For random variables $X,Y\in L^1$, if $E[X|\sigma(Y)]=Y$ and $E[Y|\sigma(X)]=X$, prove that $X=Y$ a.s.

Question

For random variables $X,Y\in L^1$, if $E[X|\sigma(Y)]=Y$ and $E[Y|\sigma(X)]=X$, prove that $X=Y$ a.s.

My attempt:

1. Note that $E[X-Y|\sigma(Y)]=E[X|\sigma(Y)]-E[Y|\sigma(Y)]=Y-Y=0$. Similarly, $E[X-Y|\sigma(X)]=0$.

2. By 1, if I could prove the following lemma, then $E[X-Y|\sigma(X,Y)]=X-Y=0$ as desired.

Lemma (not necessarily true). If $E[Z|\mathcal A]=E[Z|\mathcal B]=0$ where $\mathcal A$ and $\mathcal B$ are two $\sigma$-fields, then $E[Z|\sigma(\mathcal A,\mathcal B)]=0$.

Questions:

1. Is this "lemma" correct? Is there a counterexample?

2. If not, how should I solve the original question?

Answer 1

For the lemma, it may unfortunately not work. Let $\varepsilon$ and $\varepsilon'$ be two i.i.d. random variables such that $\varepsilon$ takes the values $-1$ and $1$ with probability $1/2$. Let $Z=\varepsilon\cdot \varepsilon'$ and $\mathcal A=\sigma(\varepsilon)$, $\mathcal B=\sigma(\varepsilon')$. The random variable $Z$ is $(\mathcal A\vee\mathcal B)$-measurable and since $\varepsilon$ is $\mathcal A$-measurable, it follows that $$ \mathbb E\left[Z\mid\mathcal A\right]=\varepsilon\mathbb E\left[\varepsilon'\mid\sigma(\varepsilon)\right]=0 $$ and similarly, $ \mathbb E\left[Z\mid\mathcal A\right]=0$.

For the second question, it was solved here.

Answer 2

I don't know your lemma is true or not but for the original question, here is my idea

$E[(X-Y)Y]=E[E(X-Y)Y|\sigma(Y)]]=E[YE[X-Y|\sigma(Y)]]=0$

Similarly, $E(X-Y)X]=0$, combine these two equations, we get $E[(X-Y)^2]=0$, which gives the desire result.