# Minimum of $F$ over Finite Perimeter Sets in $\mathbb R^N$

Problem: Let $G$ be a bounded Borel set. Let $X$ be the set of finite perimeter sets in $\mathbb R^N$ and $F: X \to \mathbb R \cup \{+\infty\}$ defined as

$F(E)= \begin{cases} Per(E) \hspace{1,5cm} \text{if} \hspace{0,2cm} \mathcal L^N(G - E)=0,\\ +\infty \hspace{2,1cm} \text{otherwise}. \end{cases}$

show that $F$ admits minumum.

Attempt: We want to use the Direct Method. We must chose the right topology on $X$. It is probably the topology induced by $L^1_{Loc}$. If $\chi_{E_n} \to \chi_{E}$ in $L^1_{Loc}$ we get $\mathcal L^N(G - E) \leq \mathcal L^N(G - E_n) + \mathcal L^N(E_n - E) = 0 + \epsilon = \epsilon$ thus the subset where $F(E)=Per(E)$ is closed for sequences. We need to show that $F\not\equiv +\infty$, $F$ is coercive, $F$ is lower semi-continuous for sequences. Pick a ball containing $G$, we get $F\not\equiv +\infty$. $Per(*)$ is lower semi-continuous thus so is $F$ thanks to the previous observation. I have some problems showing $F$ is coercive, because when $F(E) \leq \lambda$ of course $Per(E)\leq \lambda$, but in order to use the Compactness Theorem we need also $|\chi_E|_1\leq c$ but that could not be afforded I guess, because we have no control on the measure of a set covering $G$ with perimeter $\leq \lambda$.

Thanks.

The minimizer of $F$ is $L^N$-equivalent to the minimal hull containing the bounded obstacle $G$. Then the minimizer exists due to standard compactness theorem. This one-sided minimizer is also called pesudoconvex set. More references about minimal hull and pseudoconvex set include the papers by M. Miranda(1971), N. Fusco(2004) and E. Barozzi(2009).
The isoperimetric inequality should do it for you: $$Per(E)\geq C_N \mathcal{L}^N(E)^{\frac{N-1}{N}}$$ where $C_N$ depends only on $N$.