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**A First Course in Infinitesimal Calculus**

This booklet was once digitized and reprinted from the collections of the college of California Libraries. It used to be made out of electronic photos created during the libraries’ mass digitization efforts. The electronic photos have been wiped clean and ready for printing via computerized tactics. regardless of the cleansing strategy, occasional flaws should still be current that have been a part of the unique paintings itself, or brought in the course of digitization.

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**Example text**

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.