real analysis - How to show that Bernstein set is not Lebesgue measurable?

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.)

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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.