Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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. –  Martijn Jan 8 '10 at 11:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.