I've seen two kinds of demonstrations of Vitali's sets being not measurable (for example, answer number 2 here: <https://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set>, I adress it specifically because proof number one I haven't seen very often).


In both cases, the fact that the Lebesgue measure on $\mathbb{R}$ is traslation invariant is used to demonstrate that a Vitali set is not measurable:


"Given $V$ a Vitali set, define, for $k \in \mathbb{Q} \cap [-1,1]$, $V\_k = \{v + k | v \in V\}$. Then, if $\mathcal{L}(A)$ is the Lebesgue measure for $A$, it is true that $\mathcal{L}(V) = \mathcal{L}(V\_k) , \forall k \in \mathbb{Q} \cap [-1,1]$" And then this fact is used to demonstrate $V$ immesurability w.r.t Lebesgue's measure.


Since this statement does not hold for an arbitrary measure, my question is if it is or is not true that a Vitali set is not measurable for every measure $\mathcal{M}$ absolutely continuous with $\mathcal{L}$. To be more precise, $\mathcal{M}$ must be absolutely continuous with $\mathcal{L}$ when restricted to the Borel sets, $\mathcal{M}$ may be defined on a broader $\sigma$-algebra.


This can be easily answered if you don't ask $\mathcal{M}$ to be absolutely continuous w.r.t Lebesgue's measure:


Taking as probability space $(\mathbb{R},\mathcal{P(\mathbb{R})},\mathcal{I\_x})$, where $\mathcal{P(\mathbb{R})}$ is $2^\mathbb{R}$ and $\mathcal{I\_x}$ is the Dirac measure concentrated on $x$ (a pre-set, arbitrary point), then $V$ would be measurable with $\mathcal{I\_x}(V)$ being 1 or 0 depending on whether $x \in V$ or $x \notin V$.


But i cannot get a general result when I impose the absolute continuity requirement.



On the mesurability of a VItali set w.r.t. a Lebesgue absolutely continuous measure
user's answer, stated another way


Say the Vitali set $V \subseteq [0,1]$ has inner measure zero, outer measure 1. We may extend Lebesgue measure to a larger $\sigma$-algebra. Let $\mathscr F$ be the Lebesgue-measurable sets in $[0,1]$, and let the extension be
$$
\mathscr G = \{ (A\cap V) \cup (B \setminus V)\;|\; A,B \in \mathscr F\}
$$
Let $\mathcal L$ be Lebesgue measure, and define measure $\mu$ on $\mathscr G$ by
$$
\mu\big((A\cap V) \cup (B \setminus V)\big) = \mathcal L(A),
\qquad A,B \in \mathscr F .
$$
Then:  

$\mathscr G$ is a $\sigma$-algebra on $[0,1]$,  

$\mu$ is $\sigma$-additive on $\mathscr G$,  

$\mathscr G \supset \mathscr F$,  

$V \in \mathscr G$,  

$\mu(E)=\mathcal L(E)$ for $E \in \mathscr F$.


