Measuring Vitali sets

Question

If $V$ is a vector space and $m$ is the counting measure, then $$\dim(V) = \inf \{m(U) : U \subset V, \text{ span}(U) = V\}.$$

Given a measure space $(V, \mathcal M, \mu)$ such that $V$ is a vector space, have people considered the following definition? $$\dim_\mu(V) := \inf \{\mu(U) : U \in \mathcal M, \text{ span}(U) = V\}$$

In particular, considering $\mathbb R$ as a $\mathbb Q$-vector space, is there a measure $\mu$ such that $0 < \dim_\mu(\mathbb R/\mathbb Q) < \infty$ ?

Answer 1

Consider the $\mathbb Q$-vector space $\mathbb R$. Of course $1 \in \mathbb R$. Let $V \subseteq \mathbb R$ be a $\mathbb Q$-subspace complementary to $\mathbb Q\cdot 1$. (A subspace of codimension $1$. In fact, any subspace of codimension $1$ will work for this. Of course some form of AC is needed to get $V$. For example, take a Hamel base and then use the span of all but one element. The existence of $V$ may be considerably weaker than the existence of a Hamel base, though.)

Now let $\mu$ be counting measuse on the complement $\mathbb R \setminus V$. Note that every subset of $\mathbb R$ is $\mu$-measurable. Two observations: (a) $V \cup \{1\}$ is a spanning set, and has measure $1$. (b) No subset of $V$ is a spanning set, so for every spanning set $E$ we have $\mu(E) \ge 1$.

Therefore $\dim_\mu(\mathbb R / \mathbb Q) = 1$.

Answer 2

EDIT: as GEdgar pointed out, this does not answer the question.

This question is actually not very interesting; maybe it can be modified. Just let $\mu$ have weight $1$ on each of the rationals. Any Vitali set contains exactly one rational, so will have measure 1.