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Recall the result that for a nonnegative random variable $X$ on $(\Omega, \mathcal{F}, P)$, $$ E[X] = \int_0^\infty (1 - F(x)) dx, $$ where $F$ is the cdf of $X$. In many of the proofs I've seen for this we rely on Tonelli's theorem in the step $$ \int_\Omega \int_0^\infty \mathbf{I}_{[0, X(\omega))}(x) \mathrm{d}x \mathrm{d}P(\omega) = \int_0^\infty \int_\Omega \mathbf{I}_{[0, X(\omega))}(x) \mathrm{d}P(\omega) \mathrm{d}x. $$ Tonelli's theorem, however, requires that $\mathbf{I}_{[0, X(\omega))}(x)$ be $\mathcal{F} \otimes \mathcal{B}$-measurable where $\mathcal{B}$ is the Borel sigma-algebra on $\mathbb{R}$. I've spent quite a while trying to show this is the case here, but have come to believe I'm not equipped to handle showing measurability on product sigma algebras. Would anyone be able to demonstrate a method to show this?

As an aside, it seems that measurability is often implicitly assumed in applications of Tonelli and Fubini, and rather, the nonnegativity or integrability (which also implicitly assumes measurability) is focused on. Is it just that measurability is "so easy" to get that it's okay to sort of neglect?

bcf
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2 Answers2

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This might help you to handle measurability on product spaces To get the measurability consider first specific functions

$$f(x,y)= 1_A(x)1_B(y) $$

$x \mapsto \int_Y f(x,y)\, d\nu(y)= \nu(B)1_A(x)$

This is a measurable function of the first variable.

Now consider $\mathcal{C}= \big\{D \in X \times Y; x \mapsto \int_Y1_D(x,y)\, d\nu(y) \text{is measurable on } x \big\}$

note that

1) $\Omega \in \mathcal{C}$

2) $D \in \mathcal{C} \Rightarrow D^c \in \mathcal{C}$ since

$$x \mapsto \int_Y1_{D^c}(x,y)\, d\nu(y) = 1 - \int_Y1_D(x,y)\, d\nu(y)$$ is measurable on $x$

3) $D_1 \subset D_2 \subset \ldots \in \mathcal{C} \Rightarrow \cup_i D_i \in \mathcal{C}$ since

$$x \mapsto \int_Y1_{\cup_iD_i}(x,y)\, d\nu(y) = \lim_n \int_Y1_{D_n}(x,y)\, d\nu(y) $$ is measurable on $x$

To conclude note that $\mathcal{C}$ is a monotone class that contains the rectangles $A \times B$ so it contains the $\sigma$- algebra generated by the rectangles, that is the product $\sigma$- algebra.

$ f: \Omega \times \Bbb{R} \to \Bbb{R}$ $f(\omega, x) = \mathbf{I}_{[0, X(\omega))}(x) = \mathbf{I}_{[0, \infty)}(X(\omega)-x) $ note that it is the composition of measurable functions:

$(\omega , x) \mapsto (X(\omega), x) \mapsto (X(\omega)-x)$ and

$x \mapsto \mathbf{I}_{[0, \infty)}(X(\omega)-x) $

Conrado Costa
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Consider the first projection map $\pi_1: \Bbb R \times \Omega \to \Bbb R$. This map is measurable by the definition of the product sigma algebra $\mathcal B \otimes \mathcal F$.

Similarly, the second projection map $\pi_2: \Bbb R \times \Omega \to \Omega$ is measurable. Thus $X \circ \pi_2$ is measurable as a map from $\Bbb R \times \Omega \to \Bbb R$, since it is the composition of two measurable maps.

Therefore the map $Y:= \pi_1-X \circ \pi_2$ is measurable as a map from $\Bbb R \times \Omega \to \Bbb R$, since differences of measurable mappings are measurable.

It follows that $A:=Y^{-1}((-\infty,0)) \cap \pi_1^{-1}([0,\infty)) \in B \otimes \mathcal F$. Therefore, the map $(x,\omega) \mapsto I_{[0,X(\omega))}(x)$ is measurable, since it is the characteristic function of the measurable set $A$.

shalop
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