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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2What version of the theorem do you mean specifically? – 2012-03-30
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0Perhaps you're thinking of Tonelli's Theorem. In that case, it's clear why the boundedness assumption suffices. If the function is bounded below, you can assume without loss of generality you're dealing with a positive function. – 2012-03-30
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0Or Fubini's theorem for probability measures. A bounded measurable function on a probability space is automatically integrable. – 2012-03-30
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1@MichaelGreinecker Actually, OP isn't assuming bounded -- only bounded above *or* bounded below. (I think my comment mislead you. I shouldn't have called it a "boundedness assumption." Maybe a "bounded-below-ness assumption" would have been better. :) – 2012-03-30
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0Yes, my question is about probability measures (I have edited the question). – 2012-03-30
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0Though it still holds for sigma-finite measures. – 2012-03-30
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0I agree with Michael -- you really need to specify exactly which version of the theorem you're talking about. There are many different variations. (Imho, by far the most useful is the one in mama Rudin, which is a combined Fubini-Tonelli theorem that only assumes $f(x,y)$ is measurable with respect to the product $\sigma$-algebra. No integrability is assumed.) – 2012-03-30
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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)?