Hausdorff measure of intersection of a ball and a set in $\mathbb {R} ^ n$ 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))$
 A: 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, 
$$
\vert D(\chi_{E\cap B_{\rho}})\vert = \vert D\chi_E \vert \llcorner_{B_\rho} + \mathcal H^{n-1}\llcorner_{(\partial B_{\rho} \cap E)}, \qquad \mathcal L^1 \text{-a.e. } \rho >0.
$$
