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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?

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  • $\begingroup$ Dense in what? The original fat Cantor set? $\endgroup$
    – Ian
    Commented Nov 11, 2018 at 0:12
  • $\begingroup$ @Ian Yeah, dense in the fat Cantor set. I edited to clarify. $\endgroup$ Commented Nov 11, 2018 at 0:16
  • $\begingroup$ 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.) $\endgroup$
    – Ian
    Commented Nov 11, 2018 at 0:28
  • $\begingroup$ @Ian Yes, $G_\delta$ in the fat Cantor set. $\endgroup$ Commented Nov 11, 2018 at 0:42
  • $\begingroup$ @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? $\endgroup$
    – bof
    Commented Nov 11, 2018 at 1:40

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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 $$A=\bigcap_{k=1}^\infty\bigcup_{n=1}^\infty\left(d_n-2^{-n-k},d_n+2^{-n-k}\right).$$

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  • $\begingroup$ Why do $D$ and $A$ exist? $\endgroup$ Commented Nov 11, 2018 at 2:03
  • $\begingroup$ 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. $\endgroup$
    – bof
    Commented Nov 11, 2018 at 2:07
  • $\begingroup$ OK thanks for your answer. I used it to post this answer: math.stackexchange.com/a/2993377/71829 $\endgroup$ Commented Nov 11, 2018 at 2:36

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