Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set? 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?
 A: In Dorais, Filipów, and Natkaniec's paper "On Some Properties Of Hamel Bases and Their Applications to Marczewski Measurable Functions" (also available as a preprint here) the writers point that it is provable that there is a measurable Hamel basis.
They refer to "An Introduction to the Theory of Functional Equations and Inequalities: Cauchy's Equation and Jensen's Inequality" by Marek Kuczma. In a quick Google Books search, it seems that this is remarked as a corollary from previous chapters at the end of page 282.
In Corollary 11.4.3 (p.288) it is stated that non-measurable Hamel bases also exist, as well as a measurable one (whose measure has to be zero).
A: 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.
