I'm struggling with the proof for this:

Let $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ be two measurable spaces and recall the product $\sigma$-field $\mathcal{X}\otimes \mathcal{Y}$. Let $f:X\times Y\to \mathbb{R}$ be $\mathcal{X}\otimes \mathcal{Y}$-measurable function. Show that for all $x\in X$ the $f_x:Y\to \mathbb{R},f_x(y)=f(x,y)$ is $\mathcal{Y}$-measurable

My attempt, the preimage of $f_x$ is the set $M_x=\{y\in Y:f(x,y)\in X\times Y\}$, if we have shown that $M_x\in \mathcal{Y}$ then we are done, since the image of $f_x$ is equal to $f(x,y)$ which is in the product $\sigma$-Algebra.

My problem is for me it seems, that the Statement $M_x\in \mathcal{Y}$ is already coming from the Definition of $\mathcal{Y}$ since $\{y\in Y\}\in \sigma(Y)$.


Here's a sketch to get you started. You should try to fill in the details.

It suffices to show that for every $A \in \mathcal{X}\otimes \mathcal{Y}$ the set $A_x = \{y \in Y: (x,y) \in A \}$ is in $\mathcal{Y}$. And for this it suffices to show that the class $\mathcal{E}$ of all $A \in \mathcal{X}\otimes \mathcal{Y}$ for which this claim holds is a sigma-algebra that contains the measurable rectangles $B \times C$ with $B \in \mathcal{X}$, $C \in \mathcal{Y}$. It's clear that $\mathcal{E}$ contains the measurable rectangles. To see that it's a sigma-albegra note that the $(A_x)^c = (A^c)_x$ and similarly for countable unions.


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