Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set?

Question

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?

Answer 1

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).

Answer 2

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.