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I have a question concerning the excess of Caccioppoli sets: Given a Caccioppoli set $E\subset\mathbb{R}^{n}$, and let $A\in\mathbb{R}^{n}$ be open and bounded, we define according to Giusti (see http://www.springer.com/us/book/97808176315360) (Definition 5.1, page 63) the excess of $E$ in $A$ as

$$ \Psi(E,A):= |\boldsymbol{1}_{E}|_{TV(A)} - \inf\left\{|\boldsymbol{1}_{F}|_{TV(A)}: F\Delta E\Subset A \right\} \tag{*}\label{*} $$ with $|\cdot|_{TV}$ the total variation, $\Delta$ the symmetric difference and $X\Subset Y$ meaning that $\overline{X}$ compact subset of $Y$.

I have trouble in understanding this definition. For instance, pick $E=B_{1}(0)$ und $A=B_{2}(0)$, we have a set with a very smooth boundary :), however, when I compute $\Psi(B_{1}(0),B_{2}(0))$ I do not see what would prevent me from picking in the subtrahend in \eqref{*} $F=\emptyset$ resulting in $\Psi(B_{1}(0),B_{2}(0))>0$ when I was expecting that the excess of this set would be zero.

Obviously, I miss an important point or misunderstand the concept. Can you help?

Thanks in advance and best,

Alex

Edit: You can also find the definition in https://www.degruyter.com/view/j/crll.1982.issue-334/crll.1982.334.27/crll.1982.334.27.xml "Boundaries of Caccioppoli sets with Hölder continuous normal vector" by Italo Tamanini. In this publication, you can find it on the bottom of page 2 (Equation 1.1).

Alex
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  • Good point, I have changed the question accordingly. If one does not have access to the book, the paper cited below also defines the excess. Thank you very much for pointing this out! – Alex Feb 09 '18 at 22:26

1 Answers1

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I was expecting that the excess of this set would be zero.

It's not. Nowhere is it claimed that excess is zero for smooth sets. It is claimed (in Remark 5.2) that excess is zero for minimal sets. $B_1(0)$ is not minimal, its boundary has positive curvature.

In general, if $E\Subset A$, then one can take $F$ to be the empty set, and therefore $\Psi(E, A) = |1_E|_{TV(A)}$ in this case.

A set with zero excess would be one obtained by cutting across $A$ by a minimal hypersurface. If $A=B_2(0)$, take any hyperplane $P$ crossing $A$ and let $E$ be either of two parts in which $P$ cuts $A$. Then $E$ is minimal, because a perturbation of its boundary within $A$ does not decrease the size of the boundary.

As a side remark, Giusti defines $$ \psi(f, A) = \int_A|Df| - \inf\left\{\int_A |Dg| : g\in BV(A), \operatorname{spt}(g-f)\subset A\right\} $$ I understand you are interested in the case when $f$ is a characteristic function of a set, but even then the definition does not require $g$ to be a characteristic function of a set.