Show that for random variables $X,Y$ on probability space $(\Omega, \mathscr{F}, P)$, for any $\sigma$-field $\mathscr{G} \subset \mathscr{F}$ that:
\begin{align*} E[X E(Y | \mathscr{G}) ] = E[E(X | \mathscr{G}) E(Y | \mathscr{G}) ] = E[E(X | \mathscr{G}) Y ] \end{align*}
FYI, this is from Karatzas + Shreve, on page 43, in the give solution to problem 4.11.
Of course, $E[E(X | \mathscr{G})] = E[X]$. And when $X \in \mathscr{G}$, $E(X | \mathscr{G}) = X$. But in the general case, I don't see how to prove the relationship given.
$$\begin{aligned}E[XE[Y|\mathscr{G}]]&\stackrel{\textrm{TP}}{=}E[E[XE[Y|\mathscr{G}]|\mathscr{G}]]=\\ &\stackrel{\textrm{Meas.}}{=}E[E[X|\mathscr{G}]E[Y|\mathscr{G}]]=\\ &\stackrel{\textrm{Meas.}}{=}E[E[E[X|\mathscr{G}]Y|\mathscr{G]}]=\\ &\stackrel{\textrm{TP}}{=}E[E[X|\mathscr{G}]Y]\end{aligned}$$ where $\textrm{TP}$ is the tower property and $\textrm{Meas.}$ indicates the fact that if $\sigma(Z)\subseteq \mathscr{Z}$ then $E[ZH|\mathscr{Z}]=ZE[H|\mathscr{Z}]$.