Let $X,Y$ be integrable random variables such that $\mathbb E(X|\sigma(Y))=Y$ and $\mathbb E (Y|\sigma(X))=X$, where $\sigma(X)$ is the smallest sigma algebra such that $X$ is measurable. Show that $X=Y$ almost surely.
The claim seems plausible since I think of the equality $\mathbb E(X|\sigma(Y))=Y$ as $X$ being finer than $Y$. But still, I didn't manage to formally prove the claim. Any suggestions?
