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7 or any other system such that σ(G) = B(❘). e. Nn n fn = k=1 ak 1IAnk n with ak ∈ ❘ and Ank ∈ F such that fn (ω) → f (ω) for all ω ∈ Ω as n → ∞. 3 Independence Let us first start with the notion of a family of independent random variables. 1 [independence of a family of random variables] Let (Ω, F, P) be a probability space and fi : Ω → ❘, i ∈ I, be random variables where I is a non-empty index-set. , fin ∈ Bn ) = P (fi1 ∈ B1 ) · · · P (fin ∈ Bn ) . 36 CHAPTER 2. 2 [independence of a finite family of random variables] Let (Ω, F, P) be a probability space and fi : Ω → ❘, i = 1, .

1) f −1 (M ) = Ω ∈ F implies that M ∈ A. (2) If B ∈ A, then f −1 (B c ) = = = = {ω : f (ω) ∈ B c } {ω : f (ω) ∈ / B} Ω \ {ω : f (ω) ∈ B} f −1 (B)c ∈ F. (3) If B1 , B2 , · · · ∈ A, then ∞ f −1 ∞ Bi i=1 f −1 (Bi ) ∈ F. = i=1 By definition of Σ = σ(Σ0 ) this implies that Σ ⊆ A, which implies our lemma. 2. (2) =⇒ (1) follows from (a, b) ∈ B(❘) for a < b which implies that f −1 ((a, b)) ∈ F. 3 since B(❘) = σ((a, b) : −∞ < a < b < ∞). 2. 4 If f : measurable. ❘ 33 → ❘ is continuous, then f is (B(❘), B(❘))- Proof.

Using Carathe find an unique probability measure P on B(❘◆ ) such that P(B1 × B2 × · · · × Bn × ❘ × ❘ · · · ) = P1(B1) · · · Pn(Bn) for all n = 1, 2, ... , xn ∈ Bn . 8 [Realization of independent random variables] Let (❘◆ , B(❘◆ ), P) and πn : Ω → ❘ be defined as above. Then (Πn )∞ n=1 is a sequence of independent random variables such that the law of Πn is Pn , that means P(πn ∈ B) = Pn(B) for all B ∈ B(❘). Proof. , Bn ∈ B(❘). , Πn(ω) ∈ Bn}) = P(B1 × B2 × · · · × Bn × ❘ × ❘ × · · · ) = P1 (B1 ) · · · Pn (Bn ) n = P(❘ × · · · × ❘ × Bk × ❘ × · · · ) k=1 n = k=1 P({ω : Πk (ω) ∈ Bk }).