Let $(\Omega_1,\Sigma_1,\mu_1)$ and $(\Omega_2,\Sigma_2,\mu_2)$ be two totally finite measure spaces (which implies that $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras).

(As usual $\Sigma_1\times\Sigma_2$ will mean the $\sigma$-algebra generated by the set $\{ E \times F \,|\, E \in \Sigma_1 \textrm{ and } F \in \Sigma_2\}$)

Let $\mathcal{F}$ be the class of all sets $A$ such that $A\subseteq \Omega_1\times \Omega_2$ and, for all $R\in \{ E \times F \,|\, E \in \Sigma_1 \textrm{ and } F \in \Sigma_2\}$,

  1. for almost all $x\in \Omega_1$, $(A\cap R)_x=\{y\in\Omega_2 \,|\, (x,y)\in A\cap R \} \in \Sigma_2$ and, the function $x \mapsto \mu_2((A\cap R)_x)$ is $\Sigma_1$-measurable.

  2. for almost all $y\in \Omega_2$, $(A\cap R)^y=\{x\in\Omega_1 \,|\, (x,y)\in A\cap R \} \in \Sigma_1$ and the function $y \mapsto \mu_1((A\cap R)^y)$ is $\Sigma_2$-measurable.

  3. $\int_{\Omega_1} \mu_2((A\cap R)_x) d\mu_1 = \int_{\Omega_2} \mu_1((A\cap R)^y) d\mu_2$

(Note that $A$ is not supposed to be $\Sigma_1\times\Sigma_2$- measurable)

Can we prove that $\mathcal{F}$ is closed under intersection?


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