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.

There's an "accepted" answer and it's good, so normally I might not bite, but I have a slight preference for this one:
Suppose $\alpha$ is uniformly distributed in the interval $[0,2\pi)$, so that $X=(\cos\alpha,\sin\alpha)$ is uniformly distributed on the circle.
Then let $Y=(\cos(2\alpha),\sin(2\alpha))$.  Then $Y$ is also uniformly distributed on the circle.
Certainly if you know $X$ you can find $Y$, but not vice-versa.  The mapping $X\mapsto Y$ is two-to-one.
