Integration a function of a single variable over a 2-dim measure

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?

Answer

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).
$$