Show that $E(XE(Y| \mathcal{G} )) = E(YE(X | \mathcal{G}))$ We have bounded $X,Y$ and sigma field $\mathcal{G}$, $\mathcal{G}\subset \mathcal{F}$. We have to show that
$$ E(XE(Y| \mathcal{G} )) = E(YE(X | \mathcal{G}))$$
So we have
$$ E(XE(Y| \mathcal{G} )) = E\left[E(XE(Y| \mathcal{G} ))|\mathcal{G}\right] =^* E[E(Y| \mathcal{G}) \cdot E(X| \mathcal{G})] = $$
and analogically we obtain the same for
$$E(YE(X | \mathcal{G})) = E[ E(X| \mathcal{G}) \cdot E(Y| \mathcal{G})]$$
However I do not know how we got the equality with $*$.
 A: This is just the "taking out what is known" property.  Namely, if $Z$ is any integrable $\mathcal G$-measurable, then $\mathbb{E}[XZ|\mathcal G] = Z\mathbb{E}[X|\mathcal G]$.  Applying this with $Z = \mathbb{E}[Y|\mathcal G]$, which is $\mathcal G$-measurable by the definition of conditional expectation, gives the desired result.
A: Suppose that we have a Hilbert space $H$ with closed subspace $G$. Let $P:H\to G$ be the orthogonal projection onto $G$ and let $Q:H\to G^\perp$ be the orthogonal projection onto $G^\perp$. Then we have $x=Px+Qx$ for all $x\in H$. For any $x,y\in H$, since $Px\perp Qy$ and $Qx\perp Py$, $$\langle Px,y\rangle = \langle Px,Py\rangle+\langle Px,Qy\rangle= \langle Px,Py\rangle = \langle Px,Py\rangle+\langle Qx,Py\rangle =\langle x,Py\rangle.$$
Let $H=L_2(\Omega, \mathcal{F})$ and let $G$ denote the subspace of $H$ consisting of (equivalence classes) of random variables which are $\mathcal{G}$-measurable. Then $X\mapsto\mathbb{E}(X|\mathcal{G})$ is the orthogonal projection of $H$ onto $G$. In the $\mathbb{R}$ world, the inner product  is $\langle X,Y\rangle=\mathbb{E}(XY)$. Apply the previous paragraph.
$L_2(\Omega,\mathcal{F})$ is the natural setting for this problem, since otherwise $\mathbb{E}(X\mathbb{E}(Y|\mathcal{G}))$ need not be defined.
