# Combining product of conditional expectations

Let $X,Y,Z$ be real-valued random variables on some probability space $(\Omega,\mathcal{F}, P).$ Then, I am stuck on proving a seemingly simple equation: $$E\Big[1_{\{X\in A\}}|\sigma\{Y,Z\}\Big] \cdot E\Big[1_{\{Y\in B\}}|\sigma\{Z\}\Big] =E\Big[1_{\{X\in A, Y\in B\}}|\sigma\{Z\}\Big]$$ for every borel $A,B\subseteq\mathbb{R}.$

It seems to follow directly from the tower property of conditional expectations, if we could prove that $$1_{\{X\in A\}}\cdot E\Big[1_{\{Y\in B\}}|\sigma\{Z\}\Big] = E\Big[1_{\{X\in A, Y\in B\}}|\sigma\{Z\}\Big].$$

However, $1_{\{X\in A\}}$ need not be $\sigma\{Z\}$ measurable and thus the above is not necessarily true. Maybe I am failing to see something elementary, but I would appreciate any hint.

The statement is false. Take X=Y and Z independent of {X,Y}. Then $E[1_{\{X \in A\}} | \sigma \{Y,Z\}]=1_{\{X \in A\}}$, $E[1_{\{Y \in B \}} | \sigma \{Z\}]=P\{Y \in B\}$ and the right side is $P\{ X \in A \cap B\}$
Well, the strategy would be to extract the $Y{\in}B$ indicator function using the law of total expectation , but...
$${\quad\mathsf E(\mathbf 1_{X\in A, Y\in B}\mid Z)\\ = \mathsf E(\mathsf E(\mathbf 1_{X\in A}\cdot\mathbf 1_{Y\in B}\mid Y,Z)\mid Z)\\ = \mathsf E(\mathsf E(\mathbf 1_{X\in A}\mid Y,Z)\cdot\mathbf 1_{Y\in B}\mid Z)\hspace{10ex}\star\\\vdots \\ \neq \mathsf E(\mathbf 1_{X\in A}\mid Y,Z)\cdot\mathsf E(\mathbf 1_{Y\in B}\mid Z) }$$
Are you sure the target is not $\mathsf E(\mathsf E(\mathbf 1_{X\in A}\mid Y,Z)\cdot\mathbf 1_{Y\in B}\mid Z)$ ?
Because there is no reasonable way to claim $\mathsf E(g(Y,Z)h(Y)\mid Z) = g(Y,Z)\mathsf E(h(Y)\mid Z)$ would generally hold.