# Is every set with finite $\mathcal{H}^{n-1}$ measure a set of locally finite perimeter?

Given a measurable set $$E \subset \mathbb{R}^d$$, with $$\mathcal{H}^{d-1} (\partial E) < +\infty$$, is it true in general that $$E$$ is a set of locally finite perimeter? that is, is it true that $$\int_B |D \chi_E| dx$$ is finite, for every bounded ball $$B \subset \mathbb{R}^d$$?

It is well-known in geometric measure theory that, in general, the perimeter $$P(E)$$ of a measurable set $$E \subset \mathbb{R}^d$$ does not equal to $$\mathcal{H}^{d-1} (\partial E)$$; unless $$E$$ has some nice regularity properties, for example when it has $$C^2$$ boundary. In any case, by De Giorgi's structure theorem, there is a set $$\partial ^\ast E$$, called the reduced boundary of $$E$$, which the equality $$P(E)= \mathcal{H}^{d-1} (\partial^\ast E)$$ holds. Recall that by definition, $$P(E)<+\infty$$ if the characteristic function $$\chi_E$$ belongs to the space $$BV$$ of functions with bounded variation. Thus, my question is about the existence of reduced boundary $$\partial^ \ast E$$ for a set $$E$$ with with $$\mathcal{H}^{d-1} (\partial E) < +\infty$$; rather than any claim about equivalence between the two boundaries. Thus it maybe true that $$P(E)$$ exists, but $$P(E) \not = \mathcal{H}^{d-1} (\partial E)$$.

It must be said that, I guess the answer is negative, but I have no idea to prove it.

• If you mean the topological boundary of E, then the result is true, and it is not too difficult to prove. It can be found in Ambrosio, Fusco, Pallara book in chapter 3 somewhere. If instead you mean the measure theoretic boundary (points where the density either does not exist, or if it exists it is different from 0 and 1) then it is still true but much more difficult to prove (see the current answer by Leo Moos)
– Del
Apr 10, 2022 at 19:32
• @Del Thanks a lot. I think I get the point; but I still need to check reference for more details.
– XIE
Apr 14, 2022 at 18:54

The reduced boundary can be defined for just about any (measurable) subset $$E \subset \mathbf{R}^n$$, whether it is a Caccioppoli set or not.

The precise result you seem to be after should be Theorem 4.5.11 in Federer's book; in my edition this is on page 506. Let me just quickly restate it here.

Define two sets $$Q$$ and $$R \subset \mathbf{R}^n$$ respectively as containing those points where the densities of $$\mathcal{H}^n$$ restricted to $$E$$ and $$\mathbf{R}^n \setminus E$$ respectively are zero. Paraphrasing slightly, Federer's theorem states that if $$\mathcal{H}^{n-1}(K \setminus (Q \cup R)) < \infty$$ for all compact $$K \subset \mathbf{R}^n$$, then $$E$$ is a Caccioppoli set.

I haven't yet untangled the proof that Federer gives. Given that the statement seems a bit hard to find in other texts, I suspect it might be a bit technical.

Remark 1. Note that the interior of $$E$$ belongs to $$R$$ and that of its complement is in $$Q$$, so that $$\mathbf{R}^n \setminus (Q \cup R) \subset \partial E$$. If I am not mistaken this gives the result you're looking for.

Remark 2. A more precise re-telling of Federer's result would be stated in terms of what is called the $$(n-1)$$-dimensional integral-geometric measure with exponent $$1$$, which he denotes $$\mathscr{I}_1^{n-1}$$. However, this is comparable to the Hausdorff measure, up to a dimensional constant $$\beta = \beta_1(n,n-1)$$: $$\mathcal{H}^{n-1} \geq \beta \mathscr{I}_1^{n-1}$$. The immediately relevant sections in Federer's books are 2.10.5 and 2.10.6, on pages 172-174 in my edition.

• As I know the notion of reduced boundary is defined for a set of locally finite perimeter, see for example, p. 167 in chapter 15 of maggi's book "Sets of finite perimeters and geometric variational problems'; or definition 3.3 in Giusti's book "minimal surfaces and functions of bounded variation". The question can be stated a follow: if $\mathcal{H}^{d-1}(E)<+\infty$, then is it true in general, that the function $\chi_E$ belongs to the space $BV$ or not?
– XIE
Apr 8, 2022 at 20:26
• @XIE Sorry, the original answer was perhaps a bit too short. I've added more information to better explain what I was referring to. Does this answer your question now? Apr 8, 2022 at 23:49
• Thanks. It really makes sense now. It seems that you mention Federer theorem about essential boundary. Though, I should try to go into details of proof.
– XIE
Apr 14, 2022 at 18:52