# Showing "directly" that a fat Cantor set contains a non-measurable subset

Let $$L$$ be the Volterra set made by removing the middle open interval of length $$5^{-\nu}$$ at each step. In the limit, this will yield a perfect compact set of positive Lebesgue measure, i.e. a "fat Cantor set" (per a previous homework). I'm the grader in a basic measure theory class, and I'm writing a solution set for a homework problem that says to show that the fat Cantor set $$L$$ has a nonmeasurable subset. I know that in general, it's the case that any subset of $$\mathbb{R}$$ with positive measure contains a nonmeasurable subset, and that this can be shown through Steinhaus's Theorem, or at least I've seen it called that.

My question in whether this could be done in this particular setting without reference to Steinhaus's Theorem, and without too much machinery. The text covers the "construction" of the Vitali set, but does nothing else with nonmeasurable sets until this problem. If there's a solution that doesn't stray too far from what the text has already covered, I wanna use that rather than the (in my opinion not at all intuitive) Steinhaus's Theorem. I have two thoughts for possible solutions.

One thought was to devise a measure space equivalence between $$L$$ (equipped with a normalized measure) and an interval in $$\mathbb{R}$$ by factoring through $$\{ 0 , 1 \}^{\mathbb{N}}$$, pulling a Vitali set back through this mapping to a nonmeasurable subset of $$L$$, and then trying to rephrase all of this without the language of measure space isomorphism. I know this would be possible, but I feel like it'd take a lot of writing and it might not really be clear what's happening.

Another thought I had was to just build a Vitali set contained entirely in $$L$$, but I know that's not possible, since it'd imply that $$\bigcup_{q \in \mathbb{Q}} (K + q) = \mathbb{R}$$, which I don't think should be possible if $$K$$ is nowhere-dense, since that'd mean $$\mathbb{R}$$ is a countable union of nowhere-dense sets.

It's also possible that there's some other way of showing $$L$$ contains a non-measurable subset that's much simpler or intuitive than what I've considered. If so, I don't know what it'd be. Is there a more intuitive solution that I haven't considered here?

• The other well-known construction for non-measurable sets is the Bernstein set ... en.wikipedia.org/wiki/Bernstein_set Nov 22, 2019 at 1:13
• I believe that without the axiom choice, it is consistent with ZF that every set is measurable. So if you aren't using AC, there's something suspicious going on.
– zhw.
Nov 22, 2019 at 1:34
– AJY
Nov 22, 2019 at 1:38
• @zhw. That's right. Under Determinancy, every subset of $\mathbb{R}$ is Lebesgue measurable. I was using scare quotes on "construct" for this reason.
– AJY
Nov 22, 2019 at 1:38
• @bof Sorry, I misspoke. The $L$ here is made by removing the middle interval of legnth $5^{-\nu}$, not the "middle fifth". I'll correct the post.
– AJY
Nov 22, 2019 at 2:48

One method is show there exists a countable family $$F$$ with $$\cup F=\Bbb R,$$ such that for each $$f\in F,$$ the Lebesgue inner measure $$l^i(f)$$ is $$0.$$
Now if $$S\subset \Bbb R$$ is Lebesgue-measurable and not Lebesgue-null then $$\{f\cap S: f\in F\}$$ includes at least one non-measurable set.
Otherwise $$f\cap S$$ is measurable for every $$f\in F,$$ but then $$l(S),$$ the measure of $$S,$$ would satisfy $$l(S)=l(\cup_{f\in F}\,f\cap S)\le \sum_{f\in F}\,l(f\cap S)=\sum_{f\in F}\,l^i(f\cap S)\le \sum_{f\in F}\,l^i(f)=0.$$
• Consider $\Bbb R$ as a vector space over the field $\Bbb Q.$ Let $B$ be a Hamel (vector-space) basis for $\Bbb R$ with $1\in B.$ For $x\in \Bbb R$ there is a unique $\{x_b:b\in B\}\subset \Bbb Q$ such that (i) $x_b=0$ for all but finitely many $b\in B,$ and (ii) $x=\sum_{b\in B}\,bx_b.$ Now let $F=\{f_q:q\in [0,1)\cap \Bbb Q\}$ where $x\in f_q$ if $x_1-q\in \Bbb Z.$ Nov 22, 2019 at 10:06