sets with positive Lebesgue measure boundary - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:42:47Z http://mathoverflow.net/feeds/question/25993 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary sets with positive Lebesgue measure boundary coudy 2010-05-26T11:56:23Z 2010-05-26T12:58:26Z <p>Consider a compact subset $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?</p> <p>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.</p> http://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary/25997#25997 Answer by rpotrie for sets with positive Lebesgue measure boundary rpotrie 2010-05-26T12:12:10Z 2010-05-26T12:12:10Z <p><a href="http://www.jstor.org/pss/1986455" rel="nofollow">http://www.jstor.org/pss/1986455</a> Here is constructed a Jordan Curve with positive measure. This gives an example.</p> http://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary/26000#26000 Answer by KP Hart for sets with positive Lebesgue measure boundary KP Hart 2010-05-26T12:19:34Z 2010-05-26T12:19:34Z <p>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.</p> http://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary/26003#26003 Answer by Joel David Hamkins for sets with positive Lebesgue measure boundary Joel David Hamkins 2010-05-26T12:31:18Z 2010-05-26T12:40:08Z <p>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 $O$.</p>