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.



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


