Does X = Y in distribution and X being Y-measurable imply Y is X-measurable? Suppose $X,Y$ are random variables taking values in some Borel space, $X \overset {d}{=} Y$, and $X$ is $Y$-measurable.
It follows from the fact that $X$ is $Y$-measurable that there exists a measurable $f$ such that $X = f(Y)$ a.s.
Is it the case that there exists a measurable $g$ such that $Y = g(X)$ a.s.? It seems plausible, and is true for finite-valued, discrete random variables.
EDIT:
I've thought about it a bit, and one consequence is $X \overset {d}{=} f(X)$, which may or may not be helpful.

No.
Let $Y$ be uniformly distributed on $[-1,1]$, let $f(x) = 2|x|-1$ and let $X = f(Y)$.  Then $X$ is also uniformly distributed on $[-1,1]$, but $Y$ is not $X$-measurable, since, intuitively, by looking at $X$ you cannot tell the sign of $Y$.
