I'm studying measure theory, and I came across a theorem that says that, given two $\sigma$-finite measure spaces $\left(X,\mathcal M,\mu\right)$ and $\left(Y,\mathcal N,\nu\right)$, and a set $E\in \mathcal M\otimes\mathcal N$, the function $x\mapsto \nu\left(E_x\right)$ is measurable.
The definition of the product-$\sigma$-algebra $\mathcal M\otimes\mathcal N$ is the $\sigma$-algebra generated by the rectangles $A\times B$ with $A\in\mathcal M$ and $B\in\mathcal N$. Also, the $x$-section $E_x$ is defined as $E_x=\left\{y\in Y:\left(x,y\right)\in E\right\}$.
My problem is that I can't even convince myself that, given $E\in\mathcal M\otimes \mathcal N$, the sections $E_x$ are in $\mathcal N$ for all $x$. It looks like it has something to do with unitary sets being measurable (which shouldn't be necessary, I think), but I can't prove it even assuming that $\left\{x\right\}\in\mathcal M$ for all $x$.
Best Answer
Since the map $u_{x}:Y\to X\times Y$ given by $u_x(y)=(x,y)$ for a given $x$ is $\mathcal{N}$-$(\mathcal{M}\otimes\mathcal{N})$-measurable and $E_x=u_x^{-1}(E)$ it follows directly that $E_x\in\mathcal{N}$ for any $x$.
To convince yourself that $u_x$ is indeed measurable, just note that $$ u_x^{-1}(A\times B)= \begin{cases} B,\quad &\text{if }x\in A,\\ \varnothing, &\text{if } x\notin A, \end{cases} $$ which is in $\mathcal{N}$ for any $A\in\mathcal{M}$ and $B\in\mathcal{N}$.