Measurability of integral

Question

Consider a function $f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ which is continuous in the first argument, measurable in the second.

Let $m: \mathcal{B}(\mathbb{R}^m) \rightarrow [0,1]$ be a finite measure.

I am wondering if the function $F: \mathbb{R}^n \rightarrow [0,1]$ defined as $$ F(x) := m\left( \{ y \in \mathbb{R}^m \mid f(x,y) \leq 0 \} \right) $$ is measurable.

What I tried to do is to claim that $F$ is upper semicontinuous. This should imply measurability.

Michael Greinecker · Accepted Answer · 2013-09-13T12:25:39.900

2

The function $f$ is a Caratheodory function and must therefore be jointly measurable, so $A=f^{-1}\big((-\infty,0]\big)$ is a measurable subset of $\mathbb{R}^n\times\mathbb{R}^m$. Let $\nu$ be some finite measure on $\mathbb{R}^n$. The Borel-$\sigma$-algebras have the property that $\mathcal{B}(\mathbb{R}^n\times\mathbb{R}^m)=\mathcal{B}(\mathbb{R}^n)\otimes\mathcal{B}(\mathbb{R}^m)$. Let $1_A$ be the indicator function of $A$. By Fubini's theorem, $$\nu\otimes m(A)=\int_{\mathbb{R}^n}\int_{\mathbb{R}^m} 1_A(x,y)~dm(y)~d\nu(x) $$ and the function $x\mapsto\int_{\mathbb{R}^m} 1_A(x,y)~dm(y)$ is measurable. But $\int_{\mathbb{R}^m} 1_A(x,y)~dm(y)=F(x)$.

Of course, a more direct argument could be made using the essential parts of the proof of Fubini's theorem.

edited Sep 13 '13 at 12:25

answered Sep 13 '13 at 12:11

Michael Greinecker

34,865

Thanks for the comment. If we instead show that $F$ is upper semicontinuous, does this imply measurability? In this way we would not assume joint measurability. – user693 Sep 13 '13 at 12:19
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@Adam, sorry if this was not clear: A function measurable in one argument and continuous in the other is always jointly measurable. But yes, usc functions are measurable. – Michael Greinecker Sep 13 '13 at 12:24
Thanks again. By the way, is it true that $F$ is upper semicontinuous? – user693 Sep 13 '13 at 12:28
I actually don't know. – Michael Greinecker Sep 13 '13 at 12:32
A follow up. To claim joint measurability, I found that it is enough to have $f$ being a normal integrand, which is in general less strict than assuming $f$ to be Carathéodory. – user693 Sep 13 '13 at 12:51
Finally, where can I find a reference stating that analytic functions are measurable? – user693 Sep 13 '13 at 13:52
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The classic reference is Dellacherie and Meyer 1978, Probability and Potential – Michael Greinecker Sep 13 '13 at 14:28

Measurability of integral

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