How to show that Bernstein set is not Lebesgue measurable?

Question

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.

Answer 1

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 $0$.

Answer 2

Theorem: Any measurable subset of a Bernstein set must be a null set.

Proof: Let $A$ be a measurable subset of a Bernstein set $B$. Then for any $\epsilon > 0 $ there exists a closed set $F \subset A$ such that $$\lambda (A \setminus F) < \epsilon$$

$F\subset A\subset B$ implies $F\cap B^c=\emptyset $.

Hence $F$ must be countable.

Then $\lambda(A)=\lambda(F)+\lambda(A\setminus F)<\epsilon$

Since $\epsilon>0$ is arbitrary,$\lambda(A) =0$

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Claim: Bernstein set is non measurable.

Proof: Suppose a Bernstein set $B$ is measurable. Then $B$ and $B^c$ both are null sets (complement of a Bernstein set is also Bernstein). This implies the whole space has measure $0$! (contradiction).