Prove $E[XE[Y\mid\mathcal{G}]] = E[YE[X\mid\mathcal{G}]]$ Show that for bounded $X$ and $Y$ that $E[XE[Y\mid\mathcal{G}]] = E[YE[X\mid\mathcal{G}]]$.
Attempt:
Suppose that $X = _{\mathcal{X}_F}$, where $F \in \mathcal{D}$. Then for every $B \in \mathcal{D}$:
\begin{align}
& \int_B E[Y\mid\mathcal{D}\hspace{2mm} dP_{\mathcal{D}}] = \int_B  Y \, dP_{\mathcal{D}} = \int_{B \cap F}  Y \, dP_{\mathcal{D}} = \int_{B \cap F}  E[Y\mid\mathcal{D}] \, dP_{\mathcal{D}} \\[8pt]
= {} & \int_{B} E[Y\mid\mathcal{D}] \, dP_{\mathcal{D}}  = X[E[Y\mid G]]\text{ a.s.} \tag 1
\end{align}
(I'm trying to write out an $_{\mathcal{X}_F}$ infront of the conditional expectations in the last part, but it won't work)
Then we can follow the same process for $Y = {\mathcal{Y}_F}$ to complete the proof (calling it $(2)$). Then saying by both $(1)$ and $(2)$, $E[XE[Y\mid\mathcal{G}]] = E[YE[X\mid\mathcal{G}]]$.
 A: I claim that $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\big[XE[Y|\mathcal{G}]\big]=E\big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)$$
If we can prove this then we will be finished, because the right-hand side of equation $(*)$ is symmetric in $X$ and $Y$, and so the expression $E\big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]$ is also equal to $E\big[ YE[X|\mathcal{G}] \big]$, by interchanging the roles of $X$ and $Y$ in $(*)$.
In order to prove expression $(*)$, we will use the following lemma:
Lemma: Suppose $V$ and $W$ are random variables, and $W$ is $\mathcal{G}$-measurable. Then $E[VW |\mathcal{G}] = E[V|\mathcal{G}]  W$
Proof of Lemma: Prove it first when $W=\chi_{F}$ for some $F \in \mathcal{G}$ (this is easy from the definition of conditional expectation). Then extend to nonnegative $W$ by using the monotone convergence theorem, and finally extend to general $W$ by considering positive and negative parts. I'll trust you to write out the details. $\Box$
Finally, in order to see why the lemma implies equation $(*)$, let $V=X$ and let $W=E[Y|\mathcal{G}]$, so by the tower property of conditional expectation we have that $$E\big[XE[Y|\mathcal{G}]\big] = E[VW] = E\big[E[VW|\mathcal{G}]\big]=E\big[E[V|\mathcal{G}] W \big] = E \big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]$$ which proves equation $(*)$, and thus we are finished.
