A property of sets of finite perimeter

2012-02-15T21:00:05Z

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

$$P_D(A)=\sup \left \lbrace\ \int_D \chi_A {\rm div} \varphi \ dx \ : \ \varphi \in C_c^1(U;\Bbb{R}^N),\ |\varphi|\leq 1\ \right\rbrace $$

If $|A|=V$ with $V\in (0,|\Omega|)$ and $P_D(A)<\infty$ is it true that we can modify $A$ up to a set of measure zero in order to find a small ball included in $A$?

If yes, is there any reference, or easy proof for this?

Thank you.

Answer by Beni Bogosel for A property of sets of finite perimeter

2012-02-16T09:41:30Z

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.

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

$\sum_{n=1}^\infty 2\pi r_n < \infty$
$\sum_{n=1}^\infty \pi r_n^2 < \pi$

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 $Per_D(C)=Per_D(B)<\infty$.

If $C$ would contain an open ball then that ball would intersect $S \subset B$, which is not possible.

A property of sets of finite perimeter - MathOverflow

A property of sets of finite perimeter

Answer by Beni Bogosel for A property of sets of finite perimeter