Lecture 2: Measures 1 of 17 Course: Theory of Probability I Term: Fall 2013 Istructor: Gorda Zitkovic Lecture 2 Measures Measure spaces Defiitio 2.1 (Measure). Let (S, S) be a measurable space. A mappig µ : S [0, ] is called a (positive) measure if 1. µ( ) = 0, ad 2. µ( A ) = N µ(a ), for all pairwise disjoit {A } N i S. A triple (S, S, µ) cosistig of a o-empty set, a σ-algebra S o it ad a measure µ o S is called a measure space. Remark 2.2. 1. A mappig whose domai is some oempty set A of subsets of some set S is sometimes called a set fuctio. 2. If the requiremet 2. i the defiitio of the measure is weakeed so that it is oly required that µ(a 1 A ) = µ(a 1 ) + + µ(a ), for N, ad pairwise disjoit A 1,..., A, we say that the mappig µ is a fiitely-additive measure. If we wat to stress that a mappig µ satisfies the origial requiremet 2. for sequeces of sets, we say that µ is σ-additive (coutably additive). Defiitio 2.3 (Termiology). A measure µ o the measurable space (S, S) is called 1. a probability measure, if µ(s) = 1, 2. a fiite measure, if µ(s) <, 3. a σ-fiite measure, if there exists a sequece {A } N i S such that A = S ad µ(a ) <, 4. diffuse or atom-free, if µ({x}) = 0, wheever x S ad {x} S. A set N S is said to be ull if µ(n) = 0.
Lecture 2: Measures 2 of 17 Example 2.4 (Examples of measures). Let S be a o-empty set, ad let S be a σ-algebra o S. 1. Measures o coutable sets. Suppose that S is a fiite or coutable set. The each measure µ o S = 2 S is of the form µ(a) = p(x), x A for some fuctio p : S [0, ] (why?). I particular, for a fiite set S with N elemets, if p(x) = 1/N the µ is a probability measure called the uiform measure 1 o S. 2. Dirac measure. For x S, we defie the set fuctio δ x o S by 1, x A, δ x (A) = 0, x A. 1 I the fiite case, it has the well-kow property that µ(a) = #A #S, where # deotes the cardiality (umber of elemets). It is easy to check that δ x is ideed a measure o S. Alteratively, δ x is called the poit mass at x (or a atom o x, or the Dirac fuctio, eve though it is ot really a fuctio). Moreover, δ x is a probability measure ad, therefore, a fiite ad a σ-fiite measure. It is atom free oly if {x} S. 3. Coutig Measure. Defie a set fuctio µ : S [0, ] by #A, A is fiite, µ(a) =, A is ifiite, where, as above, #A deotes the umber of elemets i the set A. Agai, it is ot hard to check that µ is a measure - it is called the coutig measure. Clearly, µ is a fiite measure if ad oly is S is a fiite set. µ could be σ-fiite, though, eve without S beig fiite. Simply take S = N, S = 2 N. I that case µ(s) =, but for A = {}, N, we have µ(a ) = 1, ad S = A. Fially, µ is ever atom-free ad it is a probability measure oly if #S = 1. Example 2.5 (A fiitely-additive set fuctio which is ot a measure). Let S = N, ad S = 2 S. For A S defie µ(a) = 0 if A is fiite ad µ(a) =, otherwise. For A 1,..., A S, either 1. all A i is fiite, for i = 1,...,. The i=1 A i is also fiite ad so 0 = µ( i=1 A i) = µ(a i ), or i=1 2. at least oe A i is ifiite. The i=1 A i is also ifiite ad so = µ( i=1 A i) = µ(a i ). i=1
Lecture 2: Measures 3 of 17 O the other had, take A i = {i}, for i N. The µ(a i ) = 0, for Note: It is possible to costruct very each i N, ad, so, simple-lookig fiite-additive measures i N µ(a i ) = 0, but µ( i A i ) = µ(n) =. which are ot σ-additive. For example, there exist {0, 1}-valued fiitely-additive measures o all subsets of N, which are Propositio 2.6 (First properties of measures). Let (S, S, µ) be a measure space. 1. For A 1,..., A S with A i A j =, for i = j, we have 2. If A, B S, A B, the µ(a i ) = µ( i=1 A i) i=1 (Fiite additivity) ot σ-additive. Such objects are called ultrafilters ad their existece is equivalet to a certai versio of the Axiom of Choice. µ(a) µ(b) (Mootoicity of measures) 3. If {A } N i S is icreasig, the µ( A ) = lim µ(a ) = sup µ(a ). (Cotiuity with respect to icreasig sequeces) 4. If {A } N i S is decreasig ad µ(a 1 ) <, the µ( A ) = lim µ(a ) = if µ(a ). (Cotiuity with respect to decreasig sequeces) 5. For a sequece {A } N i S, we have µ( A ) µ(a ). N (Subadditivity) Proof. 1. Note that the sequece A 1, A 2,..., A,,,... is pairwise disjoit, ad so, by σ-additivity, µ( i=1 A i) = µ( i N A i ) = µ(a i ) = i=1 µ(a i ) + i N i=+1 µ( ) = µ(a i ). i=1 2. Write B as a disjoit uio A (B \ A) of elemets of S. By (1) above, µ(b) = µ(a) + µ(b \ A) µ(a). 3. Defie B 1 = A 1, B = A \ A 1 for > 1. The {B } N is a pairwise disjoit sequece i S with k=1 B k = A for each N (why?). By σ-additivity we have µ( A ) = µ( B ) = µ(b ) = lim µ(b k ) N k=1 = lim µ( k=1 B k) = lim µ(a ).
Lecture 2: Measures 4 of 17 4. Cosider the icreasig sequece {B } N i S give by B = A 1 \ A. By De Morga laws, fiiteess of µ(a 1 ) ad (3) above, we have µ(a 1 ) µ( A ) = µ(a 1 \ ( A )) = µ( B ) = lim µ(b ) = lim µ(a 1 \ A ) = µ(a 1 ) lim µ(a ). Subtractig both sides from µ(a 1 ) < produces the statemet. 5. We start from the observatio that for A 1, A 1 S the set A 1 A 2 ca be writte as a disjoit uio so that A 1 A 2 = (A 1 \ A 2 ) (A 2 \ A 1 ) (A 1 A 2 ), µ(a 1 A 2 ) = µ(a 1 \ A 2 ) + µ(a 2 \ A 1 ) + µ(a 1 A 2 ). O the other had, ad so µ(a 1 ) + µ(a 2 ) = (µ(a 1 \ A 2 ) + µ(a 1 A 2 )) ( ) + µ(a 2 \ A 1 ) + µ(a 1 A 2 ) = µ(a 1 \ A 2 ) + µ(a 2 \ A 1 ) + 2µ(A 1 A 2 ), µ(a 1 ) + µ(a 2 ) µ(a 1 A 2 ) = µ(a 1 A 2 ) 0. Iductio ca be used to show that µ(a 1 A ) µ(a k ). k=1 Sice all µ(a ) are oegative, we ow have µ(a 1 A ) α, for each N, where α = µ(a ). N The sequece {B } N give by B = k=1 A k is icreasig, so the cotiuity of measure with respect to icreasig sequeces implies that µ( A ) = µ( B ) = lim µ(b ) = lim µ(a 1 A ) α. Remark 2.7. The coditio µ(a 1 ) < i the part (4) of Propositio 2.6 caot be sigificatly relaxed. Ideed, let µ be the coutig measure o N, ad let A = {, + 1,... }. The µ(a ) = ad, so lim µ(a ) =. O the other had, A =, so µ( A ) = 0.
Lecture 2: Measures 5 of 17 I additio to uios ad itersectios, oe ca produce other importat ew sets from sequeces of old oes. More specifically, let {A } N be a sequece of subsets of S. The subset lim if A of S, defied by lim if A = B, where B = k A k, is called the limit iferior of the sequece A. It is also deoted by lim A or {A, ev.} (ev. stads for evetually 2 ). Similarly, the subset lim sup A of S, defied by lim sup A = B, where B = k A k, is called the limit superior of the sequece A. It is also deoted by lim A or {A, i.o.} (i.o. stads for ifiitely ofte 3 ). Clearly, we have lim if A lim sup A. 2 the reaso for the use of the word evetually is the followig: lim if A is the set of all x S which belog to A for all but fiitely may values of the idex, i.e., from some value of the idex owards. 3 i words, lim sup A is the set of all x S which belog A for ifiitely may values of. Problem 2.1. Let (S, S, µ) be a fiite measure space. Show that Hit: For the secod part, a measure space with fiite (ad small) S will do. µ(lim if A ) lim if µ(a ) lim sup µ(a ) µ(lim sup A ), for ay sequece {A } N i S. Give a example of a (sigle) sequece {A } N for which all iequalities above are strict. Propositio 2.8 (Borel-Catelli Lemma I). Let (S, S, µ) be a measure space, ad let {A } N be a sequece of sets i S with the property that N µ(a ) <. The µ(lim sup A ) = 0. Proof. Set B = k A k, so that {B } N is a decreasig sequece of sets i S with lim sup A = B, ad so µ(lim sup A ) µ(b ), for each N. Usig the subadditivity of measures of Propositio 2.6, part 5., we get µ(b ) µ(a ). (2.1) k= Sice N µ(a ) coverges, the right-had side of (2.1) ca be made arbitrarily small by choosig large eough N. Extesios of measures ad the coi-toss space Example 1.19 has itroduced the measurable space ({ 1, 1} N, S), with S = 2 { 1,1} beig the product σ-algebra o { 1, 1} N. The purpose
Lecture 2: Measures 6 of 17 of the preset sectio is to tur ({ 1, 1} N, S) ito a measure space, i.e., to defie a suitable measure o it. It is easy to costruct just ay measure o { 1, 1} N, but the oe we are after is the oe which will justify the ame coi-toss space. The ituitio we have about tossig a fair coi ifiitely may times should help us start with the defiitio of the coi-toss measure - deoted by µ C - o cyliders. Sice the coordiate spaces { 1, 1} are particularly simple, each product cylider is of the form C = { 1, 1} N or C = C 1,..., k ;b 1,...,b k, where } C 1,..., k ;b 1,...,b k = {s = (s 1, s 2,... ) { 1, 1} N : s 1 = b 1,..., s k = b k, for some k N, ad a choice of 1 1 < 2 < < k N of coordiates ad the correspodig values b 1, b 2,..., b k { 1, 1}. I the laguage of elemetary probability, each cylider correspods to the evet whe the outcome of the i -th coi is b i { 1, 1}, for i = 1,...,. The measure (probability) of this evet ca oly be give by µ C (C 1,..., k ;b 1,...,b k ) = 2 1 1 2 2 1 = 2 k. }{{} k times (2.2) The hard part is to exted this defiitio to all elemets of S, ad ot oly cyliders. For example, i order to state the law of large umbers later o, we will eed to be able to compute the measure of the set { s { 1, 1} N } 1 : lim s k = 0, k=1 which is clearly ot a cylider. Problem 1.9 states, however, that cyliders form a algebra ad geerate the σ-algebra S. Luckily, this puts us close to the coditios of the followig importat theorem of Caratheodory. Theorem 2.9 (Caratheodory s Extesio Theorem). Let S be a o- Note: I words, a σ-additive measure empty set, let A be a algebra of its subsets ad let µ : A [0, ] be a o a algebra A ca be exteded to a σ-additive measure o the σ-algebra set-fuctio with the followig properties: geerated by A. It is clear that the σ- additivity requiremet of Theorem 2.9 is 1. µ( ) = 0, ad ecessary, but it is quite surprisig that it is actually sufficiet. 2. µ(a) = =1 µ(a ), if {A } N A is a partitio of A. The, there exists a measure µ o σ(a) with the property that µ(a) = µ(a) for A A. Of Theorem 2.9. PART I. We start by defiig a measure-like object, Note: Ituitively, we try all differet called a outer measure, µ : 2 S [0, ] i the followig way: coutable covers of B with elemets of A ad miimize the total µ. { µ (B) = if µ(a ) : B =1 =1 } A, A A for all N.
Lecture 2: Measures 7 of 17 Eve though we do t expect the ifimum i the defiitio of µ to be attaied, µ has the followig properties: 1. µ ( ) = 0 (otriviality), 2. for B C, µ (B) µ (C) (mootoicity), ad 3. µ ( k B k ) k=1 µ (B k ) (subadditivity) Parts 1. ad 2. are immediately clear, while, to show 3., we pick ε > 0 ad k N ad fid a coutable cover {A k } N with elemets of A such that µ (B k ) µ(a k ) + ε. 2 k =1 Usig {A k } k N, N as a cadidate cover for k B k, we coclude that µ ( k B k ) k=1 µ (B k ) + ε. This beig true for each ε > 0 implies 3. We remark at this poit that µ ad µ coicide o A. By usig (A,,,... ) as a cadidate coutable cover of A A, we ca coclude that µ (A) µ(a), for all A A. Coversely, suppose that A N A with A A. Give that the elemets of A form a algebra, we ca assume that A A, for all N ad that {A } N are pairwise disjoit, as ay sequece coverig A ca be trasformed ito such a sequece without icreasig µ(a ). The assumed coutable Note: It is, probably, iterestig to ote additivity of µ o A ow comes ito play sice, for partitio {A that this is the oly place i the etire } N proof where the coutable additivity of of A ito elemets i A, we ecessarily have µ(a ) = µ(a), ad, µ o A is used. so µ (A) µ(a). PART II. The set-fuctio µ is, i geeral, ot a measure, but comes with the advatage of beig defied o all subsets of S. The cetral idea of the proof is to recover coutable additivity by restrictig its domai a little. We say that a subset M S is Caratheodory-measurable or µ -measurable if µ (B) = µ (B M) + µ (B M c ) for all B S, (2.3) with the family of all µ -measurable subsets of S deoted by M. We ote that, by subadditivity, the equality sig i the defiitio of the measurability ca be replaced by ; this will be used below. The first thig we eed to establish about M is that it is a algebra ad that µ is a fiitely-additive measure o M. Clearly A ad the complemet axiom follows directly from the symmetry i (2.3). Oly the closure uder fiite uios eeds some discussio, ad, by iductio, we oly eed to cosider two-elemet uios; for that, we pick M, N M, ad itroduce the followig otatio M 00 = M c N c, M 01 = M c N, M 10 = M N c M 11 = M N. (2.4)
Lecture 2: Measures 8 of 17 By the measurability of M ad N, for ay B S, we have µ (B) = µ (B N c ) + µ (B N) = µ (M 00 B) + µ (M 10 B) + µ (M 01 B) + µ (M 11 B) O the other had, M 01 M 10 M 11 = M N, so that, by subadditivity ad (2.4), we have µ ( (M N) c B ) + µ ( (M N) B ) = µ (M 00 B) + µ ( ) (M 01 B) (M 10 B) (M 11 B) µ (M 00 B) + µ (M 10 B) + µ (M 01 B) + µ (M 11 B) = µ (B), which implies that that M N M. Whe M N = a applicatio of measurability of N to B = M N yields the fiite additivity of µ o M : µ (M N) = µ ((M N) N) + µ ((M N) N c ) = µ (N) + µ (M). PART III. We ow tur to the closure of M uder coutable uios ad the σ-additive property of µ. Sice M already kow to be a algebra, it will be eough to show that it is closed uder pairwisedisjoit uios, i.e., that M M wheever {M } N are pairwise disjoit elemets i M with M = M. For N, we set L = k=1 M k so that, for B S, we have so that µ (B) = µ (B L ) + µ (B L c ) k=1 µ (B M k ) + µ (B M c ), µ (B) µ (B M c ) + k N µ (B M k ) µ (B M c ) + µ ( k (B M k )) + µ (B M c ) = µ (B M) + µ (B M c ). Sice all the iequalities above eed to be equalities, we immediately coclude that, with B = S, µ (M) = µ (M k ), k i.e., that µ is a coutably-additive measure o M. Sice A M, we have M σ(a) ad µ = µ σ(a) is the required σ-additive extesio of µ. Back to the coi-toss space. I order to apply Theorem 2.9 i our situatio, we eed to check that µ is ideed a coutably-additive measure o the algebra A of all cyliders. The followig problem will help pipoit the hard part of the argumet:
Lecture 2: Measures 9 of 17 Problem 2.2. Let A be a algebra o a o-empty set S, ad let µ : A [0, ] be a fiite (µ(s) < ) ad fiitely-additive set fuctio o S with the followig, additioal, property: lim µ(a ) = 0, wheever A. (2.5) The µ satisfies the coditios of Theorem 2.9. The part about fiite additivity is easy (perhaps a bit messy) ad we leave it to the reader: Problem 2.3. Show that the set-fuctio µ C, defied by (2.2) o the product cylliders ad exteded by additivity to the algebra A of cyliders, is fiitely additive. Lemma 2.10 (Coditios of Caratheodory s theorem). The set-fuctio µ C, defied by (2.2), ad exteede by additivity to the the algebra A of cyliders, has the property (2.5). Proof. By Problem 1.10, cyliders are closed sets, ad so {A } N is a sequece of closed sets whose itersectio is empty. The same problem states that { 1, 1} N is compact, so, by the fiite-itersectio property 4, we have A 1... A k =, for some fiite collectio 1,..., k 4 The fiite-itersectio property refers to of idices. Sice {A } N is decreasig, we must have A =, for all k, ad, cosequetly, lim µ(a ) = 0. Propositio 2.11 (Existece of the coi-toss measure). There exists a measure µ C o ({ 1, 1} N, S) with the property that (2.2) holds for all cyliders. Proof. Thaks to Lemma 2.10, Theorem 2.9 ca ow be used. I order to prove uiqueess, we will eed the celebrated π-λ Theorem of Eugee Dyki: the followig fact, familiar from real aalysis: If a family of closed sets of a compact topological space has empty itersectio, the it admits a fiite subfamily with a empty itersectio. Theorem 2.12 (Dyki s π-λ Theorem). Let P be a π-system o a oempty set S, ad let Λ be a λ-system which cotais P. The Λ also cotais the σ-algebra σ(p) geerated by P. Proof. Usig the result of part 4. of Problem 1.1, we oly eed to prove that λ(p) (where λ(p) deotes the λ-system geerated by P) is a π- system. For A S, let G A deote the family of all subsets of S whose itersectios with A are i λ(p): G A = {C S : C A λ(p)}.
Lecture 2: Measures 10 of 17 Claim: G A is a λ-system for A λ(p). Sice A λ(p), clearly S G A. For a icreasig family {C } N i G A we have ( C ) A = (C A). Each C A is i Λ, ad the family {C A} N is icreasig, so ( C ) A Λ. Fially, for C 1, C 2 G with C 1 C 2, we have (C 2 \ C 1 ) A = (C 2 A) \ (C 1 A) Λ, because C 1 A C 2 A. P is a π-system, so for ay A P, we have P G A. Therefore, λ(p) G A, because G A is a λ-system. I other words, for A P ad B λ(p), we have A B λ(p). That meas, however, that P G B, for ay B λ(p). Usig the fact that G B is a λ-system we must also have λ(p) G B, for ay B λ(p), i.e., A B λ(p), for all A, B λ(p), which shows that λ(p) is π-system. Propositio 2.13 (Measures which agree o a π-system). Let (S, S) be a measurable space, ad let P be a π-system which geerates S. Suppose that µ 1 ad µ 2 are two measures o S with the property that µ 1 (S) = µ 2 (S) < ad µ 1 (A) = µ 2 (A), for all A P. The µ 1 = µ 2, i.e., µ 1 (A) = µ 2 (A), for all A S. Proof. Let L be the family of all subsets A of S for which µ 1 (A) = µ 2 (A). Clearly P L, but L is, potetially, bigger. I fact, it follows easily from the elemetary properties of measures (see Propositio 2.6) ad the fact that µ 1 (S) = µ 2 (S) < that it ecessarily has the structure of a λ-system 5. By Theorem 2.12 (the π-λ Theorem), L cotais the σ-algebra geerated by P, i.e., S L. O the other had, by defiitio, L S ad so µ 1 = µ 2. Propositio 2.14 (Uiqueess of the coi-toss measure). The measure µ C is the uique measure o ({ 1, 1} N, S) with the property that (2.2) holds for all cyliders. 5 It seems that the structure of a λ-system is defied so that it would exactly describe the structure of the family of all sets o which two measures (with the same total mass) agree. The structure of the π-system correspods to the miimal assumptio that allows Propositio 2.13 to hold. Proof. The existece is the cotet of Propositio 2.11. To prove uiqueess, it suffices to ote that algebras are π-systems ad use Propositio 2.13. Problem 2.4. Defie D 1, D 2 { 1, 1} N by
Lecture 2: Measures 11 of 17 1. D 1 = {s { 1, 1} N : lim sup s = 1}, 2. D 2 = {s { 1, 1} N : N N, s N = s N+1 = s N+2 }. Show that D 1, D 2 S ad compute µ(d 1 ), µ(d 2 ). Our ext task is to probe the structure of the σ-algebra S o { 1, 1} N a little bit more ad show that S = 2 { 1,1}N. It is iterestig that such a result (which deals exclusively with the structure of S) requires a use of a measure i its proof. Example 2.15 (A o-measurable subset of { 1, 1} N (*)). Sice σ- algebras are closed uder coutable set operatios, ad sice the product σ-algebra S for the coi-toss space { 1, 1} N is geerated by sets obtaied by restrictig fiite collectios of coordiates, oe is tempted to thik that S cotais all subsets of { 1, 1} N. That is ot the case. We will use the axiom of choice, together with the fact that a measure µ C ca be defied o the whole of { 1, 1} N, to show to costruct a example of a o-measurable set. Let us start by costructig a relatio o { 1, 1} N i the followig 6 way: we set s 1 s 2 if ad oly if there exists N such that s 1 k = s2 k, for k (here, as always, si = (s1 i, si 2,... ), i = 1, 2). It is easy to check that is a equivalece relatio ad that it splits { 1, 1} N ito disjoit equivalece classes. Oe of the may equivalet forms of the axiom of choice states that there exists a subset N of { 1, 1} N which cotais exactly oe elemet from each of the equivalece classes. Let us suppose that N is a elemet i S ad see if we ca reach a cotradictio. For each oempty = { 1,..., k } 2 N f i, where 2N f i deotes the family of all fiite subsets of N, let us defie the mappig T : { 1, 1} N { 1, 1} N i the followig 7 maer: s l, l, T = Id ad (T (s)) l = for N. s l, l, 6 I words, s 1 ad s 2 are related if they oly differ i a fiite umber of coordiates. 7 T flips the sigs of the elemets of its argumet o the positios correspodig to. Sice is fiite, T preserves the -equivalece class of each elemet. Cosequetly (ad usig the fact that N cotais exactly oe elemet from each equivalece class) the sets N ad T (N) = {T (s) : s N} are disjoit. Similarly ad more geerally, the sets T (N) ad T (N) are also disjoit wheever =. O the other had, each s { 1, 1} N is equivalet to some ŝ N, i.e., it ca be obtaied from ŝ by flippig a fiite umber of coordiates. Therefore, the family forms a partitio of { 1, 1} N. N = {T (N) : 2 N f i }
Lecture 2: Measures 12 of 17 The mappig T has several other ice properties. First of all, it is immediate that it is ivolutory, i.e., T T = Id. To show that it is (S, S)-measurable, we eed to prove that its compositio with each projectio map π k : S { 1, 1} is measurable. This follows immediately from the fact that for k N (π k T ) 1 C k;1, k, ({1}) = C k; 1, k, where, for b { 1, 1}, we recall that C k;b = {s { 1, 1} N : s k = b} is a product cylider. If we combie the ivolutivity ad measurability of T, we immediately coclude that T (A) S for each A S. I particular, N S. I additio to preservig measurability, the map T also preserves the measure 8 the i µ C, i.e., µ C (T (A)) = µ C (A), for all A S. To prove that, let us pick F ad cosider the set-fuctio µ : S [0, 1] give by µ (A) = µ C (T (A)). It is a simple matter to show that µ is, i fact, a measure o (S, S) with µ (S) = 1. Moreover, thaks to the simple form (2.2) of the actio of the measure µ C o cyliders, it is clear that µ = µ C o the algebra of all cyliders. It suffices to ivoke Propositio 2.13 to coclude that µ = µ C o the etire S, i.e., that T preserves µ C. The above properties of the maps T, F ca imply the followig: N is a partitio of S ito coutably may measurable subsets of equal measure. Such a partitio {N 1, N 2,... } caot exist, however. Ideed if it did, oe of the followig two cases would occur: 1. µ(n 1 ) = 0. I that case µ(s) = µ( k N k ) = µ(n k ) = 0 = 0 = 1 = µ(s). 8 Actually, we say that a map f from a measure space (S, S, µ S ) to the measure space (T, T, µ T ) is measure preservig if it is measurable ad µ S ( f 1 (A)) = µ T (A), for all A T. The ivolutivity of the map T implies that this geeral defiitio agrees with our usage i this example. 2. µ(n 1 ) = α > 0. I that case µ(s) = µ( k N k ) = µ(n k ) = α = = 1 = µ(s). Therefore, the set N caot be measurable 9 i S. 9 Somewhat heavier set-theoretic machiery ca be used to prove that most of the subsets of S are ot i S, i the sese The Lebesgue measure As we shall see, the coi-toss space ca be used as a sort of a uiversal measure space i probability theory. We use it here to costruct the Lebesgue measure o [0, 1]. We start with the otio somewhat dual to the already itroduced otio of the pull-back i Defiitio 1.8. We leave it as a exercise for the reader to show that the set fuctio f µ from Defiitio 2.16 is ideed a measure. that the cardiality of the set S is strictly smaller tha the cardiality of the set 2 S of all subsets of S
Lecture 2: Measures 13 of 17 Defiitio 2.16 (Push-forwards). Let (S, S, µ) be a measure space ad let (T, T ) be a measurable space. The measure f µ o (T, T ), defied by f µ(b) = µ( f 1 (B)), for B T, is called the push-forward of the measure µ by f. Let f : { 1, 1} N [0, 1] be the mappig give by f (s) = k=1 ( ) 1+sk 2 2 k, s { 1, 1} N. The idea is to use f to establish a correspodece betwee all real umbers i [0, 1] ad their expasios i the biary system, with the codig 1 0 ad 1 1. It is iterestig to ote that f is ot oe-to-oe 10 as it, for example, maps s 1 = (1, 1, 1,... ) ad s 2 = ( 1, 1, 1,... ) ito the same value - amely 2 1. Let us show, first, that the map f is cotiuous i the metric d defied by part (1.2) of Problem 1.9. Ideed, we pick s 1 ad s 2 i { 1, 1} N ad remember that for d(s 1, s 2 ) 2, the first 1 coordiates of s 1 ad s 2 coicide. Therefore, f (s 1 ) f (s 2 ) 2 k = 2 +1 = 2d(s 1, s 2 ). k= Hece, the map f is Lipschitz ad, therefore, cotiuous. The cotiuity of f (together with the fact that S is the Borel σ- algebra for the topology iduced by the metric d) implies that f : ({ 1, 1} N, S) ([0, 1], B([0, 1])) is a measurable mappig. Therefore, the push-forward λ = f (µ) is well defied o ([0, 1], B([0, 1])), ad we call it the Lebesgue measure o [0, 1]. 10 The reaso for this is, poetically speakig, that [0, 1] is ot the Cator set. Propositio 2.17 (Ituitive properties of the Lebesgue measure). The Lebesgue measure λ o ([0, 1], B([0, 1])) satisfies λ([a, b)) = b a, λ({a}) = 0 for 0 a < b 1. (2.6) Proof. 1. Cosider a, b of the form b = 2 k ad b = k+1 2, for N ad k < 2. For such a, b we have f 1 ([a, b)) = C 1,...,;c1,c 2,...,c, where c 1 c 2... c is the base-2 expasio of k (after the recodig 1 0, 1 1). By the very defiitio of λ ad the form (2.2) of the actio of the coi-toss measure µ C o cyliders, we have ( ) ( λ [a, b) = µ C f 1( [a, b) )) = µ C (C 1,...,;c1,c 2,...,c ) = 2 = k+1 2 2 k. Therefore, (2.6) holds for a, b of the form b = 2 k ad b = 2 l, for N, k < 2 ad l = k + 1. Usig (fiite) additivity of λ, we
Lecture 2: Measures 14 of 17 immediately coclude that (2.6) holds for all k, l, i.e., that it holds for all dyadic ratioals. A geeral a (0, 1] ca be approximated by a icreasig sequece {q } N of dyadic ratioals from the left, ad the cotiuity of measures with respect to decreasig sequeces implies that ( ) ( ) ( ) λ [a, p) = λ [q, p) = lim λ [q, p) = lim(p q ) = (p a), wheever a (0, 1] ad p is a dyadic ratioal. I order to remove the dyadicity requiremet from the right limit, we approximate it from the left by a sequece {p } N of dyadic ratioals with p > a, ad use the cotiuity with respect to icreasig sequeces to get, for a < b (0, 1), ( ) ( ) ( ) λ [a, b) = λ [a, p ) = lim λ [a, p ) = lim(p a) = (b a). The Lebesgue measure has aother importat property: Problem 2.5. Show that the Lebesgue measure is traslatio ivariat. More precisely, for B B([0, 1]) ad x [0, 1), we have Hit: Use Propositio 2.13 for the secod part. 1. B + 1 x = {b + x (mod 1) : b B} is i B([0, 1]) ad 2. λ(b + 1 x) = λ(b), where, for a [0, 2), we defie a, a 1, a (mod 1) = a 1, a > 1. Geometrically, the set x + 1 B is obtaied from B by traslatig it to the right by x ad the shiftig the part that is stickig out by 1 to the left. Fially, the otio of the Lebesgue measure is just as useful o the etire R, as o its compact subset [0, 1]. For a geeral B B(R), we ca defie the Lebesgue measure of B by measurig its itersectios with all itervals of the form [, + 1), ad addig them together, i.e., λ(b) = ( (B ) ) λ [, + 1). = Note how we are overloadig the otatio ad usig the letter λ for both the Lebesgue measure o [0, 1] ad the Lebesgue measure o R. It is a quite tedious, but does ot require ay ew tools, to show that may of the properties of λ o [0, 1] trasfer to λ o R: Problem 2.6. Let λ be the Lebesgue measure o (R, B(R)). Show that 1. λ([a, b)) = b a, λ({a}) = 0 for a < b,
Lecture 2: Measures 15 of 17 2. λ is σ-fiite but ot fiite, 3. λ(b + x) = λ(b), for all B B(R) ad x R, where B + x = {b + x : b B}. Additioal Problems Problem 2.7 (Local separatio by costats). Let (S, S, µ) be a measure space ad let f, g L 0 (S, S, µ) satisfy µ ( {x S : f (x) < g(x)} ) > 0. Prove or costruct a couterexample for the followig statemet: There exist costats a, b R such that µ ( {x S : f (x) a < b g(x)} ) > 0. Problem 2.8 (A pseudometric o sets). Let (S, S, µ) be a fiite measure space. For A, B S defie d(a, B) = µ(a B), where deotes the symmetric differece: A B = (A \ B) (B \ A). Show that d is a pseudometric o S, ad for A S describe the set of all B S with d(a, B) = 0. Problem 2.9 (Complete measure spaces). A measure space (S, S, µ) is called complete if all subsets of ull sets are themselves i S. For a (possibly icomplete) measure space (S, S, µ) we defie the completio (S, S, µ ) i the followig way: S = {A N : A S ad N N for some N S with µ(n) = 0}. For B S with represetatio B = A N we set µ (B) = µ(a). 1. Show that S is a σ-algebra. Note: Let X be a oempty set. A fuctio d : X X [0, ) is called a pseudometric if 1. d(x, y) + d(y, x) d(x, z), for all x, y, z X, 2. d(x, y) = d(y, x), for all x, y X, ad 3. d(x, x) = 0, for all x X. Note how the oly differece betwee a metric ad a pseudometric is that for a metric d(x, y) = 0 implies x = y, while o such requiremet is imposed o a pseudometric. Note: Ufortuately, the same otatio µ is ofte used for the completio of the measure µ ad the outer measure associated with µ as i the proof of Theorem 2.9. Fortuately, it ca be show that these two object coicide o the domai of the completio. 2. Show that the defiitio µ (B) = µ(a) above does ot deped o the choice of the decompositio B = A N, i.e., that µ(â) = µ(a) if B =  ˆN is aother decompositio of B ito a set  i S ad a subset ˆN of a ull set i S. 3. Show that µ is a measure o (S, S ) ad that (S, S, µ ) is a complete measure space with the property that µ (A) = µ(a), for A S. Problem 2.10 (The Cator set). The Cator set is defied as the collectio of all real umbers x i [0, 1] with the represetatio x = c 3, where c {0, 2}. =1 Show that it is Borel-measurable ad compute its Lebesgue measure.
Lecture 2: Measures 16 of 17 Problem 2.11 (The uiform measure o a circle). Let S 1 be the uit circle, ad let f : [0, 1) S 1 be the widig map ( ) f (x) = cos(2πx), si(2πx), x [0, 1). 1. Show that the map f is (B([0, 1)), S 1 )-measurable, where S 1 deotes the Borel σ-algebra o S 1 (with the topology iherited from R 2 ). 2. For α (0, 2π), let R α deote the (couter-clockwise) rotatio of R 2 with ceter (0, 0) ad agle α. Show that R α (A) = {R α (x) : x A} is i S 1 if ad oly if A S 1. 3. Let µ 1 be the push-forward of the Lebesgue measure λ by the map Note: The measure µ 1 is called the uiform measure (or the uiform distribu- α (A) ). f. Show that µ 1 is rotatio-ivariat, i.e., that µ 1 (A) = µ 1( R tio) o S 1. Problem 2.12 (Asymptotic desities). We say that the subset A of N admits asymptotic desity if the limit d(a) = lim #(A {1, 2,..., }), exists (remember that # deotes the umber of elemets of a set). Let D be the collectio of all subsets of N which admit asymptotic desity. 1. Is D a algebra? A σ-algebra? 2. Is the map A d(a) fiitely-additive o D? A measure? Problem 2.13 (A subset of the coi-toss space). A elemet i { 1, 1} N (i.e., a sequece s = (s 1, s 2,... ) where s { 1, 1} for all N) is said to be evetually periodic if there exists N 0, K N such that s = s +K for all N 0. Let P { 1, 1} N be the collectio of all evetually-period sequeces. Show that P is measurable i the product σ-algebra S ad compute µ C (P). Problem 2.14 (Regular measures). The measure space (S, S, µ), where (S, d) is a metric space ad S is a σ-algebra o S which cotais the Borel σ-algebra B(d) o S is called regular if for each A S ad each ε > 0 there exist a closed set C ad a ope set O such that C A O ad µ(o \ C) < ε. 1. Suppose that (S, S, µ) is a regular measure space, ad that the measure space (S, B(d), µ B(d) ) is obtaied from (S, S, µ) by restrictig the measure µ oto the σ-algebra of Borel sets. Show that S B(d), where ( S, B(d), (µ B(d) ) ) is the completio (i the sese of Problem 2.9) of (S, B(d), µ B(d) ) 2. Suppose that (S, d) is a metric space ad that µ is a fiite measure o B(d). Show that (S, B(d), µ) is a regular measure space.
Lecture 2: Measures 17 of 17 Hit: Cosider a collectio A of subsets A of S such that for each ε > 0 there exists a closed set C ad a ope set O with C A O ad µ(o \ C) < ε. Argue that A is a σ-algebra. The show that each closed set ca be writte as a itersectio of ope sets; use (but prove, first) the fact that the map x d(x, C) = if{d(x, y) : y C}, is cotiuous o S for ay oempty C S. 3. Show that (S, B(d), µ) is regular if µ is ot ecessarily fiite, but has the property that µ(a) < wheever A B(d) is bouded, i.e., whe sup{d(x, y) : x, y A} <. Hit: Pick a poit x 0 S ad, for N, defie the family {R } N of subsets of S as follows: R 1 = {x S : d(x, x 0 ) < 2}, ad R = {x S : 1 < d(x, x 0 ) < + 1}, for > 1, as well as a sequece {µ } N of set fuctios o B(d), give by µ (A) = µ(a R ), for A B(d). Uder the right circumstaces, eve coutable uios of closed sets are closed. 4. Coclude that the Lebesgue measure o ( R, B(R ) ) is regular.