Hello,

Sorry if I explain it in my own words, I don't know if you've studied it the same way... And maybe there are small typos, but not very important.

1) We know that if Z is $\displaystyle \sigma(Y)$-measurable, then for any rv X (in L^2 I think), we have $\displaystyle E(XZ|Y)=ZE(X|Y)$

So since Y is obviously $\displaystyle \sigma(Y)$-measurable, E[E[XY|Y]]=E[YE[X|Y]] (*)

But there's something that says :

Let $\displaystyle \mathcal{B}$ be a $\displaystyle \sigma$-algebra. For any $\displaystyle \mathcal{B}$-measurable Z (positive), E[ZX]=E[Z E[X|Y]], and where X is positive. (this comes from the fact that E[X|Y] is the orthogonal projection of X over $\displaystyle L^2(\Omega,\sigma(Y),P)$, but you don't really need to know it if you haven't learnt this...)

So (*)=E[YX]

Exact same reasoning, under the condition that h is $\displaystyle \sigma(Y)$-measurable.

I hope this is clear enough

* $\displaystyle \sigma(B)$ is the smallest sigma-algebra that makes B measurable

** Note : a rv A is $\displaystyle \sigma(B)$-measurable iff there exists $\displaystyle \varphi$ which is $\displaystyle \sigma(B)$-measurable such that $\displaystyle B=\varphi\circ A$