Lebesgue measure of boundary of Caccioppoli set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:49:26Z http://mathoverflow.net/feeds/question/6253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6253/lebesgue-measure-of-boundary-of-caccioppoli-set Lebesgue measure of boundary of Caccioppoli set Martijn 2009-11-20T11:11:19Z 2010-01-05T02:10:35Z <p>Can anything be said about the measure of the <em>topological</em> boundary of a Cacciopoli set in $R^n$? Of course, the reduced boundary has finite (n-1)-dimensional Hausdorff measure, but this does not say anything about the topological boundary, for instance, points with density 0 or 1 can still be part of the boundary.</p> <p>My precise question: if $E$ is a Caccioppoli set, does there exist a measurable Cacciopoli set $F$, such that $E\triangle F$ and $\partial F$ are both Lebesgue null sets?</p> http://mathoverflow.net/questions/6253/lebesgue-measure-of-boundary-of-caccioppoli-set/10766#10766 Answer by 002 for Lebesgue measure of boundary of Caccioppoli set 002 2010-01-05T02:10:35Z 2010-01-05T02:10:35Z <p>The answer is no. Take countably many disjoint closed balls <code>$B_i$</code> contained in the square $Q=[0,1]\times [0,1]$ and such that:<br /> (i) Sum of areas of <code>$B_i$</code> is less than 1<br /> (ii) Sum of perimeters of <code>$B_i$</code> is finite<br /> (iii) <code>$\bigcup B_i$</code> is dense in $Q$<br /> Since the series <code>$\sum \chi_{B_i}$</code> converges in BV norm, the set <code>$E=Q\setminus \bigcup B_i$</code> has finite perimeter. It also has positive measure and empty interior. Any representative $F$ of the set $E$ also has empty interior and therefore $\partial F$ is not Lebesgue null.</p> <p><hr /></p> <p>By the way, any Lebesgue measurable set E has a representative F with the property </p> <p>(*) $0&lt;|F\cap B(x,r)|&lt;|B(x,r)|$ for all $x\in\partial F$ and all $r>0$. </p> <p>The proof is straightforward: add the points x for which $|E\cap B(x,r)|=|B(x,r)|$ for some r, and throw out all points x such that $|E\cap B(x,r)|=0$ for some $r$. (See Prop. 3.1 in "Minimal surfaces and functions of bounded variation" by E. Giusti.) By virtue of (*) the set $F$ has the smallest (w.r.t inclusion) topological boundary among all representatives of $E$, so if this representative doesn't help you, nothing does. </p>