By Geiss C., Geiss S.

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50 CHAPTER 3. INTEGRATION Then A g(η)dPϕ (η) = ϕ−1 (A) g(ϕ(ω))dP(ω) for all A ∈ E in the sense that if one integral exists, the other exists as well, and their values are equal. Proof. (i) Letting g(η) := 1IA (η)g(η) we have g(ϕ(ω)) = 1Iϕ−1 (A) (ω)g(ϕ(ω)) so that it is sufficient to consider the case A = Ω. Hence we have to show that E g(η)dPϕ (η) = Ω g(ϕ(ω))dP(ω). (ii) Since, for f (ω) := g(ϕ(ω)) one has that f + = g + ◦ ϕ and f − = g − ◦ ϕ it is sufficient to consider the positive part of g and its negative part separately.

N n n (2) is an exercise. (3) If ❊f − = ∞ or ❊g + = ∞, then ❊f = −∞ or ❊g = ∞ so that nothing is to prove. Hence assume that ❊f − < ∞ and ❊g + < ∞. The inequality f ≤ g gives 0 ≤ f + ≤ g + and 0 ≤ g − ≤ f − so that f and g are integrable and ❊f = ❊f + − ❊f − ≤ ❊g+ − ❊g− = ❊g. (4) Since (af + bg)+ ≤ |a||f | + |b||g| and (af + bg)− ≤ |a||f | + |b||g| we get that af + bg is integrable. The equality for the expected values follows from (1) and (2). 4 [monotone convergence] Let (Ω, F, P) be a probability space and f, f1 , f2 , ...

So we wish to extend this definition. 2 [random variables] Let (Ω, F) be a measurable space. A map f : Ω → ❘ is called random variable provided that there is a sequence (fn )∞ n=1 of measurable step-functions fn : Ω → ❘ such that f (ω) = lim fn (ω) for all ω ∈ Ω. n→∞ Does our definition give what we would like to have? 3 Let (Ω, F) be a measurable space and let f : Ω → a function. Then the following conditions are equivalent: ❘ be (1) f is a random variable. (2) For all −∞ < a < b < ∞ one has that f −1 ((a, b)) := {ω ∈ Ω : a < f (ω) < b} ∈ F.