Consider sets of Vitali's type in models of $\mathsf{ZF}+\mathsf{GCH}$ where $V \neq L$. Are there sets of Vitali's type in both $L$ and $V \backslash L$? If so, is there any way one can distinguish the constructible sets of Vitali's type from the nonconstructible sets of Vitali's type?


By a *set of Vitali's type* it is meant a subset of $\mathbb{R}$ containing exactly one element of every equivalence class of the relation $x - y \in G$, where $G$ is some fixed countable subgroup of $( \mathbb{R} , + )$.



Sets of Vitali's type in models of $\mathsf{ZF}+\mathsf{GCH}$ where $V \neq L$
There are several things to say that may answer your question.


* It is consistent with ZFC$\pm$GCH that the set of all real numbers in $L$ is countable in $V$, in which case every set of reals in $L$ has measure zero.
* It can happen that $V$ and $L$ have precisely the same sets of reals, yet $V\neq L$, because they disagree only very high up. In this case, of course there can be no set of reals in $V-L$.
* Meanwhile, if by a set of "Vitali type" you mean a set that selects one member of each equivalence class under the relation of translation by rationals, then if $L$ has such a set, then it must have all the reals (since every real differs by a rational from a member of it), and so $\mathbb{R}^V=\mathbb{R}^L$.
* Also, one can force to add a new Vitali set without adding reals, and so one can have ZFC+GCH+$V\neq L$ with Vitali sets in both $L$ and in $V-L$.


