Let $V$ be a Borel set in $\mathbb{R}^n$ such that the Lebesgue measure of $V$, $|V|$, satisfies $|V|\approx |B_1(0)|$, but $|V|\neq |B_1(0)|$ (i.e. $|V|$ is slightly greater or less than $|B_1(0)|$).

We modify the set $V$ to obtain a set $\hat{V}$ satisfying $|\hat{V}|=|B_1(0)|$ in the following way:

Fix a point $x$ on the boundary of $V$, $\partial V$, and let $r>0$ such that if we set $$\hat{V}:=\begin{cases} V\cup B_r(x),\;\text{if}\; |V|\le |B_1(0)|\\ V\setminus B_r(x),\;\text{if}\;\; |V|> |B_1(0)|\\ \end{cases} $$ we have $|\hat{V}|=|B_1(0)|$.

Let $P(E)=\sup\left\{\int_{\mathbb{R}^n}\chi_E(x) \mathrm{div}\boldsymbol{\phi}(x) \, \mathrm{d}x : \boldsymbol{\phi}\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\mathbb{R}^n)}\le 1\right\}$ is be the perimeter of a borel set $E$. See https://en.wikipedia.org/wiki/Caccioppoli_set

Why is $|P(\hat{V})-P(V)|\le P(B_r(x))$?

Heuristically, if we draw a picture, the estimate is clear, but I don't know how to prove it analytically. I tried to prove it with a case distinction, case $|V|\le |B_1(0)|$ and case $|V|> |B_1(0)|$ and using that we have the following for the characteristic functions: $\chi_{V\cup B_r(x)}=\chi_V+\chi_{B_r(x)}-\chi_{V\cap B_r(x)}$ and $\chi_{V\setminus B_r(x)}=\chi_V-\chi_{V\cap B_r(x)}$. However, I am not sure how to handle |sup.. minus sup..|calculations.


The estimate is false. Suppose $B$ is a ball with radius slightly smaller (or slightly larger) than 1, choose point $x$ on the boundary. Now you can modify your set on a small neighbourhood of $x$ so that the Lebesgue measure doesn't change much, but the perimeter increases arbitrarily (think of a comb-like shape on the boundary). That's your $V$, with $|V|\not=|B_1(0)|$. Now, the perimeter of $\hat{V}$ will be small again, as long as $B_r(x)$ contains the "comb" we created earlier. Then $$ |P(\hat{V})-P(V)| \gg P(B_r(x)),$$ with the left hand side arbitrarily large.

In general, for $E$ and $F$ sets of finite perimeter, you have $$P(E\cup F) + P(A\cap F) \le P(E)+P(F),$$ and $$P(E\setminus F)\le P(E)+P(F),$$ see the book of Maggi on sets of finite perimeter, (12.17) and (12.19).


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