Independence of variables implies no functional representation? My question's pretty simple, I just thought the title phrases it pretty well.. Anyway, the Doob-Dynkin lemma says that $X$ is $\sigma (Y)$-measurable iff there's a measurable $f$ such that $X=f(Y)$ (this is what I refer to as a functional representation, henceforth "f.r"). Intuitively, If two random variables are independent I don't except them to have a f.r of each other. Then again my intuition sucks.. So the question is ( _edit: for not a.s constant $X,Y$_ )$$\sigma(X)\perp\sigma(Y)\overset{?}{\implies}\text{no f.r}$$I don't know how to connect measurability and independence.. _Update: found and posted solution from stats.stack_

Well, I've found a solution (I didn't think of it myself) so I'm posting it. I'm not sure what the protocol is for answering your own question.

Under the assumption $X,Y$ are not a.s constant, the first answer to this question proves the contrapositive.