Exercise 1.4.36 (Linearity in measure)

Let (X,,μ) be a measure space, and let f : X [0,+] be measurable. Show that

(i)
For every c [0,+] we have Xfd() = c ×Xfdμ.

(ii)
If (μ)n is a sequence of measures on (X,) then Xfd i=1μ n = i=1Xfdμn.

Answers

We have to demonstrate that

sup {SimpXhd ( i=1cμ n) : h is simple h f } = c× i=1sup {SimpXhdμ : h is simple h f }

An arbitrary simple integral with respect to the linear combination of measures is contained in the set on the left-hand side

SimpXhd ( i=1cμ n) = i=1ka i ( i=1cμ n(f1(a i))) = c i=1 i=1ka iμn(f1(a i)) = c i=1SimpXhdμn

if and only if its product with 1c and each summand is contained in the set on the right-hand side.

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2020-07-20 00:00
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