measure theory - The infimum of a measurable $f:Xtimes Yto[-infty,infty]$ over $X$ is measurable $bar f:Yto[-infty,infty]$

Let $(X,\mathscr M)$ and $(Y,\mathscr N)$ be measurable spaces, and let $f:X\times Y\to[-\infty,\infty]$ be some measurable function. Then we may define $\bar f:Y\to[-\infty,\infty]$ by $$\bar f(y)=\inf_{x\in X}f(x,y)$$ but is this function $\mathscr N$-measurable? I suspect the answer is yes, though I don't see why.

Edit: Note that the statement that the projection map $\pi:X\times Y\to Y$ takes measurable sets to measurable sets is false.

Answer

The statement is in general false: Let $X= Y= \mathbb{R}$ and $\mathcal{M}$ the Lebesgue-$\sigma$-algebra and $\mathcal{N}$ the Borel-$\sigma$-algebra. Now take any set $E \subset \mathbb{R}$ which is Lebesgue-measurable, but not Borel.

Define $$f(x,y) = \begin{cases} 0 & \text{if } x=y \\ 1 & \text{else} \end{cases}.$$
Of course, this map is measurable according to $\mathcal{M} \otimes \mathcal{N}$, because $\{(x,y) \in \mathbb{R} : x=y\}$ is closed and thus a Borel-set. On the other side $$\inf_{x \in E} f(x,y) = 1-1_{E}(y)$$
is not Borel-measurable.