Prove that from $\mathbb E(X\mid Y) \ge Y, \mathbb E(Y\mid X) \ge X$ follows that $X = Y$.

Question

Let $X$ and $Y$ discrete random variables. Prove that from $\mathbb E(X\mid Y) \ge Y, \mathbb E(Y\mid X) \ge X$ follows that $X = Y$.

I tried to prove it by contradiction. If the statement is not true, then without loss of generality $X \lt Y.$ Then from $X \le \mathbb E(Y\mid X)$ we have that $Y \lt \mathbb E(Y\mid X)$.

As a result, we have $Y \lt \mathbb E(Y\mid X)$ and $Y \le \mathbb E(X\mid Y)$, what can we do next?

Answer 1

"If the statement is not true, then without loss of generality $X < Y$."

This is wrong. What you can say is that without loss of generality $\mathbb P(X < Y) > 0$.