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This is a problem from Billingsley's text Probability and measure.

Suppose that $f$ is nonnegative on a $\sigma$-finite measure space $(\Omega, \mathscr{F}, \mu)$. Show that $$\int_\Omega f d \mu = (\mu \times \lambda)[(\omega, y) \in \Omega \times \mathbb{R}^1: 0 \leq y \leq f(\omega)]. \tag{1}$$ Prove that the set on the right is measurable.

My Attempt:

Denote the set on the right by $G$. First, by condition, the product measure $\mu \times \lambda$ exists on the $\sigma$-field $\mathscr{F} \times \mathscr{R}^1$. I am able to show that $G$ is measurable and $(1)$ holds if $f$ are simple functions. It looks naturally that for general nonnegative $f$, let $\{f_n\}$ be a sequence of simple functions such that $f_n \uparrow f$. Then define $G_n = [(\omega, y) \in \Omega \times \mathbb{R}^1: 0 \leq y \leq f_n(\omega)]$ accordingly for every $n$. It is expected that $G_n \uparrow G$ so that the measurability of $G$ follows from the measurability of $G_n$. However, this seems unjustified, since in fact $$G_n \uparrow [(\omega, y) \in \Omega \times \mathbb{R}^1: 0 \leq y \color{red}{<} f(\omega)].$$ So we have to show that $$[(\omega, y) \in \Omega \times \mathbb{R}^1: f(\omega) = y, y \geq 0] = [(\omega, f(\omega)): f(\omega) \geq 0] \tag{2}$$ is measurable. And here is where I got stuck. Can someone show me why $(2)$ is measurable or to show the assertion through other ways? Thank you very much.

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First of all you need to assume that $f$ is $(\cal F, B)$-measurable, where $\cal B$ is the Borel sigma algebra on $\Bbb R$.

Define a map $g: \Omega \times \Bbb R \to \Bbb R$ as $(\omega, y) \mapsto f(\omega)-y$. Note that $g$ is $(\cal F \otimes B, B)$-measurable, since it is the difference of two $(\cal F \otimes B, B)$-measurable functions, namely $h_1(\omega, y) := f(\omega)$ and $h_2(\omega, y) := y$.

It follows that the set $A:=g^{-1}[0,\infty)$ is measurable. Hence $A \cap B$ is measurable, where $B=\Omega \times [0,\infty) \in \cal F \otimes B$. But $A \cap B$ is precisely the set you are trying to prove is measurable.

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  • $\begingroup$ This looks good to me, thank you. But in your way, is it also easy to show that $(1)$ also holds? $\endgroup$ – Zhanxiong Oct 24 '15 at 21:02
  • $\begingroup$ I got it, $(1)$ then follows from the Fubini's theorem. $\endgroup$ – Zhanxiong Oct 24 '15 at 21:05
  • $\begingroup$ Exactly, just let $D$ be that set. And then notice that $f(\omega) = \int_{\Bbb R} 1_D(\omega,y) dy$ and integrate both sides w.r.t $\mu$. Then apply Fubini. $\endgroup$ – Shalop Oct 24 '15 at 21:11

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