conditional mean / integration Consider a $\sigma$-algebra $\mathcal{G}$ and a sub-$\sigma$-algebra $\mathcal{f}$ For any $A \in \mathcal{f}$ holds: $$\int_A E(E(X|G)|F)dP = \int_A E(X|G) dP$$ Why is this the case?

Let $Y=E(X|\mathcal G)$. We are asked to show that $\int_A E(Y|\mathcal F)dP=\int_A Y dP$ for $A \in \mathcal F$. This follows from the definition.