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N (iii) The properties limx→−∞ F (x) = 0 and limx→∞ F (x) = 1 are an exercise. 10 Assume that µ1 and µ2 are probability measures on B(❘) and F1 and F2 are the corresponding distribution functions. Then the following assertions are equivalent: (1) µ1 = µ2 . (2) F1 (x) = µ1 ((−∞, x]) = µ2 ((−∞, x]) = F2 (x) for all x ∈ ❘. 3. INDEPENDENCE 35 Proof. (1) ⇒ (2) is of course trivial. We consider (2) ⇒ (1): For sets of type A := (a1 , b1 ] ∪ · · · ∪ (an , bn ], where the intervals are disjoint, one can show that n n (F1 (bi ) − (F1 (ai )) = µ1 (A) = µ2 (A) = i=1 (F2 (bi ) − (F2 (ai )).

E. 5. ) if P-almost surely or {ω ∈ Ω : P(ω) holds} belongs to F and is of measure one. Let us start with some first properties of the expected value. 1 Assume a probability space (Ω, F, P) and random variables f, g : Ω → ❘. (1) If 0 ≤ f (ω) ≤ g(ω), then 0 ≤ ❊f ≤ ❊g. (2) The random variable f is integrable if and only if |f | is integrable. In this case one has |❊f | ≤ ❊|f |. ❊f = 0. s. s. s. and ❊f exists, then ❊g exists and ❊f = ❊g. , then (4) (5) Proof. (1) follows directly from the definition.

3 Let (Ω, F) and (M, Σ) be measurable spaces and let f : Ω → M . Assume that Σ0 ⊆ Σ is a system of subsets such that σ(Σ0 ) = Σ. If f −1 (B) ∈ F for all B ∈ Σ0 , f −1 (B) ∈ F for all B ∈ Σ. then Proof. Define A := B ⊆ M : f −1 (B) ∈ F . Obviously, Σ0 ⊆ A. We show that A is a σ–algebra. (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.