Proof expectation $E[(X-Y)1_{\{X>a\}}]=E[(X-Y)1_{\{Y>a\}}]=0$

Question

I need to show: $$E[(X-Y)1_{\{X>a\}}]=E[(X-Y)1_{\{Y>a\}}]=0$$

Given are two random variables $X,Y$ on a probability space $(\Omega,\mathcal{F},P)$ with $E[|X|]<\infty$ and $E[|Y|]<\infty$ as well as $E[X|\sigma(Y)]=Y$ and $E[Y|\sigma(X)]=X$.

Given as a hint is that $1_{\{X>a\}}$ is $\sigma(X)$-measurable. Now I know that $1_{\{B\}}$ is $\sigma(X)$-measurable because $\forall B \in\sigma(X) \Rightarrow (1_{\{B\}})^{c}\in\sigma(X)$, but I cant see where this helps me.

Can anyone please help?

Answer 1

Hint: Try calculating $\mathrm{E}[(X-Y)\mathbf{1}_{X>a}]$ as $\mathrm{E}[\mathrm{E}[(X-Y)\mathbf{1}_{X>a}\mid \sigma(X)]]$.