Coniditional expectation for bounded random variables I need to show that for bounded random variables $X$ and $Y$ we have 
${\rm{E}}[X{\rm{E}}(Y||\mathcal{F})]={\rm{E}}[Y{\rm{E}}(X||\mathcal{F})]$.
The only property of conditional expectation I am aware of that requires boundedness is the Jensen's Inequality. So, I have tried to consider a function $\phi(X)=XY$, and since ${\rm{E}}(|XY|)<\infty$, we have ${\rm{E}}(XY||\mathcal{F})= Y{\rm{E}}(X||\mathcal{F})$ since $\phi(X)$ is linear in $X$, and hence convex and concave. Using same argument for $\phi(Y)=XY$, we obtain he result. I am not sure though if we can apply jensen's inequality on this $\phi$. Is this correct? If not, Any ideas?

$$
\mathbb E[X~\mathbb E[Y~|~\mathcal F]]=\mathbb E[\mathbb E[X~\mathbb E[Y~|~\mathcal F]~|~\mathcal F]] = \mathbb E[\mathbb E[X~|~\mathcal F]\mathbb E[Y~|~\mathcal F] ]=\mathbb E[\mathbb E[Y~\mathbb E[X~|~\mathcal F]~|~\mathcal F] ]=\mathbb E[Y~\mathbb E[X~|~\mathcal F] ]
$$

Boundedness:
The boundedness of the random variables guarantees the boundedness of the (conditional) expectations above. Since $|X|\leqslant M$ for some $M\geqslant 0$, we have
$$-M=\mathbb E[-M~|~\mathcal F]\leqslant \mathbb E[X~|~\mathcal F]\leqslant \mathbb E[M~|~\mathcal F]= M\tag{1}
$$
In addition, since $|Y|\leqslant N$ for some $N\geqslant 0$, we also have from (1)
$$
-\max\{M^2,N^2\}\leqslant\mathbb E[Y~\mathbb E[X~|~\mathcal F]]\leqslant\max\{M^2,N^2\}
$$
Similar arguments can be made for the remaining (conditional) expectations.
