Let $A \subset \mathbb{R}$ be sets of positive Lebesgue measure. Let $\Gamma$ be a countable dense subgroup of the additive group $\mathbb{R}$. Consider the partition of $\mathbb{R}$ canonically associated with the equivalence relation $x \in \mathbb{R} \wedge y \in \mathbb{R} \wedge x - y \in \Gamma$. Let $X$ be any selector of this partition. We define an $A, \Gamma$ Vitali set to be $V_{A, \Gamma} = X \cap A$.
My question is can any non-null set be expressed as a countable union of $A, \Gamma$ Vitali sets. i.e. given a set $Y \subset{R}$ such that the Lebesgue measure of $Y$ is not zero, does there exist $A_{i}, \Gamma_{i}$ such that $Y = \bigcup_{i \in \mathbb{N}} V_{A_{i}, \Gamma{i}}$?
Ok I'll have a go at proving this myself.
I think the answer is no.
Proof:
We make use of the following Lemma whose proof can be found here: There exists a partition of the real line into continuum many Bernstein sets.
Consider one of these Bernstein sets, $B$. Then a countable union of translations of $B$ is also non-measurable.
Now suppose $B = \bigcup_{n \in \mathbb{N}} V_{A_{n}, \Gamma_{n}}$ as in the above question.
Then $\bigcup_{\gamma \in \bigcup_{n \in \mathbb{N}}\Gamma_{n}} B \oplus \gamma$ is a countable union of translations of $B$ with infinite measure contradicting the earlier statement.
QED