Why are the conditions of Fubini's theorem satisfied for probability measures, provided that the function is measurable and bounded above or below? More specifically, Let $\mu$ be a probability measure on $\mathcal{X}$ and $\nu$ a probability measure on $\mathcal{Y}$, and let $\mu\times\nu$ be product measure on $\mathcal{X}\times\mathcal{Y}$. If $f:\mathcal{X}\times\mathcal{Y}\rightarrow\mathbb{R}$ is measurable with respect to $\mu\times\nu$, then $\int_{\mathcal{X}\times\mathcal{Y}}fd(\mu\times\nu)=\int_{\mathcal{X}}\left(\int_{\mathcal{Y}}f(x,y)\nu(dy)\right)\mu(dx)=\int_{\mathcal{Y}}\left(\int_{\mathcal{X}}f(x,y)\mu(dx)\right)\nu(dy)$ provided that $f\geq C>-\infty$ or $f\leq C<\infty.$
Fubini's Theorem and bounded functions
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0Sure, I added more details. Sorry about that. – 2012-03-30
1 Answers
Below is my favorite version of the Fubini-Tonelli Theorem. It's almost straight out of Rudin's "Real and Complex Analysis." I believe it answers your question because all that is assumed in the first part is that the function is non-negative and measurable. If your function is bounded below, then of course you can simply add a large enough constant and make it non-negative. (I've added a more precise statement of this below.)
Added
If you take this version as "The Fubini Theorem," then the solution to your problem would be as follows:
"If $f\geq C>-\infty$, then the function $g=f+C$ satisfies the hypotheses of the Fubini Theorem. Therefore, the conclusion holds for $g$, so it holds for $f$ as well."
Follow up question: Does the claim "so it holds for $f$ as well" depend on your measure spaces being finite (e.g. probability spaces)?