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Let $(X,\mathfrak B(X))$ and $(Y,\mathfrak B(Y))$ be two measurable spaces and let $\mu$ be a finite measure on the product $\sigma$-algebra $\mathfrak B(X)\otimes \mathfrak B(Y)$. Let $f:X\to\Bbb R$ be a bounded $\mathfrak B(X)$-measurable function and define $\nu(A) = \mu(A\times Y)$ to be a measure on $\mathfrak B(X)$. How to show that $$ \int_{X\times Y}f(x)\mu(\mathrm dx\times \mathrm dy) = \int_Xf(x)\nu(\mathrm dx). $$ I guess, this is completely trivial, however I couldn't come with a completely formal solution. Perhaps, that shall follow from some change of variables equality and the fact that $\nu$ is a pushforward measure of $\mu$ under the projection $\pi_X$? Or from the fact that $f:X\times Y\to\Bbb R$ is $\mathfrak B(X)\otimes \{\emptyset,Y\}$-measurable function and both measures coincide on this $\sigma$-algebra?

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Can't be it done via a standard argument, i.e. 1) Show that it holds for indicator functions $1_A$ for $A\in\mathfrak{B}(X)$. 2) Show that if it holds for two measurable, bounded functions $f,g$ then it also holds for $f+g$. 3) Show that if it holds for a sequence $(f_n)_{n\geq 1}$ if measurable, bounded functions, then it also holds for $f=\lim_n f_n$. – Stefan Hansen Jan 20 '13 at 11:49
@StefanHansen: like monotone class, or like simple functions arguments? I guess, it shall work but I hoped, it shall be more direct. – Ilya Jan 20 '13 at 11:51
This is a simple functions argument. The 3 steps above shows that it holds for any bounded measurable function. I'll think about the problem, and see if it can be done in a more direct way. – Stefan Hansen Jan 20 '13 at 11:52
@StefanHansen: thanks! – Ilya Jan 20 '13 at 11:55
up vote 1 down vote accepted

Let $\pi_X:X\times Y\to X$ be the projection, i.e. $\pi_X((x,y))=x$ for $(x,y)\in X\times Y$. Then $\nu$ is the image measure of $\mu$ under $\pi_X$, i.e. $$ \nu(A)=\mu(A\times Y)=\mu(\pi_X^{-1}(A)),\quad A\in\mathfrak{B}(X). $$ Integration with respect to $\nu$ can be described via integration with respect to $\mu$ (but this result relies on a standard argument as described in the comments, see e.g. this answer): If $f$ is bounded and $\mathfrak{B}(X)$-meaurable, then $$ \int_X f(x)\,\nu(\mathrm dx)=\int_X f\,\mathrm d\nu=\int_{X\times Y}f\circ\pi_X\,\mathrm d\mu=\int_{X\times Y}f(x)\,\mu(\mathrm dx,\mathrm dy). $$

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Oh, so one shall go from the RHS to the LHS in the equality in OP. That's nice! Do you have any book source for the pass from $\nu$ to $\mu$ via the projection? – Ilya Jan 20 '13 at 12:05
My immediate thought is that you can't go from the LHS to the RHS, because you can't describe $\mu$ uniquely by $\nu$ (i.e. $\pi_X$ has no inverse function). I'm not sure if you can do something clever to be able to go from the LHS to the RHS though. You can take a look at Measures, Integrals and Martingales by R. L. Schilling. He treats the idea of integration with respect to image measures. Not sure if he treats the explicit case where the mapping is the projection though. – Stefan Hansen Jan 20 '13 at 12:08
I see, but that doesn't seem to be necessary - I liked the way you presented the solution. I'm interested though in some book sources on the passing from one integral to another in a general form given for a function $\varphi$ in your linked post - are you aware of any? Most of these results are not hard to proof, but sometimes it's better just to have a proper reference. – Ilya Jan 20 '13 at 12:14
Yes, chapter 14 in the before-mentioned book (maybe I was unclear, but it treats exactly the kind of passing from one integral to another as described in $(*)$ in the linked answer). – Stefan Hansen Jan 20 '13 at 12:18
@Ilya A change of variable result that one can use here is given as Theorem 3.6.1 in Bogachev. – Michael Greinecker Jan 20 '13 at 15:34

For a rigorous proof, the question has to be slightly modified since $f$ is not defined on $X\times Y$. So let $\tilde{f}:X\times Y\to\mathbb{R}$. we want to show that $$\int_{X\times Y}\tilde{f}(x)\mu(\mathrm dx\times \mathrm dy) = \int_Xf(x)\nu(\mathrm dx).$$ Now, two functions with the same distribution have the same integral. This is obviously satisfied here, since $\mu\big(\tilde{f}^{-1}(B)\big)=\mu \big(f^{-1}(B)\times Y\big)=\nu\big(f^{-1}(B)\big)$.

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Yeah, you're right. I made the identification that $\int_{X\times Y} f\,\mathrm d\mu$ should mean $\int_{X\times Y}f\circ \pi_X\,\mathrm d\mu$ – Stefan Hansen Jan 20 '13 at 15:51

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