TITLE: Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set?
QUESTION [16 upvotes]: The existence of subsets of the real line which are not Lebesgue measurable can be argued using the Axiom of Choice. For example, define an equivalence relation on $[0, 1]$ by
$a \thicksim b$ if and only if $a - b \in \mathbb{Q}$
and let $S \subset [0, 1]$ contain exactly one representative from each class; $S$ is not Lebesgue measurable.
Now, the Axiom of Choice also gives us that every vector space has a basis, and in particular $\mathbb{R}$ has a basis over the field $\mathbb{Q}$. Can a similar (but assumedly much more involved) argument show that every such basis is a nonmeasurable set?

REPLY [5 votes]: Here are some short proofs: An easy way to see that there is a null basis of $\mathbb{R}$ over $\mathbb{Q}$ is by noting that $C + C = [0, 2]$ where $C$ is the usual Cantor set. The reason why a Hamel basis cannot be analytic is also easy: If $B$ is an analytic Hamel basis, then the $\mathbb{Q}$-linear span of $B$ without one element is analytic and non measurable which is impossible. Finally, by transfinite induction one can construct a Hamel basis $B$ which meets every perfect set. But then $B$ has full outer measure and zero inner measure so it non measurable.