Let $U$ be an open set in $\mathbb{R}^2$. How to prove that the boundary of the CLOSURE of $U$ has Lebesgue measure 0 ? Thanks.


It appears that the statement in your question is not always true. See https://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary for excellent answers. For related details, see also Regular open set whose boundary has nonzero volume. and https://mathoverflow.net/questions/24264/a-question-about-the-osgood-curve.

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    $\begingroup$ In fact, it's not hard to find examples where the measure of $U$ is arbitrarily small while the closure is the entire ${\bf R}^2$, just consider small balls around a countable dense subset. $\endgroup$ – tomasz Jul 19 '13 at 8:49
  • $\begingroup$ @tomasz I agree; however, the boundary of $\mathbb{R}^2$ is the empty set and thus has measure zero. So, I don't think it's a counterexample to the OP's claim. $\endgroup$ – Amitesh Datta Jul 19 '13 at 9:05
  • $\begingroup$ Thanks, I understand. Your links are helpful. I am preparing for my qualifying exam, and this problem is in the collection of past exams, so I didn't expect it to be wrong. Thank you very much for your help. $\endgroup$ – Boyu Zhang Jul 19 '13 at 9:24

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