Let $X$, $Y$, and $Z$ be three random variables in $L^{1}(\Omega, \mathcal{F}, P)$. Suppose that we have $E(X|Y)=Z$, $E(Y|Z)=X$, and $E(Z|X)=Y$. Show that $X=Y=Z$ a.s.
I am able to show this in $L^{2}$; in fact, there, it's a very easy proof. I'm having trouble proving it for $L^{1}$, though, because in $L^{1}$, conditional expectations are defined in terms of sub $\sigma$-algebras of $\mathcal{F}$, and that's not the way they're expressed here. I suppose my main problem is expressing the conditional expectations given in the problem as ones I can use results for $L^{1}$ on. Can somebody please help me do that?
$E(X\mid Y) = Z$ implies that $Z$ is $\sigma(Y)$-measurable and in the same way $E(Z \mid X) = Y$ implies that $Y$ is $\sigma(X)$-measurable. By transitivity, this proves that $Z$ is actually $\sigma(X)$ measurable, so that $E(Z\mid X)=Z = Y$.
Et caetera.