One question about conditional expectation

Question

Let $X,Y\in L^1(\Omega,\mathfrak{F}, P)$, where $\mathfrak{F}$ is a $\sigma$-algebra and $P$ is the probability measure on $\Omega$. If $E[X|Y]=Y$ a.s. $E[Y|X]=X$ a.s., then prove that $X=Y$ a.s..

Now I can prove it is true for $X,Y$ are simple function, but I stuck at how to pass it to general measurable functions.

possible dublicate of https://math.stackexchange.com/questions/666843/if-exy-y-almost-surely-and-eyx-x-almost-surely-then-x-y-almost-surel — Shashi, Dec 27 '17 at 08:31
@E-A, To use monotone convergence theorem, we should find pairs of simple functions $(X_n,Y_n)$ satisfying the above conditions. I failed to find them. — aurora_borealis, Dec 27 '17 at 08:34
@aurora_borealis, yeah I thought you proved it simple functions; either way; this question as Shashi pointed out, is a duplicate — E-A, Dec 27 '17 at 08:35

score 2 · Accepted Answer · answered Dec 27 '17 at 08:26

Hints:

Show that $$\mathbb{E}[(X-Y) 1_{\{X>q\} \cap \{Y \leq q\}}] =0$$ for any $q \in \mathbb{R}$.
Prove that $$\{X>Y\} = \bigcup_{q \in \mathbb{Q}} \{Y \leq q\} \cap \{X>q\}$$
Combine Step 1 and 2 to conclude that $\mathbb{P}(X>Y)=0$.
Interchange the roles of $X$ and $Y$ to get $\mathbb{P}(Y>X)=0$. Thus $$\mathbb{P}(X \neq Y) = \mathbb{P}(X>Y)+\mathbb{P}(Y>X)=0.$$

One question about conditional expectation

1 Answers1