Let $A$ a subset of $\mathbb R ^n$, $B=B(x,r) \subset \mathbb {R} ^n$ an open ball, and denote the $(n-1)$-dimensional Hausdorff measure in $\mathbb R ^n$ by $\mathcal H^{n-1}$. Also assume that $\mathcal H^{n-1} (\partial A) < + \infty$ (One can assume that $A$ is a set of finite perimeter in necessary).

In this case, is it the following identity holds?

$ \mathcal H^{n-1}(\partial (A \cap B))= \mathcal H^{n-1}((\partial A) \cap B) + \mathcal H^{n-1}( A\cap ( \partial B))$


Let me state and prove the following:

Proposition. Let $E \subset \mathbb R^n$ be a set of finite perimeter. For $\mathcal L^1$-a.e. $\rho>0$ the following equality holds: $$ P(E \cap B_{\rho}) = P(E, B_{\rho}) + \mathcal H^{n-1}(E \cap \partial B_{\rho}). $$

Proof. Let $u \in BV(\mathbb R^n)$ and set $v_\rho:=u \chi_{B_{\rho}}$, being $\chi_A$ the characteristic function of a set $A$. Then we have $$\tag{0} Dv_{\rho} = (Du)\llcorner_{B_\rho} - \gamma_{\rho}(u) \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}} $$ where $\gamma_{\rho}(u)$ is the trace of $u$ on $B_\rho$. By a well-known characterization of the trace operator, we know that if $x \in \partial B_\rho$ $$\tag{1} \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_\rho \cap B_r(x)} \vert u(y)-\gamma_{\rho}(u)(x) \vert \, dy \to 0, \qquad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho. $$ On the other hand, we know that $\mathcal L^n$-a.e. $x$ is a Lebesgue point for $u$, i.e. $$\tag{2} \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x)} \vert u(y)- u(x) \vert \, dy \to 0, \qquad \mathcal L^{n}\text{-a.e. } x \in \mathbb R^n. $$ Furthermore, by Coarea Formula, we know that if $f$ is a Lipschitz map $$ \vert \nabla f \vert \mathcal L^n = \mathcal H^{n-1}\llcorner_{\{f=t\}} \otimes \mathcal L^1. $$ Now take $f=\vert \cdot \vert$ and hence \begin{equation*} \mathcal L^n = \mathcal H^{n-1}\llcorner_{\partial B_{\rho}} \otimes \mathcal L^1, \end{equation*} (which is the well-known formula for the polar change of coordinates). From this it follows that (2) is equivalent to $$ \text{for } \mathcal L^1\text{-a.e. } \rho>0, \, \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x)} \vert u(y)- u(x) \vert \, dy \to 0, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho. $$ Hence, being the integrand non negative, we also deduce
$$\tag{3} \text{for } \mathcal L^1\text{- a.e. } \rho>0, \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x) \cap B_\rho} \vert u(y)- u(x) \vert \, dy \to 0, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho. $$ By comparison between (1) and (3) we get \begin{equation*} \text{for } \mathcal L^1\text{ - a.e. } \rho>0, \quad \gamma_\rho(u) = u \vert_{\partial B \rho}, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho. \end{equation*} Now it is easy to conclude: from (0) we have $$ Dv_{\rho} = (Du)\llcorner_{B_\rho} - u \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}}, \qquad \mathcal L^1 \text{-a.e. } \rho >0 $$ which is $$ D(\chi_{E\cap B_{\rho}})= (D\chi_E)\llcorner_{B_\rho} - \chi_E \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}}, \qquad \mathcal L^1 \text{-a.e. } \rho >0 $$ and since the two measures are mutually singular we deduce, taking total variations, $$ \vert D(\chi_{E\cap B_{\rho}})\vert = \vert D\chi_E \vert \llcorner_{B_\rho} + \mathcal H^{n-1}\llcorner_{(\partial B_{\rho} \cap E)}, \qquad \mathcal L^1 \text{-a.e. } \rho >0. $$

  • $\begingroup$ Thanks! It really make sense. $\endgroup$ – Humed Jan 24 at 15:18
  • 1
    $\begingroup$ @HamedPourmohammad You are welcome, I am glad to help when I can. $\endgroup$ – Romeo Jan 24 at 15:56
  • $\begingroup$ just two (maybe stupid) questions: 1- $E$ is a set of finite perimeter, then by De Giorgi's structure theorem, $P(E)=\mathcal H ^{n-1}(\partial ^\ast E)$. Am I right? 2- I understand that your statement is true for a.e. $\rho$. But fix a ball $B=B_{\rho}$ and let $E= \mathbb {R} ^ n \setminus B$. Then for this $E$ and $B$ the both statements are false. Is it possible to improve it? $\endgroup$ – Humed Jan 28 at 11:48
  • $\begingroup$ Question 1. Yes. Question 2. I have not understood which are the statements you are referring to. $\endgroup$ – Romeo Jan 28 at 13:01

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