# A perfect nowhere dense set which intersects every open set with positive measure?

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.