Let's view $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and pick some basis $(v_\alpha)_{0 \leq \alpha < \mathfrak{c}}$ of it. We can then consider the subspace $L$ spanned by $(v_\alpha)_{0 < \alpha < \mathfrak{c}}$ (ie leaving out one vector from the basis). Given the horrible way we have built $L$, I don't suppose there is much a priory reason for $L$ to be measurable. However, I am wondering whether we can say something about the outer measure of $L$.

1$\begingroup$ This is the Vitali set, which is not Lebesgue measurable, and in fact has infinite outer measure and zero inner measure $\endgroup$ – Pietro Majer Apr 8 at 15:54

1$\begingroup$ Vitali sets can be found in arbitrarily small intervals, if my memory serves me right. $\endgroup$ – Asaf Karagila Apr 8 at 16:02

$\begingroup$ @AsafKaragila Correct; in fact a nonLebesgue measurable set can be found even in any Lebesgue set of positive measure. $\endgroup$ – Pietro Majer Apr 8 at 16:05

$\begingroup$ @Pietro: Just to clarify on your comment which could be read to imply somehow that Vitali sets are the only examples of nonmeasurable sets, which is of course not true. We have many different examples of nonmeasurable sets! $\endgroup$ – Asaf Karagila Apr 8 at 17:15
This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:
$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.
On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $SS$ is a nbd of $0$. Therefore $V=VV$, which is nowhere dense, contains no measurable set of positive measure.

$\begingroup$ It's not a Vitali set, since it is closed under sums and products by a rational scalar, and a Vitali set is not (a priori) closed under such sum and products. Moreover, Vitali sets can be taken from arbitrarily small intervals so they can have arbitrarily small outer measure. $\endgroup$ – Asaf Karagila Apr 8 at 16:05

$\begingroup$ That's depend on your definition of "Vitali set"; of course there are non measurable sets even into any measurable set with positive measure. A standard way to exhibit a nonLebesgue measurable set is, a rational hyperplane of $\mathbb{R}$, which is essentially the original quotient construction. $\endgroup$ – Pietro Majer Apr 8 at 16:18

2$\begingroup$ Nobody imposed a set of representatives to be bounded. I am simply claiming that there is a bounded one. And I'd be happy to see any reference to Vitali set meaning anything other than the set of representatives. $\endgroup$ – Asaf Karagila Apr 8 at 16:42

4$\begingroup$ @AsafKaragila If the omitted basis vector, called $v_0$ in the problem, happens to be a rational number, then the subspace spanned by the other vectors is a set of representatives of $\mathcal R/\mathcal Q$, a Vitali set. If the omitted basis vector is something else, then you get a set of representatives of $\mathcal R/(v_0\mathcal Q)$, which I think still qualifies as a version of the Vitali set. Shrink it by a factor $v_0$ and it becomes a genuine Vitali set again. $\endgroup$ – Andreas Blass Apr 8 at 17:12

1$\begingroup$ @Skeeve since you made me think, I got a simpler reason (added) $\endgroup$ – Pietro Majer Apr 8 at 18:12