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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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up vote 21 down vote accepted

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.

share|cite|improve this answer Here is constructed a Jordan Curve with positive measure. This gives an example.

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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. – Harald Hanche-Olsen May 26 '10 at 12:21
It does the trick. The Jordan curve in that paper is a "thinned out" variant of the Peano curve. – Xandi Tuni May 26 '10 at 12:24

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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