Exercise 1.4.48 (Measures induced by measurable functions)

$$ The function ${\mu }_g$ is non-negative, since $g$ is unsigned. We verify the measure axioms (Definition 1.4.27).

(i)

We have ${\mu }_g(\varnothing )=1_{\varnothing }g=0=0$.

(ii)

Let $E_1,E_2,\dots \in B$ be a countable sequence of disjoint measurable sets. Then $${\mu }_g\left(\bigcup _{n=1}^{\infty }{E}_n\right)=\left(1_{\bigcup _{n=1}^{\infty }{E}_n}\times g\right)=\left(\sum _{n=1}^{\infty }{1}_{E_n}g\right)=\sum _{n=1}^{\infty }\left(1_{E_n}g\right)=\sum _{n=1}^{\infty }{\mu }_g(E_n)$$

since for disjoint sets ${(E_n)}_{n\in \mathbb{N}}$ the indicator function is additive with respect to the union. The crucial step, exchanging sums and integrals, follows by Tonelli’s theorem for sums and integrals (Corollary 1.4.45).