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 $L$.



Measure of rational hyperplanes of $\mathbb{R}$
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, $S-S$ is a nbd of $0$. Therefore $V=V-V$, which is nowhere dense, contains no measurable set of positive measure.


