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In $\mathbb{R}^n$, let $E \subset \mathbb{R}^n$ such that $E$ has measure zero. Prove that $\bar{E}$ and $\partial E$ need not have measure zero.

I think I have a poor understanding of this. I know that $\bar{E} = int E \cup \partial E$

I am thinking that $E$ must be an open set. Because if $E$ were closed, then we get $\partial E\subset E$ and this implies $\partial E$ must have measure $0$. But now I have strayed away from the problem completely.

Should I start with a cover on $\partial E$ and deduce that the total volume over the cover is not necessarily less than $\epsilon$?

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If $E$ is a finite set, then $E$ has measure zero and $\partial E=\bar{E}=E$... – Brandon Apr 20 '14 at 2:19
How do we know $E$ is finite? – Hawk Apr 20 '14 at 2:20
Branden's point is that the assertion to be proved is misstated. It should say that the closure and boundary might not have measure 0, because in some cases they do have measure 0. – Andreas Blass Apr 20 '14 at 2:21
up vote 4 down vote accepted

You're right that $E$ shouldn't be closed, but it doesn't follow that it should be open. The only open set of measure 0 is the empty set. You could take $E$ to be the set of rational numbers (in $\mathbb R$).

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Why can we take $E = \mathbb{Q}$? What if $\mathbb{Q} \subset E$? – Hawk Apr 20 '14 at 2:24
$\mathbb Q$ has measure 0 because it's countable, and its closure and boundary are both $\mathbb R$, which does not have measure 0. – Andreas Blass Apr 20 '14 at 2:25
But how do you know $E$ is the rationals? – Hawk Apr 20 '14 at 2:26
You are allowed to pick what $E$ is. – Ben Blum-Smith Apr 20 '14 at 2:29
I don't know that $E$ is the rationals; I'm giving one example. See Brandon's and my comments on the question on the question for why one couldn't possibly prove the claim for arbitrary $E$. – Andreas Blass Apr 20 '14 at 2:30

Consider $\Bbb Q$. The boundary and the closure of $\Bbb Q$ is $\Bbb R$.

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