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

Solution 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).

Solution 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.