So, it's pretty clear that for independent $X,Y\in L_1(P)$ (with $E(X|Y)=E(X|\sigma(Y))$), we have $E(X|Y)=E(X)$. It is also quite easy to construct an example (for instance, $X=Y=1$) which shows that $E(X|Y)=E(X)$, does not imply independence of $X$ and $Y$.

However, since $X,Y$ are independent iff $E(f(X)g(Y))=E(f(X))E(g(Y))$ for all bounded Borel functions $f,g:\mathbb{R}\to\mathbb{R}$, does it not make sense that $X$ and $Y$ are independent iff $E(f(X)|Y)=E(f(X))$, again for all bounded Borel functions $f:\mathbb{R}\to\mathbb{R}$?

Clearly, one of the implications hold (since $f(X)$ will be independent of $Y$), but what about the other way around? Am I completely wrong in my assessment?