# Measure of rational hyperplanes of $\mathbb{R}$

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 $$L$$.

• This is the Vitali set, which is not Lebesgue measurable, and in fact has infinite outer measure and zero inner measure – Pietro Majer Apr 8 at 15:54
• Vitali sets can be found in arbitrarily small intervals, if my memory serves me right. – Asaf Karagila Apr 8 at 16:02
• @AsafKaragila Correct; in fact a non-Lebesgue measurable set can be found even in any Lebesgue set of positive measure. – Pietro Majer Apr 8 at 16:05
• @Pietro: Just to clarify on your comment which could be read to imply somehow that Vitali sets are the only examples of non-measurable sets, which is of course not true. We have many different examples of non-measurable sets! – Asaf Karagila Apr 8 at 17:15

## 1 Answer

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.

• 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. – Asaf Karagila Apr 8 at 16:05
• 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 non-Lebesgue measurable set is, a rational hyperplane of $\mathbb{R}$, which is essentially the original quotient construction. – Pietro Majer Apr 8 at 16:18
• 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. – Asaf Karagila Apr 8 at 16:42
• @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. – Andreas Blass Apr 8 at 17:12
• @Skeeve since you made me think, I got a simpler reason (added) – Pietro Majer Apr 8 at 18:12