Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
- $(E_i,\mathcal E_i)$ be a measurable space
- $X_i$ be an $(E_i,\mathcal E_i)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$
- $f:E_1\times E_2\to E_3$ be $(\mathcal E_1\otimes\mathcal E_2,\mathcal E_3)$-measurable
- $X_3:=f(X_1,X_2)$
Assuming $X_2$ is independent of $\mathcal F$, are we able to show that $X_3$ is conditionally independent of $\mathcal F$ given $X_1$, i.e. $$\operatorname P\left[X_3\in B_3,F\mid X_1\right]=\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\;\;\;\text{almost surely}\tag1$$ for all $B_3\in\mathcal E_3$ and $F\in\mathcal F$?
Let $B_3\in\mathcal E_3$ and $F\in\mathcal F$. We need to prove that $$\operatorname P\left[X_1\in B_1,X_3\in B_3,F\right]=\operatorname E\left[1_{\{\:X_1\:\in\:A\:\}}\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\right]\tag2.$$ What's the easiest way to show $(2)$? Maybe we are able to reduce the problem to the case $f^{-1}(B_3)=A_1\times A_2$ for some $A_i\in\mathcal E_i$, but I'm missing the right argument for that.
EDIT: If necessary, feel free to impose a stronger notion of measurability on $f$.