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.