Proof random inequal variables equal union of disjunct sets on Q with density argument

Question

I need to proof

$$\{X\neq Y\}=\bigcup_{q\in\mathbb{Q}}\{X>q,Y\leq q\}\cup\{X\leq q,Y>q\}$$

and also show $X=Y\ \ \ \ P-a.s.$

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

Now, my approach would be to proof both sides (like '$\subseteq$' first and then '$\supseteq$') and in the hints there is stated to use a density argument i.e. $\mathbb{Q}$ is dense in $\mathbb{R}$.

However I have no idea how to start and make the proof...can someone please help?

Answer 1

Update: Sorry, made a mistake, but try to look at the complement of $\{X\neq Y\}$, and I think you should be fine ;)

Update2: To refine my hint: \begin{align} &\left(\bigcup_{q\in\mathbb{Q}}\{X>q,Y\leq q\} \cup \{X\leq q,Y> q\}\right)^c\\ &= \bigcap_{q\in\mathbb{Q}}\{X>q,Y\leq q\}^c \cap \{X\leq q,Y> q\}^c\\ &= \bigcap_{q\in\mathbb{Q}}(\{X\leq q\}\cup\{Y>q\}) \cap (\{X> q\}\cup\{Y\leq q\})\\ &= \bigcap_{q\in\mathbb{Q}}(\{X\leq q\}\cup\{Y>q\}) \cap (\{X> q\}\cup\{Y\leq q\})\\ &= \bigcap_{q\in\mathbb{Q}}(\{q<X\leq q\}\cup \{q<Y\leq q\}\cup \{X,Y>q\}\cup \{X,Y\leq q\}) \end{align} Now it sould be easy to conclude the equality to $\{X=Y\}=\{X\neq Y\}^c$.

Update3: Since $\{q<X\leq q\} = \{q<Y\leq q\} = \emptyset$, we have $$\left(\bigcup_{q\in\mathbb{Q}}\{X>q,Y\leq q\} \cup \{X\leq q,Y> q\}\right)^c = \bigcap_{q\in\mathbb{Q}}(\{X,Y>q\}\cup \{X,Y\leq q\}).$$ Now you can observe that the right side consists of all Points $p$, such that for all $q\in\mathbb{Q}$ holds $X(p),Y(p)>q$ or $X(p),Y(p)\leq q$. Since $\mathbb{Q}$ is dense, it must be $X(p)=Y(p)$. (The converse inclusion is trivial.)

Answer 2

The equality of two sets is clear. To show $X=Y$ a.s. with conditions you mentioned, first we assume that $X, Y\in L^2$. Then, the conditions follow that $E(XY|Y)=Y^2, E(XY|X)=X^2$, this implies $EX^2=EXY=EY^2$. Hence, $$E(X-Y)^2 = EX^2-2EXY + EY^2 = 0,$$ and $X=Y$ a.s. Next we consider $X, Y \in L^1$. Denote $X_n=X\wedge n = \min\{X, n\}, Y_n=Y\wedge n=\min\{Y, n\}$. Using the fact that if $U\leqslant V$ a.s. and $EU =EV$, then $U=V$ a.s., we get $E(X_n|Y)= Y_n, E(Y_n|X)= X_n$ for all $n$. Next, denote $X_n'=X_n\vee (-n) = \max \{X_n, -n\}$ and $Y_n'=Y_n\vee (-n) = \max \{Y_n, -n\}$, by the same arguments, $E(X_n'|Y)=Y'_n, E(Y_n'|X)=X'_n$ for all $n$. Observer that $X'_n, Y'_n \in L^2$ for each $n$, and they satisfy the first case, we get $X'_n= Y'_n$ a.s. for all $n$. Hence, $X=Y$ a.s.