By Daniel A. Murray

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Proof Fix n > 0 and let Xn = X ∧ n. Then Xn is bounded, hence in Lp . If we have proved our inequality for Y and Xn , the MCT shows that it also holds for Y and X = limn↑∞ Xn . We can thus take X ∈ Lp without loss. z Since for any z ≥ 0, p {z≥x} xp−1 dx = p 0 xp−1 dx = z p , we have, integrating z = X(ω), over Ω and using Fubini X(ω) X p dP = Ω xp−1 dx P (dω) p Ω 0 ∞ =p 0 ∞ =p 0 Ω 1{X(ω)≥x} P (dω) xp−1 dx xp−1 P (X ≥ x)dx. But xP (X ≥ x) ≤ {X≥x} Y dP by hypothesis, so (using Fubini again) the RHS is no greater than ∞ p ∞ xp−2 Y dP dx = p 0 With 1 p {X≥x} + 1 q Y Ω 0 xp−2 1{X≥x} dxdP.

E f dμ is Proof If a, b ∈ R and f, g ∈ L1 , then |af + bg| ≤ |a| |f | + |b| |g|, so that E |af + bg| dμ is finite. Hence L1 is a vector space. 15 the integral is positive-linear on M+ (F), and consider the difference of h1 , h2 in M+ (F): we have h1 + (h1 − h2 )− = (h1 − h2 )+ + h2 and both sides are in M+ (F), so that we can apply the additivity of the integral on M+ (F) to conclude that (h1 − h2 )− dμ = h1 dμ + E E (h1 − h2 )+ dμ + E h2 dμ. E Rearrange to obtain (h1 − h2 )dμ = E h1 dμ − E h2 dμ.

F For the opposite inequality, take simple functions φ ≤ f on E and ψ ≤ f on F to construct a simple function θ = φ1E + ψ1F ≤ f on E ∪ F , so that φdμ + E θdμ ≤ ψdμ = F E∪F f dμ. E∪F The RHS is an upper bound independent of the choices of φ, ψ. Fix ψ and take the sup over Y (E, f ) then take the sup over Y (F, f ) to conclude that f dμ ≥ E∪F f dμ + E f dμ. 8 On [0, 1] with Lebesgue measure m, define the function f by setting f = 0 on the Cantor set C and for each k ≥ 1, let f (x) = k on each of the intervals of length 3−k removed from [0, 1] in the construction of C.

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