Let $f :\mathbb R^{2} \to \mathbb R$ be bounded an measurable, and $\mathcal{G}\subseteq \mathcal{B}(\mathbb R^{2})$ a sub-sigma-algebra such that random variable $X$ is $\mathcal{G}$ measurable while $Y$ is independent of $\mathcal{G}$.
Show that $E[f(X,Y)\lvert \mathcal{G}]= E[f(X,Y)]$ a.s.
It is clear that we will have to use elementary tools such as the Dynkin Lemma as well as the simple functions but there is an issue that confuses me. $X$ is a real random variable, i.e. $X: \Omega \to \mathbb R$ so it can only be measurable with respect to $\mathcal{G}$ in the case that $\Omega\subseteq \mathbb R^{2}$, right? Because otherwise we would have mismatching dimensions?
In this case, we have that $X(\omega,\overline{\omega})$ is well-defined for $(\omega,\overline{\omega})\in \Omega$
My ideas for the proof so far:
Define $f = 1_{A\times B}:\mathbb R^{2} \to \mathbb R$ with $A, B \in \mathcal{B}(\mathbb R)$, then
$E[f(X,Y)\lvert \mathcal{G}]=E[1_{A\times B}(X,Y)\lvert \mathcal{G}]=E[1_{A}(X)1_{B}(Y)\lvert \mathcal{G}]=1_{A}(X)E[1_{B}(Y)\lvert \mathcal{G}]=1_{A}(X)E[1_{B}(Y)]$. I do not see how to proceed.
Of course I could use some generator $\epsilon:=\{B_{1}\times B_{2} \in \mathcal{B}(\mathbb R)\times \mathcal{B}(\mathbb R) \}\subseteq \mathcal{B}(\mathbb R^{2})$ of $\mathcal{G}$ but I do not see how this would help me.
Let $f(x,y)=x$. Then your equation becomes $E(X|\mathcal G)=EX$. This is false since LHS is $X$, not $EX$.