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Short Recap: Σ-Algebras Generated by Maps

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Short recap: σ-algebras generated by maps 1) Let (X; A) and (Y; B) be measurable spaces and f : X ! Y be a map. Then we define the pull back σ-algebra as f −1(B) := ff −1(B): B 2 Bg This is the smallest σ-algebra on X such that f is measurable. In particular, f is (A; B)- measurable if and only if f −1(B) ⊆ A. 2) If (fi)i2I is a family of maps fi : X ! Yi, then [ −1 σ(fi; i 2 I) := σ fi (Bi) i2I S −1 is the σ-algebra generated by (fi)i2I . Please note that i2I fi (Bi) is not a σ-algebra in general, and as in 1) we can easily prove that σ(fi; i 2 I) is the smallest σ-algebra on X such that all fi are measurable. Exercise 4 (σ(X)-measurable random variables) . Let (Ω; A) be a measurable space. a) Let a1; : : : ; an 2 R be pairwise distinct and A1;:::;An 2 A be pairwise disjoint with Sn Pn 1 Ω = j=1 Aj. Let X := j=1 aj Aj and σ(X) be the σ-algebra generated by X. i) Determine σ(X). ii) Let Y be a σ(X)-measurable random variable. Show that Y is constant on each set Aj, j = 1; : : : ; n. iii) Show that Y can be written as a function of X. b) Let X :Ω ! R be an arbitrary random variable and let Y :Ω ! R be σ(X)- measurable. Show that there exists a function Φ such that Y = Φ(X): Hint: algebraic induction. Solution: a) i) In this particular case we obtain −1 n[ o σ(X) = X (B(R)) = σ fA1;:::;Ang = Aj : J ⊆ f1; : : : ; ng : j2J −1 −1 a) ii) We have Y (B) 2 σ(X) for all B 2 B(R), in particular Y (fzg) 2 σ(X) for −1 all z 2 R. Now assume that Y (fzg) 6= ; for some z 2 R. Then there exists a set −1 S A 2 σ(X) n f;g such that Y (fzg) = A = j2J Aj for some index set J ⊆ f1; : : : ; ng, i.e. Y (!) = z for all ! 2 A. This means, that Y is constant on each set Aj, j = 1; : : : ; n. a) iii) By part ii) we can find z1; : : : ; zn 2 R such that n n X 1 X 1 Y = zi Ai = zi fX=aig: i=1 i=1 1 Pn 1 By defining the map g : R ! R by g(x) = i=1 zi faig(x) we finally obtain n X 1 g(X(!)) = zi fX=aig(!) = Y (!): i=1 Pn 1 b) 1) Let Y be a simple function, i.e Y = i=1 zi Ai for some (zi) ⊆ R and (Ai) ⊆ σ(X). −1 Then there exist sets Bi 2 B(R) such that Ai = X (Bi), i = 1; : : : ; n. Now define n X 1 '(x) := zi Bi (x); x 2 R: i=1 Then we obtain n n X 1 X 1 '(X(!)) = zi Bi (X(!)) = zi Ai = Y: i=1 | {z } i=1 =1 −1 (!) X (Bi) 2) Let Y :Ω ! R be a positive function. Then there exists a sequence of simple functions Yn such that 0 ≤ Yn % Y pointwise, in particular Y = supn Yn. By part 1) we can find a sequence of functions 'n such that 'n(X) = Yn, n 2 N. Now take ' := supn2N 'n. Then we get '(X) = sup 'n(X) = sup Yn = Y: n n 3) Let Y be a general measurable map. Then Y = Y+ − Y− with Y+;Y− ≥ 0. By part 2) we can construct functions '+ and '− such that Y+ = '+(X) and Y− = '−(X). Defining ' := '+ − '− we finally get '(X) = '+(X) − '−(X) = Y+ − Y− = Y: Moreover, we obtain the following result: Lemma T2. Let fi : (Ω; A) ! (Ωi; Bi), i 2 I, and ': (Ω0; A0) ! (Ω; A) be maps. Then we have ' is (A0; σ(fi; i 2 I))-measurable () fi ◦ ' is (A0; Bi)-measurable for all i 2 I: Proof: First assume the left-hand side. Since ' is (A0; σ(fi))-measurable and fi is (σ(fi); Bi)-measurable for any fixed i 2 I, we trivially obtain that fi ◦ ' is (A0; Bi)- measurable for all i 2 I. S −1 Now assume the converse. Let M := i2I fi (Bi) be the generating system of σ(fi; i 2 I) −1 and A 2 M be any set. Then there exists an i 2 I and Bi 2 Bi with A = fi (Bi) satisfying −1 −1 −1 −1 ' (A) = ' (fi (Bi)) = (fi ◦ ') (Bi) 2 A0: Since A 2 M was arbitrary, the claim follows. 2 Exercise 5 (Product probability spaces) . Let (Ωj; Aj; Pj), j 2 N, be probability spaces, 1 1 O Ω := × Ωj; and A := Aj := σ(πj; j 2 N) j=1 j=1 the product σ-algebra generated by the projections πj :Ω ! Ωj (πj(!) = !j, j 2 N). Show that there exists a uniquely determined probability measure P on A satisfying 1 n Y P A1 × ::: × An × × Ωj = Pj(Aj) j=n+1 j=1 for any sets Aj 2 Aj, j = 1; : : : ; n. N1 N1 1 N1 N1 Notation: P = j=1 Pj and j=1(Ωj; Aj; Pj) := (×j=1 Ωj; j=1 Aj; j=1 Pj). Hints for the proof: 1) For n 2 N define 1 Fn := Bn × × Ωj : Bn 2 A1 ⊗ ::: ⊗ An j=n+1 and show that this is an ascending sequence of σ-algebras. S1 2) For F := n=1 Fn show that F ⊂ A and σ(F) = A. 3) On Fn define the probability measure 1 µn Bn × × Ωj := P1 ⊗ ::: ⊗ Pn(Bn) j=n+1 and on F the map P by P(A) := µn(A) if A 2 Fn: 4) Show that P(Ω) = 1 and P is σ-additive. 5) Finally, use Caratheodory's extension theorem as well as the measure uniqueness theorem to conclude the proof. Before we start with the proof of this result we shortly recall Caratheodory's extension theorem and the measure uniqueness theorem: Caratheodory's extension theorem. Let R be a semi-ring, µ: R! [0; 1] be a set function satisfying 1) µ(;) = 0 2) µ is σ-additive Then there exists a measureµ ~ on σ(R) with µ(A) =µ ~(A) for all A 2 R. 3 Measure uniqueness theorem. Let M ⊆ P(Ω) be an intersection stable generator of the σ-algebra A and let µ1, µ2 be measures on A such that µ1(A) = µ2(A) for all A 2 M: S If there exists an exhausting sequence of sets (An)n≥1 ⊆ M with Ω = n≥1 An with µ1(An) = µ2(An) < 1 for all n 2 N, then µ1 = µ2. 1 Solution: 1) Let Fn := Bn × ×j=n+1 Ωj : Bn 2 A1 ⊗ ::: ⊗ An , n 2 N. Using that n ⊗j=1Aj is a σ-algebra, it can be easily verified that Fn is a σ-algebra as well. Moreover, n+1 n since Bn × Ωn+1 2 ⊗j=1 Aj for any set Bn 2 ⊗j=1Aj we trivially obtain Fn ⊆ Fn+1. 2) Now define F := S F (which is not a σ-algebra in general). Then F is an algebra: n2N n i) Since Ω 2 Fn for all n 2 N, we obtain Ω 2 F. ii) Let A 2 F. Then there is an n 2 N such that A 2 Fn. Using that Fn is a σ-algebra C C we get A 2 Fn, i.e. A 2 F. iii) Let A; B 2 F. Then there are m; n 2 N such that A 2 Fn and B 2 Fm. Without loss of generality we may assume that n < m. Then A; B 2 Fm by part 1). Using again that Fm is a σ-algebra, we get A [ B 2 Fm, and this implies A [ B 2 F by definition. In particular, F is an intersection stable semi-ring. Consider now the maps pn :Ω ! Ω1 × ::: × Ωn; pn(!) = (!1;:::;!n): n Then pn is (A; ⊗i=1Ai)-measurable. This can be seen by considering the maps n n n πj : × Ωi ! Ωj; πj ((!1;:::;!n)) = !j; j = 1; : : : ; n i=1 n which has the property πj ◦ pn = πj, and this is (A; Aj)-measurable for each j = 1; : : : ; n. n By Lemma T2 pn is (A; σ(πj ; j = 1; : : : ; n))-measurable. | n{z } =⊗j=1Aj 1 For any set A 2 F we can find an n 2 N such that A = Bn × ×j=n+1 Ωj for some n Bn 2 ⊗j=1Aj. Using the measurability of pn above we obtain −1 A = pn (Bn) 2 A: This implies F ⊆ A. 1 Define the sets Rn := fA1 × :::An × ×j=n+1 Ωj : Aj 2 Ajg, n 2 N. Then we have 1 1 [ −1 [ πj (Aj) ⊆ Rj ⊆ F ⊆ A; j=1 j=1 which implies that σ(F) = A. 3) On Fn we define the probability measure µn by 1 n µn Bn × × Ωj := ⊗j=1Pj(Bn) j=n+1 4 and the set function P on F by P(A) := µn(A) if A 2 Fn: 4) By construction P is well-defined and it holds that P(Ω) = µ1(Ω1) = P1(Ω1) = 1: Now let A1;:::;An 2 F be disjoint sets. Then there is an m 2 N such that A1;:::;An 2 Fm. This implies n n n n [ [ X X P Aj = µm Aj = µm(Aj) = P(Aj): j=1 j=1 j=1 j=1 which means that P is finitely additive.
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