Say we have $Y \in L^2(\Omega,\mathcal{A},P)$ and that $E(Y|X) = X, E(Y^2|X) = X^2$. Then show that $Y=X$ a.s
My approach: Define $\mathcal{C} = \{\omega : X(\omega) = Y(\omega)\}$. Then $Y = X1_{\mathcal{C}} + Z1_{\mathcal{C}^c}$ for some Z. Then
$E(Y|X) = X \implies XE(1_{\mathcal{C}} | X) + E(Z1_{\mathcal{C}^c} | X) = X$ and $E(Y^2|X) = X^2 \implies X^2E(1_{\mathcal{C}} | X) + E(Z^21_{\mathcal{C}^c} | X) = X^2$ together which $\implies XE(Z1_{\mathcal{C}^c} | X) = E(Z^21_{\mathcal{C}^c} | X)$ and I am unable to get any further.
I do however know that we must have $E(YW) = E(XW) \ \forall W \in L(\Omega,\sigma(X),P)$ from the definition of conditional expectation.
Can you guys help me out here?
Hint: the assumptions imply that $X$ is square integrable. Compute $\mathbb E[(X-Y)^2\mid\ X]$.