A property of sets of finite perimeter - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:01:43Z http://mathoverflow.net/feeds/question/88560 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88560/a-property-of-sets-of-finite-perimeter A property of sets of finite perimeter Beni Bogosel 2012-02-15T21:00:05Z 2012-03-01T10:22:12Z <p>I have a question regarding sets of finite perimeter. I feel that it should be true, but I didn't manage to prove or find a reference about it. Suppose $D$ is an open, bounded subset of $\Bbb{R}^n$, and define the perimeter of a measurable set $A \subset D$ as </p> <p>$$P_D(A)=\sup \left \lbrace\ \int_D \chi_A {\rm div} \varphi \ dx \ : \ \varphi \in C_c^1(U;\Bbb{R}^N),\ |\varphi|\leq 1\ \right\rbrace$$</p> <blockquote> <p>If $|A|=V$ with $V\in (0,|\Omega|)$ and $P_D(A)&lt;\infty$ is it true that we can modify $A$ up to a set of measure zero in order to find a small ball included in $A$?</p> </blockquote> <p>If yes, is there any reference, or easy proof for this?</p> <p>Thank you. </p> http://mathoverflow.net/questions/88560/a-property-of-sets-of-finite-perimeter/88617#88617 Answer by Beni Bogosel for A property of sets of finite perimeter Beni Bogosel 2012-02-16T09:41:30Z 2012-02-16T09:41:30Z <p>I'm sorry that I answer my own question, but I found out the answer this morning from my teacher. There are examples of sets of finite perimeter with positive measure, which do not contain any open ball.</p> <p>For example, take $D=B(0,1)$, the unit ball in $\Bbb{R}^2$ and denote $S=D \cap \Bbb{Q}^2=(x_n)_{n \geq 0}$. Then, we can find a sequence of positive real numbers $(r_n)$ such that</p> <ul> <li>$\sum_{n=1}^\infty 2\pi r_n &lt; \infty$</li> <li>$\sum_{n=1}^\infty \pi r_n^2 &lt; \pi$</li> </ul> <p>Take $B=\bigcup_{n=1}^\infty B(x_n,r_n)$. Then $C=D\setminus B$ has the desired property. Indeed, $|C|=|D|-|B|>0$, and $Per_D(C)=Per_D(B)&lt;\infty$. </p> <p>If $C$ would contain an open ball then that ball would intersect $S \subset B$, which is not possible.</p>