Exercise 3, § 8

Let T be a locally compact space, $\mu$ a positive measure on T, $f$ a numerical function $\geq 0$ defined on T. In the product space $T \times \mathbf{R}$, denote by $D_f$ the set of points $(t,x)$ such that $0 \leq x \leq f(t)$; on the other hand let $\nu$ be Lebesgue measure on $\mathbf{R}$.

a) Show that for $f$ to be $\mu$-measurable, it is necessary and sufficient that $D_f$ be a measurable set for the product measure $\lambda = \mu \otimes \nu$. (To see that the condition is necessary, prove that if $f$ is $\mu$-measurable, then for every compact subset K of T, $D_f \cap (K \times \mathbf{R}_+)$ is the union of a $\lambda$-negligible set and a countable family of sets of the form $A \times I$, where A is $\mu$-measurable and I is an interval of $\mathbf{R}_+$. To show that the condition is sufficient, prove that if it is satisfied then, for every compact subset K of T, there exists a dense set $H \subset \mathbf{R}_+$ such that, for every $\alpha \in H$, $K \cap f([\alpha,+\infty])$ is $\mu$-measurable, using Cor. 2 of Prop. 7.)

b) Show that, for $f$ to be $\mu$-integrable, it is necessary and sufficient that $D_f$ be $\lambda$-integrable, and one then has $\lambda(D_f) = \int f d\mu$. Moreover, denoting by $g$ the decreasing numerical function on $\mathbf{R}_+$ defined by $g(t) = \mu(f([t,+\infty]))$ (possibly equal to $+\infty$ for $t = 0$ (Ch. IV, §5, Exer. 29)), one has $\int f d\mu = \int_0^{+\infty} g(t) dt$.