Is zero-measured set in $\mathbb{R}^n$ always closed? Or can you please give an example of zero-measured non-closed set in $\mathbb{R}^n$? Here, measure means Lebesgue measure.
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5It cannot be open. But it could fail to be closed, eg ${1/n \ | \ n \in \mathbb{N}}$. – Chris Jan 21 '21 at 00:30
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5Think about $\mathbb{Q}$ – Alexandre Sallinen Jan 21 '21 at 00:31
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The comments already give two examples, but I'm going to chime in here so this question stops showing up as unanswered. Both comments give examples of countable sets that are not closed. Every countable set has Lebesgue measure $0$, so any non-closed countable set will do.
There's also non-closed uncountable sets of measure $0$. Let $C$ be the standard Cantor set in $[0, 1]$. It is an uncountable closed set of measure $0$. The union of two sets of measure $0$ still has measure $0$, so $\mathbb{Q} \cup C$ has measure $0$, and it's definitely not closed.
Even better, this link shows that there are plenty of Lebesgue measure $0$ sets that aren't even Borel.
Michael Jesurum
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