Theory of the Integral


 Mavis Miles
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1 Spring 2012 Theory of the Integrl Author: Todd Gugler Professor: Dr. Drgomir Sric My 10, 2012
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3 Contents 1 Introduction Office Hours nd Contct Informtion Set Theory: Intersections, Unions, nd Complements Homework Algebrs of Sets The Axiom of Choice Countble Sets Reltions nd quivlences Homework 2/7/ Prtil Orderings nd the Mximl Principle Husdorff Mximum Principle Well ordering The wellordering Principle The Rel Numbers The Completeness Axiom The Nturl Numbers Axiom of Archimetes The xtended Rel numbers Sequences of Rel Numbers Homework Open nd Closed Set of Rel Numbers Continuous Functions Borel Sets Lebesgue Mesure Outer Mesure Review Questions Outer Mesure Mesurble Sets nd Lebesgue Mesure xm Mesurble Sets (continued) Mesurble Functions Homework Continued
4 CONTNTS 1.24 Littlewood s Three Principles Homework Continued The Lebesgue Integrl The Riemnn Integrl Step Functions Lebesgue Integrtion Additionl homework The Lebesgue Integrl of Bounded Function over set of Finite Mesure The Integrl of Nonnegtive Function Review for Midterm Solution to (1) Solution to (2) Solution to (3) Solution to (1) Solution to the other (2) Post xm, Integrl of Nonnegtive Function The Generl Lebesgue Integrl The Lebesgue Convergence Theorem Bck from the brek, Quick Review The Generl Lebesgue Integrl The Lebesgue Convergence Theorem Differentition of Monotone functions Homework Continued Functions of Bounded Vritions Differentition of n Integrl Homework The Differentition of n Integrl Absolute Continuity Convex Functions nd of Clss, L P Spces Minkowski s Inequlity Minkowski s Inequlity for 0 < p < Hölder Inequlity Convergence nd Completeness Approximtions in L p Bounded Liner Functionls on the L p Spces
5 Chpter 1 Introduction The book we will be using for this clss is Rel Anlysis by H.L. Royden.We re using the 3rd edition, or the 4th, depending on your interests Office Hours nd Contct Informtion Dr. Sric s office is in KY 405, nd office hours will be Tuesdy/Thursdy from 23 p.m. My emil is In cse you need it, my office phone is The grding policy is s follows: 1. Homework (30 %) 2. One inclss xm (30 %) 3. A (cumultive) Finl xm (40 %) The generl pln of our course is to cover prt I sections 25 in our textbook. Also we will jump to chpters or so, to go through some mesure theory. 1.1 Set Theory: Intersections, Unions, nd Complements Suppose we hve set, X. All other sets will be subsets of X. P(X) is the set of ll subsets of X, nd is clled the power set of X. xmple. X = {1, 2, 3} So, P(x) = {{}, {1}, {2}, {3}, {1, 2}{1, 3}, {2, 3}, {1, 2, 3}} Notice tht if X = n, then P(X) = 2 n. Definition. If A, B X (or A, B P(X)) Then we hve: A B = {x : x A nd x B} 5
6 CHAPTR 1. INTRODUCTION We hve the following fcts: 1. A B = B A 2. A B A 3. A B = A A B 4. (A B) C = A (B C) Proof of 3. Let us ssume tht A B = A. By property two, we see tht A B B. Since A B = A, we know tht A B. Let us ssume tht A B. We wish to show tht A B = A, which we cn do by showing tht A B A nd tht A A B. By 2, we hve tht A B A. Now let x A. By A B, we hve tht x B s well. Then, x A nd x inb, which equivlent to sying tht x A B. This shows tht A A B, so we hve tht A B = A. Definition. We hve the following definition for union: A B = {x : x A or x B} We hve the following fcts bout union: 1. A B = B A 2. A (B C) = (A B) C = A B C 3. A A B 4. A = A B B A The following fcts re known bout both union nd intersection: 1. A (B C) = (A B) (A C) 2. A (B C) = (A B) (A C) Proof of (2). First let us try to show the following: A (B C) (A B) (A C) Tke n element x A (B C). By definition, this mens tht x A or X (B C), which is equivlent to sying tht tht x B nd x C. We hve two cses to del with now: Notice tht if x A, then clerly x must be in A B nd A C. Then x (A B) (A C). Notice lso tht if x B nd x C, then x x A B nd A C, from which it follows tht x (A B) (A C). So, we re done. The proof of (1) will be left to you. Definition. The empty set, denoted, is the set tht hs no elements. 6
7 1.1. ST THORY: INTRSCTIONS, UNIONS, AND COMPLMNTS We hve the bsic fcts bout the empty set: 1. A = A 2. A = Definition. If A X, then the complement A C of A (reltive to X ) is s follows: A C = {x X : x / A} xmple. If X = {1, 2, 3} nd A = {1}, then A C = {2, 3}. Notice tht C = X nd tht X C =. We hve the following fcts: 1. (A C ) C = A 2. A A C = X 3. A A C = 4. A B B C A C Definition. DeMorgn s Lws: 1. (A B) C = A C B C 2. (A B) C = A C B C (1) follows from noticing tht x (A B) C is the sme s sying tht x / (A B) nd this sys tht x / A nd x / B. From this it follows tht x A C B c. Going the other wy, sying tht x A C B C, this tells us tht x A C nd X B C, from which we see tht x / A nd x / B, so x / (A B), telling us tht x (A B) C. Definition. We hve the following definition of the difference, or reltive complement of two sets A nd B: let A, B X. Then B A = {x X : x B nd x / A} notice tht the following is true: B A = B A C Definition. A B is clled the symmetric difference of A nd B, nd is defined s follows: A B = (B A) (A B) = A B (A B) Definition. If A B =, then we sy tht A nd B re disjoint sets. Definition. A collection C of sets is disjoint collection or collection of pirwise disjoint sets if ny two sets in C re disjoint. Definition. The intersection of the collection C is the set of ll elements of X tht belong to ech member of C. This cn be denoted s follows: A C A or {A : A C} 7
8 CHAPTR 1. INTRODUCTION We hve tht: A C A = {x X : ( A)(A C x A} We lso hve the union of n rbitrry collection of set, A C Agin, we hve DeMorgn s lws, telling us tht: = {x X : ( A)(A C nd x A} ( ) C A = ( ) C A C A = A C A C A C A C A C In trying to prove the rightmost identity, we hve the following: Proof. Let x ( A C A) C. This tells us tht x / A CA, which tells us tht ( A)(A C x / A) which tells us tht ( A)(A C x A C ) x A C A C. We hve the following distributive lws: 1. B [ A C A] = A C[B A] 2. B [ A C A] = A C[B A] Also, the union of the empty collection of sets is empty, nd the intersection of n empty collection of set is X, since by definition since the collection is empty there is nothing to check, nd we wind up with ll x X. If collection is given in sequence, C = {A i } then A i = A, A C A i = A A C Definition. {A i }, i N is clled index set. We cn lso tlk bout {A λ : λ λ}, nd we cn tlk bout their unions nd intersections: Let λ A λ A A A = {x : λ A nd x A λ } A λ = {x : ( λ)(λ A) x A λ } f : X X be function. Then let {A λ } λ A be collection of subsets of X, in which cse: f[ λ A f[ λ A A λ] = λ A A λ] 8 λ A f(a λ ) f(a λ )
9 1.2. HOMWORK Now we ssume tht {B λ } λ A is n indexed collection of subsets of Y, then we hve tht f 1 [ λ B λ ] = λ f 1 (B λ ) nd tht f 1 [ λ B λ ] = λ f 1 (B λ ) Notice tht we hve the following chin of events: x λ f 1 (B λ ) λ A, f(x) B λ f(x) λ B λ x f 1 ( λ B λ ) We lso hve tht f 1 [B C ] (f 1 [B]) C, f(f 1 [B]) = B f 1 (f[a]) A 1.2 Homework Pge 17 questions (16, 17) nd Pge 19 question (19). 1.3 Algebrs of Sets Definition. A collection A of subsets is clled n lgebr of sets (or boolen lgebr), if: 1. For ll A, B A, then A B A 2. For ever A A, A C A. Given A, B A, we would like to show tht A B is in A. Bsed on wht we know (long with DeMorgn s lws) we hve: (A B) C = A C B C A C B C A (A C B C ) C A A B = (A C B C ) C A Finite unions nd finite intersections re lso in the lgebr. Proposition 1. Given ny collection C of subsets of X, there is smllest lgebr A which contins C. Proof. P(X) is n lgebr, nd it contins C Consider the fmily F of ll lgebrs tht contin C. Then, A = {B : B F} C Now we just need to verify tht A is n lgebr. This follows from seeing the following: let A, B A. Since both A, B re in A, this implies tht A, B B, for ll B F. Since B is n lgebr, this implies tht A B B for ll B F, from which we see tht A B A. Definition. The smllest lgebr contining C is clled n lgebr generted by C. 9
10 CHAPTR 1. INTRODUCTION Proposition 2. Let A be n lgebr of subsets of X, nd {A i } is sequence of sets in A.Then, there exists sequence {B i } of sets in A such tht B n B m = for n m, nd B i = A i. Proof. Define the following: B 1 = A 1. B 2 = A 2 A 1 = A 2 A C 1. Generlly, for n > 1, we define: B n = A n (A 1 A 2... A n 1 ) = A n A C 1 A c 2... A C n 1 So, B n A, nd B n A n. Given some m < n, we cn look t B m B n. Well, B m B n A m B n = A m (A n A C 1 A C 2... A C n 1) = since m < n, nd since you ll hve m some where in tht chin of complements, so you ll hit the empty set somewhere when you tke the intersection of A m with A C m. We lso know tht B i A i B i A i. So, let x A i. Then, x belongs to t lest one A i. Let i 0 be the smllest such index such tht x A i0. Then, x B i0 = A i0 (A 1 A 2... A i0 1) So this implies tht A i B i A i = B i Definition. An lgebr A of sets is clled σlgebr or Borel field, if every union of countble collection of sets in A is lso in A. Any σlgebr contins countble intersections, which follows gin from union. A i A ( A i ) C = A i A Proposition 3. Given ny collection C of subsets of X, there is smllest σlgebr tht contins C (this is homework question) 1.4 The Axiom of Choice Let C be ny collection of nonempty sets. Then, there is function F defined on C which ssigns to ech set A C n element F (A) in A. The function F is clled choicefunction. If our collection C = {X λ } λ Λ Notice tht A B = {(, b) : A, b B}. How to we tlk bout the direct product of n infinite collection, C = {X λ } λ Λ? Well, it is collection of ll sets {x λ } λ Λ indexed by Λ such tht x λ X λ. Remrk. If one X λ is the empty set, then 10
11 1.5. COUNTABL STS is empty. If ll X λ, then X λ λ Λ X λ λ Λ 1.5 Countble Sets Definition. A set is countble if it is the rnge of some sequence. Definition. A set is finite if it is either empty or the rnge of finite sequence. A set is countbly infinite if it cn be put in oneone correspondence with the nturl numbers N. Proposition 4. very subset of countble set is countble. Proof. Let = {x n } be countble set, nd A. If A is empty, then we re done by definition. Otherwise, let x A. Define sequence {y n } s follows: if x n A, then y n = x n. If x n / A, then y n = x. We now hve sequence {y n }, nd its rnge is A thus it is countble. Proposition 5. Let A be countble set. Then the set of ll finite sequences of A is countble. Proof. We estblish oneone correspondence between the set of ll finite sequences of elements of A nd N. Since A is oneone with N, let us look t sequences of the following type: Now, Notice tht < 2, 3, 5, 7,..., P n k,... > n = 2 x1 3 x 2... P x k k f : n (x 1, x 2,..., x k ) which is finite sequence in N {0}. The rnge of f re ll finite sequences in N {0}, which contins ll finite sequences in N nd thus is countble. Proposition 6. The set of ll rtionl numbers is countble they mp into ll sequences of length 2. Proposition 7. The union of countble collection of countble sets is countble. Proof. C, {A n }. ch A n is set {x m n } m=1, indexed by the nturl numbers. Thus, A n is indexed by sequences of length 2, {m, n} 11
12 CHAPTR 1. INTRODUCTION 1.6 Reltions nd quivlences R is reltion on set X is subset of X X. We write the following: (x, y) R xry Some exmples of reltions on the set R re =, or. We denote them in the following wy: = {(x, x) : x R {(x, y) : x, y R, x y} Definition. A reltion R is trnsitive on X if xry, xrz xrz Notice tht the exmples bove re trnsitive reltions. Definition. A reltion R is symmetric if xry yrx Notice tht = is symmetric, but is not symmetric reltion. Definition. A reltion R on X is reflexive if xrx For ll x X. Agin, = is n exmple of reflexive reltion. Notice tht < is not reflexive on R. Definition. A reltion R on X is n equivlence reltion if it is reflexive, symmetric, nd trnsitive. For exmple, = is n equivlence reltion on R. However, is not n equivlence reltion. Suppose tht is n equivlence reltion on some set X. Then, we cn define: x = {y X : x y} If y, z x, then x y, nd x z.then, y z. This set x is clled n equivlence clss. Thus, either x = y or x y =. One nice construction of such equivlence reltions is prtition on set, where ech pir of sets in your prtition re either disjoint or the sme. So, we get this: X/ = { x : x X} = A Collection of quivlence Clsses Since x x, then x for ll x X. Definition. A binry opertion on X is mpping mp : X X X. One exmple is the ddition of rel numbers. An equivlence reltion is comptible with binry opertion + if x x nd y y x + y x + y If + is comptible with, then + defines new binry opertion on the set Q = X/. For exmple, x/, ; y/ Q, (x/ ) + (y/ ) = (x + y/ ) One good exmple is the ddition of rtionl numbers. 12
13 1.7. HOMWORK 2/7/ Homework 2/7/2012 On pge 23, do questions 24 nd Prtil Orderings nd the Mximl Principle Definition. A reltion R on set X is ntisymmetric if xry nd yrx x = y One exmple would be the reltion on R. Similrly, on P(X) is n ntisymmetric reltion. Definition. A reltion < is prtil ordering on X if it is trnsitive nd ntisymmetric. Some exmples of such reltion include on R, nd on P(X). Definition. A prtil ordering < on set X is liner ordering if for ll x, y X either x < y or y < x. One such exmple is on R. However, on P(X) is not liner ordering if X hs more thn one element (for exmple you could pick two disjoint subsets of X, neither of which will then be contined in ech other). We red < b As precedes b, or less thn b. Alterntively, we cn sy tht b is greter thn. Definition. If X, n element is the smllest element in if for every x, if x then < x. It is very simple to show tht this smllest element must be unique. Definition. A miniml element of is n element such tht there is no x with x nd x <. xmple. X = {1, 2, 3, 4} C = {{1}, {1, 2}, {3}} Notice tht with the reltion, C hs two miniml elements: {1}, {3}. But, there is no smllest element. Fct. The smllest element is miniml element. This cn be shown from their definitions. Definition. If for ll x X, x < x then < is reflexive prtil ordering. If it is never true tht x < x, then < is clled strict prtil ordering. 1.9 Husdorff Mximum Principle Let < be prtil ordering on set X. Then, there exists mximl linerly ordered subset S of X. Remrk. X is mximl with property M if ny S S does not hve property M Well ordering A strict liner ordering on set X is clled wellordering if every nonempty subset of X contins the smllest element. 13
14 CHAPTR 1. INTRODUCTION xmple. Tke the set N with the reltion <. This is wellordered. Alterntively, if we tke R nd the reltion <, it is not wellordered. This follows from tking n open intervl on R, which does not hve smllest element The wellordering Principle very set X cn be wellordered. Proposition 8. There is n uncountble set X tht is wellordered by reltion < in the following wy: 1. There is the lrgest element Ω in X. 2. If x X, nd x Ω, then is countble. {x X : y < x} Proof. (i) Let Y be n uncountble set. By the wellordering principle, Y hs wellordering <. If Y does not hve the lrgest element, then we introduce Z = Y {α}, where α / Y. Ordering on Z extends the ordering < on Y by sying tht for ll y Y, y < α. Then, α is the lrgest element of Z. (ii) The set of ll y Z such tht {x Z : x < y} is countble is nonempty, becuse α is in this set. Let Ω be the smllest element in this set. Let Then, X stisfies (i) nd (ii). X = {x Z : x < Ω or x = Ω} Ω is clled the first uncountble ordinl, nd nything less tht Ω is clled countble ordinl The Rel Numbers Definition. The set R is set tht consists of two binry opertions + nd, nd cn be constructed from the following xioms: 1. (R, +, ) is field, or in other words, (R, +) is commuttive group (for ll x, y R, x+y R, ddition is commuttive, there exists n identity element (0), nd there exists n dditive inverse for ll elements in R) nd (R, ) for ll x, y R we hve tht x y R, is ssocitive, there exists multiplictive identity (1) in R, nd for ll the nonzero elements in R there exists nother element in R tht is its multiplictive identity, multipliction is commuttive, nd multipliction distributes over ddition. We hve the xioms of order: there is set P of positive rel numbers tht stisfies: 1. For ll x, y P, x + y P 2. For ll x, y P, x y P 14
15 1.11. TH RAL NUMBRS 3. For ll x P, x / P 4. For every x R, x = 0 or x P or x P Thus, our set is clled n ordered field. Such exmples of n ordered field include Q, nd of course R. We cn introduce the following reltion by definition: Similrly, by definition x < y y x P x y y < x or y = x Definition. Let S R. The number b is n upper bound for S if for every x S, x b. Definition. The lest upper bound for S is number b such tht if c is n upper bound of S, then b c. Sometimes this is denoted: sup(s) = b And clled the supremum of S Definition. A lower bound for S is if for ll x S, x. Similrly, the gretest lower bound for Sis if for ny lower bound d for S, we hve tht d. Sometimes this is denoted imf(s) = Pronounced infimum. xmple. Let S = { 1 n : n N} Notice tht 2 is n upper bound for S. However, lso notice tht sup(s) = 1. Similrly,  6 is lower bound for S, however the imf(s) = The Completeness Axiom very nonempty set of rel numbers which hs n upper bound hs lest upper bound The Nturl Numbers We hve the following definition: Notice tht N R. N = {1, 2, 3, 4,...} Axiom of Archimetes For ll x R, there exists n N such tht x < n. This xiom isn t prticulrly helpful, but it does give us the following: Corollry 9. Between ny two rel numbers, there is rtionl number. 15
16 CHAPTR 1. INTRODUCTION The xtended Rel numbers The extended rel numbers re the set R, long with the two numbers +,. Notice tht And tht < x < x + = x = x = if x is positive, + = = (± ) = (± ) = undefined (± ) = 0 = 0 by convention Definition. If S hs no upper bound, then sup(s) =. If S hs no lower bound then imf(s) = Sequences of Rel Numbers A sequence is mp f : N R, f(n) = x n, denoted {x n }, or {x n } nd we hve tht lim x n = x n if for ll ɛ > 0 there exists N N such tht for ll n N, x n x < ɛ Definition. {x n } is Cuchy sequence if for ll ɛ > 0 there exists N such tht for ll n, m N we hve tht x n x m < ɛ Fct. If sequence hs limit, then it is unique. Fct. If sequence of rel numbers is Cuchy, then it hs limit. This follows from the Completeness xiom. Definition. x is cluster point of sequence {x n } if every neighborhood of x contins infinitely mny elements of the sequence. Definition. lim n x n = If for ll C > 0, there exists N(C) C such tht if m N(C) then x n > C. Similrly, lim x n = n If for ll D R, there exists N(D) N such tht if n N(D), then x n < D. xmple. x n = n!. Then lim n x n = Also, if x n = n then lim n x n = 16
17 1.11. TH RAL NUMBRS Definition. Let {x n } be sequence of rel numbers. The limit superior of {x n } is defined by lim x n = inf n sup k n x k = lim sup k n x k n n xmple. We cn show tht the limit superior of the following sequence is 0: Definition. The limit inferior of {x n } is: x n = ( 1)n n n = 0 lim n x n = sup n inf k n x k Notice tht unlike wit the limit superior, inf k n x k is monotoniclly incresing. Fct. x = lim n x n if nd only if it stisfies: 1. For ll ɛ > 0, there exists n such tht x k < l + ɛ for ll k n 2. For ll ɛ > 0 nd for ll n, there exists k n such tht x k > l ɛ. xmple. x n = sin( nπ ) + 1. Notice tht 2 lim x n = 2, n n lim x n = 0 Fct. nd lim ( x n) = lim x n n n lim x n lim n x x n lim x n + lim y n lim(x n + y n ) lim x n + lim y n lim(x n + y n ) lim x n + lim y n Let us prove tht lim( x n ) = lim x n. Proof. We hve tht lim ( x n) = inf n sup k n ( x k ) = inf n ( inf k n x k ) = sup n (inf k n x k ) n We hve the following fct: So, This follows from seeing tht A = { : A} sup( A) = infa sup( A) = infa So, letting b = sup( A) we notice tht b is n upper bound for A, nmely for ll A we hve tht b. From this, b, for ll A. Thus, b is lower bound for A. Let c be lower bound for A, so c for ll A. In this cse, c for ll ]ina, so c is n upper bound for the set A. Since b is the lest upper bound, we hve tht b c, or sup( A) = b C. This implies tht sup( A) is the gretest lower bound, nd is the infinitum of A. 17
18 CHAPTR 1. INTRODUCTION Homework Prove the lst two of the bove clims tht ren t finished Open nd Closed Set of Rel Numbers We hve the stndrd open intervl on R, nd we hve which is n infinite open intervl, similrly (, b) = {x : < x < b} (, ){x; x > } (, b) = {x : x < b} which is lso n infinite open intervl. Alterntively, [, b] = {x : x b} which is closed intervl. Notice tht [, b) nd (, b] re hlfopen nd hlfclosed intervls. Definition. A set O of rel numbers is clled open if for every x O, there exists some δ > 0 such tht ech y with y x < δ belongs to O. Another wy to sy this, is to sy tht (x δ, x + δ) O Open sets, for exmple, include things like open intervls, R, Proposition 10. The intersection of two open sets O 1 nd O 2 is n open set. Proof. Let x O 1 O 2. We know tht there exists some δ 1 > 0 such tht (x δ 1, x + δ 1 ) O, becuse O 1 is open. Similrly, O 2 hs some δ 2, so defining δ = min{δ 1, δ 2 }, we see tht So, O 1 O 2 must be open. (x δ, x + δ) O 1 O 2 As corollry, this implies tht the finite intersection of open sets is open. Proposition 11. The union of ny collection C of open sets is open. Proof. Suppose we took U = O C O. If x U, this implies tht there exists some O 1 in our collection such tht x O 1, nd since O 1 is open there must exist some δ > 0 such tht the intervl round x of rdius δ is open. xmple. ( 1 n, 1 n ) = {0} Notice tht this infinite intersection of open sets is {0} which isn t open. 18
19 1.12. OPN AND CLOSD ST OF RAL NUMBRS Proposition 12. very open set of rel numbers is the union of countble collection of disjoint open intervls. Proof. Let O be n open set. For ll x O, there exists y > x such tht (x, y) O. There lso exists some z < x such tht (z, x) O. Let nd b = sup{y : (x, y) O} = inf{z : (z, x) O} This implies tht < x < b, lso, I x = (, b). Our first clim is tht I x O. Let w I x. We cn sy tht x < w < b. Then, there exists some y such tht y > w nd (x, y) O. This implies tht w O. Also, b / O. If b O, then you know tht there would hve to exist some δ > 0 such tht (b δ, b + δ is subset of O. But if this were true, then b would not be the supremum of ll {y : (x, y) O}, since (x, b + δ ) O. This would imply tht (x, b + δ) O), nd then b is not 2 the supremum. There is similr proof tht / O. It cn follows tht x O I x = O. If we hve to intervls (, b) nd (c, d) tht re elements of {I x { x O nd (, b) (c, d), then we wnt to show tht (, b) = (c, d). If their intersection ws nonempty, then c < b, nd lso, < d. Since c / O, then c. Since / O, then we know lso tht / (c, d), which implies tht c. Putting these two together, we hve tht = c. Similrly, we get tht b = d. Thus, (, b) = (c, d). Thus, {I x } x O is disjoint fmily of open sets. ch open intervl contins rtionl number, since the intervls re disjoint this gives welldefined function from Q {I x } x O, thus {I x } x O is countble. Proposition 13. Let C be collection of open sets of rel numbers. Then there exists countble subcollection {O i } u=1 such tht O C O = O i Proof. Let U = O C O. Let x U. This implies tht there exists some O C such tht x O. Since O is open, there exists I x such tht I x (n open intervl) such tht x I x O. Suppose tht I x = (, b). This implies tht there exists J x, n open intervl with rtionl endpoints such tht x J x nd J x I x. The collection of ll intervls with rtionl endpoints is countble set. Thus {J x } x U is countble collection, nd x U J x = U, becuse x J x. For ech J x, choose one O C such tht J X O we get countble collection. Definition. A rel number x is point of closure of if for ll δ > 0, there exists some y such tht x y < δ. Remrk. very point of is point of closure of. Definition. is the set of points of closure of. Notice tht. Proposition 14. If A B, then A B, nd lso A B = A B. Proof. Since A A B, then A, B (A B), so A B (A B) 19
20 CHAPTR 1. INTRODUCTION Let x / A B. Then there exists δ 1 > 0 such tht no y A stisfies x y < δ 1. Similrly there exists δ 2 > 0 such tht no y B stisfies x y < δ 2. Simply, tke δ = min(δ 1, δ 2 ) > 0. If x y < δ, then y / A, nd y / B, which implies tht y / A B. This mens tht x / A B. Definition. A set F is closed if F = F. xmple. Some exmples of closed sets include closed intervls, R, nd. Also, [, ) nd [, b) re closed. Proposition 15. For ny set, the set is closed. It is lso true tht =. Proof. Let x. Given δ > 0, there exists y such tht x y < δ. Since y, there then 2 exists some z tht y z < δ. Finlly, we get tht x z x y + y z < δ + δ = δ, which lso implies tht x. Proposition 16. If F 1 nd F 2 re closed, then F 1 F 2 is closed. Proof. Bsed on wht we proved before, which must be closed. F 1 F 2 = F 1 F 2 = F 1 F 2 Corollry 17. The union of finitely mny closed sets is closed. xmple. Look t [ 1, 2 1 ]. Tking their union, we ll get (0, 2), which is open despite the n n fct tht ech one of these sets is closed. Proposition 18. The intersection of ny collection F of closed sets is closed. Proof. Let x ( F F F ). Then for ny δ > 0 there exists some y F F F such tht x y < δ. Then y is in ech F F. Then, x F. But since F = F (remember, F is closed) we thus hve tht x F F F. So, F F F is closed, since ny point in the closure of this intersection is in the intersection. Proposition 19. The complement of ny open set is closed, nd the complement of ny closed set is open. Proof. Let O be open. If x O, then there exists δ > 0 such tht if x y < δ, then y O. Then, x / O C. Thus, O C = O C, from which we gther tht O C is closed. Let F be closed, nd sy tht x F C. Then x is not point of closure of F C, which implies tht there exists δ > 0 such tht if x y < δ, then y F C. Thus, F C is open. Definition. A collection C of sets covers set F is F {O : O C}. If ech O is open, we sy tht C is n open cover. If C is finite, then C is finite cover. Theorem 20. HeineBorel: Let F be closed nd bounded set of rel numbers. Then ech open cover of F hs finite subcover. Proof. Assume covering C of F is given. First, we ssume tht F = [, b]. We define = {x : x b such tht for [, x] there is choice of finite subcover of C } First, we observe tht. In prticulr, this tells us tht. is bounded, let c = sup() R. Since b is n upper bound for c, we hve tht c b. We need to show tht 20
21 1.13. CONTINUOUS FUNCTIONS c = b. Since c [, b], there exists some open set O C such tht there exists some ɛ such tht (c ɛ, c + ɛ) O. Since c = sup(), this implies tht there exists some x nd x > c ɛ. This implies tht there exists some finite subcollection {O 1, O 2,.., O n } of C tht covers [, x]. Then {O 1,..., O n } covers [, c ɛ]. c + ɛ/2 [ ( ] ] ) ] c ɛ x c c + ɛ Thus c is not the supremum of unless c = b. This finishes our first cse, when F = [, b]. Now let F be ny closed bounded set of the rels. Then, there exists some [, b] F. Then we define the following: C := C F C which is lso n open cover of [, b]. By the first cse, there exists some {O 1,..., O n, F C }, subcollection of C which covers [, b]. Restricting this set to {O 1,..., O n }, we hve subcollection of C, nd covers F. Proposition 21. Let C be collection of closed sets with the property tht every finite subcollection of C hs nonempty intersection nd suppose tht one of the sets in C is bounded. Then, F C F This will be left s n exercise, the ide is to pply the HeineBorel Theorem. b 1.13 Continuous Functions Suppose we hve R nd we hve f : R. f is continuous t x if for ll ɛ > 0 there exists some δ > 0 such tht for ll y if x y < δ then f(x) f(y) < ɛ. Definition. f is continuous on A if f is continuous t ech point of A. Proposition 22. Let f : F R be continuous function, nd F R is closed nd bounded. Then f is bounded on F, nd it ssumes it is mximl nd miniml on F. Proof. For ll x F. there exists n intervl I x such tht x I x where if y I x F then f(x) f(y) < 1. This implies tht f(y) < f(x) + 1 for ll y I X F. Then, {I x } x F is n open cover of F. By the HeineBorel theorem, we hve tht there exists finite subcover {I x1...i xk } of F. Let M = mx{ f(x 1 ) + 1,..., f(x k ) + 1} Let y F. Then there exists some intervl I xi tht contins y. Then f(y) < f(x i ) + 1 M. Thus f is bounded on F by the constnt M. 21
22 CHAPTR 1. INTRODUCTION Since f is bounded, m = sup x F f(x) is finite. We need t o show tht there exists some x 1 F. such tht f(x 1 ) = m. Suppose such x! does not exist. Then f(x) < m for ech x F. Then by the continuity of f, for ech x F there exists n intervl I x contining x such tht for ll y I x we hve 1 2 f(y) f(y) = f(y) < 1 (f(x) + m) 2 (we bndoned the bove proof to do the following) There exists n intervl I x such tht x I x. For ll x I x F, we hve so f(y) f(x) < 1 (m f(x)) 2 f(y) f(x) < 1 (m f(x)) 2 f(y) < 1 (m + f(x)) 2 Well, {I x } x F is n open cover of F. By the HeineBorel theorem gin, there exits finite subcover {I x1,..., I xn } of F. Let = mx{f(x 1 ),..., f(x n )} < m. For y F, there exists I x contining y, nd we hve f(y) < 1 2 (f(x i) + m) 1 2 ( + m) < 1 (m + m) = m 2 This implies tht m is not the lest upper bound becuse 1 ( + m) is n upper bound which is 2 less thn m. This is contrdiction, thus f must chieve tis mximum. Proposition 23. Let f : (, ) R Then f is continuous if nd only if f 1 [O] is open for ech open O R. Proof. For ll O open subsets of R, we wnt to show tht f 1 [O] is open. Given x R, for ll ɛ > 0 we hve I = (f(x) ɛ, f(x) + ɛ) R nd is open, nd x f 1 (I). This implies tht there exists some δ > 0 such tht (x δ, x + δ) f 1 (I) so f((x δ, x + δ)) (f(x) ɛ, f(x) + ɛ) Working in the other direction, let f be continuous function nd let O R tht is open. Let x f 1 (O). Then f(x) O, O is open implies tht there exists some ɛ > 0 such tht There then exists some δ > 0 such tht which in turn implies tht which implies tht f 1 (O) is open. (f(x) ɛ, f(x) + ɛ) O f((x δ, x + δ)) (f(x) ɛ, f(x) + ɛ) (x δ, x+]delt) f 1 ((f(x) ɛ, f(x) + ɛ)) f 1 (O) 22
23 1.13. CONTINUOUS FUNCTIONS Definition. A relvlued function f on set is uniformly continuous if for ll ɛ > 0 there exists δ > 0 such tht if x, y nd x y < δ then f(x) f(y) < ɛ xmple. Uniformly continuous functions re lso continuous. However, it is not true tht continuous functions re uniformly continuous. For exmple, f(x) = 1 x, x > 0 is continuous, but not uniformly continuous. This follows from the fct tht the function f(x) s x 0 gets so steep so quickly, the δ s dispper s you choose ɛ. Another exmple is f(x) = x 2 Proposition 24. If function f is continuous function defined on closed nd bounded set F, then f is uniformly continuous. Proof. Given ɛ > 0, x F (where f : F R), then there exists δ x > 0 such tht if x y < δ x nd y F then f(x) f(x) < 1 2 ɛ. Let I x = (x 1 2 δ x, x δ x), x I x, is n open cover of F s x F. By HeineBorel, there exists finite subcover {I x1,..., I xn }. Let δ = min 1 2 δ x 1, δ x n } > 0 R. We let y, z F, nd ssume tht y z < δ. There exists some I xi contining y nd I xj contining z, which implies tht z x i z y + y x i < δ δ x i < δ xi This implies tht f(z) f(x i ) < 1ɛ. It is lso true tht f(y) f(x 2 i) < 1 ɛ. We now hve tht 2 so we re done. f(z) f(y) f(z) f(x i ) + f(x i ) f(y) < 1 2 ɛ ɛ = ɛ Definition. A sequence {f n } of functions defined on set converges pointwise on to function f if for ech x we hve tht lim f n(x) = f(x) n Definition. A sequence of functions {f n } on uniformly converges on to function f if for ll ɛ > 0 there exists some N N such tht for every x we hve tht f n (X) f(x) < ɛ for ll n N. xmple. Suppose tht f n (X) = x n for 0 x 1. In this cse, f n converges pointwise to f(x) = 0 for ll 0 x < 1 nd 1 for x = 1. However, f n does not converge uniformly to f. This ide follows from wht hppens to the distnce t the point (1, 0): the functions f n do not get closer to f nicely. 23
24 CHAPTR 1. INTRODUCTION 1.14 Borel Sets Remrk. A countble union of closed sets is not necessrily closed. Definition. The collection B of Borel Sets is the smllest σlgebr which contins ll of the open sets. It is certin tht B is generted by ll closed sets. It is lso true tht B is generted by ll open intervls. A set is F σ if it is countble union of closed sets. Notice tht every closed set is F σ. Also, notice tht every countble set F σ. An exmple of F σ set tht is not countble is the following set: N, which follows directly from our lst exmple. Alterntively, { 1 n : n N } is nother exmple. Definition. A set is G δ if it is countble intersection of open sets. An open intervl is F σ, for one could do the following: (, b) = [ + 1 n, b 1 ] n 1.15 Lebesgue Mesure Tking the intervl I = [, b], the length l of l(i) of I is l(i) = b. The length is n exmple of set function ( it ssocites rel number to ech set in collection subsets of set). l is set function on the collection of intervls in R. The length of n open set is defined s follows: the sum of the lengths of the open intervls of which it is composed. Given collection M of sets, with m set function on M such tht for every M, we would like to define m such tht m() R 0 { } nd such tht m stisfies our wishlist of properties: 1. m() defined for ll R. 2. For every intervl I, we would like to hve tht m(i) = l(i). 3. If { n } is disjoint sequence then m( n n ) = N m( n ) 4. m( + y) = m() for ll nd for ech y R. In other words, we wnt to be invrint under trnsltion. With intervls, we cn notice tht which preserves distnce trivilly. [, b] + y = [ + y, b + y] 24
25 1.15. LBSGU MASUR Unfortuntely, it is not possible to stisfy ll four properties for ny set function m. Wht we will end up doing is wekening (1): m() will be defined on s mny sets s possible, nd we will keep properties 24. We will require tht M is σlgebr. Any such M hs to contin Borrel σlgebr, by property number 2 (since the Borel σlgebr is the smllest σlgebr which contins ll of the open sets in some collection) Outer Mesure For ny subset A R, we will consider countble collection {I n } of open intervls which covers A; in other words A I n n Definition. Lebesgue outer mesure of A is s follows: m (A) = inf l(i n ) A ni n From the definition, it is immedite tht m ( ) = 0. Also, if A B, then m (A) m (B). It lso follows tht m {x} = 0. One wy to see this would be to let I n = ( x ɛ 2 n + 1, x + ɛ ) 2 n + 1 in which cse, ( ) 1 l(i n ) = ɛ = ɛ 2 n nd since our choice of ɛ ws rbitrry, we let ɛ 0. Proposition 25. The outer mesure of n intervl is its length. Proof. Notice tht is considered n open intervl. We hve [, b], nd tht for ll ɛ > 0, tht ( ɛ, b + ɛ) [, b]. This tells us tht the outer mesure Where we define m ([, b]) b + 2ɛ I n = ( ɛ, b + ɛ), I n =, n 1 since this holds for ll ɛ > 0,it follows tht m ([, b]) b. It remins to be shown tht m ([, b]) b. Let {I n } be n rbitrry collection of open intervls tht covers [, b]. By the HeineBorel Theorem, {I n } hs finite subcover Since [, b], there exists n intervl ( 1, b 1 ) contining which is in tht finite subcover. If b 1 < b, let ( 2, b 2 ) be n intervl in the finite subcover which contins b 1. Then n 2 < b 1 < b 2 We cn continue this process, were we would then sy tht 3 < b 2 < b 3 25
26 CHAPTR 1. INTRODUCTION nd we cn continue this process until b is in n intervl, which must hppen (this process stops becuse the collection is finite, which we hve from the HeineBorel theorem). We hve now tht i < b i 1 < b i So, k l(i n ) l(( i, b i )) = b k k + b k k b 1 1 = b k ( k b k 1 )... ( 2 b 1 ) 1 Notice tht ech term k b k 1 is negtive, so we cn remove them: b k 1 b m [, b] b from which we conclude tht m [, b] = b = l([, b]) If I is ny finite intervl, then given ɛ > 0 there exists closed intervl J I such tht l(j) > l(i) ɛ Since l(i) ɛ < l(j) = m (J) m (I) m (I) = l(i) = l(i) This implies tht s ɛ 0, then l(i) = m (I) For homework, prove this when I is n intervl: if I is ny infinite intervl, then given some ɛ > 0 there exists closed intervl J I such tht l(j) > l(i) ɛ 1.16 Review Questions Notice tht [, b) Is Borel set, since it is F σ set. This cn be shown by looking t the union: [, b 1 n ] of closed sets. Notice tht f n (x) = x n 0 x
27 1.17. OUTR MASUR uniformly converges to f(x) = 0. This tkes some work to show: given ɛ > 0, we need to find N such tht if n N then x n 0 < ɛ First notice tht x n 0 = x n = x n 1 2 n < ɛ for every x in the intervl x [0, 1 ]. We wnt to find n from the following: 2 So, 1 2 n = ɛ n log 1 2 = log ɛ N = [ log ɛ log 1 ] Outer Mesure If A R then m (A) = inf l(in ) A n In This definition hs some consequences: for exmple, if A B, then m (A) m (B). This follows directly from the fct tht A hs more covers thn B does. Proposition 26. If I is n intervl, then m (I) is the length of I, l(i). Proof. First, look t closed intervl [, b]. We hve to show tht m [, b] b nd tht m [, b] b. The first one follows quickly, choosing I 1 = [ ɛ, b + ɛ] nd for ll others to be I n =, n = 2, 3,... In this cse, the sum of the lengths of I n is b + 2ɛ. Notice tht m [, b] l(i n ) = b + 2ɛ m [, b] b Since we let ɛ to go zero. For the second cse, we hve to work bit hrder  but we cn us the HeineBorel theorem. Suppose tht we hve cover {I n }. There must exists I 1,..I k, finite subcover for [, b]. Bsed on how this intervl should look, it hs to look like ( 1, b 1 ), contining. If this set doesn t contin b, we look t point b 1. Since we hve cover, we hve to dd nother set tht contins b 1. This process continues k times, since we hve finite cover. We cn then show tht (thnks to the overlps of these sets) tht the sums of this lterntive cover is greter thn or equl to b. If I is finite, in this cse there exists closed set J such tht J I, nd tht l(j) l(i) ɛ (there is trick tht you cn do to mke this work) given some ɛ > 0. So, m (J) = l(j) since J is closed intervl. Since J I, this implies tht But we lso note tht I I. Then, m (J) m (I) m (I) m (I) = l(i) = l(i) 27
28 CHAPTR 1. INTRODUCTION Thus wht we hve is Which ultimtely tells us tht m (I) = l(i). l(i) ɛ m (J) = l(j) m (I) m (I) l(i) We now pproch the cse in which I is infinite. Suppose we hd something like (, ) or (, b). Well suppose hd some infinite intervl: in ny cse, given some M > 0 it is possible to find some closed set J I such tht l(j) = m (J) M This then implies tht m (I) m (J) > M for ll M > 0. This then implies tht m (I) =, which is the length of I, l(i). Proposition 27. Let {A n } be countble collection of sets of rel numbers. Then In other words, m is countbly subdditive. m ( A n ) m (A n ) Proof. For A n, there exists cover {I n, i} by open intervls such tht m (A n ) i l(i n, i) ɛ 2 n So Since m A n ( l(i n, i) ɛ ) ( ) = l(i n i 2 n n, i) ɛ m ( A n ) ɛ n i n ɛ 2 n = ɛ Becuse {I n, i},, is cover of the union of ll A n. Notice tht N N is countble, therefore our set {I n, i},, is countble. Let ɛ 0, in which cse you get the conclusion we wnted. Just s remrk, we need to consider the cse when one m (A n ) = seprtely. However, it is trivil cse, we don t hve nything to check. Corollry 28. If A is countble, then m (A) = 0. This follows from seeing tht A is the countble union of singletons, nd bsed on bove, we hve the fct tht m (A) = 0. Corollry 29. The set [0, 1] is not countble. If it were, then m [0, 1] would be 0 however, we know bsed on the fct tht it is closed intervl tht m [0, 1] = 1. Proposition 30. Given ny set A nd ny ɛ > 0, there exists n open set O such tht A O nd m (O) m (A) + ɛ. There is set G which is G δ such tht A G nd Proof. For homework! m (A) = m (G) 28
29 1.18. MASURABL STS AND LBSGU MASUR 1.18 Mesurble Sets nd Lebesgue Mesure Outer mesure is countble subdditive, s we showed, but not countbly dditive (in generl). We restrict ourselves to smller fmily of subsets of R in order to get mesure. Definition. A set is sid to be mesurble if for ech set A we hve tht m (A) = m (A ) + m (A C ) Remrk. It is lwys true tht m (A) m (A ) + m (A C ). This follows directly from the fct tht our mesure is countbly subdditive. This tells us tht is mesurble if nd only if m (A) m (A ) + m (A C ) for ll A R. Fct. If is mesurble, then C is mesurble. This follows from the fct tht ( C ) C =, nd tht ddition is commuttive. Fct. The, R re both mesurble. This follows simply from plugging in these sets for, nd recognizing tht t lest one of the terms in our sum of m is zero. Lemm 31. If m () = 0, then is mesurble. Proof. We tke A R. In this cse, A = A C A So, m (A ) + m (A C ) m () + m (A) = m (A) Using the fct tht m () = 0. This implies tht is mesurble. Lemm 32. If 1 nd 2 re mesurble, then 1 2 is mesurble. Proof. Tke A R, some rbitrry set. Since both 2 is mesurble, we hve tht m (A 1 C ) = m (A 1 C 2 ) + m (A 1 C 2 C ) Just s remrk: notice tht A 1 C 2 C = A ( 1 2 ) C Now looking t A ( 1 2 ), we hve tht A ( 1 2 ) = (A 1 ) (A 2 ) = (A 1 ) (A 2 1 C ) So, m (A [ 1 2 ]) m (A 1 ) + m (A 2 1 C ) (by the dditivity of countble sets). Looking t the whole thing, m (A [ 1 2 ]) + m (A [ 1 2 ] C ) m (A 1 ) + m (A 1 C 2 ) + m (A 1 C 2 C ) 29
30 CHAPTR 1. INTRODUCTION Now, wht we relly hve is = m (A 1 ) + m (A C 1 ) becuse 2 is mesurble. Now using the fct tht 1 is mesurble, This then implies tht 1 2 is mesurble. = m (A) Corollry 33. The fmily M of ll mesurble sets is n lgebr. Lemm 34. Let A R be ny set nd let 1,..., n be finite sequence of disjoint mesurble sets. Then, n n m (A [ i ]) = m (A i ) Proof. The poof follows by induction on n. It is true for n = 1, nd we ssume tht it is correct for n 1, we need to show tht it is true for n. Well we hve n A [ i ] n = A n Becuse i s re ll disjoint. Another thing tht is true is s follows: n A [ n 1 i ] n C = A [ By ssumption n is mesurble, which implies tht n m (A [ i ] n 1 i ]) = m (A n ) + m (A [ i ]) Using our inductive hypothesis, n = m (A 1 ) Recll from the lst clss tht we hd the following Lemm: Lemm 35. Given 1, 2 mesurble, 1 2 is mesurble. As corollry to this, we hve Corollry 36. The collection M of ll such sets is n lgebr. We lso hd the following: Lemm 37. If A is ny set, 1,..., n re mesurble nd disjoint sets, then n m (A [ i ]) = n m (A i ) We now hve the following: Theorem 38. The collection M of mesurble sets is σ lgebr. 30
31 1.18. MASURABL STS AND LBSGU MASUR Proof. Since M is n lgebr, it is enough to show tht countble union of elements of M is gin n element of M. For exmple suppose tht {A n } is in M. We wnt to show tht their union is in M, tht is: A n = where is mesurble. We need to show tht m (A) = m (A ) + m (A C ) for ny set A such tht A R. Since is the union of our collection of A s, nd since ll such A M ( nd M is n lgebr ) then = n Such tht ech F n is pirwise disjoint, nd in M. This cn be constructed in the following wy: 1 = A 1, 2 = A 2 1, 3 = A 3 ( 1 2 ),... Now, let We lso know tht F C n n F n = i M contins C. From this we cn write the following: m (A) = m (F n A) + m (Fn C A) By the definition of mesurble set. notice tht this line, m (A) = m (F n A) + m (Fn C A) m (F n A) + m (A C ) But notice tht n m (F n A) + m (A C ) = m ([ n ] A) + m (A C ) Using our lemm bove, we hve tht = which is true for ll n N, tht n m (A i ) + m (A C ) n m (A) m (A i ) + m (A C ) Wht we do is let n, from which we get m (A) m (A i ) + m (A C ) Now, by countble subdditivity, we hve tht m (A) m (A ) + m (A C ) This implies tht is mesurble. 31
32 CHAPTR 1. INTRODUCTION Lemm 39. The intervl (, ) is mesurble. Proof. Let A R. Let A 1 = A (, ) nd let A 2 = A (, ) C = A (, ]. We need to show tht m (A) m (A 1 ) + m (A 2 ) Of m (A) =, then we re done. If m (A), then given ɛ > 0 there exists some {I n } cover of A by open intervls such tht ( ) m (A) l(i n ) ɛ Let I n = I n (, ) nd let I n = I n (, ]. If we hve tht then Similrly, tells us tht Thus, Notice tht A 1 I n m (A 1 ) m ( I n) A 2 m (A 2 ) m (A 1 ) + m (A 2 ) Since both I n nd I n re intervls. Thus, I n m (I n ) m (I n) [m (I n) + m (I n)] m (I n) = l(i n), m (I n) = l(i n) [m (I n) + m (I n)] m (I n ) = l(i n ) m (A) + ɛ since our choice of ɛ ws rbitrry, we let ɛ 0, in which cse m (A 1 ) + m (A 2 ) m (A) The theorem tht follows from this lemm is s follows: Theorem 40. very Borel set is mesurble. In prticulr, ech open nd ech closed set is mesurble. 32
33 1.18. MASURABL STS AND LBSGU MASUR Proof. Recll tht M is σlgebr. Since (, ) re mesurble, then (, ] re mesurble. From this, it follows tht sets of the form (, b) re mesurble s well. This follows from seeing tht (, b] = (, b) {b} nd since both (, b] nd {b} re mesurble, thus {b} C is mesurble, nd since our set is mesurble. More directly, notice tht (, b) = (, b] {b} c (, b) = (, b 1 n ) nd ech of these sets is mesurble. From this it would follows tht (, b) is mesurble. We see tht s long s < b, tht (, b) (, ) = (, b) It then follows tht (, b) M, nd tht M, σlgebr, contins ll open intervls, nd implies tht M contins ll Borel sets. Definition. If is mesurble set, then the Lebesgue mesure m() is the outer mesure m (). Proposition 41. Let { i } be sequence of mesurble sets. Then, If the i s re pirwise disjoint, then This is the property of countble dditivity. m( i ) m( i ) m( i ) m( i ) Proof. The first prt of our proposition follows from countble subdditivity of the outer mesure. Suppose we hve the sequence { i }. Then n m (A [ i ]) = n m (A i ) A = R tells us tht So n m ( ) = m ( i ) n i i 33
34 CHAPTR 1. INTRODUCTION Thus From this we conclude tht Letting n, we hve tht n m ( i ) m ( i ) = m( i ) n m( i ) m( i ) m( i ) n m ( i ) Which follows from seeing tht this is n infinite sequence (incresing) tht is bounded bove, so it must hve limit. Proposition 42. Let { n } be infinite decresing sequence of mesurble sets. Let m( i ) <. Then, m( i ) = n lim m( n ) We hve tht { n } is decresing if n n+1 for ll n. Proof. We cll Let We hve the following: = i F i = i i+1 1 = F i which re pirwise disjoint. Thus, ccording to countble dditivity, Notice tht And tht these two re disjoint. We get m( 1 ) = m(f i ) = m( i i+1 ) i = i+1 ( i i+1 ), i+1 1 m( i ) = m( i+1 ) + m( i i+1 ) Since we re under the ssumption tht m( i ) is finite for ll i, we hve tht So, m( i i+1 ) = m( i ) m( i+1 ) n m( i i+1 = (m( i ) m( i+1 )) = n lim [m( i ) m( i+1 )] = n lim [m( 1 ) m( n+1 )] 34
35 1.19. XAM So, Thus = m( 1 ) lim n m( n+1 ) m( 1 ) = m( 1 ) lim n m( n ) m( 1 ) m( 1 ) = m() = lim n m( n ) 1.19 xm The exm will be on Thursdy, the 29 th Mesurble Sets (continued) A set is mesurble if for ll A R m (A) = m (A ) + m (A C ) We sid tht if is mesurble, then the Lebesgue mesure of is given s m(e) = m () We sid tht if we hve { n } where n n+1 where ech n is mesurble, then lim m( n) = m( n ) n This ws ll shown in our lst clss. We hve the following now: Proposition 43. Let be ny set. The following re equivlent: 1. is mesurble 2. Given ɛ > 0, there exists n open set O such tht m (O ) < ɛ 3. Given ɛ > 0 there exists closed set F such tht m ( F ) < ɛ 4. There exists G in G δ with G nd m (G ) = 0 35
36 CHAPTR 1. INTRODUCTION 5. There exists F in F σ with F nd m ( F ) = 0 If m () <, the bove re equivlent to number (6): 6. Given ɛ > 0, there exists finite union U of open intervls such tht m (U ) < ɛ Proof. Assume first tht m () < ɛ. We first wnt to show tht (1) (2). We know tht m () is finite, given ɛ > 0 we cn conclude tht there exists cover {I n } such tht Since by definition, Let l(i n ) > m () + ɛ m () = inf l(in ) In O = I n, which is n open set. O is mesurble becuse ech I n is mesurble (open sets re mesurble). We lso know tht O. Then, m(o) m () + ɛ Which is true becuse {I n } is cover of O, so Since is mesurble, we hve tht But this is the sme s sying Since O. From bove, we hve tht Cnceling out since m() is finite, we hve tht m(o) = m (O) l(i n ) m () + ɛ m(o) = m (O) = m (O ) + m (O C ) = m() + m(o ) m() + m(o ) m() + ɛ m (O ) < ɛ Now we wnt to show tht (2) (4). This is the symmetric difference of sets 36
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