Let $(X,\mathcal X, \eta)$ and $(Y,\mathcal Y, \nu)$ be measure spaces and denote $(X\times Y, \mathcal X \otimes \mathcal Y, \eta \times \nu)$ be the usual product measure space. I assume that they are NOT $\sigma$-finite measure spaces.
Let $E \in \mathcal X \otimes \mathcal Y$. I would like to prove that the map $x\mapsto \nu(E_x)$ is measurable (from $X$ to $[0,\infty]$), where $E_x:= \{y \in Y: (x,y) \in E \}$ is a section set which I know is measurable. In the $\sigma$-finite measure case, I know how to do it using Monotone Class Theorem.
Question: Is is still true that $x\mapsto \nu(E_x)$ is still measurable in general measure spaces? If not, could we provide a counter-example?
Any hints are highly appreciated since I don't know where to start.
Best Answer
You asked: Is it still true that $x\mapsto \nu(E_x)$ is still measurable in general measure spaces?
Answer: No. Here is a counterexample.
Consider $([0,1],\Sigma, \lambda)$ where $\Sigma$ is Lebesgue $\sigma$-algebra and $\lambda$ is the Lebesgue measure. Let $V \subseteq [0,1]$ be a set that is NOT Lebesgue measurable. Consider also $([0,1],\Sigma, \nu)$ where $\nu$ is the measure defined by $\nu(\{a\}) = 1$ if $a\notin V$ and $\nu(\{a\}) = 2$ if $a\in V$. Clearly, $\nu$ is not $\sigma$-finite.
Let $E = \{(x,x) : x \in [0,1]\}$, the diagonal of $[0,1] \times [0,1]$. Clearly $E \in \Sigma \otimes \Sigma$ and, we have that, each section $E_x:= \{y \in Y: (x,y) \in E \}$ is measurable.
However, the map $x\mapsto \nu(E_x)$ (from $[0,1]$ to $[0,\infty]$) is NOT measurable. This maps coincides with $\chi_V +1$ and we know $\chi_V$ is not (Lebesgue) measurable ($\chi_V$ is the indicator function of $V$).