Conditional Expectation for independent sigma-algebras

Question

Let $X,Y,Z$ be random variables such that $\sigma(X,Y)$ is independent of $Z$.

How can I prove that for any measurable $A \subset \mathbb{R}$ we have: $$E[\mathbb{1}_{A}(X)|\sigma(Y,Z)] = E[\mathbb{1}_{A}(X)|\sigma(Y)]$$

This seems clear intuitively, but I have failed to find a formal proof.

Answer 1

Note that $\sigma(Y,Z)$ is generated by $\{A\cap B: A\in\sigma(Y),B\in\sigma(Z)\}$. Also for any $A\in\sigma(Y)$ and $B\in\sigma(Z)$ one has $$ \mathsf{E}[f(X) 1_{A\cap B}]=\mathsf{E}[f(X) 1_A] \mathsf{P}(B) =\mathsf{E}[\mathsf{E}[f(X)\mid \sigma(Y)] 1_{A\cap B}], $$ where $f$ is any suitable function.