Let $(E_j)_j\subset \mathbb{R}^n$ be a sequence of Borel sets with $P(E_j)<\infty$ for all $j$, where $P(E)=\sup\left\{\int_{\mathbb{R}^n}\chi_E(x) \mathrm{div}\boldsymbol{\phi}(x) \, \mathrm{d}x : \boldsymbol{\phi}\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\mathbb{R}^n)}\le 1\right\}$ is the perimeter of a borel set $E$. Let $\lambda(E_j)=\lambda(B_1(0))$ for all $j$, where $\lambda$ is the Lebesgue measure and $B_1(0)$ is the open unit ball in $\mathbb{R}^n$. If $\lambda(E_j \triangle B_1(0))\to 0$ for $j\to\infty$, is then $P(E_j)=P(B_1(0))$ for $j$ large enough?
$\triangle$ denotes the symmetric difference. I'm not sure if the condition $\lambda(E_j \triangle B_1(0))\to 0$ for $j\to\infty$ is needed, but I think so. I am asking this becuase I need something like this to understand a technical proof in a paper. However, I have no idea how to prove it and whether the answer is yes or no. I appreciate any kind of help.
Best Answer
I think $P(E_j)$ can actually go to infinity. Let $E_j$ be $B_1(0)$ except we delete a small arc around $(0,1)$ and replace is by a triangle with height $\sqrt{j}$ and base length $\frac{1}{j}$. Then the areas of $E_j$ do converge to the area of $B_1(0)$ but the perimeters go to infinity.