Martingale Theory Problem set 1, with solutions Measure and integration 1.1 Let be a measurable space. Prove that if , , then . (Ω; F) An 2 F n 2 N \n2NAn 2 F HINT FOR SOLUTION: Apply repeatedly De Morgan's identities: \ [ An = Ω n (Ω n An): n2N n2N 1.2 Let (Ω; F) be a measurable space and Ak 2 F, k 2 N an innite sequence of events. Prove that for all ! 2 Ω 11\n[m≥nAm (!) = lim 11An (!); 11[n\m≥nAm (!) = lim 11An (!): n!1 n!1 HINT FOR SOLUTION: Note that ( if 0 #fn 2 N : ! 2 Ang < 1; lim 11A (!) = n!1 n if 1 #fn 2 N : ! 2 Ang = 1: ( 0 if #fm 2 : !2 = A g = 1; lim 11 (!) = N m An if n!1 1 #fm 2 N : !2 = Amg < 1: 1.3HW (a) Let Ω be a set and Fα ⊂ P(Ω), α 2 I, an arbitrary collection of σ-algebras on Ω. We assume I 6= ;, otherwise we don't make any assumption about the index set I. Prove that \ F := Fα α2I is a σ-algebra. (b) Let C ⊂ P(Ω) be an arbitrary collection of subsets of Ω. Prove that there exists a unique 1 smallest σ-algebra σ(C) ⊂ P(Ω) containing C. (We call σ(C) the σ-algebra generated by the collection C.) (c) Let (Ω; F) and (Ξ; G) be measurable spaces where G = σ(C) is the σ-algebra generated by the collection of subsets C ⊂ P(Ω). Prove that the map T :Ω ! Ξ is measurable if and only if for any A 2 C, T −1(A) 2 F. Hint for (c): Prove that fA ⊂ Ξ: T −1(A) 2 Fg is a σ-algebra. SOLUTION: (a) Check the axioms of σ-algebra for F: (i) \ (8α 2 I):Ω 2 Fα =) Ω 2 Fα α2I (ii) \ (8n 2 N): An 2 Fα =) (8α 2 I)(8n 2 N): An 2 Fα α2I [ =) (8α 2 I): An 2 Fα n2N [ \ =) An 2 Fα n2N α2I (b) Denote I(C) := fF ⊂ P(Ω) : F is a σ-algebra and C ⊂ Fg: Since P(Ω) 2 I(C), I(C) 6= ;. Hence, by applying (a) \ σ(C) := F; F2I(C) is a σ-algebra which contains C as subset. By construction it is the smallest such object. (c) We prove that H := fA ⊂ Ξ: T −1(A) 2 Fg ⊂ P(Ξ) is a σ-algebra. Indeed: (i) T −1(Ξ) = Ω 2 F =) Ξ 2 H: (ii) −1 (8n 2 N): An 2 H =) (8n 2 N): T (An) 2 F [ −1 −1 [ =) T (An) = T ( An) 2 F n2N n2N [ =) An 2 H: n2N 2 By assumption, C ⊂ H and thus due to (b) G = σ(C) ⊂ H. −1 1.4 (a) Let f : R ! R and assume that for any a 2 R, f ((−∞; a)) 2 B, where B denotes the σ-algebra of Borel-measurable subsets of R. Prove that f is Borel-measurable, i.e. for any A 2 B, f −1(A) 2 B. (b) Let f : R ! R and g : R ! R be Borel-measurable functions. Prove that f ◦ g : R ! R is also Borel-measurable. (c) Let f : R ! R be piece-wise monotone function. Prove that f is Borel-measurable. HINT FOR SOLUTION: (a) This is special case of part (c) of problem 3, with the particular choice: (Ω; F) = (Ξ; G) = (R; B) (B denotes the Borel-sigma-algebra on R, generated by the topol- ogy.), and C = f(−∞; a): a 2 Rg ⊂ P(R). Note that B = σ(C). (b) Let A ⊂ R, then (f ◦g)−1(A) = g−1(f −1(A)). Hence, since both functions f and g are assumed to be Borel-measurable A 2 B =) f −1(A) 2 B =) g−1(f −1(A)) 2 B. (c) Prove rst that if f : R ! R is piecewise monotone then inverse images of intervals are countable unions of intervals. 1.5HW Let Ω = f1; 2; 3; 4g and F := f;; f1g; f3g; f1; 3g; f2; 4g; f1; 2; 4g; f2; 3; 4g; f1; 2; 3; 4gg G := f;; f1g; f2g; f1; 3g; f3; 4g; f1; 3; 4g; f2; 3; 4g; f1; 2; 3; 4gg H := f;; f1g; f4g; f1; 4g; f2; 3g; f1; 2; 3g; f2; 3; 4g; f1; 2; 3; 4gg (a) Decide, which of the collections F, G and/or H are σ-algebras and which are not. n (b) Let f :Ω ! R be dened as f(n) := (−1) . Decide whether f is measurable or not with respect to the σ-algebras identied in question (a). SOLUTION: (a) F and H are σ-algebras. G is not a σ-algebra. (b) f is F-measurable but not H-measurable. 1.6 Let Ω = N, F := P(N) and dene µ : F! [0; 1] as follows: ( 0 if jAj < 1; µ(A) = 1 if jAj = 1: Prove that µ is an additive but not a σ-additive measure on (N; P(N)). 3 HINT FOR SOLUTION: Finite additivity follows from the fact that nite union of nite sets is nite. However -additivity doesn't hold. Indeed , but σ N = [n2Nfng X µ(N) = 1 6= 0 = µ(fng): n2N 1.7 Bonus Let Ω = N and #A \ [0; n] C := fA ⊂ N : lim =: ρ(A) exists g: n!1 n For A 2 C we call the number ρ(A) 2 [0; 1] the Césaro density of the set A. The Césaro density measures in a sense the relative weight of the subset A within N. Unfortunately, the collection C ⊂ P(N) is not even an algebra of subsets, and thus the Césaro density can not serve as a decent measure. Give an example of two sets A; B 2 C for which A \ B 62 C. HINT FOR SOLUTION: Let 2k 2k E := f2 + 2l : k 2 N; 0 ≤ l < 2 g; 2k+1 2k+1 F := f2 + 2l : k 2 N; 0 ≤ l < 2 g; 2k+1 2k+1 G := f2 + 2l + 1 : k 2 N; 0 ≤ l < 2 g: In plain words: - E is the set of all even numbers in intervals of the form [22k; 22k+1), k 2 N. - F is the set of all even numbers in intervals of the form [22k+1; 22(k+1)), k 2 N. - G is the set of all odd numbers in intervals of the form [22k+1; 22(k+1)), k 2 N. These are clearly disjoint sets. Dene A := E [ F , B := E [ G. Then check that #A \ [0; n] #B \ [0; n] 1 lim = lim = n!1 n n!1 n 2 while #E \ [0; n] 1 2 #E \ [0; n] lim = 6= = lim : n!1 n 3 3 n!1 n 1.8 Bonus Construction of the Vitali set example of a subset of [0; 1) which can't be Lebesgue 4 measurable. Let Ω := [0; 1) and dene on Ω the following equivalence relation: x ∼ y if and only if x − y 2 Q: Let V ⊂ [0; 1) consist of exactly one representative element from each equivalence class according to ∼. (Note, that this construction relies on the Axiom of Choice.) For q 2 Q \ [0; 1) denote Vq := fy = x + q (mod 1) : x 2 V g: Prove that 0 0 (i) The sets Vq, q 2 Q \ [0; 1), are congruent: for any q; q 2 Q \ [0; 1), Vq0 = (q − q) + Vq (mod 1). 0 0 (ii) For any q; q 2 Q \ [0; 1), if q 6= q then Vq \ Vq0 = ;. (iii) S V = [0; 1). q2Q\[0;1) q Conclude that the Vitali set V can not be Lebesgue measurable. HINT SOLUTION: (i) By construction, for all , is congruent with . So, they are q 2 [0; 1) \ Q Vq V0 = V all congruent between them. 0 0 (ii) Assume that for q 6= q there is x 2 Vq \ Vq0 . Then there are y; y 2 V , so that x = y+q and x = y0 +q0 (mod 1), and hence y−y0 = q0 −q 6= 0. But, by construction, V contains one single representative from each class of equivalence, so can't contain two dierent elements whose dierence is non-zero rational. (iii) Let x 2 [0; 1). Denote by x∗ the representative of the class [x] := fy 2 [0; 1) : ∗ y ∼ xg in V and let q = x − x . Then clearly x 2 Vq. Assume now, that V is assigned Lebesgue measure λ(V ) 2 [0; 1]. Then, since Vq-s are all congruent with V , all must have the same Lebesgue measure: for all q 2 [0; 1)\Q λ(Vq) = λ(V ): On the other hand, [ [0; 1) = Vq; q2[0;1)\Q where the sets Vq are pairwise disjoint and they are countably many. By σ-additivity of measure we must have X λ(Vq) = λ([0; 1)) = 1: q2[0;1)\Q Now, assuming λ(V ) = 0 we get λ([0; 1)) = 0, assuming λ(V ) > 0 we get λ([0; 1)) = 1. Both possible assumptions lead to contradiction. The case is that the set V is not measurable and there is no way to assign Lebesgue measure to it. 5 1.9HW (a) Let r; rn 2 R, n 2 N and assume limn!1 rn = r. Prove that r = sup inf rn = inf sup rn : m n≥m m n≥m (b) Let (Ω; F) be a measurable space, fn :Ω ! R a sequence of real valued functions and f :Ω ! R, dened as f(!) := infn fn(!).