4th Preparation Sheet - Solutions

Prof. Dr. Rainer Dahlhaus Probability Theory Summer term 017 4th Preparation Sheet - Solutions Remark: Throughout the exercise sheet we use the two equivalent definitions of separability of a metric space (X, d): The topology of (X, d) is generated by countably many open sets There exists a countable dense subset A X of (X, d). Task P14 (Separability). Solve the following tasks. (a) Let L 1 (R) := {f : R R f is measurable and f(x) dx < }, (b) Let and let f 1 := f(x) dx be the L 1 -norm. Show that (L 1 (R), 1 ) is separable. Hint: Approximate f by simple functions. Use Theorem 3.4(i) of the lecture and the separability of R. BV (R) := {f : R R f has bounded variation}, where f BV := sup n N sup <t0 <...<t n< n i=1 f(t i) f(t i 1 ) is the variation of f. Show that (BV (R), BV ) is not separable. (c) Let l 0 (R) := {a = (a n ) n N R (a n ) n N is convergent to 0}, and define a := sup n N a n. Show that (l 0 (R), ) is separable. (d) Let l (R) := {a = (a n ) n N R (a n ) n N is a bounded sequence}, and define a := sup n N a n. Show that (l (R), ) is not separable. Hint: You may use without proof that M := {0, 1} N := {a = (a n ) n N : a n {0, 1}} is uncountable which directly follows from the fact that this set can be identified with the binary representation of all real numbers in [0, 1]. Solution: (a) Let (D n ) n N be a countable basis of the topology of (R, ). Let M := n N i 1,...,i { n n N D i k } which is still a countable basis of the topology. This modification allows us to assume that finite unions of elements of M are in M again. We propose the set of primitive functions A := { n y k 1 Bk : n N, y k Q, B k M} = n N 1 y 1,...,y n Q B 1,...,B n Q { n y k 1 Bk }

which is countable as a union of countable sets. Reduction to nonnegative functions: By the decomposition f = f + f into positive and negative parts and the inequality f g 1 f + g + 1 + f g 1, we can assume w.l.o.g. that f 0. Let > 0. Reduction to primitive functions: By results from Probability Theory 1 we can find a sequence of nonnegative primitive functions φ n : R R such that φ n f. By definition of the measure integral, we have (λ is the Lebesgue measure) 0 φ n dλ f dλ, i.e. f φ n dλ 0. Thus we can find a nonnegative primitive function φ N = M N x k1 Ak with A k A (w.l.o.g. λ(a k ) > 0), x k [0, ) (w.l.o.g. x k > 0) such that f φ N 1. 4 Reduction to rational primitive functions: Since Q is dense in R we can find y k Q (0, ) such that x k y k, K = 1,..., M 4M N λ(a k ) N. Put ψ N := M N y k1 Ak, then M N φ N ψ N 1 x k y k λ(a k ) 4. Reduction to functions out of A: By Theorem 3.4(i) of the lecture applied to (R,, λ) there exist open sets O k A k such that λ(o k \A k ) 4M N y k. Since (R, ) is separable, for each k N we can find (C k,i ) i N M (M is the countable basis of the topology of (R, )) such that O k = i N C k,i. Defining the disjoint sets D k,i := C k,i \ i 1 j=1 C k,j, we see that λ(o k ) = i N λ(d k,i ), therefore there have to exist m k N such that Define M N ψ N := 4M N y k > λ(o k ) m k i=1 λ(d k,i) = λ ( O k \ m k i=1 C k,i). y k 1 m k i=1 C k,i. Since m k i=1 C k,i M (see the motivation at the beginning), we have ψ N A. Furthermore, which shows in total that ψ N ψ M N N 1 y k 1 Ak 1 m k i=1 C dλ < k,i 4 + 4, λ(o k \A k )+λ(o k \ m k i=1 C k,i) f ψ N 1 f φ N 1 + φ N ψ N 1 + ψ N ψ N 1 <. (b) For x R, define f x : R R, f x (y) := 1 (,x] (y). Obviously it holds that f x BV = 1 (f x has only one jump of height 1). For x 1, x [0, 1], x 1 < x we have f x f x1 = 1 (x1,x ], thus f x1 f x BV = (since f x f x1 has two jumps of height 1). Suppose that there exists a countably dense subset A BV (R) of (BV (R), BV ). Because

A is dense in (BV (R), BV ), for each x R we can find an a x A with f x a x < 1. In total, with x 1, x R, x 1 x : = f x f x1 f x1 a x1 + a x1 a x + a x f x 1 = 1 1 < 1 f x 1 a x1 a x f x a x1 a x, which shows a x1 a x. This is a contradiction to the property of A to be countable, because is uncountable. (c) Define {a x : x R} A A := m N{b = (b n ) n N : b n Q for 1 n m, b n = 0 for n > m} = m N Q m {(0) n N } which is countable as a countable union of countable sets. Obviously A consists of sequences converging to 0, thus A l 0 (R). We now show that A is dense in (l 0 (R), ). Let (a n ) n N l 0 (R) be arbitrary. Let > 0. Since a n 0, there exists N N such that for all n > N, a n <. Since Q is dense in R, for each 1 n N there exists b n Q such that a n b n. We now choose b = (b n) n N A, where b n = 0 for n > N. Then {, n = 1,..., N, a n b n a n, n > N, thus sup n N a n b n <. (d) Note that for all m M (see the hint for the definition of M), m = 1 <, thus m l (N). For m, m M, m 1 m there has to exist at least one index n N with m n m n, and thus m m = sup n N m n m n 1. Suppose that there exists a countably dense subset A l (N) of (l (N), ). Because A is dense in (l (N), ), for each m M we can find a m A such that m a m < 1 4. In total, with m 1, m M, m 1 m : 1 m 1 m m 1 a m1 + a m1 a m + a m m 1 = 1 1 4 1 4 < 1 m 1 a m1 a m m a m1 a m, which shows a m1 a m. This is a contradiction to the property of A to be countable, because is uncountable. {a m : m M} A Task P15 (Polish spaces). Solve the following tasks. (a) Let (X, d) be a Polish space and U X an open, nonempty subset of X. Show that (U, d) is again Polish. 1 Hint: Define the metric ρ(x, y) := d(x, y)+ f(x) f(y) on U, where f(x) :=. inf{d(x,z):z X\U} For separability, show that ρ induces the same topology as d on U. 3

(b) Let C[0, ) := {f : [0, ) R f is continuous}. Define the metric d(f, g) := k sup f(x) g(x) x [0,k] on C[0, ). Assume that it is already known that (C[0, k], ) is Polish for k N. Show that (C[0, ), d) is Polish. Solution: (a) Note that the topology of (U, d) is the subspace topology {A U : A X open in (X, d)}. In the following we will show that g : (U, d) ((0, ), ), x inf{d(x, z) : z X\U} is continuous. g is well-defined since for each fixed x U, we can find > 0 such that B (x) = {z X : d(x, z) < } U which implies g(x) /. Since d(x, z) d(x, y) + d(y, z) inf d(x, z) inf d(y, z) d(x, y) z X\U z X\U it follows that (swapping x, y in the above inequality for the same result with interchanged signs) inf d(x, z) inf d(y, z) d(x, y). z X\U z X\U This finishes the proof of the continuity of g. Since 1 : (0, ) (0, ) is continuous, we x conclude that f : (U, d) ((0, ), ) is continuous as a composition (*). We first show that ρ generates the same topology: Let A U be open in (U, d), and let x A. Then there exists > 0 such that {z U : d(x, z) < } A. Since d ρ, we have {z U : ρ(x, z) < } {z U : d(x, z) < } A, thus A is open in (U, ρ). Let A U be open in (U, ρ), and let x A. Then there exists > 0 such that {z U : ρ(x, z) < } A. Since f : (U, d) ((0, ), ) is continuous (cf. (*)), there exists δ > 0 such that d(x, z) < δ implies f(x) f(z) /. Choose := min{δ, /}. Then {z U : d(x, z) < } {z U : ρ(x, z) < } A, thus A is open in (U, d). Separability: Let A X be a countable dense subset of (X, d). Then A := A U is still countable. Let x U be arbitrary. Since U is open, there exists > 0 such that {z X : d(x, z) < } U. Since A is dense in (X, d), we find some y A with d(x, y) < /. This implies that y U, i.e. y A U = A. This shows that A is a countable dense subset of U. Completeness: Let (x n ) n N U be a Cauchy sequence in (U, ρ). Since d ρ, it follows that (x n ) n N is also Cauchy in (X, d). Since (X, d) is complete, there exists a limit x X such that d(x n, x) 0. Obviously, it holds that x U U (where U is the boundary in (X, d)). Assume that x U X\U. Then inf{d(x n, z) : z X\U} 0. This yields f(x n ) as well as f(x m ) (n, m ) and is therefore a contradiction the fact that (x n ) n N is a Cauchy sequence in (U, ρ) (since ρ(x, y) = d(x, y) + f(x) f(y) ). Thus x U. Since f is continuous on U (see (*)) and d(x n, x) 0 with x U, we obtain that f(x n ) f(x) 0. Thus ρ(x n, x) 0, i.e. (x n ) n N is convergent in (U, ρ). (b) Separability: For k N, let A k be the countable dense set in (C[0, k], ). Define A := {f C[0, ) : f [0,k] A k, f(x) = f(k) for all x > k}. k N =:B k 4

Note that there exists a one-to-one mapping from B k to A k since the domain of definition of the functions of A k is only extended to [0, ) glueing a constant line on these functions. Thus A is countable as a countable union of countable sets. Fix some f C[0, ). Let > 0. For N N large enough it holds that k=n+1 k <. Since f [0,N] C[0, N], there exists some g A N C[0, N] such that f [0,N] g <. The N corresponding element g B N A then fulfills d(f, g) = k sup f(x) g(x) x [0,k] N sup x [0,N] f(x) g(x) + k=n+1 k < N N + =. Remark: A direct proof of separability in the manner of task 14 from exercise sheet 4 without using prior information on C[0, k] is also possible. Completeness: Let (f n ) n N be an arbitrary Cauchy sequence in (C[0, ), d). Since d(f n, f m ) = k sup x [0,k] f n (x) f m (x) 0 for m n, each summand has to converge to 0, i.e. for all k N it holds that sup f n (x) f m (x) 0 (m n ). x [0,k] This shows that for all k N, (f n [0,k] ) n N is a Cauchy sequence in (C[0, k], ). Since (C[0, k], ) is complete, there exists a limit f k C[0, k] with f n [0,k] f k 0. Define f(x) := f k (x) for x [0, k]. Since the uniform limits are unique, this definition is welldefined. We conclude that f C[0, ). Furthermore, f n [0,k] f [0,k] 0. Now let > 0 be arbitrarily chosen. Let K N be large enough such that k=k+1 k <. Choose N N large enough such that for all n N, sup x [0,K] f n (x) f(x) <. Then we have for n N: K d(f n, f) K sup x [0,K] f n (x) f(x) + k=k+1 k < K K + =, showing that d(f n, f) 0. Task P16 (The set of cylinder sets). Let (S, A) be a measurable space and T an arbitrary index set. For index sets J I T, π I,J : S I S J, (x i ) i I (x j ) j J are the projections. Recall that B(A) T := σ(π T,{t} : t T ) is the smallest σ-algebra which makes the one-dimensional projections measurable. Put C := (a) Show that C is an algebra over S T. (b) Show that σ(c) = A T. I T finite π 1 T,I (AI ) Remark: In the lecture, one would often use (S, A) = (R, B(R)) and T = [0, ). Then C = n N 0 t 1 <t <...<t n π T,{t1,...,t n}(b(r) n ). Solution: (a) In the following we show the three properties of X to be an algebra: (i) Choose I = {t} T. Because S A, we have S T π 1 T,{t} (A), thus ST C. (ii) Let B 1, B C. Then there exist finite I 1, I T and A 1 A I 1, A A I such that B 1 = π 1 T,I 1 (A 1 ), B = π 1 T,I (A ). 5

Define I := I 1 I and A := π 1 I,I 1 (A 1 ) π 1 I,I (A ). Then we have (by the stability of the preimage w.r.t. unions) B 1 B = (π I,I1 π T,I ) 1 (A 1 ) (π I,I π T,I ) 1 (A ) = π 1 T,I (A). Because the projections π I,I1 : (S I, A I ) (S I 1, A I 1 ), π I,I : (S I, A I ) (S I, A I) are measurable (*), we have A A I and thus B 1 B π 1 T,I (AI ) C. Proof of (*): It is well-known that E := { i I 1 A i : i I 1 : A i A i } generates the σ- algebra A I 1, i.e. σ(e) = A I 1. For i I 1 A i = E E we have π 1 I,I 1 (E) = i I 1 π 1 I,{i} (A i) A I (since A I is the smallest σ-algebra which makes the projections to one component measurable). Therefore, π 1 I,I 1 (E) A I. In total, π 1 I,I 1 (A I 1 ) A I. (iii) Let B C. Then there exists a finite I T and A A I such that B = π 1 T,I (A). Because A I is a σ-algebra, we have A c A I. Thus B c = π 1 T,I (Ac ) π 1 T,I (AI ) C. (b) Because A T = σ(π 1 T,{t} (A) : t T ) and π 1 T,{t} (A) C, we have AT σ(c). Let I T be a finite index set. It is well-known that the set E := { i I A i i I : A i A} generates the product σ-algebra A I. It is therefore enough to show that π 1 T,I (E) AT (*), because then we obtain π 1 T,I (AI ) = π 1 T,I (σ(e)) well-known rule = σ(π 1 T,I (E)) AT σ-algebra A T. As a consequence, C A T (note that I T finite is a possibly uncountable union, but we do not unite sets, but collections of sets!). Proof of (*): If A = i I A i with A i A i, then π 1 T,I (A) = i I π 1 T,{i} (A i) A T, because it is the finite intersection of sets that are in A T. Task P17 (Measurability and continuity). Let X = (X t ) t 0 be a real-values stochastic process on (Ω, A, P), i.e. for each t 0, X t : Ω R is (A, B(R))-measurable. (a) Show that the above measurability condition is equivalent to the fact that X : Ω R [0, ) is A-B(R) [0, ) -measurable (see the definition in Task P16). From now on assume that X is right-continuous (i.e. t X t (ω) is continuous for all ω Ω). (b) Show that X is product measurable, i.e. X : Ω [0, ) R is ( A B([0, )) ) -B(R)- measurable. (c) Show that the following subsets of Ω are in A: M 1 := { t 0 : X t 0}, M := { lim X t = 0}, t M 3 := {t X t is Lipschitz continuous with constant L}, L > 0. Solution: (a) We have to show the equivalence [ ] t 0 : Xt 1 (B(R)) A X 1 (B(R) [0, ) ) A. 6

Since B(R) [0, ) = σ(e) with E = t 0 {π 1 [0, ),{t}(b(r))}, we have X 1 (B(R) [0, ) ) = X 1 (σ(e)) ( = σ (X 1 ( = σ t 0 t 0 well-known rule = σ(x 1 (E)) )) ( π 1 [0, ),{t} (B(R)) = σ Xt 1 (B(R)) ) = σ(x t : t 0). t 0 X 1 (π 1 [0, ),{t} (B(R))) ) Now, " "is obvious. For " ", note that by assumption t 0 X 1 t (B(R)) A (note that we build the union of sets of sets and thus this is still a subset of A), thus X 1 (B(R) [0, ) ) = σ( t 0 X 1 t (B(R))) A. (b) Define the stochastic process X (n) t := X nt /n for t 0, where denotes flooring. Since nt /n t for n and t X t is continuous, it holds that X (n) X on Ω [0, ) (indeed, for each t 0 and ω Ω we have X (n) t (ω) = X nt /n (ω) X t (ω)). We now show that X (n) is A B([0, ))-B(R)-measurable, then it follows that X is also measurable with respect to the same σ-algebras as limit of X (n). We only have to show measurability on the generating system {[0, c] : c > 0} of B(R). It holds that = = {(ω, t) Ω [0, ) : X (n) t (ω) c} {(ω, t) : X nt /n (ω) c, t [ k n, k + 1 n )} k=0 ( {ω Ω : X k/n (ω) c} [ k n, k + 1 ) n ) A B([0, )), k=0 A B([0, )) since {A B : A A, B B([0, ))} A B([0, )) and σ-algebras are closed with respect to countable unions. (c) M 1 : By continuity we have that M 1 ( ) = {t [0, ) Q : X t 0} = t [0, ) Q {X t 0} A A as a countable union of elements of A. (*) holds since for arbitrary t [0, ) we can find a sequence t n [0, ) Q with t n t. Since X has continuous paths, 0 X tn X t, thus X t 0 (this implies {t [0, ) Q : X t 0} M 1, the other relation is obvious). For the other two cases M and M 3, these facts are similarly proved and we will therefore abbreviate the argumentations. M : It holds that M = { k N : N N : t [N, ) : X t 1 k } X = cont. k N N N t [N, ) Q { X t 1 k } A, A since countable unions and intersections of elements of σ-algebras are again elements of the σ-algebra. M 3 : Here we have M 3 = { s, t [0, ) : X t X s L s t } X cont. = 7 s,t [0, ) Q { X t X s L s t } A. A

Submission: No submission! This sheet will (partly) be discussed in the exercise groups. Homepage of the lecture: http://math.uni-heidelberg.de/stat/studinfo/teaching_stat/wt_ss017/index.html 8