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 $L$.

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    $\begingroup$ 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
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    $\begingroup$ 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 non-Lebesgue 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 non-measurable sets, which is of course not true. We have many different examples of non-measurable 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, $S-S$ is a nbd of $0$. Therefore $V=V-V$, 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 non-Lebesgue measurable set is, a rational hyperplane of $\mathbb{R}$, which is essentially the original quotient construction. $\endgroup$ – Pietro Majer Apr 8 at 16:18
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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @Skeeve since you made me think, I got a simpler reason (added) $\endgroup$ – Pietro Majer Apr 8 at 18:12

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