Are all random variables conditional expectations? Let $Z$ be a random variable on a probability space. Is it true that we can always find $X,Y$ two random variables on the same space such that $E[Y|X]=Z$ almost surely ?
What about this weaker (is it really weaker ? It feels so, but I am not sure) statement: Let $\mu$ be a probability measure on $\mathbb R$. Is it true that we can always find $X,Y$ such that $E[Y|X]\sim \mu$ ?
Any solution is welcome, but extra points if you can "build" the random variables $X,Y$.
 A: Let $(X, \Omega, \mu) \in \mathcal{P}$ a probability space,  $\mathcal{B} \subseteq \Omega$ a sub $\sigma$-field of $\Omega$, $Y : X \rightarrow \mathbb{R}$ a $\Omega$-measurable function of finite expectation. Conditional expectation naturally extends itself to the question of existance of a $\mathcal{B} $-measurable function $Z$ on $X$ such that :
$$ \forall A \in \mathcal{B}, \int_A ZdP = \int_A Y dP  $$
If $Y$ has a finite expectation (and we need it for the very definition of conditional expectation) such a random variable $Z$ does exists (We define $m$ a measure on $\mathcal{B}$ by $ \forall A \in \mathcal{B}, m(A) = \int_A YdP $ and apply Radon-Nykodym theorem to get a $\mathcal{B} $-measurable function $Z$ such that $ \forall A \in \mathcal{B}, m(A) = \int_A ZdP $)
A good intuition on this for $\mathcal{L}^2$ functions is that expectation is just the (linear) projection from $\Omega$-measurable functions onto $\mathcal{B} $-measurable function.
Now for the second question, by definition of $Z$ you have that the answer is positive.
Now if we drop the assumption of finite expectation it becomes a bit more complex as (  R. E. Strauch. "Conditional Expectations of Random Variables Without Expectations." Ann. Math. Statist. 36 (5) 1556 - 1559, October, 1965. ) in short, you could define a new conditional expectation to be
$$ E(Y^+ \mid \mathcal{B}) - E(Y^- \mid \mathcal{B}) $$
provided this difference is defined almost everywhere. The paper goes on about pathologies of such a definition.
