Vitali sets are nonmeasurable and in particular are not null sets. But all Vitali sets are in some sense small, as described below. Let $V$ be any Vitali set and let $k \in \mathbb{N}$. For each $i \in \mathbb{Z}$, consider the set $A_i = V \cap [\frac{i}{k}, \frac{i+1}{k} )$ and let $B_i = A_i - \frac{i}{k} = \{x-\frac{i}{k} : x \in A_i\}$. Since the difference between two different elements of $V$ is irrational, the sets $B_i$ for $i \in \mathbb{Z}$, are pairwise disjoint. The union of the $B_i$ is a subset of the open interval $[0,\frac{1}{k})$. This shows that given any interval, no matter how small, it is possible to partition any Vitali set $V$, by intersecting $V$ with half-open intervals, into countably many pieces and then translate those pieces so that they all fit, without overlap, inside the interval.
Here is the formalization I have in mind. We say $A \subseteq \mathbb{R}$ is "small" if for all $\epsilon>0$ there exists a sequence of half-open intervals $I_n = [a_n, b_n)$ which together partition $\mathbb{R}$ and a sequence of real numbers $t_n$ such that the sequence $(A \cap I_n) - t_n$ is a pairwise disjoint sequence of sets whose union is contained in the interval $[0,\epsilon)$.
Is there a natural finitely additive measure on $\mathbb{R}$ whose null sets are exactly the sets that are small in the above sense? Can Lebesgue measure be extended to a finitely (but not countably!) additive measure so that the sets that are small in the sense above are precisely the null sets?
