Solution. 1 Solutions of Homework 1. Sangchul Lee. October 27, Problem 1.1

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1 Solutio Sagchul Lee October 7, Solutios of Homework 1 Problem 1.1 Let Ω,F,P) be a probability space. Show that if {A : N} F such that A := lim A exists, the PA) = lim PA ). Proof. Usig the cotiuity of P alog mootoe sequeces of measurable sets see [Dur10, Theorem 1.1.1]), we have ) ) lim P A m = PA), lim P A m = PA). ) A m = P m= m= ) A m = P m= m= O the other had, by the mootoicity of P we have m= A m A m= A m ad hece ) P A m PA ) P A m ). m= m= Therefore the claim follows from the squeezig lemma. Problem 1. [Dur10, Exercise 1.1.] Let Ω = R, F = all subsets so that A or A c is coutable, PA) = 0 i the first case ad = 1 i the secod. Show that Ω,F,P) is a probability space. Proof. Step 1. We first prove that F is a σ-algebra. By the costructio it is clear that F ad F is closed uder complemets. Moreover, if {A } F the either all of them are coutable so that A is also coutable, or A c is coutable for at least oe N so that A c ) c = A c is coutable. Therefore F is closed uder coutable uios ad hece the claim follows. Step. Next we show that P is a probability measure. Agai, it is immediate from defiitio that 0 = P ) PA) for A F. Now let {A } F be a family of disjoit sets i F. If either all of them are coutable, the their uio is also coutable ad ) P A = 0 = PA ). 1

2 O the other had, if A c is coutable for some N the for all other m we have A m A c ad hece A m is coutable. So there is exactly oe idex for which A is cocoutable. So ) P A = 1 = PA ). Sice the coutable additivity holds i ay case, we are doe. Problem 1.3 [Dur10, Exercise 1.1.3] Recall the defiitio of S d from Example Show that σs d ) = BR d ), the Borel subsets of R d. Proof. A stadard approach to this type of problem is to show that oe family is cotaied i the other. This is ofte doe by showig that oe family cotais all the geerators of the other. First we show the easier part, σs d ) BR d ). Notice that ay geerator d a i,b i ] of S d ca be writte as d a i,b i ] = a i,b i + 1 ] BR d ). d This shows that S d BR d ) ad therefore σs d ) BR d ) by the miimality of σs d ). Next we show the reverse, BR d ) σs d ). Agai, it suffices to prove that σs d ) cotais all the ope sets of R d. To this ed, let U be ay ope set. We also itroduce families of cubes { d ki C =, k ] } i + 1 : k i Z S d, C = D. 1.1) Sice each C is coutable, the same is true for C. The the set Ũ = C C C d C U satisfies Ũ U ad Ũ σs d ). But sice ay poit of U is a iterior poit, it should lie i some C C satisfyig C U. Therefore U = Ũ ad the claim follows. Problem 1.4 [Dur10, Exercise 1.1.4] A σ-field F is said to be coutably geerated if there is a coutable collectio C F so that σc) = F. Show that BR d ) is coutably geerated. Proof. Aalyzig the previous solutio, it follows that C defied i 1.1) is coutable ad geerates BR d ). Problem 1.5 Let M be a compact metric space ad CM) the space of cotiuous fuctios o M edowed with the supremum orm. For BCM)) deotig the Borel subsets of CM), prove that BCM)) = σ { f CM) : f t) a,b]} : t M, a b ) 1.)

3 Proof. As usual, we prove that oe cotais all the geeratig elemets of the other. We also itroduce otatios which will come hady for our proof. E t B) = { f CM) : f t) B}, B f,r) = {g CM) : f g sup < r} Let C = {E t a,b]) : t M, a b } deote the geeratig set of the right-had side of 1.). Sice the projectio map f f t) for each t M is cotiuous o CM), we kow that E t a,b + )) is ope. The E t a,b]) = E t a,b + ) ) BCM)) ad hece σc) BCM)). To show the other directio, we recall the fact that CM) is secod coutable. That is, there exists a coutable basis {B f i,r i )} for the topology of CM). Sice every ope set of CM) is writte as a coutable uio of this basis elemets, it suffices to show that each basis elemet is a elemet of σc). To this ed, we pick a coutable dese subset D of M by ivokig the fact that M is separable. The by cotiuity, for each f CM) ad r > 0, we have B f,r) = {g CM) : f g sup r} = E t [ f t) r, f t) + r]) σc) ad hece B f,r) = This completes the proof that BCM)) σc). t D B f, 1 r) σc). Problem 1.6 [Dur10, Exercise 1.1.5] i) Show that if F 1 F are σ-algebras, the i F i is a algebra. ii) Give a example to show that i F i eed ot be a σ-algebra. Proof. i) We write F = i F i for simplicity. Let A F. The there exists a idex i such that A F i ad hece A c F i F. This shows that F is closed uder complemets. Now let A,B F. The there exist i ad i such that A F i ad B F i. The A B F i i F ad hece F is closed uder fiite uios. ii) Cosider Ω = N with F i = σ{1},,{i}). The F = i F i cotais all sigletos of N ad hece σf) = N. O the other had, every elemet of F is either fiite or cofiite. For istace, {,4, } / F. Problem 1.7 [Dur10, Exercise 1.1.6] A set A {1,, } is said to have asymptotic desity θ if A {1,,,} lim = θ Let A be the collectio of sets for which the asymptotic desity exists. Is A a σ-algebra? a algebra? Proof. The aswer is that A is ot eve a algebra. To show this, we costruct a couter-example as follows: Pick a icreasig sequece k ) k=1 of positive itegers such that k/ k+1 0 as k. We costruct two sets A ad B as 3

4 follows: ) ) A = N, B = N k, k+1 ] N 1) k 1, k ] k=1 That is, B is costructed by cocateatig suitably chose subsets of eve itegers ad odd itegers alteratigly. Writig [] = {1,,, } for simplicity, it is clear that 1 A [], 1 k=1 B [] + 1 ad hece both A ad B have the asymptotic desity 1. O the other had, we also have ad similarly A B) [ k+1 ] = A B) [ k ] = k N j, j+1 ] k + N j 1, j ] = k+1 + k 1 k j j 1 ) k+1 + k N j, j+1 ] k + N j 1, j ] k 1 = k Combiig two estimates together, it follows that A B) [] lim sup j+1 j 1 ) k k 1. A B) [] = 1, limif 1. Therefore A B has o asymptotic desity ad hece the claim follows. Problem 1.8 [Dur10, Exercise 1..1] Suppose X ad Y are radom variables o Ω,F,P) ad let A F. Show that if we let Zω) = Xω) for ω A ad Zω) = Y ω) for ω A c, the Z is a radom variable. Proof. This immediately follows by writig Z 1 B) = A X 1 B)) A c Y 1 B)). If B BR), the all the sets ivolved i the right-had side are F-measurable ad hece Z 1 B) F. Problem 1.9 [Dur10, Exercise 1..3] Show that a distributio fuctio has at most coutably may discotiuities. Before the proof, let us recall some facts about mootoe fuctios. Sice F is mootoe icreasig, both Fx ) = lim y x Fy) ad Fx + ) = lim y x Fy) exist ad satisfy Fx ) Fx + ). So the oly possibly type of discotiuity of F is the jump discotiuity with positive icremet. 1 st Proof. Defie the fuctio h : 0,1) Q R by hr) = sup{x R : Fx) r}. 4

5 If D deotes the set of discotiuity of F, the for each x D we have 0 Fx ) < Fx + ) 1. So there exists r 0,1) Q such that Fx ) < r < Fx + ), which implies that hr) = x. Therefore But sice Rageh) is coutable, the same is true for D. D Rageh) = {hr) : r 0,1) Q} d Proof. Let D ε = {x R : Fx + ) Fx ) ε} be the set of poits with jump size ε. It is clear that D = D 1/ is the set of discotiuity of F. Now we claim that Fx + ) Fx ) ) 1 1.3) x D ε holds for ay ε > 0. Ideed, cosider a arbitrary fiite subset F D ε ad eumerate the elemets of F i icreasig order F = {x 1 < x < < x }. The for ay sufficietly small δ > 0 we have x F Fx + δ) Fx δ)) = Fx i + δ) Fx i δ)) 1 Fx i + δ) Fx i δ)) + Fx i+1 δ) Fx i + δ)) = Fx + δ) Fx 1 δ) 1 ad takig δ 0 shows that x F Fx + ) Fx )) 1. Takig supremum over F proves 1.3). Boudig each term of 1.3) from below by ε, we have ε D ε 1, or equivaletly, D ε ε 1. So D ε is fiite. The D is a coutable uio of fiite sets ad hece is coutable. Problem 1.10 [Dur10, Exercise 1..4] Show that if Fx) = PX x) is cotiuous the Y = FX) has a uiform distributio o 0,1), that is, if y [0,1], PY y) = y. Proof. We first prove the claim for y 0,1). Cosider 0 < y < y < y + < 1. By the itermediate value theorem, there exist x ± such that Fx ) = y ad Fx + ) = y +. The we have {X x } {FX) y} {X x + } ad hece y = Fx ) = PX x ) PY y) PX x + ) = Fx + ) = y +. Takig limit as y y ad y + y proves the claim for y 0,1). Whe y = 0, we utilize cotiuity of P to obtai PY 0) = lim PY ) = lim = 0. Fially, PY 1) = 1 holds trivially ad therefore the proof is complete. Problem 1.11 [Dur10, Exercise A.1.1] Give a example of two probability measures µ ν o F = all subsets of {1,,3,4} that agree o a collectio of sets C with σc) = F, i.e., the smallest σ-algebra cotaiig C is F. Proof. Cosider C = {{1,},{,3},{3,4},{4,1}} with µ{1}) = µ{3}) = 1, µ{}) = µ{4}) = 0, ν{1}) = ν{3}) = 0, ν{}) = ν{4}) = 1. 5

6 Refereces [Dur10] R. Durrett. Probability: Theory ad Examples. Cambridge Series i Statistical ad Probabilistic Mathematics. Cambridge Uiversity Press,