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Let $X,Y \in \mathcal{L}^1(\Omega,\mathcal{F},\mathbb{P})$ and assume that $E[X|Y]=Y$ and $\mathbb{E}[Y|X]=X$.

I want to show that $\mathbb{P}(X=Y)=1$.

I started working on \begin{align} \mathbb{E}[(X-Y)\mathbb{1}_{\{X>z,Y\leq z\}}] + \mathbb{E}[(X-Y)\mathbb{1}_{\{X\leq z,Y\leq z\}}] \end{align} for arbitrary $z \in \mathbb{R}$. However, I do not know how to continue. Any help is appreciated.

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    **Hint:** Let $h:\Bbb R\to(0,1)$ be continuous and strictly increasing. Show that $\Bbb E[(X-Y)\cdot(h(X)-h(Y))]=0$. Notice that $(X-Y)\cdot(h(X)-h(Y))\ge 0$ with equality only if $X=Y$. – 2017-01-02

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