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!!

  • $\begingroup$ 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. $\endgroup$ – Jihlbert Jan 6 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.