# Is measure of $\bar{E}$ always equal to zero, where $E$ has measure $0$ and nowhere dense in $\mathbb{R}$

let E is subset of $$\mathbb{R}$$ ,where E is nowhere dense and outer measure of $$E=0$$. Then outer measure of $$\bar{E}=0$$? I think is there exist a subset of $$\mathbb{R}$$ that is nowhere dense set and has measure zero whose closure is equal to generalised Cantor set having some positive measure. Any Hint..

This is false. Let $$C$$ be a fat Cantor set (i.e. Cantor set of positive measure). Let $$E$$ be a countbale dense set in $$C$$. Then $$E$$ is nowhere dense because its closure $$C$$ has no interior. $$E$$ has measure $$0$$ but its closure has positive measure.
• Can we take intersection of all rational number between $[0,1]$ with fat Cantor set having some positive measure as a countable dense set?
• @Tony that depends on the construction of $C$, which may have no rational elements - see math.stackexchange.com/a/133236/6460 Jun 13, 2022 at 11:56
• No you cannot do that. But every subset of $\mathbb R$ (and every subset of any separable metric space) is separable. There is alwasy a countable dense set in any subset of $\mathbb R$. @Tony Jun 13, 2022 at 11:56
• @Tony ... "Can we take intersection of all rational number between [0,1] with fat Cantor set" That depends on the fact Cantor set $C$. If you construct $C$ so that the endpoints of the intervals removed are all rational, then yes. Jun 13, 2022 at 11:57