Conditional Expectation: $E(X|Y) \geq Y$ and $E(Y|X) \geq X$ implies $X=Y$ a.e. Suppose for random variables $X$ and $Y$ with finite expectation we he have $E(X|Y) \geq Y$ and $E(Y|X) \geq X$.  We want to show that $X=Y$ almost everywhere.
My attempt:
Let $A \in \sigma(X)$ and $B \in \sigma(Y)$.  Then we have
$\int_B X = \int_B E(X | Y) \geq \int_B Y$ and $\int_A Y = \int_B E(Y | X) \geq \int_A X$.
My idea now is to consider sets of the form $\{X>a>Y\}$ for $a$ rational but I'm having trouble with getting the other variable involved.  The issue seems to be that $\{X>a\}$ is in $\sigma(X)$ but not necessarily in $\sigma(Y)$ and it seems like we cannot get the other inequality working.  Any help would be appreciated.
 A: Rewriting what you have in the question body, we have:
$\int_B X - Y \geq 0$ and $\int_A X - Y \leq 0$ for $B \in \sigma(Y)$ and $A \in \sigma(X)$. We need to show that these are actually equalities, so let us just do one. Let $B$ be an arbitrary set, and note that $\int_{B^c} X - Y \geq 0$ since $B^c$ is in $\sigma(Y)$ as well. We then have (by some abuse of notation) $\int_{B^c} + \int_B = \int_\Omega$. So, we overall have
$$0 \leq \int_B X - Y = \int_\Omega X - Y - \int_{B^c} X - Y \leq 0$$
since the first term is negative ($\Omega \in \sigma(X)$) and the second one is positive. Thus, for any $B \in \sigma(Y)$, $\int_B X - Y = 0$. You can do the same argument for all $A$ in $\sigma(X)$ as well, so we have that $X = Y$ for all sets $A \in \sigma(X) \cup \sigma(Y)$. Now, if you want the full sigma algebra $\sigma(X, Y)$, let us note that $1_{A \cap B} = 1_{A} + 1_{B} - 1_{A \cup B}$, so you can show that $X = Y$ in an algebra generated by the two sigma algebras. You can then use monotone convergence to go to a monotone class containing these to show closure under countable unions and intersections. Then, monotone class theorem for sets should give you the overall result.
A: 
Let $(\Omega, \Sigma, P)$ be a probability space.
Suppose for random variables $X$ and $Y$ with finite expectation we he have $E(X|Y) \geq Y$ and $E(Y|X) \geq X$.  We want to show that $X=Y$ almost everywhere.

For any  $A \in \sigma(X)$ and any $B \in \sigma(Y)$. We have
$\int_B X = \int_B E(X | Y) \geq \int_B Y$ and $\int_A Y = \int_A E(Y | X) \geq \int_A X$.
Since $X$ and $Y$ have finite expectation, we can conclude that, for any  $A \in \sigma(X)$ and any $B \in \sigma(Y)$,
$$\int_B (X -  Y)  = \int_B X - \int_B Y \geq 0$$ and $$\int_A (X - Y)= -\left( \int_A Y - \int_A X\right) \leq 0$$
Since $\Omega \in \sigma(X) \cap \sigma(Y)$, it follows immediately that $\int_\Omega (X -  Y) =0$.
Since for all  $B \in \sigma(Y)$,  $B^c \in \sigma(Y)$, we have
$\int_B (X -  Y) \geq 0$, $\int_{B^c} (X -  Y) \geq 0$ and
$$ \int_B (X -  Y)  + \int_{B^c} (X -  Y) = \int_\Omega (X -  Y) =0 $$
So for all $B \in \sigma(Y)$ we have
$$\int_B (X -  Y) = 0  \tag{1}$$
In a similar way, since for all  $A \in \sigma(X)$,  $A^c \in \sigma(X)$, we have
$\int_A (X -  Y) \leq 0$, $\int_{A^c} (X -  Y) \leq 0$ and
$$ \int_A (X -  Y)  + \int_{A^c} (X -  Y) = \int_\Omega (X -  Y) =0 $$
So for all $A \in \sigma(X)$ we have
$$\int_A (X -  Y) = 0 \tag{2} $$
From $(1)$ and $(2)$, we have that for all $C \in  \sigma(X) \cup  \sigma(Y)$,
$\int_C (X -  Y) = 0 $.
Now, note that $\Gamma = \{ D \in \Sigma : \int_D (X -  Y) = 0 \}$ is a $\sigma$-algebra and, since $\sigma(X) \cup  \sigma(Y) \subseteq \Gamma$, we have that $\sigma( \sigma(X) \cup  \sigma(Y)) \subseteq \Gamma$.
So, for all $C \in \sigma(X, Y) = \sigma( \sigma(X) \cup  \sigma(Y))$, we have
$\int_C (X -  Y) = 0 $.
Now, note that $[X \geq Y]= \{ w \in \Omega : X(w) \geq  Y(w)\} \in \sigma(X, Y)$ and
$[X < Y]= \{ w \in \Omega : X(w) <  Y(w)\} \in \sigma(X, Y)$. So
$$\int_\Omega |X-Y| = \int_{[X \geq Y]} (X-Y) + \int_{[X < Y]} (-(X-Y)) = \int_{[X \geq Y]} (X-Y) - \int_{[X < Y]} (X-Y)  =0$$
So, $X=Y$ almost everywhere.
