# Every convex set is of locally finite perimeter

I need to prove that every convex subset of $$\mathbb{R}^n$$ is of locally finite perimeter.

$$E$$ is of locally finite perimeter if there exists a vector-valued Radon measure $$\mu_E$$ s.t. the Gauss Green theorem holds: that is for each compactly supported vector field $$T$$ $$\int_{E}div(T)=\int_{\mathbb{R}^n}T\cdot d\mu_E.$$

Moreover the perimeter of $$E$$ is defined as the total variation of $$\mu_E$$, that is $$P(E;A):=|\mu_E|(A)$$.

Let me state the following lemma which the book suggests to use: Let $$H_t=\{x : e\cdot x for $$t\in \mathbb{R}$$ and $$e\in S^{n-1}$$ (an half space) and $$E$$ a set of locally finite perimeter with $$|E|<\infty$$. Then $$\mu_{E\cap H_t}= (\mu_E)_{|_{H_t}}+ eH^{n-1}_{|_{E\cap H_t}}.$$ From this follow that $$H^{n-1}(E\cap \partial H_t)\leq P(E; H_t)$$, $$P(E\cap H_t)\leq P(E)=P(E;\mathbb{R}^n)$$.

Let now $$C$$ be a convex set, this happens if and only if $$\bar{C}=\bigcap_n H_n$$ where $$H_n$$ are closed half spaces. The suggestion from the book is the following: first prove that if $$E$$ is of finite perimeter and $$C$$ is convex, then $$P(E\cap C)\leq P(E)$$ (which is an easy consequence of the second inequality in the last result and of the fact that $$C$$ is a countable intersection of half spaces). Then refine this argument to prove that every convex set if of locally finite perimeter.

I didn't get the suggestion of "refine" the argument, and so i am asking for help. Thanks to everyone who will use time to respond me

• A. Ninno can you specify the book? In particular I need the formula above for the intersection between $E \cap H_t$. Is Maggi's book? If yes, in the book the author wrote this formula for a set of finite perimeter in $\mathbb{R}^n$ and you for a locally finite perimeter with $|E| < \infty$. Why? I would need your version :)
– ty88
May 9, 2020 at 14:52

First assume that $$E$$ is compact. Then, your inequality says that you can approximate it from above by a sequence $$E_n$$ of convex polytopes with decreasing perimeters. Then, the sequence $$\mu_{E_n}$$ is weak*-precompact by Banach$$-$$Alaoglu theorem, so we can assume by passing to a subsequence that $$\mu_{E_n}\to \mu$$. Hence $$\int_E \mathrm{div}\,T=\lim_{n\to\infty} \int_{E_n} \mathrm{div}\,T=\lim_{n\to\infty} \int Td\mu_{E_n}=\int Td\mu,$$ and we are done.
If $$E$$ is not compact, intersect it with balls of large radii and and pass to a limit.
• Thank you very much! For convenience of future readers i add that the same argument that you presented works with the compactness of the sets of finite perimeter: that is If $\{ E_h\}_h$ are of finite perimeter and they satisfies $sup_h P(E_h)<\infty$ and $E_h\subset B_R$ for a certain R Then there exists $E$ a set of finite perimeter s.t. $E_h\to E$ and $\mu_{E_h}\to^{\star}\mu_E$. Mar 25, 2020 at 20:10