# Prove that two random variables are equal almost surely

This is a problem from Durret's probability textbook. Show that if $$X,Y$$ are random variables with $$E(Y|\mathcal{G})=X$$ and $$EY^2=EX^2\leq\infty$$, then $$X=Y$$ a.s.

My question is, by the definition of conditional expectation, $$\forall A\in \mathcal{F}$$, $$\int_{A} YdP = \int_{A} E(Y|\mathcal{G})dP=\int_{A} XdP$$ Doesn't this already imply that $$X=Y$$ a.s? Why do we need the condition that $$EY^2=EX^2\leq\infty$$?

The equation $$\int_A XdP=\int_AYDP$$ holds only for $$A \in \mathcal G$$ and not for all $$A \in \mathcal F$$. So we cannot conlude that $$X=Y$$ a.s. using just the definition.
We have $$EY^{2}=EX^{2}=E(E(Y|\mathcal G))^{2}\leq E E(Y^{2}|\mathcal G)) =EY^{2}$$. Hence, we must have equality throughout. And the fact that $$EE(Y|\mathcal G)^{2}=EE(Y^{2}|\mathcal G)$$ a.s implies that $$E(Y-E(Y\mathcal G))^{2}=0$$ as seen by expanding the square. It follows now that $$Y$$ is (a.s equal to a r.v. which is) $$\mathcal G$$ measurable and hence $$X=E(Y|\mathcal G)=Y$$ a.s.