By Geiss C., Geiss S.

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S. s. and ❊|f g| = 0. Hence we may set f˜ := f (❊|f |p ) 1 p and g˜ := g (❊|g|q ) q 1 . We notice that xa y b ≤ ax + by for x, y ≥ 0 and positive a, b with a + b = 1, which follows from the concavity of the logarithm (we can assume for a moment that x, y > 0) ln(ax + by) ≥ a ln x + b ln y = ln xa + ln y b = ln xa y b . Setting x := |f˜|p , y := |˜ g |q , a := p1 , and b := 1q , we get 1 1 q g| |f˜g˜| = xa y b ≤ ax + by = |f˜|p + |˜ p q 60 CHAPTER 3. INTEGRATION and ❊|f˜g˜| ≤ p1 ❊|f˜|p + 1q ❊|˜g|q = p1 + 1q = 1.

45 ❘, then af + bg is integrable and Proof. (1) We only consider the case that ❊f + + ❊g + < ∞. Because of (f + g)+ ≤ f + + g + one gets that ❊(f + g)+ < ∞. Moreover, one quickly checks that (f + g)+ + f − + g − = f + + g + + (f + g)− so that ❊f − + ❊g − = ∞ if and only if ❊(f + g)− = ∞ if and only if ❊f + ❊g = ❊(f + g) = −∞. 1) which implies that ❊(f + g) = ❊f + ❊g. 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.

Fn ∈ Bn) = P (f1 ∈ B1) · · · P (fn ∈ Bn) . 9 what does it mean that a sequence of events is independent. Now we rephrase this definition for arbitrary families. 3 [independence of a family of events] Let (Ω, F, P) be a probability space and I be a non-empty index-set. , one has that P (Ai1 ∩ · · · ∩ Ain ) = P (Ai1 ) · · · P (Ain ) . 4 Let (Ω, F, P) be a probability space and fi : Ω → ❘, i ∈ I, be random variables where I is a non-empty index-set. Then the following assertions are equivalent.