# A dense measure $0$ $G_\delta$ subset of the Fat Cantor set?

The fat Cantor set is a nowhere dense subset of $$\mathbb{R}$$ with positive Lebesgue measure. My question is, does there exist a $$G_\delta$$ set dense in the fat Cantor set with Lebesgue measure $$0$$?

If such a set does exist, is it possible to produce an actual example of it?

• Dense in what? The original fat Cantor set? – Ian Nov 11 '18 at 0:12
• @Ian Yeah, dense in the fat Cantor set. I edited to clarify. – Keshav Srinivasan Nov 11 '18 at 0:16
• And it is also $G_\delta$ in the fat Cantor set, or is it $G_\delta$ in $\mathbb{R}$? (The two aren't homeomorphic so it matters.) – Ian Nov 11 '18 at 0:28
• @Ian Yes, $G_\delta$ in the fat Cantor set. – Keshav Srinivasan Nov 11 '18 at 0:42
• @Ian Why does it matter? The fat Cantor set is a closed set, so it';s a $G_\delta$ set, so any relative $G_\delta$ set in the fat Cantor set is a real $G_\delta$ set, right? – bof Nov 11 '18 at 1:40

## 1 Answer

Your fat Cantor set is closed, so it's a $$G_\delta$$ set. If $$C$$ is any $$G_\delta$$ subset of $$\mathbb R$$, there is a $$G_\delta$$ subset of $$C$$ which is dense in $$C$$ and has measure zero. Namely, let $$D$$ be a countable dense subset of $$C$$, and let $$A$$ be a $$G_\delta$$ set of measure zero containing $$D$$. Then $$A\cap C$$ is a $$G_\delta$$ set of measure zero and is dense in $$C$$.

To give an explicit example, you would start by defining an explicit fat Cantor set. Next, you need a countable dense subset $$D$$; you can do that by taking, for each interval $$[a,b]$$ with rational endpoints such that $$[a,b]\cap C\ne\emptyset$$, the least element of $$[a,b]\cap C$$. Next, you need to specify an enumeration of $$D$$; that is easily obtained an enumeration of the rational intervals $$[a,b]$$. So now we have $$D=\{d_n:n\in\mathbb N\}$$, a countable dense subset of $$C$$. Finally, define $$A=\bigcap_{k=1}^\infty\bigcup_{n=1}^\infty\left(d_n-2^{-n-k},d_n+2^{-n-k}\right).$$

• Why do $D$ and $A$ exist? – Keshav Srinivasan Nov 11 '18 at 2:03
• Every subset of $\mathbb R$ (or any separable metric space) has a countable dense subset. Every countable set has Lebesgue measure zero. Every set of Lebesgue measure zero is contained in a $G_\delta$ set of measure zero. – bof Nov 11 '18 at 2:07
• OK thanks for your answer. I used it to post this answer: math.stackexchange.com/a/2993377/71829 – Keshav Srinivasan Nov 11 '18 at 2:36