# Proof of a convergence of sets in the context of Finite Perimeter sets

Let $$E \subset \mathbb{R}^n$$ be a set of finite perimeter that satisfies $$\mathcal{L}^n (E) < \infty$$. Assume that $$E$$ is symmetric with respect to the hyperplane $$\{x_n = 0\}$$. We know that there exists a sequence $$(E_h)_h$$ of bounded open sets having polyhedral boundary and that are symmetric with respect to the hyperplane $$\{x_n = 0\}$$ such that $$E_h \to E \quad \text{as } h \to \infty \quad \text{and} \quad \lim_{h \to \infty}P(E_h) =P(E).$$ and $$\lim_{h \to \infty} P(E_h; F) = P(E; F) \text{ for all } F\subset \mathbb{R}^n \text{ that satisfies } P(E; \partial F )=0.$$ For all $$h \in \mathbb{N}$$ and $$z \in \mathbb{R}^{n-1}$$ let us set $$\begin{gather*} (E_h)_z := \{ t \in \mathbb{R} : (z,t ) \in E_h \}, \quad (E)_z := \{ t \in \mathbb{R} : (z,t ) \in E \} \\ m_h (z) := \mathcal{L}^1 ((E_h)_z) \text{ with }z \in \mathbb{R}^{n-1} , \quad G_h := \{ z \in \mathbb{R}^{n-1} : m_h (z) > 0 \} \\ m (z) := \mathcal{L}^1 (E_z) \text{ with }z \in \mathbb{R}^{n-1} , \quad G := \{ z \in \mathbb{R}^{n-1} : m(z) > 0 \} . \end{gather*}$$ I have to prove that (up to a subsequence) $$G_h \to G$$ as $$h \to \infty$$ (which means $$\lim_{h \to \infty} \mathcal{L}^{n-1} (G_h \triangle G)= 0$$).

My attempt: if we prove that $$\lim_{h \to \infty} \mathcal{L}^n ( G \setminus G_h) =0 \text{ and } \lim_{h \to \infty} \mathcal{L}^n ( G_h \setminus G) =0$$ then I have finished.

By Fubini's Theorem (and using the fact that the vertical slices of these sets are intervals) it can be showed that $$\begin{equation*} \begin{split} \mathcal{L}^n( & E_h \triangle E) = \int_{\mathbb{R}^{n-1}} \mathcal{L}^1 \bigl( (E_h \triangle E)_z \bigr) \, dz = \int_{\mathbb{R}^{n-1}} \mathcal{L}^1 \bigl( {E_h}_z \triangle {E}_z \bigr) \, dz = \\ & = \int_{\mathbb{R}^{n-1}} \left| \mathcal{L}^1 ({E_h}_z) - \mathcal{L}^1 ({E}_z) \right| \, dz = \int_{\mathbb{R}^{n-1}} \left| m_h(z) - m(z) \right| \, dz. \end{split} \end{equation*}$$ Therefore passing to the limit as $$h \to \infty$$ we get $$0 = \lim_{h \to \infty} \mathcal{L}^n ( {E_h} \triangle E) = \lim_{h \to \infty} \int_{\mathbb{R}^{n-1}} \left| m_h(z) - m(z) \right| \, dz.$$ Up to a subsequence, for a.e $$z \in \mathbb{R}^{n-1}$$, $$m_h(z) \to m(z)$$. This proves that $$\lim_{h \to \infty} \mathcal{L}^n ( G \setminus G_h) =0,$$ Indeed for a.e. $$z \in G$$, since $$m_h(z) \to m(z)$$ and, by definition of $$G$$, $$m(z) >0$$ , we have that $$m_h (z) > 0$$ if $$h$$ is great enough, hence $$z \in G_h$$ if $$h$$ is great enough.

But we I don't know how to prove that $$\lim_{h \to \infty} \mathcal{L}^n ( G_h \setminus G) =0$$. The above argument can't be adapted to this case because I think there could exist $$z \notin G$$ (ie $$m(z)=0$$) such that $$m_h (z) \to m(z)$$ but $$m_h(z) >0$$ for all $$h$$ (thus $$z \in G_h$$ for all $$h$$). I've been trying for hours but I can't conclude!

Thank you very much for any advice!!

• Vertical slices of these sets aren't necessary intervals on $\mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $\left(E_h\right)_z$ an interval. – Jihlbert Jan 6 at 17:18