An elementary proof that finite perimeter implies finite volume (up to complementation) While I was studying some geometric measure theory I came across the following fact:
Let $E \subset \mathbb{R}^n $ be a set of finite perimeter $P(E, \mathbb{R}^n)<+\infty$, prove that $E$ or $E^c$ have finite volume.
This interesting, and at first sight innocuous, proposition is a byproduct of the isoperimetric inequality for sets of finite perimeter when one does not require (as in the classical isoperimetric inequality) that $E$ is bounded or regular. But this requires some fine tools in analysis and geometric analysis, such as sets of finite perimeter, some Sobolev's inequality (that is equivalent to the isoperimetric inequality), and a lot of technicalities. Moreover, using the distributional definition of the perimeter one treats $\chi_E$ as an element of $BV(\mathbb{R}^n)$ , and in some references, I think they implicitly assume that $\chi_E \in BV(\mathbb{R}^n)$ that is, in particular, $\| \chi_E \|_{L^1} = |E| <+\infty $, so there is some sort of circular reasoning going on if one then wants to use this isoperimetric inequality to prove that $|E|<+\infty $. I think that with some care this approach works but it is quite technical.
Nevertheless, it seems to me that such a clean property should have a nice proof for regular sets that only uses the classical isoperimetric inequality for regular bounded sets and some sort of coarea formula at most. I think that, if this has a simple proof in the sense described above, this could be a very nice exercise to give to good students in the future, but I am stuck in the attempt of an "elementary" proof of this property. I restate here what I want to prove and briefly sketch below my attempts.
Fact: Let $E \subset \mathbb{R}^n $ be a set with $\partial E$ smooth and $\mathcal{H}^{n-1}(E)<+\infty$, prove that $E$ or $E^c$ have finite volume.
Proof attempts: We denote by $ \lesssim $ inequalities that are satisfied up to absolute constants depending on $n$ and $\mathcal{H}^{n-1}(\partial E)$, which we know is finite, and we let $B_R$ be the ball of radius $R\ge 0$ centered at the origin. Lastly, we call $w_n=|B_1(0)|$ and $\alpha_n = \mathcal{H}^{n-1}(\partial B_1(0))$ and assume $n \ge 3$. Consider the sets $E \cap B_R$ and apply the isoperimetric inequality to get
$$|E \cap B_R|^{\frac{n-1}{n}} \lesssim \mathcal{H}^{n-1}(\partial (E\cap B_R)) = \mathcal{H}^{n-1}(\partial E \cap B_R) + \mathcal{H}^{n-1}(E \cap \partial B_R) \lesssim 1+\mathcal{H}^{n-1}(E \cap \partial B_R) ,$$
where we have used that $\mathcal{H}^{n-1}(\partial E) < +\infty $. From this inequality I see essentially two ways to continue the argument:

*

*First: note that $\frac{d}{dR}|E \cap B_R| = \mathcal{H}^{n-1}(E \cap \partial B_R)$, hence for $g(R):= |E \cap B_R| $ we can rewrite the above inequality as
$$ g^{\frac{n-1}{n}} \lesssim 1+g' , \tag{1}$$
and we also know that $g(0)=0$ and that $g$ is nondecreasing. Since the exact same argument hold for $E^c$ we know that the same inequality holds for the function $\tilde{g}(R)=w_nR^n-g(R)$, from here we would like to conclude that one between $g$ and $\tilde{g}$ must be bounded. Just comparing the size of both sides one (morally) knows that if $g$ grows like $R^{\theta}$ for some $\theta \in [0,n]$ then necessarily $\theta =0$ or $\theta =n$. This seems to be pointing in the right direction since, if $E$ is the one between $E$ and $E^c$ with finite volume, then $g$ grows like $R^0$ and $\tilde{g}$ like $w_n R^n$. One cannot conclude just by analytic arguments, because one can have $g(R)=t R^n$ and $\tilde{g}(R)=(w_n-t)R^{n}$ (for some $t\in (0,w_n)$) and they both satisfy $(1)$. Nevertheless, in my mind this growth should clearly imply that $\mathcal{H}^{n-1}(\partial E) = +\infty $.


*Second: note that by the above inequality if we can prove that $ f(R) := \mathcal{H}^{n-1}(E \cap \partial B_R)$ is bounded in $R$ (for $E$ or for $E^c$) then we are done. This function $f$ was $g'$ before, but the argument that we want to use to bound $f$ is different from what we did above, so I think the two attempts are not the same. Consider the function $u:=|\cdot| : \partial E \to \mathbb{R}^+ $ (the restriction of the norm to $\partial E$), since $|\nabla_{\partial E} u | \le |\nabla u| =1 $ on $\partial E$ by the coarea formula we have
$$ \int_{0}^\infty \mathcal{H}^{n-2}(\partial E \cap \partial B_R) dR = \int_{\partial E} |\nabla_{\partial E} u| \le \mathcal{H}^{n-1}(\partial E) <+\infty . \tag{2}$$
This encodes the information that the perimeter is finite, it is saying that even if both $E$ and $E^c$ are unbounded (which may be the case) the quantity $\mathcal{H}^{n-2}(\partial E \cap \partial B_r)$ becomes uniformly very small, in the sense that it is integrable. Moreover, say one gives for granted the following version of the isoperimetric inequality for Riemannian manifolds (does someone has a suggestion on how to avoid this?): let $M \subset \mathbb{R}^n$ be a smooth manifold of dimension $m$ and $\vec{H}:M\to \mathbb{R}^n$ the mean curvature vector, then
$$ |M| \lesssim \left( \mathcal{H}^{n-1}(\partial M) + \int_{M} |\vec{H}|\right)^{\frac{m}{m-1}}. $$
Since the mean curvature of the sphere $\partial B_R$ is $|\vec{H}|=\frac{1}{R}$, using this inequality with $M=E\cap\partial B_R$ gives
$$f(R)^{\frac{n-2}{n-1}} = \mathcal{H}^{n-1}(E \cap \partial B_R)^{\frac{n-2}{n-1}} \lesssim \mathcal{H}^{n-2}( \partial E \cap \partial B_R) + \frac{1}{R} f(R) , $$
integrating this inequality on $[0,R]$ and using $(2)$ we get
$$\int_{0}^R f(R)^{\frac{n-2}{n-1}} dR \lesssim 1 + \int_{0}^R \frac{f(R)}{R} dR ,$$
again with $f(0)=0$ and $f$ nondecreasing. Analougusly to above the same inequality holds for $\tilde{f}(R)=\alpha_n R^{n-1}-f(R)$ , and we would like to conclude that one between $f$ and $\tilde{f}$ must be bounded. This inequality seems far stronger than $(1)$ looking at the sizes for both sides, can one conclude from this stronger inequality?
By far mine may not be the best approach to the problem, I would be happy to see something different. To me, there are few elementary tools to compare the volume and the perimeter of a set: the isoperimetric inequality, some sort of calibration/divergence theorem, and the coarea formula/Fubini. Can someone see an elementary way to do this?
Edit: as pointed out below in the answers, assuming the relative isoperimetric inequality
$$ \min(|E\cap B_R|, |E^c \cap B_R|) \le C(n)P(E, B_R)^{\frac{n}{n-1}}$$
the statements follows, I sketch here my approach. By the relative isoperimetric inequality, since $P(E)<+\infty$ we have
$$ \min(|E\cap B_R|, |E^c \cap B_R|) =: \phi(R) \le C ,$$
and $\phi$ is nondecreasing. Now we consider the sequence $(\phi(k))_k$. For each $k \ge 1$ the minimum in the definition of $\phi(k)$ is attained by $E$ or $E^c$ (intersected with $B_{k}$), we claim that definitely the minimum is attained by $E$ or $E^c$, that is this choice cannot oscillate indefinitely. Indeed, suppose there exists a subsequence of indices $\Lambda \subset \mathbb{N}$ such that, for all $k\in \Lambda$:
$$ \phi(k) = |E\cap B_{k}| \,\,\,\,\, \text{and} \,\,\,\,\, \phi(k+1) = |E^c\cap B_{{k+1}}| \,,$$
or the other way around swapping $E$ and $E^c$. Then this would imply
$$|B_{k}| = |E\cap B_{k}| + |E^c \cap B_{k}| \le |E\cap B_{k}| + |E^c \cap B_{{k+1}}| \le 2C \,,$$
letting $ k \to \infty $ with $k\in \Lambda$ we obtain a contradiction. So definitely say $E$ (or $E^c$) attains the minimum in $\phi(k)$, that is
$$ \phi(k) = |E \cap B_k| \le C \,, $$
for all $k$ sufficiently large, letting $k \to \infty$ we conclude $E$ (or $E^c$) must have finite volume.
Final question: what is the easiest way of deducing the relative isoperimetric inequality from the classical one?
 A: This is a classical argument that you can find in Ambrosio, Fusco, Pallara's book, Theorem 3.46. By the local isoperimetric inequality in the unit cube and a scaling argument,
$$c(n)\frac{1}{L^{n-1}}P(E,Q_{L}(x))\geq  \min\left\{\frac{|Q_L(x)\cap E|}{|Q_L(x)|},\frac{|Q_L(x)\setminus E|}{|Q_L(x)|}\right\},$$
where $Q_L(x)$ is any cube of side $L$ centered at $x$.
If $P(E)<\infty$, then you can choose $L$ big enough, say $L=L_0=(4c(n) P(E))^\frac{1}{n-1}$, so that the LHS is $<\frac12$. As a consequence, one of the two terms at the RHS is necessarily $<\frac12$.
Now here's the trick: by continuity in $x$ of the two fractions at the RHS, and since the two terms sum to $1$, the term that is $<\frac12$ must be the same for every cube $Q_{L_0}(x)$! Assume for instance that it's the first one, so that, using $|Q_{L_0}(x)|={L_0}^n$, we obtain
$$c(n){L_0}P(E,Q_{L_0}(x))\geq {|Q_{L_0}(x)\cap E|}$$
for every $x$. Then it is sufficient to cover almost all $\mathbb{R}^n$ with disjoint cubes $Q_{L_0}(x_i)$ and sum the inequality among them to deduce
$$P(E)\geq \sum_i P(E,Q_{L_0}(x_i))\geq \frac{1}{c(n)L_0}\sum_i|Q_{L_0}(x_i)\cap E|=\frac{1}{c(n)L_0}|E|.$$
In particular, $E$ has finite volume, and substituting the value for $L_0$ you also obtain the isoperimetric inequality
$$P(E)\geq c'(n)|E|^\frac{n-1}{n}.$$
If the term that is $<\frac12$ is the other one, then you obtain the same for $\mathbb{R}^n\setminus E$.
