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 w...
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?
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.
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.
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).
Somehow I've seen before only the 2nd edition of your previous comment, not the last one - sorry. The case when we have a projection I guess is a special case, so of course it's not necessary to have an explicit reference to that particular instance
Dudley's "Real Analysis and Probability" have very little about this kind of integration. Just stating the theorem and giving a 5 line proof using that standard argument
@StefanHansen that's true. In our case that were lecture notes of the lecturer, as dry as you cannot even imagine. I learnt only later that being formal is not necessary being dry
@Stefan: I don't have Dudley now - does he discuss it for general pushfowards?
there are a lot of nice discussions (mostly non-math) going on there and nice people hanging out
@StefanHansen: and finally, your comment here shall be an answer. I think, you got the question in a right way - though maybe it's worth waiting for clarifications
Discussion between Ilya and Stefan Ha…
Discussion between Ilya and Stefan Ha…