Distribution function and conditional expectation This might be a strange question and I'm completely confused.
Let $(\Omega,\mathscr{F},P)$ be a probability space and $\xi:\Omega\rightarrow\mathbb{R}$ be a random variable. Let $F:\mathbb{R}\rightarrow\mathbb{R}$ be the c.d.f. of $\xi$, i.e., $F(x)=P(\xi\leq x)$ for all $x\in \mathbb{R}$. My question is whether $E(\xi|F(\xi))=\xi$, where $F(\xi)$ is the composition of $F$ and $\xi$. Of course the question boils down to whether $\xi$ is measurable with respect to $F(\xi)$.
What makes me confused is the following example. Consider $\Omega=\mathbb{R}$ and $\mathscr{F}$ is the Borel $\sigma$-algebra. Let $P$ be the uniform distribution over $[0,1]$, and $\xi:\mathbb{R}\rightarrow\mathbb{R}$ be such that $\xi(x)=x$. Then clearly $\xi$ is not measurable w.r.t. $F(\xi)$ and hence $E(\xi|F(\xi))\neq \xi$. But if we restrict attention to $\Omega=[0,1]$, $\mathscr{F}=\mathscr{B}([0,1])$, then $\xi$ is measurable w.r.t. $F(\xi)$ and hence $E(\xi|F(\xi))=\xi$.
It seems to me $F(\xi)$ contains all relevant information about $\xi$, but $\xi$ is not measurable w.r.t $F(\xi)$ because some measure 0 sets. Is there any general statement about $E(\xi|F(\xi))=\xi$?
 A: The most general statement you can make in your counterexample is $E(\xi\mid F(\xi))=\xi$ almost surely. 
Conditional expectations are equal except on a set of measure zero. That is, if $Y$ and $Y'$ both satisfy the requirements for $E(X\mid\mathscr G)$, then $P(Y=Y')=1$. In your counterexample it is true that $\xi$ is not measurable with respect to $F(\xi)$ when $\Omega=\mathbb R$; however, there is a version $\xi'$ that equals $\xi$ with prob 1 such that $\xi'$ is measurable and $E(\xi\mid F(\xi))=\xi'$, namely $\xi':=F(\xi)$. 
The result $E(\xi\mid F(\xi))=\xi$ a.s. is true in general. Write $\mathscr G:=\sigma(F(\xi))$ and let $\mathscr G^*$ be the completion of $\mathscr G$, i.e., $\mathscr G^*$ is the sigma-algebra consisting of all $A\cup Z$ where $A\in\mathscr G$ and $Z$ is a subset of a set of probability zero. Argue that $\xi$ is measurable with respect to $\mathscr G^*$, and therefore $\xi$ is a version of $E(\xi\mid\mathscr G^*)$. But any version of $E(\xi\mid\mathscr G)$ is also a version of $E(\xi\mid\mathscr G^*)$, and therefore equals $\xi$ almost surely.
