De Giorgi's structure theorem for reduced boundary - why compact submanifolds?

158 Views Asked by At

De Giorgi's structure theorem states the following (Theorem 15.9 from Maggi's Sets of Finite Perimeter and Geometric Variational Problems, or e.g. Theorem 4.3 of Giusti's Minimal Surfaces and Functions of Bounded Variation).


If $E$ is a set of locally finite perimeter in $\mathbb{R}^n$, then [...] there exist countably many $C^1$ hypersurfaces $M_h$ in $\mathbb{R}^n$, compact sets $K_h \subset M_h$, and a Borel set $S$ with $\mathcal{H}^{n-1}(F)=0$, such that $$\partial^* E=F \cup \bigcup_{h \in \mathbb{N}} K_h.$$


I don't understand how the sets $K_h$ can be taken to be compact if $E$ has vertices. For example, the reduced boundary of a square consists of the square's boundary minus the square's four vertices (Maggi's Example 15.4). How can we write that set as a union of compact submanifolds and a set of measure zero?

1

There are 1 best solutions below

0
On BEST ANSWER

I worked it out. The theorem only requires a countable union of compact sets. We can take a sequence of compact subsets of the form $K_h:=\{x \in \partial E : \operatorname{dist}(x,\partial E \backslash \partial^* E)\ge\frac{1}{h}\}$.