I have a homework problem where I'm asked to show that if $X$ and $Y$ are random variables with $E[X|\mathcal{G}] = Y$ and $E[X^2] = E[Y^2]$ then $X = Y$ a.s.
My approach is to use Chebyshev's inequality on $X - Y$ to show that the probability that the two differ is zero, i.e.,
$$P[|X - Y| \geq \epsilon] \leq \frac{Var(X - Y)}{\epsilon^2}$$
where $\epsilon$ is arbitrary.
I also can tell that $E[E[X|\mathcal{G}]] = E[X] = E[Y]$, so the two expectations are equal in addition to the second moment.
So, to show that the right side goes to zero, I use the fact that $Var(X - Y) \leq E[(X - Y)^2]$ and expand the expression in the expectation to find $Var(X - Y) \leq E[X^2] + E[Y^2] - 2 E[X Y]$. Once I show that the correlation term equals $E[X^2]$ I am done, though I'm not sure how to get there. I feel like I'm missing something obvious.
Any suggestions, hints, or alternative approaches would be appreciated. Thanks.