Proof of $E[XY|\mathscr{H}]=XE[Y|\mathscr{H}]$ for $\mathscr{H}$-measurable $X$

Question

I reckon this is a pretty trivial result in the context of elementary probability, but how would one show it in the context of measure-theoretic probability theory? That is, that $E[XY|\mathscr{H}]=XE[Y|\mathscr{H}]$ for $\mathscr{H}$-measurable $X$ where $X$ and $Y$ are random variables.

Answer 1

When $X=I_A$ for some measurable set $A$ (w.r.t. $\mathcal H$) this follows from definition of conditional expectation. Now go to simple functions and then take limits. This gives the result for $X$ non-negative. Now write $X$ as $X^{+}-X^{-}$. You have to assume that $E|XY| <\infty$ for the result to be true.

Answer 2

Let $$ Z=E[XY\mid \mathscr{H}]-XE[Y\mid \mathscr{H}]=E[XY\mid \mathscr{H}]-XY+X[Y-E[Y\mid \mathscr{H}]] $$ and note that $Z$ is a $\mathscr{H}$ measurable function. Further for $H\in \mathscr{H}$, we have that $$ EZI_H=E(E[XY\mid \mathscr{H}]-XY)I_H+E(Y-E[Y\mid \mathscr{H}])XI_H=0 $$ by the definition of conditional expectation where we use the fact that $XI_{H}$ is $\mathscr{H}$ measurable. Taking $H=I(Z>0)$ and $H=I(Z<0)$ it follows that $$ Z=0 $$ with probability one from which the claim follows.