# Is a Jordan curve encircling a finite-perimeter set rectifiable?

Let $\gamma:[0,1]\rightarrow \mathbb R^2$ be a (continuous) simple closed curve (Jordan curve). The curve is not assumed to be rectifiable, i.e. we don't assume a priori that the length of the curve $$\textrm{Len}(\gamma):=\sup\left\{\sum_{i=1}^n\left|\gamma(t_{i-1})-\gamma(t_i)\right|\; :\; 0\leq t_0<\ldots<t_n\leq 1,\;n\in\mathbb N\right\}\in[0,+\infty]$$ is finite. Let $E$ be the internal bounded set (from Jordan curve theorem). Assume that $E$ has finite perimeter $\textrm{Per}(E)$ in the sense of the Caccioppoli sets i.e., if $\partial^* E$ is the reduced boundary, then $\textrm{Per}(E)=\mathcal H^1(\partial^* E)<+\infty$. Is it true that the curve $\gamma$ is rectifiable, that is $\textrm{Len}(\gamma)<+\infty$? (In which case $\textrm{Len}(\gamma)= \textrm{Per}(E)$)

The viceversa ($\gamma$ rectifiable hence $E$ of finite perimeter) is easy to find in the literature.

• There is an interesting example in these notes by Nicola Fusco (Example 1.4). Take $\{q_i\}$ a dense set in $\mathbb{R}^n$ and consider $E=\bigcup_{i=1}^{\infty} B_{2^{-i}}(q_i)$. Then $|\partial E|= \infty$, but $E$ has finite perimeter. This is not a counterexample to the question, as $E$ is not a simple curve, but possibly still interesting. May 12, 2016 at 18:47
• The problem still appears difficult to me. I thought at some point that the answer could be affirmative by using the representation theorem for indecomposable finite perimeter set on the plane (see e.g. uam.es/becarios/aferrier/Ferriero-Fusco_8.pdf , Theorem 3). However, it is difficult to get rid of the infinite familiy of curves ${C_i}^-$. May 15, 2016 at 15:43

Yes. The representation theorem for indecomposable finite perimeter sets in the plane cited in @guestDiego's comment (Ferriero and Fusco, 'A note on the convex hull of sets of finite perimeter in the plane', http://www.aimsciences.org/article/doi/10.3934/dcdsb.2009.11.103, Theorem 3) implies that $$\partial E$$ is a union of rectifiable curves whenever $$E$$ is a connected open set of finite perimeter. In this case, since $$E$$ is the interior of $$\gamma$$, we have that $$\partial E=\gamma$$ is a single rectifiable curve.