# Fubini's Theorem and bounded functions

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 $f\leq C<\infty.$

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What version of the theorem do you mean specifically? –  Michael Greinecker Mar 30 '12 at 14:30
Perhaps 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. –  William DeMeo Mar 30 '12 at 14:41
Or Fubini's theorem for probability measures. A bounded measurable function on a probability space is automatically integrable. –  Michael Greinecker Mar 30 '12 at 14:45
@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. :) –  William DeMeo Mar 30 '12 at 14:50
Yes, my question is about probability measures (I have edited the question). –  V.A. Mar 30 '12 at 14:57

"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)?