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))$



Hausdorff measure of intersection of a ball and a set in $\mathbb {R} ^ n$
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, 
$$
\vert D(\chi\_{E\cap B\_{\rho}})\vert = \vert D\chi\_E \vert \llcorner\_{B\_\rho} + \mathcal H^{n-1}\llcorner\_{(\partial B\_{\rho} \cap E)}, \qquad \mathcal L^1 \text{-a.e. } \rho >0.
$$


