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..
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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.
answered Jun 13, 2022 at 11:47
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$\begingroup$ Can we take intersection of all rational number between $[0,1]$ with fat Cantor set having some positive measure as a countable dense set? $\endgroup$– TonyJun 13, 2022 at 11:53
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$\begingroup$ @Tony that depends on the construction of $C$, which may have no rational elements - see math.stackexchange.com/a/133236/6460 $\endgroup$– HenryJun 13, 2022 at 11:56
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$\begingroup$ 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 $\endgroup$ Jun 13, 2022 at 11:56
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$\begingroup$ @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. $\endgroup$– GEdgarJun 13, 2022 at 11:57
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