;;; x ~ 1. [Kanpur 1972] 7. Let < fn > be a sequence of measurable functions defined on a set E of finite measure, and suppose that there is a real number M such that I f,,(x) I .;:; ; M for all n and all x. lim . If f(x) = fn(x) for each x In E, then prove that n

11-00

I

E

f- lim

-11_ 00

J J, E

n'

[Kanpur 1970]

8. Define the Lebesgue-integral of a functionf, with respect to the measure Over a given set. Let < fn > be a sequence of measurable functions defined on the p.-rneasurable space E wch that ~ fl (x) ~ f2 (x) E;;; ... ... (x E E). lim . If f,,(x)-;..f(x), then show that

°

11-;" 00

f fn dp. J.f d/-', (n.-~CXj). -?

n}

[Kanpur 1969]

193

THEOREMS ON CONVERGENCE OF SEQUENCES 00

~'

show that the series

p. (En) converges.

[Banaras III 701

n=1

10. State and prove Beppo-Levi's theorem. [Banaras IV 70] 11. State and prove Lebesgue's dominated ~onvergence theorem. [Banaras 1965, IV 70) 12. Let < In > be a sequence of real valued measurable functions on (-00, 00) and -let 00

1/" (x)

1:

I=~ (x) E

V (-00, 00).

n=1

Prove that

f: L~)

In

(x)

dX=n~

S: I.

(x) dx

provided that either side of the equality exists. [Banaras IV 70] Hint. r/> (x) E V (-00, (0) => ~ (x) is measurablo and ~ (x) is L-intep·able on ( - 00, (0). 13. Let < I .. > be a monotonic non-decreasing sequence of integrable functions over E and let prove that

f 1=

lim

J

Lim

n-»oo

In i,

1ft.

[Banaras 69] 14. State and prove the theorem of dominated convergence. [Banaras 67] 15. Let fbe a (finite) rea) valued Lebesgue measurable function on [0, 1] and let E .. ={x E [0, I]: n-l 0, 3 3 > 0 s.t. I f(t) - f(x) ! < e whenever I t-x I Using this in (3), we get IF(x+hl=-F(X) -f(x}

or

197

\< I ~ I J:

IF(X+~)-F(X) -f(x} \
f(x) Ml' g(x) M 2 • Of x E [a, b], Ml and M2 are upper bounds of f(x) and g(x) respectively in [a, b].


0 S.t.

I g(x) I ;:.

0

>of x E [a, b].

To prove that !Ix) is absolutely continuous over [a, b]. gt X)

207

ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINlTE INl EGRALS

i: 1_1___I _1 =

i; I g(bt)-g(at) I

1

k=1

g(ilk )

g(ak)

g(b k ) (a k ) I

k=l'

0

I

0 S.t.

~

1c~1

\ f(h k) - ftat) \ ,


bI , a 2, bz, ...... a,., b,. where a=a1 < bi ~ (12 < b2 ~ ...... ~ an < bn=b. Again divide the closed interval [a, b] by means of points a=co -< C1 < C2 < ..... c =b no in no parts s.t. CkH -CI: < 8. Consequently for any subdivision of [CTc,

E I f(XHl)-f(x.)

•

I

CkH]

XI E

[CTc, CI:+1]

V~~+1 (f) ~ 1.

i.e.

It follows that

f ...... +V C"o Vba (f )=V C1 (f)+V c. ()+ Co

~

C"O_l

(f)

1+1+ ...... =n o

~!

i.e.

C1

( f)


3

7],

h

0 s.t. If(x+~~ - f(x) I < 'I

>

=> If(x+h)-f(x)

I
Qcorrespor.Jing to each E in the definition of absolutely continuity of f If we suppose X k ~ Xk+l' then a=yo o m(Ek)=O.

,Lim f To prove that k_do '£ J<x) dx=O. ;

Ie

Case I. ~henJ(x) is Throughout the discussion we shall suppose that

E" C E ¥ kEN. Slncej(x) is bounded over E and hence we can take

« ~ J(x) ~ (3 on the set Ell.

Using the first mean value theorem, we get

Making k-+oo and observing (1) we get

f Lim f o ~ k-+oo E"

Lim «.0 ~ k-oo EIcJ(x) ~x ~ ~.O or This

J(x) dx ~ 0

~ k~:fE".f(X) dx=O

• •. (1)

211

ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS

Case II. When f(x) is unbounded. Suppose, without losing generality, thatf(x) ~ O. For if the result is applicable to positive functions, it is equally applicable to negative functions and therefore it is so by addition in general case. Let mEN be arbitrary.

Define [f(x)]", as follows: when /(x) ~ m whenf(x) > m.

[/(x)]",= {f{X)

m

Then [.r(x)]", is L-integrable and we define

J Ii

.r(x) dx= Lim

m-.oo

Observe that [.r(x)] ...

f

Ii

L.r(X)].n dx.

. .. (2)

< f(x)

and therefore JEk [f(x)]", dx
~

for k

... (8)

ko

Adding (4) and (8), we get fEk {JiX)-[ f(x)

lnJ dx+IEt [f(X) ]nodx < ,'+,'.

Setting e=E'+e, we get JEtf(X) dx .. I(x)

~

o.

:. JEk If(x) I dx < e Now, or

IIEk/(X) dX\

I

JEkf(X) dx

This

=>

< II

I
F'(x)--J(x)=O ~ F'(x)=/(x) a.e. Remark, If we write g'(t)=Ft1). then

li~ ~ [+'- F(I)

df=lim }

1:+'- [~l get)

I [

=Iim /, g (x)

] dt

Jc-c+'-

C _ ' ( ) - 1:'( ) -I' - 1m j?'(c-l-h)-gi II - )--g c -C' C ,

Theorem 17. III: [a, b] .... R is all increojillg lUi cticn (nondecreasing). then I I : [a. b] .... R is L-integrable and

I:

I '(x) dx

~I

Proof. Consider a sequence tions I. : [a, bJ~R defined as

(b)-I (a).

[Punjab 65, 67; Banaras 69, 71J of non-negative func-

< In >

In(x)=n [ I (x+}) ---/(x)

1'"

x E [a, b].

218

ABSOLUTE CONTINUOUS FUNCTIONS, INDEfiNITE IN1EGKALS

Since I ..(x) is an increasing function and hence it is L-integrable. The sequence < In > converges to I (x) a.e. I

f'(x) =

i.e.

Lim n-+oo

.

I .. (x) a.e.

Applying Fatou's lemma. we get

I

b

..

f'(x) dx

~

Lim inf

n-+ 00

fib

I ..(x) dX}

... (1)

Q

Extend the definition of the function I by setting l(x)-4(b) If b oo n by remark of Theorem 16. P. 217. =/tb) - I( '1). Finally, Lim inf

n-+

00

Jil /J

I,,(x)

dx=/tb)-/~a).

..

ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS

219

Using this in (I), we obtain

J:

rex) dx :S;" f(b)--f(a).

Proved. Theorem 18. Fundamental theorem of integral calculus : If f is an integrable function its indefinite integral F(x) has almost everywhere a finite differential coefficient or Iff is an integrable function on [a, bJ and if F(x) = then

1:

F(t) dt+F(a).

for almost everywhere on la, b]. [Banaras 1967] Letfbe an integrable function on [a, b] and let

F'(x)=f(x)

Proof.

F(X)=jlDa f(t) dt+F(a),

... ( I)

1'(0) being any constant.

To prove that F (x)=f(x) a.e. 011 [a, b). Without loss of generality, we may suppose that f(x) ~ 0 tv x. Let m be a positive integer and define a function [f (Xj]". as follows

If(X)Jm={f~X) i.ff(x) ~ m m Iff~x) > m Evidently [f(x)l", ~f(x) in every case Ax) - [f~x)]m ~ O. so that Consequently

1: l f(l)-[/(I)].,.] dt=G",(x)

__ (2)

is an increasing function of x. As a result of which Gm{x) has non-negative differential coefficient. Since [fV)]". is bounded and measurable and so

~x By (2),

1:

[f(t)Jm dt=[f(x)]". ... (3), by theorem 16.

J: f(t) dt=G".(x) +J: [f(l)],,, dt, F(x)-F(a)":'Gm(x)+ J: [f(t)l",

dl, by (I).

i.e.

Differentiating w.r.t. x, F'(x)--O=G'.,.(x)+[f(x)]m, by (3) ~ [f(x)].,. as G'".(x);;;;r o. :. FJx) ~ [f\x)]m a.e. But m i, arbitrary and hence m~ OJ, give~ ':F'(x)'~ f(Jc}

... (4)

220

ABSOLUIE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS

1: J:

But Also

f(x) dx=F(b)-F(a), by (1) r(x) dx=F(b)-F(a).

The last two equations show that

J: I:

or

r(x) dx=

J:

f(x) dx

[r(x)-f(x)] dx=O.

Also F'(x)-f(x) ~ 0 a.e. by (4). Then we must have F'(x) - f(x)=O a.e. or F'(x}-f(x)

a.e.

Problems related to Lebesgue point of a function. Theorem 19. Let x be a Lebesgue point of afunctionf(t). Then show that the indefinite integral F(x) =f(a)

1:

+

f(t) dt

is aijj'erelltiable at cach point x and r(x)=f(x).

Let x be a Lebesgue point of the functionf(t), so

Proof. that

Lim I J",+h h-+O 11", I f(t)-f(x) Also let

F(x)=fw)+J"'o /(t) dt

To prove that Since

I dt=O

... (1)

... (2)

F'(x) =f(x).

F (x) =Lim F(x+h) -F(x) h ~O It

Hence if we show that Lim \ F(x+h~ -F(x) _ (x)

11--+ }

"

.

1=0,

... (3)

the result will follow. From (2). F(x+h)--F(x)= ",+A a f(f) dt- I'"0 /(t) dt

J

=.j

1·+" /\t) dt

",+A j. t) dt+ JOII f(t) dt= "'

221

ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS

:. !I~±h)h =- ~i~) = h~ I"'+"

f(t) dt

:II

•••

J,"+II f(x) dt=h-1 f(x)"J,"+h

1 Also 11:11 or

1 JGI+" II 81 f(x)

or

1 JGI+" f(x)=h 81 f(x) dt

I dt=h f(x)

(4)

[1'"+" t '" =f(x)

dt=f(x) . (5)

Subtracting (5) from (4),

IF (X+~-F (x) -- f(x) \=1 ~f:+h [[(t) - f(x)ldt \ 1 ~ h-I., 'f(t)-f(x)! dt. £+TI

•

Making h-+O and using (I), we get

I

I

~im F(x+h~--F(x) -f(x) ~ 0

,1-+0

But

Lim

h

~

... (6)

I~(x+h)-F(x) -f(x) I ~ 0 hi?'

... (7)

h ...O,

Since modulus of any quantity is always non-negative. Combining (6) and (7), we get the required result. Theorem 20. Every point of continuity of a summable function f(t) is a Lebesgue point off(t). Proof. Letf(t) be sum mabie over the closed interval [a, b]. Also letftt) be continuous. at t=xl • To prove that Xl is a Lebesgue point of the functionf(t). Recall thatftt) is summable over [a, b] if

L b

•

f(t) dt=a finite quantity

f(t) is continuous at f=X l implies given e > 0, there exists 8 > 0 such that 'f(t) -/(.'"1) , < t whenever I t From which we get

I

Xl+h

Xl

or

1

II

\

I fll)--/tXl) I dt < •

fXl+h Xl

I f(l) -f(x!) I dl
0, i.e.

J:

I f II' dx
O.

Sometimes we denote the class of such functions by the symbol LP in case the mention of interval is not necessary. Then [1 or simply L denote be a sequence of functions belonging to £Pspace. The sequence is said to converge in mean with index p, if given II> 0,3 no E N s.t. m, n ~ no => 111m (x)-In (x) lip < • or if

f

I I,,(x)-/m(x) IP

dx~O as m-+oo and n~oo.

In this ca'le the sequence is called meanlundamental sequence

Or Cauchy sequence. The sequ~nce is said to converge in mean with index p to a function I(x) if given I! > 03 no. s.t. n ~ no => If/-I,. lip < 6 or if

II

eN

1-/n'IP

dx~(} as n_oo.

12·S. Definition. Cauch; sequence. Let < lro > be a sequence of functions belonging to a L'P space, This sequence is said to be a Fundamental sequence or Cauchy sequence if given II> 0, 3, no E N s.t. m, n ;iii no :0> iif.. -/.... \lp < •. Later on we shall show that every convergent sequence is a. Cauchy sequence. (Refer Theorem 13, Page 241). 11'6. Definition. Completeness of L'P·space. An LP-space is said to be complete if every Cauchy sequence in tne space is convergent at some point of the space,' i.e. for every Cauchy sequence < I .. > in the space, there is an element lin the space s.t. /.,-+1. A complete normed linear space is called Banach space. Theorem I. II IE Lp and g ~ I, then g E V,. Proof. Let IE Lp (a, b), so that (i) I is measurable over (a, b), (ii) I/I P is integrable i.e.,

Also let

J: I/P'dx
g(x) is measurable over (a, b). Hence the result (iii). •• g(x) ~ I(x)

. J:gig ;" or

)t

If)

dx ~ •

If 121 dx
I. To prove that IE L (a, b), it is enough to prove that (i) {is measurable over (a, b) (ii) 1 I I is integrable over (a, b).

By hypothesis, (iii) I is measurable over (a. b). (iv) I 1/1' is integrable over (a, b). Let E=(a, b). Clearly (iii) => 0). Let A=E( III ~ I), B=E( Then E=A U B, A n B=0.

III
1, then i.e.

11/+g I,

~ II

111,+11 g 1121'

(Burdblwan 1990,Meerut 1988, 87 ; Banaras 70, 68; Indore 78J Proof. Let/, g E £P (a, b), p > 1. To show that 0) I+g E Lp (a, b) (ii)

11/+g II,

< 111/1.+11 gil,·

Let q be a number conjugate to p, then

L +!. = 1. q

p

I, g E £P (a, b) .,. I+g E LP (a, b) Refer Theorem 3. g E LP (a, b) => I+g E LV (a, b) • (/+g)plq E Lq (a, b). Now 1 E £P (a, b), (f+g)Plq E U (a, b).

and J,

On applying Holder's inequality to these functions, we get

I: or

If 1·1 J~ g I plq dx ~

(J: III" dx rIP U: 'f+g

f:lfl.lf+glp,qdX.~ O:I/PJdx

I(P/q)q

f'p u: I/+g Iv

dx )"'1

dxf/'l ... (1)

Since /, g E LV (a, b) and therefore interchanging f and g in the last inequality, we get

J: I g I·I[+g

Iv/ q

dx ~

(J: I g Iv dxf'V U: If+g IV dx/,q

... (2)

Adding (I) and (2), we get

J: If ,.,

[+ g IV;q dx

+J: I g 1·lf+g Iv/q dx

232

OJ-SPACES

~

[(J"OI/IJ>dx )1/11+ (Jbo,gl1l dX )1/11] ((bJ.I/+gI1ldx )1111 ... (3)

..• p~+~=l :• q So tong

l+I!.._p q .

as I/+g I'=I/+g 1.I/+g

111«

1

~ (1/1+lg 1).I/+g 111111 =1/1.I/+g 1"}I1+1 g '.I/+g Ipla.

Integrating, we get

J: II+gIPdx~ J: 1/1.I/+gllllqdx+J: Igl·l/fglPlqdx J: I I+g P' dx " [0: I I I" dx rIP +0: 1 I" dx tv J(J: if+g dx r On using (3), we get

Dividing by

0: or

(J: 11+

g

Ip dx

til

ll

IP

g

and observing that

1 1 1--=-, we get q ,p

I/+g!, dx

r'p~(J: 1/111 dx Y'' +0: I

g

I" dx

f'P

II/+g II, ~ 11/11,,+ II gil,,· Ex. State and prove Minkowski's inequality lor V'-spaces. [Banaras M.Sc. Final IV 1970] Theorem 6. Schwarz's inequality. II f, g E L2 (a, b); then Ig eLand IIlg 112 ~ 11/1/2 II g 112' Proof. Suppose I, g E LI (a, b) so that (i) f and g are measurable over (a, b) Oi) 1I 12, I g 12 are integrable over (a, b), a.e.,

J: I j(x) dx < f: g(x) dx < J: I j(x) I~ dx+ I: 1g(x) Ii dx < 12

00,

I

This =>

To prove that Ig E L(a, b) (iv) IIlg II ~ 11/112' II g 112 (i) => fg is measurable over (a, b)

12

00.

00.

. .. (1)

(iii)

... (2)

233

O'-SPACES

Suppose x E (a, b) is arbitrary. Then obviously ( I f(x) /-1 g(x) I )2 ~ O. This gives 2 I f(x) 1.1 g(x) I Et;; I f(x) 12+1 g(x) 12

or

n

f(x) g(x) I dx

~!

By virtue of (1), this

U: ~

'f(x) 12 dx+

I:,

J: I

g(x)

f(x) g(x) I dx
'f(x) g(x) , is integrable over (a, b) From (2) and (3), the result (iii) follows. Let at be any real number. Then (II If'

or or

at 2

oe 2

2 +211 I fl· If 1

~ 0

I g I + I g 12

J: If 12 dx+2:x J: Ifg I dx+j:

where p=J:

I.flla dx,

If p=O, i.e, if

0

I g 12 dx ~ 0

J:

q=2

If /2

J:

Ifg

... (4)

I dx,

~x=o then

r=J:

Ig

12 dx

If 12 =0

/(x)=O a.e. on (a, b) In this case both the sides of the inequality namely

I:

are Zero.

Ifg

I dx

~

U: 'f 12 fa 0: I dx

or

Set

f1.=~q

q2

q2

in (4). Then ~!:+

Putting the values of p, q, r we get

2

g

/2 dx

t2

Excluding this trivial case we suppose that p:;toO.

----+r ~ 0 or -. q2+4pr 4p 2 p ' q ~ 2pl'2 r1/2.

or

or

~

a 2p+aq+r ~ 0

or

or

+ I g I )2

... (3)

(

~

;pq) q+r ;.. 0

0

c?'

i

J: IlgldX~2(J: Ifl t U: Igl~dx fa J: I I ~ (1: If t2 0: / t2 2

fg

dx

dx

2

12 dx

g 12 dx

This is also expressed by writing IIjg II ~ II I liz II gl!z.

Hence the result (iy).

.(5)

234 Remark. This theorem IS also expressed by saying that: Let f and g be measurable functions on [a, b]. If f and g are square integrable, then fg E L [a, b] and

Solution.

1:

Then

[Banaras 1971] First write the proof of the previous Theorem 6. 1 fg I

~(J:

dx

If 12 dx

r 0: /2

I

g

12 dx r~

Since If 12-P and het'lce the last gives

J:

I/g

1

~

dx

IJ: fg dx I~ I:

But

Ex. t.

IIJ .. fg dx b

lJo1f(x)

Solution. we get or

I L

I~ [j1 0

Proved .

(J

I]. show that

I f(x) 12 dx

)1/2 .

[Banaras 1971]

If f.g E V (0, 1], then by Schwarz's inequality, IIfg II ~ 11/.1'2 II g 112

1: I fg 1dx ~ U:

J: ! f I [For

or

by (6)

[

If 12

dx

Taking g(x) = 1 V x, we get

or

... (6)

I fg I dx

{J

IflE L2 [0, dx

dxtlO: g2 dXY'2

[ jbf = dx ]I/B[Jb.. 154 dx ]1/1 , ~ b12 dx )1/2[Jb g2 dx ]1/2 . ~

...

(J:r

dx

I:

~

[1: II

Igl2 dx

IJ: f dx I ~ J:

12 dx

rIll: I

I' d.~J/I

r /2

=(t dX=1]

I f I dx

~

U: I

If: I~ U: I f r'2. f dx

g

f

II dx Tla

12 dx

Ex. 2. Iff and g are Jquare integrable in the Lehesge sense, then prove that 1+ g is also square integrable in the Lebesgue sense Qnd [Meer ut 1971J

235

f and

Solution. sense so that

f,

g are square integrable in the Lebesgue g E L2 [a, b].

(I +g)

To prove

E L2 [a, b]

li/+g 112 ~ 11l112+11 gilM

and

By Schwarz's inequality, f, geL2[a, b] ~ fgE L[a, b] ..• (1) Alsof, gEU[a, b) ~ p, g2EL[a, b] ... (2) (1) and (2) :::> P+g2+2/gEL[a, b] ~ (/+g)2EL[a, b] ~ l+gE Uta, b]. This proves the first required result I1+ g 12 ~ ( I f I + I g 1)2

= 1/1 2 + IgI2+2Ifl·lgl·

= I/I!+/g

. J: I f+g 12 ~ J:

12

+21lg

If !I+

1

C

I

g i2+2

1: Ilg I .

By Schwarz's inequality, this gives

f: I

f+ g

12

J: I11 + J: I

~

1

+2[1:

g

12

T/2 U:

If 12

I g 12

J/

2

=[0: I t2 +0: I \2 fa r (J: I/+gI2t2~(j: 1/12 f2+(J: I t2 fl2

g

Taking positive square root

g,a

II 1+ g 112

or

~

'I f

II~+II

g 112·

Tbeorme 7. Let < fn > be a St quence of integrable functiuns which converge in mean to a IUllction f 1 hell the sequence converges in mea,mre to f Proof. See theorem I, chapter 10, page 177. Theorme 8. Let I (x) ~ 0 be an integrable lunction a e. 011 a measurable set E. Then! E

I

(x) dx=O iff f (x)=O a.e. on E.

The proof of this theorem is left .as an exercise for students,

236

LP-SPACES

Therome 9. Show that (U', d) is a metric space. [Kolhapur 1968J Proof.

In, f,,,

Let

We define d

(In,

E LP (a, b) be arbitrary.

f,,,)= II In-1m

111'=0: lin-I", Ip dx t

P

We have to prove that (£1', d) is a metric space. (i)

:.

i.e. (ii) d

(I,

...

I in

-1m (x) I ~ 0 I In f.,. I p dx II/n--Im II p=d (In.!m) (x)

0:

t

g)=O II

~

;;;.

0

O.

I -g II p=O

*S: I I-g

P

121 dx=O

11- g 121 =0

a.e., By Theorem 8. a.e., I =g a.e. • d (.f. g)=O ifl=g a.e. (iii) d(/, g)=d(g,/) For II--gi = Ig-II (iv) IIf-g II (f--h)+(h-g) II" where!, g, hEU'. On applying Minkowski's inequality, we get IIf-gll,,= II (f-h)+(It--g) II" E;;; II f-/H,,+ II h-g lip =d (f, h)+d (h, g)

-=- 1/- g 1 =0 11

11,,=




is enough to show

\I 1m --In III'
is a sequence of functions belonging to an LP-space. 1/ this sequence is convergent, then it is a Cauchy sequence. [Meerut 1990] Of A sequence 0/ functions belonging to an LP-space is a Cauchy sequence if this sequence has its limit in the space. Proof. Let the sequence < /n > of functions belong to an £1'space S.t. this is a convergent sequence so that it must converge to a functionf E LP-space (say) i.e., 3f E £P (a, b) S.t.

Lim fn=f

a.e.

n~oo

To prove that (1) ~ given E

>.;f /I

IIfm

~ no:;:>

11/- f"

-In II ,,=[1 (." -/+f f"

II

,< 11/", -/11 + 11/-/","

,

~

..( I)

< {" > is a Cauchy sequence. > 0, there exists a positive integer II

E

II

< '2

no S.t.

¥ m, n ~ nil

p

hy Minkowsk:'s inequality

p.

is a Cauchy sequence. Ex. If < /.. > is a Cauchy sequence, then it has a limit. Hint Refer to first part of Theorem 12, the result£ at ollce follows. Theorem 14. A sequence

0/ functions

belonging to an

Prooi. If po~slhle let /n~/, rr.~g. Then the identity .f g=/ /n +f,.-g gives ii/ g II ~ :: / -Iii Ii + II I .. - g n . This follows from Minkowski in~quality. Since -.-/:;:> !If,,-/il =0 as n~oo Ift-+g ~ II/.. -g II =0 as ,,--,00. Now (I) shows that IIf- g II ~ 0

r.,

. .. (1)

242

IIf-g II

But

~

0 is always true.

Combining the last two,

IIf-gil=o a.e. => f==g

This

Proved.

Approximation by continuous functions. Theorem 15. Iff is a bounded measurable function on [a, b] thenf.r a given ~ > 0, a a continuous function g on [a, b] s.t.

IIf-g 112 < •. Proof. Suppose f is a bounded closed interval [a, b] and let

mea~urable

function over the

F(X)=S: f(t)dt,x E [a;b]

... (1)

... (2)

Then F' (x)=f(x) a.e. on [a, b]. Let M be an upper bound of I/(x) 1in [a, b]. Then.

I f(x) I ~ M V x

E [a, b].

If x, x+h E [a, b], then F(x+h) -F(x)=

[+11 f(t) dt -

1:

f(t) dt

[+II f(t) dt+ 1: f(/) dl= [+II J(t) dl F(x+h)-F(x)= j f(t) dt =

Gl+1I

:.

01

... (3)

From which, 1 F(x+h)--F(x)

i=/ (+TIf(t) dt I ~ J:+h If(l) J dt=Mh. ~

I dt

GI+II 01

M

1 F(xth)-f~x) 1 ~ Mh.

< 8 and Mh < El , we get I F(x+h)-F(x) I < tl' whenever

Taking h

h

< 8.

This => F(x) is continuous in [a, b]. For each natural number n, we define Gn(x)=n

r

Gl+(l/"'

J

,

/(t) dt, x E [a, b]

.•• (4)

V'-SPACES

243 = n[ F

(X +~) - F( x) ]

G,.(x)=n [ F(X+!

)-F(X)]

... (5)

Also F(x) is continuous in [a, b]. Consequently G.(x) is continuous in [a, b]. From (5),

(X+n!-.) --F(x)

/ F Lim

Lim ~

n-+oo u,,(x)= n-+oo - - - lIn - - -

_ Lim F(x+h) -F(.~'Le=F'(X) = r.(x) by (2)

-h-+O

h

Lim

,

Gn(x)=/(x) a.e. on [a, b]

n-?oo

From (4),

J~

I G.. (x) I = I n !

~

...

j"'t1'" .f(t) dl \ ~ n 1.,+1/11

",+l/..

11M J

1/(1)

11M dt=- =M.

n

x

I G,,(x) I ~ M I G,,-/12 ~ (I ~

¥ n E N and V x E [a, bJ

Gn 4M2,

I + III y.

)2 ~ (M+M)2=4M2

n.

By virtue of (6), Lim n-~

0Ci

G..(x)-/(x)=O

or

Lim

n->oo

- 0 I Gn-11 2 .

Hence, by Lebesgue's bounded corivergence theorem, Lim

J: I G.. -/12 :.

r

x=

Lim n ... oo This

1: Lim I G.. -/12 dx= J:

Lim

n-+oo

(fb

fb I Gil-liZ dx=O CI

I Gra -/12 dx )1 12=0.

CI

'* Lim II G.. -/1I2=0 '* given > 0, 3 positive integer no S.t. fi

V n ~

110

I dt

(;C

(l:

\

G•• -/12

... (6)

=> II/-G.. 11

~


a g(x) ~ f(x) g(x) ~ f3.g(x) as g ~ 0 -

IX

f Ii g(x) dx

~

f Ii f(x) g(x) dx

~ (3 IE

g(x) dx

J

[asf, g are integrable => fg is integrable]

Jrg

f~.fg. dX=A ..

=>

where q. Corollalt'y.

dx

... (1)

~ i\ ~

To prove t f g

13.

dx=/(~)

Ie

g dx.

f is integrable

=> f is continuous => 3 x, y E E S.t. f(x)=rz,J(Y)=fJ and/(x) takes every value between a and (J

=> 3 , E E s.t. f(~)=A where

:. t

fg

dx=f(~)

t

IX

~

).

~

p.

g dx, by (I).

Theorem 3. Second mean value theore... If f( x) is integrable over (a, b), and g(x) is positive, bounded, and non-increasing, then f/(X) g(x) dx=g (a+O)f:f(X) dx where

~

is some number between a and b.

Fix s > 0 S.t. e < g (a+O) -g (b -0). This => 3 a point Xl s.t. Proof.

g(a+O)-g(~)

< e

LEBE~GUE

fURTHER TREOREMS ON

249

INTEGRATION

Similarly there are points x 2 , xa," ... , x" s.t. g(xr - 1 +O)-g(x)


(S)

J:

Theorem 6.

..

1:

I:

I g'(x) I dx

obtain

f{x) g'(x) dx

~~~ I

Therefore This

1:

i E;; ~ (L)

f(x) g'(x) dx /=0

f(x) g'(x) dx

f(x) dg (x)=(L)

J:

fiX) g'(x) dx by (I).

Proved. Helly's second theorem. Let f(x) be a continuous

function defined 011 the interval (a, b) and < gn(x) > be a sequence of functions which converges to a finite fUllctkm g(X) at erery point of [a, b]. If

V/) (gn) a




(g(X;,

1

t>

g(x)

In this event (5) and (6) taken together imply that

lI~ I a f dgm

faa r dg < I

Ii

1;I < ~I-J

254

FVRrHER THEOREMS ON l.EBESGUE INTEGRATION

Thi

~

I

b

Lim

In_CO

Lim

n_oo

a

Jba

I(x) dgm (X)=Jb I(x) dg (x) a

I{x) dgn (X)=J b I(x) dg (x). a

7. Definition. Cumulative distribution function. Let B be a family of Borel sets, each of which is a subset of R. A measure IIdefined 011 B is called Borel measure on the real line. If II- is finite Borel measure, therl we associate to it a function F known as its cumulath'c distribution function by setting ... ( I) F(x)=f1o (-00, x] ¥ x E R It is easy to verify that F is a real valued monotone increasing function Also ~, (a, b]=F(b)-F(a). ...(2) F(b)-F(o)=p (-00, D]_.p (-00, aJ=f' (a, b]. For Since

n (a, b+~] n

(a,

b]= ,,=1

Hence it (a, b]= Lim f' ( a, .

n-+oo

b+J..] n

[Refer to Theorem 17, Chapter 7, page 100] Consequently F(b)= Lim F (b+~)=F(b+). n_co n This proves, by definition, that F(x) is continuous on the right. Furthermore p(b}= Lim p (b- ~,b ] n_oo n =

n~:

[

F(b)

-F (b- !)],

=F(b) -

by (2)

!.)

Lim F (b=F(b)-F(b-). n_oo n If p{b} were zero, i.e., Borel measure of a singleton set {b} is zero, then F(b)=F(b- ,. This shows that F(x) is continuous on the left. have shown that F(b)=F(b+ )=F(b-). This shows that F(x) is ccntinuous if p{b} =0 00

Since

n (- ,-nl=0,

n~l

m(0)=0.

Finally we

255

FURTHER THEOREMS ON LEBESGUE INTEGRATION

Therefore

Lim

n -+--00

F(n)=O

or

Lim

x-+-oo

F(x) =0.

Thus we have shown that 'if po is Borel measure on the real line, then the cumulative distribution function F is a monotone increasing bounded function which is continuous on the right hand and Lim F(x) =0.

x-.-oo Definition. Lebesgue-Stieltjeg integral. Let f be a non-negative Borel measurable function and F a monotone increasing function which is continuous on the right. Then the Lebesgue-Stieltjes integral off w.r.t. F is defined as f

f

dF=f f dpo,

where po is a finite Borel measure and F its cumulative distribution function. : Let f be any Borel measurable function, then we define f+(X)~-{ f(x) if f(x) ~ 0 o if f(x) < 0 J 0 if f(x) ~ 0 f-(x) = 1 -f(x) if f(x) 271f. => M

=> 2nJiI-

M =- 2n~i-+E
0, there is a polynomial P (x) s.t. I /(x)-P(x) f < • yo iC E (0, 1].

Solution. Here write proof of Theorem 1. In the end of the proof put B,.(x)=P(x).

260

SEMI CONTINUOUS FUNCTIONS

Problem 2. For ellery continuous/unction/(x) on (0, I), there is a sequence 0/ polynomials < Pn(x» S.t. < Pn{x) > converges uni/crmly to J(x) w.,..t. x E LO, 1] as n~tX). Solution. Here write proof of theorem I. In the end of the proof put B,.(x)=Pn(x).

.. C.x~.(t _x)n-k (k- nx)1 ~ (n/4).

Problem 3. Prove that E

1>:=0 I>:

Solution. Here write the proof of equation (7) Theorem 1. Now we shall prove a very important theorem known as Weierstrass approximation theorem of real analysis. Theorem 2. (Weierstrass Theorem). Let lex) be a continuous /ullction defined on the closed interval [a, b J. For every 6 > 0, there exists a polynomial P(x) S.t., I/(x) -P(x)

1
s.t. < p,.(x) > converges uniformly to lex) w.r.t. x E [0, I] as n~OO. The proof of the lemma starts [For proof refer to Theorem I]. This completes the proof of the lemma. Going back to the proof of the actual theorem, suppose that / is a continuous function on [a, b]. If [a, b]=[O, I), then theorem is proved. Now we suppose (a, b]:;e[O, 1]. Define g(y)=/[a+(b-a) y] Then g(O) =/(a) , g(l)=/tb). Also/is continuc.us on La, b]. This leads to the conclusion that g(y) is continuous on (0. 1]. Hence, by the lemma, just proved, :I a polynomial Q(y) s.t. I g(y)-Q(y) ! < 6, ..... (1), 0 ~ y ~ 1. If x E [a, bJ a.nd y E [0, I], then set' x a y= -_._-- so that boa

'61

SEMI CONTiNUOUS FUNCTIONS

g(y)~g (:=:)=1 [a+(b-a) (~=:)]=f(X). Then we have j!(X)-Q

(:=:) I
fies our requirements.

Theory and problems related to semi-continuous runctio~s. 14'2. Definition. Limit superior and Limit inferior. Let a function f be defined on a set E and let Xu be a limit point of E. Consider an arbitrary real number ~ > O. Write M,,(xu)=sup {[(x)}, m.,(xo)= inf {f(x)}. x E (xo 8, x o +3) n E. The quantity M~(xo) does not increase and M~(xo) does not decrease with decreasing 8. Hence the (finite or infinite) limits Lim Lim M(xo) = 8_0 M.,(x o), m(xo)=8_0 m.,(xo) exi!,t. M(xo) is called limit superior and m(xu) is called limit inferior. We also write Lim M(xo) =

x-+xo

[(x),

If Xo E E, then we have the relation m(x,,) E;;, I(xo> C; M(xo)· Notice that the function may not be defined at Xo itself. 14'3. Definition. Semi· continuous function A function I, defined on a closed intervallO, Ij, is said to be lower semi·continuous or upper semi-continuous at the point Xo according as Lim - - (x)-/(xo)

X-Xo'

or

Lim x-+xu

r

.

J(X)=j x I 0

Iff is continuous at the point x~, then it will be both upper and lower continuous. F·or I"IS contmuous at Xo => Ji1,x)= Lim

X-Xo

ji(x)= Lim r- -- J{x)

X--'o

262'

SEMI CONTINUOUS FUN'!TIONS

Conversely if1 is lower and upper both continuous at the point Xo then 1 is continuous at XII provided 1 is a finite function. Then lower semi-continuity of a fmlction at a point Xo is equivalent to upper semi-continuity of the function - I at the same point xo' 14'4. Definition. A function I is said to be lower semi continuous on the segment [a, b) if it is defined and lower semi continuous at each point of this seginent. In a similar fashion we definf upper semi-continuity of a function on a segmen~ [a, b]. Theorem 3. Let a lunction I be defined in a closed interval [a, b] and let Xo E [a, b]. Also let < x" > be an arbitrary sequence 01 points in [a, bJ s.t. < x" > converges to xI) Then a necessary anc1 sufficient condition that 1 be lower semi continuous at Xo is that Lim I(x.) , n_oo I(x,,) Proof. Lim X,,=X o => 3 a subsequence < X n-+oo n" Lim Lim k-oo l(x/1,) = /1.-+ oc I(x,,)

;

kEN> s.t.

... (1)

Making use of the result : The limit superior of the function I at a point Xu is the greatest of the numbers which are limits of the sequence of the for~ l(x1),f(X 2),f(xa), ... where x" E [a, hI and x,,~Xo We see that Lim I(x) ~ Lim /(x ) n_ CI) 11/: x ... Xo . . Lim Lim Usmg (1), thIS becomes '.-- f(x) ::;;;; - . (x,,) fI-+CIJ

" - Xo

Let f be lower semi-continuous at ;r 0 so that Lim f(xo) = ":::':;Yo f(x). Lim To prove thatf(xo)

f

..• (5)

Urn- ji(X ) = ji( xo)' X-Xo is lower semi continuous at Xo.

Theorem 4. Let a function f be gefilled on a closed interval [a, b], Xo E [a, b] and f(x o) >- 00. A necessary and sufficient condition that I be lower semi continuous at x, is that 'to every I(xo} > A, there corresponds a 0 > 0 such that I(x) > A provided I x-xo I < b, X E [a, b]. Proof. [a, b] and

(i)

where

Xo

Suppose a functionf is defined on a closed interval E [a, bJ. Also suppose thatf(xo} > - co.

Let f be lower semi continuous so that Lim f(x o) = - - f(x) 3 8> 0 s.t. )C -+Xo Lim Lim ' f(x)=m(xo) = --~ m.l (x..) x ~Xo x-)oxo

..• ( I)

... (2)

m.,(xo)=inf {I(x) : x E (xo-8, xo+o) n [a, b]}. Also let f(x o' > A. To prove that 3 XE [a, b] s.t.f(x) > A whenever I x -Xo I < O. From (J) and (2), it follows that 3 x E (xo-'S, x o+8)nla, b]s.t. m~(xo) > A .•. (3) Since f(x} ~ ma(xo) for I x - Xo I < 8. On using (3),j(x) > A for I x-xo I < 8. Conversely, suppose that/(xo) > A. Now we can find a 8 > 0 corresponding to this A s.t. ma(xo) ~ A, which in turn implies that m(xo) ;;;;. A.

with

Increasing A and taking the limit as we obtain m(xo) d: I(xo)' m(xo) is always true.

A - /(xo) ,

But

f(x o) ~

264

SEMI CONTINUOUS FUNCTIONS

Combining the last two inequalities, we get Lim f(x o}= m(xo) = f(x o}· X-+Xo

This proves that/ex) is lower semi

contin~ous

at

Xo·

Remark. If (x.) is a finite number, then the above theorem can also be restated in the following form: A necessary and sufficient condition that /tx) be lower semicontinuous at the point Xo is that to every to> 0, 3 a number ~>O s.t. f(x o) - E < f(x) provided I x-xo I < Ii and x E (a, b]. The theorem gives a relation between semi continuity and continuity. Theorem S. Suppose that the function f and g are defined on the segment [a, b] and also suppOle that f and g both are lower semi continuous at x=rxo. If the sumi+g is defined on [a, bJ, then it is lower ~emi continuous at xO. Solutiod. We may suppose thatf(xo) +g(xo) > - 00, hut such of the numbersf(xo} and g(x o) is distinct from -d:). Consider an arbitrary number A s.t. /(x..,}+g(x o) > A. Then there exist numbers Band C s.t. f(x o} > B, g(xll ) > C so that B+C > A. Also f(x) and g(x) both are semi continuous at xu. Therefore a a number ~ > 0 s.t. !(x»B, g(x»C where xe[a, b) and I x-xo I < 8. lRefer to Theorem 4J For this x, j(x) > B, g(x) > C .. f(x)+g(x) > B+C > A => f(x)+g(x) > A. This proves that f{x) + g(x) is lower semi continuous at xo. (Refer to Theorem 4] Theorem 6. A necessary and sufficient condition that a fUIlClioll f defined on the segment F- [a, b] be lower semi cofltinuous 011 E is that the set E (f ~ c:) be closed for arbitary real c. Proof. Let a function/be defined on the segment £=(a, b1. Also suppose that c is any real number. (i) Let f be lower semi continuous on E. Let A ~E (f ~ c). To ~rove that 4 is closed.

SEMI

CONrI~UOLJ~,

265

HIN('noNS

If the points of sequence <x,,> belong to A and if x,,~xo' then lower sl'mi continuity off implies that f (xo ) ~ -Lim.) --- J~xn . [Refer to Theorem 3] I1~OO

Xn

E A

~

f

(x,,) E;

Lim

~n~oo'

C

r( ) X,.

=> ,(xo)

~

=> f (xu)

~ C

c'

..... C

Lim --. f(xn}

I1~OO

~ C

=> xoEA.

Thus x,,-Xo E A. This => limit point of the set {xn: 11=1, 2, 3, ...... } C A belongs to A => A is closed. (ii) Conversely suppoxo

If possible, let us suppose that f (xo) > B. Our assumption implies that 3 a sequence


v(A)+ ... (4)

Since v assumes atmost one of values - co, and co, and veE) is finite, therefore, (4) says that v( A) is finite and the series 00

E

1'~1

From (4),

1 . - IS convergent.

n_

~'(A)- 00 in (5), we get vCR) O.


()

A is a positive 'iet with veAl > O. This concludes the problem. Theorem 5. (Hahn Decomposition Theorem). Suppose r is a signed measure on a measurable space (X, A). Then :I a positil' . set P and a negatii'e set Qs.t. P n Q=0, X=P U Q A being a-a/~ebra of subsets of X. rBurdbw an 199.0, Punjab 1967•. 65 j Proof. Let A be a-algebra of subsets of X. Let v be a signed measure on a measurable space (X, A). Since v assumes atmost one of the values - co and + 00. Without loss of generality we can suppose that v does not take - co. Consider the family B of all negative subsets of X and let Con~equcntly

A=inf [I'(E) : E E B}. . B s.t. Lim v( En) Th en:J a sequence <En> In

n--* CO

\

-n.

B is a family of negative sets

::;. <Eft> is a

~equence

of negative sets

oc

=> U E t is a negative set, by i=l

Theorem 2.

=> Q is a negative set on taking 00

Q= U E,. ;=1

17hus Q is a negative subset of X. Then, acc.Jrding to (1), Next we consider the subset Q-E.. of Q. Q=(Q- En) U En

I'(Q) ~.\.


.\>-00 Next our aim is to show that P=X-Q is a positive subset ofX. Suppose the contrary. Then P is not positive and so P is negative. Hence by definition. for every measurable set E C p. vee) < O. Now E is a measurable subset of X with negative measure. Making use of the result : If E is a measurable set of finite negative measure. i.e., - 0 0 < veE) < 0, then E contains a set A with yeA) < 0, we obtain a negative set ACE s.t. YeA) < O. Since A and Q both are disjoint negative subsets of X and their union A U Q is also negative. Refer to Theorem 2. Consequently v(A U Q) ~ .\, by virtue of (1). But ,\ ~ v(A U Q)'= v(A) + v(Q) = v(A)+ >... or .\ ..; v(A)+.\ or veAl ;;. O. Contrary to the fact that YeA) < O. Hence our assumption is wrong, i.e.," P is not positive" is wrong. Therefore P is positive. Thus P=X - Q is positive and Q is negative. :. X=p U Q, P n Q=0· A decomposition of a measurable space X into t~ subsets P and Q s.t. X=P U Q, P n Q=0, where P and Q are positive and negative sets respectively relative to. the signed measure v, is called Hahn decomposition for the signed measure v. Also P and Q are called positive and negative components of X. Remark. The Hahn decomposition is not unique. If {Po Q} is a Hahn decomposition for v, then we may define two signed measures ~+ and v- s.t. y=v+-v- by setting v+(E)=v(E n P) v-(E)=-v(E Q).

n

IS·3. Singular measures. Two measures VI and V2 are said to be mutually singular (in symbol "I .L v2 ) if 3 a measurable set A C Xs.t. v1(A)=0=Vi(X -A).

For example, the measures v" and mutuaIly singular.

v-, defined

as above, are

274

SIGNED MEASURE

15'4. Definition. Jordan decomposition. [Punjab 67,65] Let v be a signed measure on a measurable space (X, A). By Jordan decomposition of v, we mean a pair (VI' Va) of mutually singular measures VI and VB satisfying the condition V=V1-V I • For example a pair of measures (v+, v-) defined as above, is a Jordan decomposition of v. 15'5. Definition. Absolutely continuous measure function. Let v and ~ be measure functions defined on a space (X, A). The measure v is said to be absolutely continuous w.r.t. p. iff p.(A)=O ¥ A E A ~ v(A)=O and is denoted by V ~ ".. In case ~ is a-finite, the converse is also true. Theorem 6. (Radon Nikodyn Theorem). Let (X, A,,...) be a a-!inite· measure space. LeI v'be a measure defined on A s.l. v is absolutely. continuous w.r.t. p.. TlJen there exists a non-negative measurable function f s.t. V(E)=isfdf' ¥ Ee A. The function f is uniqne in the sense that if g is any measurable functioll with this property, then g(x)-=f(x) almost everywhere in X w.r t. ".. {BuJ'dbw,an,1990;,Kanpur Statistics 1977, 76; Punjab 66, 65; Banaras 64] Proof. To establis,h the existence of the functionf, let us assume that p. is finite. ~ is finite *v -fJ'" is a signed measure on A for each rational number «.

Let T denote the set of all non-negative rational number. For every t e T, let (Ph Q,) be a Hahn decomposilion for the signed measure V-Ip.. If at ;;. t, then Q, is negative for v -11.".. For if E is a measurable subset of Q" then v(E)-Glf' (E) "I dl'= 00. [For /(x)=O Y x E P, hence/: E ... [ -00, 00] is not integrable] Hence

V(E)=ff/df'

This establishes the equality in case I' (E n P) > O. Absolute continuity of v w.r.t. I' implies that v(Enp)-O. Select a positive integer n and set .

Et:={X

E E:

~~1 < lex)

!:., we have n

E1c C{x EX :f(x) ~ n}CS.= n{Q,: y t < q}. Also Q, is a negative set for the signed measure v-ql' and Elt i~ a sub~et of Q,. Therefore E" is a negative set. (Refer Theorem 2). This => (v-ql') (Et ) ~ 0 => v(Ek ) ~ qp.(Ek ) k

=> v(Ek) ~ -n '" (E,,). On the other hand, since

.... (1)

E" C { x E X :f 11 f c

Pet-t,/ft

1) '" ] (E,,) ~ O. For P(k-t,/ft is positive,

v(Ek )

~

(

k ~ 1) 1'(Ek ).

• ••

(2)

From (I) and (2),

k-l) ",(E (-n

Ie )

~ v(E/,) ~

k Ii

",(E~).

. .. (J)

Since E is a disjoint u"ion of sets En P and E". Hence ",(Enp)='=v(Enp). It follows from coulltahle additive prop .• y of integral that

ff Ii

d",=

~ froLt f

11=0

d"" v(E) =

E v (Et )

k-l

By definition of E •• k-l -

k

~f(x)~-on

n

E".

"

Applying first mean va.lue theorem,

n (k-l)

f , co] be any measurable function satisfying

the condition For each n

l'(E)==

e:

f,. g

d". ¥ E

e:

A.

N, let

A .. ={x E X :J(x)-J(x) ~ J/n} E A and Bn={x e: x: g(x)-/(x) ~ lIn} E A. By definition of An, I(x) -g(x) ;;t JIn on A...

Applying first mean value theorem, we get fA .. (I-g) d".

or

~ ~ I£(A

fA .. Id".-fAn gdfl

II )

~ ~"'(A.)

v(A..)-v(A,,) ~ (lIn) fl(A.) (lIn) ".(A .. ) ~ 0 or I£(An) ~ O.

or or

But I£(A,.) ;;;;. 0 is always true :. ".(An) =0. Similarly we also have I'(B,.)=O. Let

C={x

e:

-

X : J(x) #g(x)} = U (A,.UB .. ). n_l

."

Then p(C)=1: p(A,.)+}J ,,(B..)==1: 0+1: 0=0.

"

..

Then p(C)=O. This => f-g a.e., on X w.r.t. p. Exactly in a similar way the theorem can be proved in case ~ is a - infinite. Remark. The functionJ: X-+[ -00,00], defined as above, is called Radon Nikodom derivativ~ of the measure v w.r.t. 1£. In symbol,

f= dv. dp.

Theorem 7. (Lebesgue Decomposition Theorem). [Punjab 69, 66. 65] Lft (X, A. ".) be a a-finite measure space end v a.-finite measure defined on A. Then 3 two uniquely determined measllres \I. and 1'1 s.t. V= VO+V1 • Va 1. "., VI « ",.

278

SIGNED MEASURE

Proof. >'=I-'+v. p. and v are G-finite implies that>. is a-finite.

" < ~ and v ..

p. . "I E E A.

'E

Let A={x EX: f(x) Then X

E

>

OJ, B={x EX: f(x)=O}.

··~UB, A()B=0.

Furthermore p.(B)=LfdA=O:

Defir ~ two functions Vo, VI: A-.[oo, 00] by requiring that Vo (E)=v (EnB), VI (E)=v (EnA) ¥ EEA. Then Vo and VI are measures on A and satisfy the condition V=VO+Vl'

v. (A)=v(AnB)=v (0)=0, or Vo (A)=O. Thus p(B)=O=vo (A)=vo (X -B) i.e. p.(B)=O==vo (X -B) This ~ Vo is mutually singular to p. ;;:> vol-I-'· To show that VI ~ p.. For this let EEA be arbitrary s.t. I-'(E)=O.

Then

Isf dA=p(E)=O

fE/ d)..=O.

or

Also /(x)

~0

a.e. on E relative to ~. on A()E and hence VI (E)==v (EnA) by definition of VJ ~" (EnA)=O. :. VI (E) ~ O. But VI (E) ~ o. Combining these two results, VI (E)=O. Thus we have shown that ,.(£)=0 =>

¥

xE E.

This ~ f=O Since f>O

Th~

To prove the uniqueness of

VI

(E)=-O

.~ 2

b"=2 • X

SIR

JI

Ans.

2 • cos (n7tx) +mc T dx ]

.•. (1)

FOURIER SERIES

291

4 cos nlt+-.2 2 (. = [ -sm -n1tX)2] = -4- (-1)". nrc nrc nrc 2, nrc Hence (I) becomes

- -4 (- I)" sm . (nrcx) 1: -2 "=1 nrc

X=

a I(x)=f+

:1a" 01)

cos (nrcx) -,-

... (2)

Here 1=2.

(b)

Let

a,,=

f- t I(x) cos (n~x) dX=~ f: I(x) cos (n;x) dx

=

.

r x cos (nltx) dx= (2X sin n1tX)2 _~ II sin ~ dIC 2 nit 2 . nrc, 2

J8

4 ( nrcx)! 4 =0+ T2 cos -2 =""""iZ

nlt

=-iz [( -I)" --I] jf n Tt

:.

an =

nit

4)

nrc-I]

n:;t:O.

~.~ jf n= 1, 3,

nit

r

LCOS

5, 7, ...... and a,.=O if n=2, 4 .... .

=1- J: J{x) dx=~ r: x dx=2.

ao

01)

nltx

4

Now (2) becomes, f(x) = I +}; 22 [( -I)" -I] cos -2 "=1 n 7t

f(x)=l-~[~ cos (r.X)+~ cos (3nX) n: P 2.:z 2

or

+~ cos(5~X)+ ...... ] Problem 3. cosine series. . Solution.

where

Expand f(x)==sin x, 0

We have

I=!Jit rc 0

n: ,

x

in a Fourier

,

sin x cos

I'IX

[sin (I-n)

dx.

Here /=rc.

dx

... (1)

x+sin (l+n) xl dx

=_! [COS (I-II) X + cos (n+ 1) rc

< x

a ':-'1 a. cos (nrex) I(x)=t+ -,-

a..=~ f~ Itx) cos (";X, a.. _3


and < bn > are FOl/ri('r c()efficip.llt.~ of lJ functior lim lim f , then n-?oo 0 11 =0, n-+oo b,,-=O.

295

FOURIER SERIES

Proof. Fix Ii > O. Define an unbounded measurable function g on [ -n, 1t] s.t.

J:"

Then g E 1..2

1 f{x)--g(x) I

dx

j!.


7t

J'It sill (n+l) u ---. - '(-I .')- [nf(x+u)+f(x-u)] duo 0

Sill

nU

[Banars 1961, 63] Proof.

U

m

Sm (x)= ~-+ 1: (a" cos -

"~l

flX-t

btl sin

I1X).

. _.( I)

296

FOURIER SERIES

Q,.=!KJo(2" f(t)

1

nt dt, b,.=! 2• f(t} sin nt dt n 0 Putting these values in (1),

with

1t

Sm=--K1 Ill.. f(t) 0

COS

[ i+

m1:

(cos nt cos nx+ sin nt sin nx) ] dt

n-l

r.Sm=J:' f(t} [~ +n~l cos n (t-x) ]dt

or

Put 1t

t-x=u. Then

Sm= J2"-« f(x+u) -~

[I-2+

m ~

cos nu ] dll

n-l

. "! sin (m+i) u Smce j+ 1: cos nu= - 2 . (i) ,.~l SIn u

Dm(u), say.

Hem:e 1t,Sm=f_r"II-1I f(xtU) Dm (ll) du.

. .. (2)

Evidently Dm (u) is a periodic function of period 2/1 and feu)

is given to be periodic fUI)ction of period 21t. Hence the interval [-x, 21t-x] can be replaced by another interval of length 2 .. namely [0, 21tJ. Thus Tt

Sm= J•2" l(x+u) Dm (u) du=

I" 0

f(x+u) Dm (u) du

+J:" f(x+ u) Dm (u) du Putting u= -t in the second integral, 71:

Sm=I' f(u+x) Dm (u) du+ o

-=

1:

f(u+x) Dm (u) du+

r-

2

..

J-1I"

r:

flx--t) Dm (-t) (-dt)

I(x -I) D", (t) dt as Dm (/)=Dm (- 1)

=[ f(x+u) Dm (u) du+[ I(x-t) Dm

. . 1: or

SIII=l

1t

(I)

dt,

by the

same reasoning discussed above [/(x+IIJ+/(x-U) Dm (1I) du

J*' [f(x+u)+f(x-u)] 0

Dm (u) du

297

FOURIER SERIES

or

S,,= ~ 11:

J" [!(x+u)+!(x-u)] Dn (u) duo 0

17'7. Summation of series by arithmetic means. [Burdbwan 1990] Let E an be a series of real numbers with partial sums

If l: an is not convergent, i.e., if lim S" does not tend to a limit, then it is sometimes possible to associate with the series a 'sum' in a less direct way. This simplest method is known as 'Summation by arithmetic means.' . W rIte a,,= Sl +S2+n "+S,, - .

Then a .. is the arithmetic mean of the partial sums. If S" -+ S, then also (1,,- S. But an may tend to a limit even though lim Sf! does not exist. A series for which a" tend to It limit, is said to be summable by arithmetic means, or by (C, 1). Hence C stands for Cesaro's fur first order. Examples (i) E( - 1)" is summable (C, 1) to the sum i. (ii) I+O-I+I+O-l+ ............ is !>ummablc (C, I) to the sum 2/3. Theorem 6. Fejer's integral. Establish rh,' illtegral

-_ 1

0,. ~2--

nn

10

sin 2 (.nll) --'~(l

Sin

211)

, • lflX+U)+!(X-U)] duo .

Proof. Let S,. denote the 11th partial sum of the Fourier series of f. Then

S,,=! R

Wh ere

j"

l.!(x+u) +/(x-u)j Dn

(U)

dx.

0

~~

l. D,. (U)=2+

~

f)bl

... (3)

sin(11+.) U cos mu .... -,-- - -2 Sill (iU)

[Prove this as in Theorem 5]. Write then, by (3), a .. _1-. nn

J" [/(x+u)+f(x-II)) [ 0

l·

",_1

Em (u) ]dU

298

1:

nut

1

Dm (u)=

m~l

m",l

sin ('!z+i)!~ = sin2 (/~'!.!.3l. 2 sm (iu) 2 Sll1~ (~U)

. 1 sin 2 (nuI2) Taking Kn (u) =2. 2( ) ,t

r1n=~ f1t n

or

r1n

Sill

tu

,

we get

£!(x+u)+f(x-u)] Kn (u) du

... (4~

0

=-21-

Jao [ f(x+u)+f(x-u) ] sin 2(nuI2) 12 duo 0 SID (U ) 2

n!;

This is known as Fejer's integral. Remark. It was Fejer who found that the operation of' "summation by arithmetic means" is applicable to the Fourier, series

178.

Sumtnability of Fourier series. Here we shall discuss the case, for what values of x the Fourier seri,!s of f e L [- Jr, Jr] is (C, 1) summable to f(x) ? i.e., lim for what x e [-Jr, n], a.. (x)=f(x). n-.oo 1

Here

all (x)=n (1n

(x)=.!n

,.

Sm (x). We have se~n that

E m=l

I'"a

[!(x+t)+f(x- I)] K,. (t) dt

... (4)

[This is equation (4) of Theorem 6] tn a gpecial case if l= I, then Sn(x)=1 V Il so that all (x)=-=(nln) = 1. Now (4) gives 2 ft 2 Kn(t)dt or /lx)=f(;'(.)K.. (t)dt (5)

1=-J 1t

jft

no

0

1

.. ·

U +fl .\ -- t) -·f(x) Kn (t) df (4) - (5) gives c. (x) l(x);..+..;2 J1t0 ({(X+ '--~2'--

If 2 then -;-

lim lim. a" (X)=/(x), i.e., n .... oo n-oo [a .. (x)-j(x)]=O

JIt [!(X+l) -I) --J~x) ----2+f(x '... --0

Theorem 7.

~CL+ :2

£

.. _1

1

K.. (1)=0 as Il~OO.

1// is continuous and Ie L t --tt, ttl. then

(a .. cos nx+bn

Sin

I/x)=f(x), (c, I).

FOURIER SERII:.. S·

Proof.

299 1

We have seen that 1 ~ 'It [f(x+t)+I(x-t)] Dn (t) dt Sn (x)=0

1t

a,.

(x)=~

; Sm(X)=-~ Jft n m~l nr. " 1

[f(x+t)+/(x-~t)]

J

·n

11

.>Jut - X Dm (/)-- E n 111_1 n 111-1

sin

(m+~) t

2 f>in ( "2

2n sin 2

1t

) =Kn (I)

~ '2 )

11

J l/(x+t)-t-/(x--l)] Kn

(t) dt

... (I)

0

1=:= 1,

In a special case if 011

E

Dm (t) ] dl m=l"

-~--'---t-"

= sin:! (n+.n/ 1 Hence u,. (x)= -

[

then S .. (x)= 1 ¥

n so

that

(X)= (njn) = 1. Then (1) ~

2

11

It

0

2

10"

1=-j

or

I(x)=

1:

Kn(t)dt

... (2)

I(x) K" (f) dl

... (3)

(1)-(3) gives

I

CI"

(x)-fvc)

< !- lir 1:

:.

0

I"'" ! 11: [J(x+tH f(x .-. t)-2/(.t)] K .. (I) dr\

[I /(.t+t)-/(x) 1+1 (X--'I) -f(x) I ] I K" (t) I dl ... ( 4', f is continuO\JI; ~ I l(x1) - f(xz) I -< 6 whenever I X 1 -X2 I au E V (scalar multipltication law). This law must satisfy the following conditions: (6) a(u+v)=au+av. (7) (a+b) ll=au+bu. (8) (ab) u=a (bu) (9) For the unity element I EF lu=u where II, V, W E Vand a, b E F. Instead of saying "V is a vector space over a field F", we always say "V(F) is a vector space." The elements of V are called vectors and clements of Fare called scalars. When F=R the field of real numbers, then VCR) is called real vectors space. Similaqy V(C) is called cpmplex vector space.

302

BANACH SPACE

18'2. Subspace. Let V(F) be a vector space. A non-empty subset Wof V is called a subspace of V if W itself is a vector space over F w.r 1. the operations of vector addition and scalar multiplication in V. The spaces to} and V(F) are called trivial subspaces or improper sub O. Theorem 1. Let N be a normed linear space. Let d be J function on N defined as d(x, y)= 1\ x-y 1\ ¥ x, YEN. Then dis a metric on N Proof. (I) x, YEN => x-y E N. Since II x-y II ;a. 0, by (n I ). Hence d(x, y) ~ O. (2) d(x, Y)= II X-Y II = II (-1) (y-x) II = I (-1) I .11 y-X II = II y-x U =d(y, x) :. d(x, y)=d(y, x) (3) d(x, z)= /I x-zlI=lI{x-y)+(y-z) II ~ II x-y II lIy-zll· by (114) =d(x, y)+d (y, z) or d(x, z) ~ d(x, y)+d(y, z). Thus d is a metric on N. Remark. Thus we have seen that a normed linear space can be treated as a metric space. Thus we have given a topological structure to normed linear space. Hence we can define closed sphere, limiting point etc. in a normed linear space. 18'14. Banach space. [Kanpur 1984]. Any sequence <Xn> in a normcd linear space (N, II II) is said tv converge to Xo E N, i.e., xn-+xo if given E > 0, 3 positive integer no S.t. n ;;? no ~ II Xn - Xo II < E. It follows that lim 0 'f n-+oo II xn-xo II = I xn~xo· A sequence <XII> in N is called Caucby sequeD~e if given • > 0, 11 positive integer 110 s.t. m, n ;;> no ~ II X",-x,. II < E.

+

305

BANACH SPACE

A normed linear space is said to be complete if every Cauchy sequence in N is convergent in N. A normed linear space which is complete as a metric space, is called a Banach Space or B space. [Kanpur M.Sc. F. 1983]

Examples. (i) The vector space C [0, I] with the norm II x II =max { I XCI) /, t E [0, I]} is a Banch space. (ii) The vector sp8.G,e C. of convergent sequence wiih the II x II =sup { I x-I : n=I, 2... } [Kanpur 1989] norm is a Banch space (iii) The vector space 11" p ~ I of sequ,ences for which

.!

I x.. \p
eN. xn .... x.

T(x.)-y

~

x E N, where y= T(x). 18'21. Some elem~ntary defil1itions. (iJ Let (X, d) be'a metric space and A C X. A point Xo E X is called limit point or

limiting point or cluster point or accumulation point if every open sphere S~ (xo) contains points of A other than xo, i.e., (Sr (xo)-{xol) n A~ fl) S,(xo) = open sphere of radius r and centre Xo ={x EX: d(x. Xo) < r}. (ii) The set of all limit points of A is denoted by D (A) and is called derived set of A. (iii) A sequence < Xn > in a metric space (X. d) is called a Cauchy sequeuce if given c > 0, 3 positive integer no S.t. m, n ~

110 ~

d(x,., xm)


0, :I no E N s.t. II Tn (u) - T (u) 1\ < Ii ¥ n ~ no· (ii) Uniform Convergence. The sequence Tn E B(N, N/) is said to converge uniformly t~ T if given fi > 0,:1 no EN s.t.1\ Tn--TII '-A!).-l=A (l-t A- 1 ) Also 0 < A < I ;> I-A> 0

••• (1)

< A
. (1- 1).-])=0 or t=l as A < 1

or

Now At t=I.0