Can anything be said about the measure of the topological 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.

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?


1 Answer 1


The answer is no. Take countably many disjoint closed balls $B_i$ contained in the square $Q=[0,1]\times [0,1]$ and such that:
(i) Sum of areas of $B_i$ is less than 1
(ii) Sum of perimeters of $B_i$ is finite
(iii) $\bigcup B_i$ is dense in $Q$
Since the series $\sum \chi_{B_i}$ converges in BV norm, the set $E=Q\setminus \bigcup B_i$ 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.

By the way, any Lebesgue measurable set E has a representative F with the property

(*) $0<|F\cap B(x,r)|<|B(x,r)|$ for all $x\in\partial F$ and all $r>0$.

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.

  • $\begingroup$ Thanks! This was the kind of answer I was looking for. My question actually came from the result of E. Giusti you mention. I was hoping to sharpen it, but clearly this is not possible. $\endgroup$
    – Martijn
    Jan 8, 2010 at 11:45

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