A mapping $f$ from $\mathbb R$ to $\mathbb R$ is called a choice function if, for any $x, y \ {\rm in}\ \mathbb R$, $f(x)-x \in\mathbb Q$ and $f(x)=f(y)$ whenever $x-y$ is rational.
My questions is: Is there a Lebesgue-measurable choice function?
Note: Here I use the equivalence relation for the construction of Vitali sets: x and y are in the same equivalence class iff x-y is rational. So choice functions pick up one element from each equivalence class as its representative value.
On
There is at least no Borel measurable function with this property. Suppose there were a Borel measurable function $f$ with this property. Then it would have a Borel measurable graph $G=\{(x,y):y=f(x)\}$. Now the projection of $G$ on its second coordinate is an analytic set and hence Lebesgue measurable. But it is a Vitali set, contradicting its measurability.
I don't know how to extend this to Lebesgue measurable functions.
Suppose $f$ is measurable. Then $V = \{x \in \mathbb{R} \,:\,f(x) = x\}$ is a measurable representative system for the equivalence relation $x \sim y$ if and only if $x - y \in \mathbb{Q}$. That is to say, $V$ is a measurable Vitali set, but such a set doesn't exist.