If $EX^2<\infty$ and $E(X|\mathcal G)$ is $\mathcal F$-measurable then $E(X|\mathcal G)=EX$
There is one step in the proof which I don't understand, set $Y=E(X|\mathcal G)$ and then why is $E((X-Y)Y)=0$, from a theorem I know that $Y\in\mathcal F$ is such that $E(X-Y)^2$ is minimal
\begin{align} E[(X-Y)Y]&=E[E((X-Y)Y|\mathcal{G})]\\&=E[YE(X-Y|\mathcal{G})]\\&=E[Y(E(X|\mathcal{G})-E(X|\mathcal{G}))]\\&=0. \end{align} The first equality uses the Law of Iterated Expectations, the second "factoring out what is known," and the third linearity plus another application of "factoring out what is known."