By Geiss

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**Example text**

1) we assume random variables ϕ, ψ : Ω → ❘ such that ϕ ≥ 0 and ψ ≥ 0. We find ∞ measurable step functions (ϕn )∞ n=1 and (ψn )n=1 with 0 ≤ ϕn (ω) ↑ ϕ(ω) and 0 ≤ ψn (ω) ↑ ψ(ω) for all ω ∈ Ω. 3, and ϕn (ω) + ψn (ω) ↑ ϕ(ω) + ψ(ω) give that ❊ϕ + ❊ψ = lim ❊ϕn + lim ❊ψn = lim ❊(ϕn + ψn) = ❊(ϕ + ψ). 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.

K=1 Fubini’s Theorem In this section we consider iterated integrals, as they appear very often in applications, and show in Fubini’s Theorem that integrals with respect to product measures can be written as iterated integrals and that one can change the order of integration in these iterated integrals. In many cases this provides an appropriate tool for the computation of integrals. Before we start with Fubini’s Theorem we need some preparations. First we recall the notion of a vector space. 1 [vector space] A set L equipped with operations + : L × L → L and · : ❘ × L → L is called vector space over ❘ if the following conditions are satisfied: (1) x + y = y + x for all x, y ∈ L.

4 (1) The function g(x) := |x| is convex so that, for any integrable f , |❊f | ≤ ❊|f |. (2) For 1 ≤ p < ∞ the function g(x) := |x|p is convex, so that Jensen’s inequality applied to |f | gives that (❊|f |)p ≤ ❊|f |p . For the second case in the example above there is another way we can go. It ¨ lder-inequality. 5 [Ho (Ω, F, P) and random variables f, g : Ω → ❘. If 1 < p, q < ∞ with p1 + 1q = 1, then ❊|f g| ≤ (❊|f |p) p1 (❊|g|q ) 1q . Proof. We can assume that ❊|f |p > 0 and ❊|g|q > 0. s. s. and ❊|f g| = 0.