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

Question

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.

Answer 1

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.