A perfect set is a closed set with no isolated points. A nowhere dense set is a set whose closure has empty interior. My question is, what is an example of a nonempty perfect nowhere dense subset of $[0,1]$ such that every intersection of the set with an open set is either empty or has positive Lebesgue measure?

Is the Fat Cantor Set an example of such a set? Or does no such set exist?


Any closed set $F\subseteq\mathbb R$ admits a decomposition $F=N\cup F_0$ where $N$ has Lebesgue measure zero and $F_0$ is a closed set whose intersection with every open set is either empty or has positive measure. Of course $F_0$ can have no isolated points. Moreover, if $F$ is nowhere dense then $F_0$ is nowhere dense, and if $F$ has positive measure then $F_0$ is nonempty. Therefore, any nowhere dense closed set $F$ of positive measure will contain a set $F_0$ with the properties you want. Also, if $F$ arises from the usual construction of a "fat Cantor set", then $N=\emptyset$ and $F_0=F$.

P.S. Given a closed set $F$, let $\mathcal U$ be the collection of all open intervals $I$ with rational endpoints such that $F\cap I$ has measure $0$. Then $U=\bigcup\mathcal U$ is an open set and $N=F\cap U$ has measure 0. Let $F_0=F\setminus N=F\setminus U$. Then $F_0$ is a closed set, and the intersection of $F_0$ with any open set, if nonempty, has positive measure.

  • $\begingroup$ How do you prove that any closer set has such a decomposition? $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 13:21
  • $\begingroup$ The decomposition can be proved in a way analogous to the "condensation point" method for proving the Cantor-Bendixson theorem. A couple of possibly useful search phrases are "measure dense set" and "metrically dense". $\endgroup$ – Dave L. Renfro Nov 11 '18 at 13:27

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