Fractal set $E$ such that the indicator function $\mathbf{1}_E$ is BV

Question

Is there a "fractal" set $E \subset \mathbb R^2$ such that the indicator function $\mathbf{1}_E$ is in $BV(\mathbb R^2)$?

Answer 1

An open subset $E$ of $\mathbb R^{N}$ such that $\mathbf 1_E$ belongs to $BV$ is said to be with "finite perimeter" and this implies that $$ \mathcal H^{N-1}(\partial E)<+\infty, $$ where $\mathcal H^{N-1}$ stands for the $N-1$ Hausdorff measure. Gauss-Green formula holds true for such open sets. As a result, we have in general $$ 0<\mathcal H^{N-1}(\partial E)<+\infty, \text{ which implies for $\epsilon >0$, } \mathcal H^{N-1+\epsilon}(\partial E)=0, \mathcal H^{N-1-\epsilon}(\partial E)=+\infty. $$