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Consider a compact subset $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?

I guess it's wrong, because the topological assumption is invariant w.r.t homeomorphism, in contrast to being of zero Lebesgue measure. But I don't see any simple counterexample.

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    $\begingroup$ For completeness, let me link to a fully detailed proof of the fact that a regular open subset of $\mathbb{R}$ can have boundary with positive measure, which I posted some time ago on MSE. $\endgroup$
    – Gro-Tsen
    Mar 28 at 12:33

3 Answers 3

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Construct a Cantor set of positive measure in much the same way as you make the `standard' Cantor set but make sure the lengths of the deleted intervals add up to 1/2, say. Let $U$ be the union of the intervals that are deleted at the even-numbered steps and let $V$ be the union of the intervals deleted at the odd-numbered steps. The Cantor set is the common boundary of $U$ and $V$; their closures are as required.

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http://www.jstor.org/pss/1986455 Here is constructed a Jordan Curve with positive measure. This gives an example.

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    $\begingroup$ Ah! You beat me to it. I'll delete my own answer, which is a duplicate of yours (and you need the rep more than I do). Let me just add that this curve (the Osgood curve) has been mentioned here on MO before. The search box will find it for you. $\endgroup$ May 26, 2010 at 12:21
  • $\begingroup$ It does the trick. The Jordan curve in that paper is a "thinned out" variant of the Peano curve. $\endgroup$
    – Xandi Tuni
    May 26, 2010 at 12:24
  • $\begingroup$ Let me mention another reference in book form where the construction of a Jordan curve of positive area is done: Hans Sagan, Space filling curves, chapter 8. $\endgroup$
    – coudy
    Mar 28 at 10:09
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Let $D_0,D_1,\ldots$ enumerate a sequence of disjoint intervals in the unit interval with $\bigcup_n D_n$ open dense and having measure less than $1$. For example, place a very tiny interval around each rational number, so that the sum of the intervals is less than $1$. Now, let $E=\bigcup_n D_{2n}$ be the union of the even intervals and $O=\bigcup_n D_{2n+1}$, the union of the odd intervals. The entire interval is the union of $E$, $O$ and their boundaries, so one of these boundaries must have positive measure. So we may take $K$ to be the closure of $E$ or $O$.

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