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A Bernstein set is a subset of the real line that meets every uncountable closed subset of the real line but that contains none of them. (They can be easily generalized to $\mathbb R^n$, but for the sake of simplicity we might stick with $\mathbb R$.) See also this post for some more details: What's application of Bernstein Set? (A proof of existence of Bernstein sets is given there - it is based on transfinite induction and well-ordering theorem. Also various properties of Bernstein sets are mentioned there, including the fact that existence of Bernstein sets cannot be shown in ZF.)

My questions are:

  • How can we show that a Bernstein set is not Lebesgue measurable?
  • Can these proofs be generalized to other measures?

By the latter I mean whether something like this can be said about some of the proofs: "We have shown that Bernstein set is not Lebesgue measurable. But the same proof works for any translation-invariant measure such that all closed sets are measurable and bounded sets have finite measure." (This is just a hypothetical example to make a bit clearer what I mean by the second question.)


When I was thinking about this problem, I thought that one way to go would be using regularity of Lebesgue measure. Let $B$ be a Bernstein set. If $C\subseteq B$ is compact, then it has to be countable and thus $\mu(C)=0$. Similarly, if $B\subseteq U$ then $B$ does not intersect the closed set $\mathbb R\setminus U$, hence $U$ is complement of countable set and $\mu(U)=\infty$. So from regularity of Lebesgue measure we get that $B$ is not measurable.

I have considered also posting my attempt sketched in the previous paragraph as an answer. But I decided not to do so - maybe somebody who knows more about this will be able to expand on this or add some other related results and useful observations.

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  • $\begingroup$ Well, the outer measure of a Bernstein set is full. The inner measure, as you point out, is null. $\endgroup$ – Asaf Karagila Dec 10 '17 at 6:58
  • $\begingroup$ @AsafKaragila I am not sure whether the rhyme was intentional or just a lucky accident, but this reminded my of Yemon Choi's poems on meta/tea. $\endgroup$ – Martin Sleziak Dec 10 '17 at 7:03
  • $\begingroup$ While I am no Fezzik, I do enjoy the occasional rhyme (and probably more than just some other next guy). The one above was by pure luck, not unlike a chicken and duck. (Sorry, I've been marathoning Friends recently...) $\endgroup$ – Asaf Karagila Dec 10 '17 at 7:08
  • $\begingroup$ @MartinSleziak: They don't actually rhyme. "Full" has the vowel of "foot" and "null" has the vowel of "nut". $\endgroup$ – Eric Wofsey Dec 10 '17 at 7:08
  • $\begingroup$ @EricWofsey Well, you're certainly right. That's the problem when I try to discuss rhymes or puns in English. (I am not a native English speaker.) $\endgroup$ – Martin Sleziak Dec 10 '17 at 7:11
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Here is one quick way of doing this: The complement of a Bernstein set is a Bernstein set.

Now, it is enough to argue that a Bernstein set has inner measure $0$. That implies that also its complement—another Bernstein set—has inner measure $0$, and that is certainly impossible if the sets were measurable.

And indeed, every compact subset of a Bernstein set is countable. So the inner measure is $0$.

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  • $\begingroup$ how can we prove that the complement of a Bernstein set is Bernstein? $\endgroup$ – Pol Sep 18 '18 at 20:59
  • $\begingroup$ Well. Can the Complement contain a perfect subset? Can it be disjoint from one? $\endgroup$ – Asaf Karagila Sep 18 '18 at 21:11

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