I have $E(X|Y,Z)=0$, $X$ independent of $Y$ and of $Z$ and I want to conclude that $E(X)=0$ ($X,Y,Z$ are real-valued random variables). Okay it seems quite obvious, but if I try to make a strict argument I find a missing step.
I tried to approach this from both sides:
From the independence follows, that $X$ is independent of $\sigma(Y) \cup \sigma(Z)$.
The first expression is, more exactly $=E(X| \sigma(Y,Z))$. $\sigma(Y,Z)=\sigma(\sigma(Y) \cup \sigma(Z))$.
It is known, that in general $\sigma(\mathcal{A} \cup \mathcal{B}) \neq \mathcal{A} \cup \mathcal{B} (*)$ for $\sigma$-algebras $\mathcal{A}, \mathcal{B}$. Question: How can I conclude $E(X)=0$ anyway?
Maybe for independent random variables the equality $(*)$ does hold? Or there is m maybe an alternative, more convenient expression for conditional expectations on more than one random variables?