I would like to show the following statement using the general definition of conditional expectation. I believe it is true as it was also pointed out in other posts.
Let $X,Y$ be conditionally independent random variables w.r.t a sigma algebra $\mathcal{G}$. Furthermore the expected values of $X,Y$ and $XY$ exist.Then $$\mathbb{E}(XY\mid \mathcal{G})= \mathbb{E}(X\mid \mathcal{G})\mathbb{E}(Y\mid \mathcal{G}).$$
Using the definition of conditional expectation that is used on wikipedia it suffices to show that $(1)$ $\mathbb{E}(X\mid \mathcal{G})\mathbb{E}(Y\mid \mathcal{G})$ is $\mathcal{G}$-measurable and $(2)$ that $\mathbb{E}(\mathbb{E}(X\mid \mathcal{G})\mathbb{E}(Y\mid \mathcal{G})\mathbf{1}_A)=\mathbb{E}(XY\mathbf{1}_A)$ for all $A \in \mathcal{G}$. $(1)$ is clear as the product of two $\mathcal{G}$-measurable functions is $\mathcal{G}$-measurable. For the second part I am able to deduce $\mathbb{E}(XY\mathbf{1}_A)=\mathbb{E}(X\mathbf{1}_A)\mathbb{E}(Y\mathbf{1}_A)/\mathbb{P}(A)=\mathbb{E}(\mathbb{E}(X\mid \mathcal{G})\mathbf{1}_A)\mathbb{E}(\mathbb{E}(Y\mid \mathcal{G})\mathbf{1}_A)/\mathbb{P}(A)$ using conditional independence and the definition of conditional expectation. But then I am lacking any idea to show that the latter equals the LHS in $(2)$. Do you have any ideas how to proceed?