Let $(E,\mathcal{E})$, $(\Omega,\mathcal{F})$ and $(G,\mathcal{G})$ be measurable spaces and consider a measurable function $X:(E\times \Omega, \,\mathcal{E}\otimes\mathcal{F})\to (G,\,\mathcal{G})$ such that X depends only on its second argument, i.e. $X(y,\omega)=X(\omega),\forall y\in E\,\forall \omega\in\Omega$.
Fix now $y\in E$ and define $\tilde{X}:\Omega\to G,\,\omega\mapsto X(y,\omega)$.
Show that $\tilde{X}$ is measurable wrt. $\mathcal{E}$.
My attempt: Fix $A\in\mathcal{G}$ then $\tilde{X}^{-1}(a)=\{\omega\in\Omega:X(y,\omega)\in A\}$. We know that $X^{-1}(A)\in \mathcal{E}\otimes\mathcal{F}$. I have then tried writing $\tilde{X}^{-1}(A)=X^{-1}(A)\cap\{ (x,\omega)\in E\times\Omega: x=y\}$ but that's where I got stuck. I would somehow like to go from general elements of $\mathcal{E}\otimes \mathcal{F}$ to those of $\mathcal{E}\times \mathcal{F}$ in order to conclude.
Check that $E \times \overline X^{-1} (A) =X^{-1} (A)$. Can you finish?