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.