If $X$ is independent of $Y$, are there conditions under which $X \perp Y \mid f(Y)$ for measurable functions $f$? Suppose we have r.v.s $X,Y$ such that $X$ is independent of $Y$. Would this automatically imply that $X \perp Y \mid f(Y)$ holds for all measurable functions $f$? What conditions might permit this to be true?
 A: By "$\ X\perp Y\ |\ f(Y)\ $" I presume you mean that the independence condition
$$
P\big(X\in A, Y\in B\ |\ f(Y)\big)= P\big(X\in A\ |\ f(Y)\big) P\big(Y\in B\ |\ f(Y)\big)
$$
is satisfied for all measurable $\ A\ $ and $\ B\ $.  This is indeed the case if $\ X\ $ and $\ Y\ $ are independent.
If $\ X\ $ and $\ Y\ $ are independent, then $\ P\big(X\in A\ |\ f(Y)\big)= P(X\in A)\ $ for all measurable $\ A\ $, so the condition reduces to
$$
P\big(X\in A, Y\in B\ |\ f(Y)\big)= P\big(X\in A\big) P\big(Y\in B\ |\ f(Y)\big)\ .
$$
By definition, $\ P\big(X\in A, Y\in B\ |\ f(Y)\big)\ $ is the (almost surely) unique random variable $\ Z_{A,B}\ $that satisfies the equation
$$
P\big(X\in A, Y\in B, f(Y)\in C\big)=\int_{Y^{-1}( f^{-1}(C))}Z_{A,B}(\omega)dP(\omega)
$$
for all measurable $\ C\ $, and $\ P\big(Y\in B\ |\ f(Y)\big)\ $ is similarly the random variable $\ Z_B\ $ satisfying
$$
P\big(Y\in B, f(Y)\in C \big)=\int_{Y^{-1}( f^{-1}(C))}Z_B(\omega)dP(\omega)
$$
for all measurable $\ C\ $. Multiplying this last equation by $\ P(X\in A)\ $, we get
\begin{align}
\int_{Y^{-1}(f^{-1}(C))}P(X\in A) & \overbrace{ P\big(Y\in B\ |\ f(Y)\big)}^{Z_B(\omega)}dP(\omega)\\
&= P(X\in A) P\big(Y\in B, f(Y)\in C \big)\\
&= P(X\in A) P\big(Y\in B\cap f^{-1}(C)\big)\\
&= P(X\in A, Y\in B\cap f^{-1}(C)\big)\ \text{, by independence}\\
&= P\big(X\in A, Y\in B, f(Y)\in C\big)
\end{align}
for all measurable $\ C\ $. Comparing this with the equation defining $\ P\big(X\in A, Y\in B\ |\ f(Y)\big)\ $ above, we must have
$$
P\big(X\in A\big) P\big(Y\in B\ |\ f(Y)\big)= P\big(X\in A, Y\in B\ |\ f(Y)\big)
$$
almost surely, by the uniqueness of conditional probability.
