Let $(X,Y)$ and $Z$ be independent random variables. Show $E(X|Y) = E(X|Y,Z)$. No other information is given.
First off, if $(X,Y)$ and $Z$ are independent, are $X$ and $Z$ (and $Y$ and $Z$) also independent?
To show $E(X|Y) = E(X|Y,Z)$, I only considered measurable sets of the form $F \cap G$ for $F \in \sigma(Y)$, $G \in \sigma(Z)$, and it's not quite clear that this exhausts $\sigma(Y,Z)$. In any case, conditional on this, I argued as follows:
$E(E(X|Y)\chi_{F\cap G}) = E(E(X\chi_{F\cap G}|Y)) = E(E(X\chi_{F}\chi_{G}|Y)) = P(G)E(E(X|Y)\chi_{F}) = P(G)E(X\chi_{F})$
and
$E(E(X|Y,Z)\chi_{F\cap G}) = E(X\chi_{F\cap G}) = E(X\chi_{F}\chi_{G}) = P(G)E(X\chi_{F})$
Comments are welcome.
I suppose your random variables are $\mathbb{R}$-valued.
If $(X,Y)$ and $Z$ are independent, then $X$(respectively $Y$) and $Z$ are independent. This is because
\begin{align} P(X\in A, Z \in B) &= P((X,Y) \in A \times \mathbb{R}, Z \in B) \\ &= P((X,Y) \in A \times \mathbb{R})P(Z \in B)\\ &= P(X \in A)P(Z \in B) \end{align}
To show $E(X|Y) = E(X|Y,Z)$, since $E(X|Y)$ is $\sigma(Y)$-measurable thus $\sigma(Y, Z)$-measurable. we only need to show for any measurable and bounded $f : \mathbb{R}^2 \to \mathbb{R}$, $$E(E(X|Y)f(Y,Z)) = E(Xf(Y,Z))$$
Similarly to what you wrote, we can prove that for $f \in S =\{\chi_{F\times G} : F \text{ and } G \text{ are measurable subsets of } \mathbb{R}\}$, since \begin{align} E(E(X|Y)\chi_{F\times G}(Y, Z)) &= E(E(X\chi_{F}(Y)|Y)\chi_{G}(Z)) \\ &= E(E(X\chi_{F}(Y)|Y))E(\chi_{G}(Z)) \\ &= E(X\chi_{F}(Y))E(\chi_{G}(Z)) \\ &= E(X\chi_{F}(Y)\chi_{G}(Z)) \end{align}
Then we can use the functional version of monotone class theorem to conclude.