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?

• There are measurable bases, of measure $0$. I believe that under CH, and possibly without it, there are no measurable bases with measure $\gt 0$. – André Nicolas Jun 4 '13 at 16:55
• Sierpinski showed in 1920 that there exist such a basis that has Lebesgue measure zero and F. B. Jones showed in 1942 that no such basis can be an analytic set (hence, no such basis can be a Borel set). For references and more information, see these old sci.math posts. – Dave L. Renfro Jun 4 '13 at 16:56
• @AndréNicolas: Due to Steinhaus theorem, there is no measurable basis with positive measure. – 23rd Jun 4 '13 at 16:58
• So far, the bases are possibly measure zero, definitely not positive measure, and definitely not Borel. Is it easy to show a nommeasurable basis exists? (I assume it is, but I didn't find explicit mention in the references.) – Ben Passer Jun 4 '13 at 17:05

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.
• Could somebody please check that the reasoning in the following comments is correct? $C + C = [0, 2]$ implies that the $\mathbb{Q}$-span of $C$ is all of $\mathbb{R}$, i.e. $C$ is a generating set. A Zorn's lemma argument (with reverse inclusion) shows that $C$ can be reduced to a basis, which must be measurable and null as $C$ is null. – Ben Passer Jun 26 '13 at 15:53
• As for $B$, if $B$ is an analytic basis, then write $1$ as a finite $\mathbb{Q}$ combination of basis elements. We may replace those finitely many elements so that $1$ is an element of the basis, and the new $B$ remains analytic. The $\mathbb{Q}$ span of $B \setminus \{1\}$ is analytic and contains one member of each class (for the relation defined in the question, applied to $\mathbb{R}$ instead of $[0, 1]$), so it is nonmeasurable (a contradiction). – Ben Passer Jun 26 '13 at 15:58