Given $\Omega\subset \Bbb R^N$ is open bounded, we say $E\subset \Omega$ has finite perimeter in $\Omega$ if $\chi_E\in \operatorname{BV}(\Omega)$.
Follows from Evans & Gariepy's book, we write for any $\varphi\in C_c^1(\Omega,R^N)$, if $E$ has finite perimeter then we have $$\int_E \text{div }\varphi dx =\int_\Omega \varphi\cdot \gamma\,\, d\|\partial E\|$$ where $\|\partial E\|$ denote the Radon measure which is the distribution derivative of $\chi_E$.
Question:
Prove that the measure $\|\partial E\|$ is supported inside $\partial E$.
(This comes from Evans & Gariepy, and actually, from Ambrosio, Fusco, Pallara as well.)
My effort:
Pick up any ball $B$ such that $B\cap \partial E=\varnothing$ and try to show that $\|\partial E\|(B)=0$. But it does not really work if I only know the definition I wrote above...
Any help would be really appreciate!!
Here I discovered my answer... it is simple... really...
The key idea is what do we mean by $\|\partial E\|$ is supported inside $\partial E$. We need to show that for any open side $E$ such that $E\cap\partial E = \varnothing$ then we have $\|\partial E\|(B)=0$.
Next, what is $\|\partial E\|(B)$? It is $$\sup\left\{\int_E \text{div} \varphi\,dx,\,\,\varphi\in C_c^1(B),\,\,\|\varphi\|_{L^\infty}\leq 1 \right\} $$
Therefore, if $B$ is outside $E$, we done; if it is inside $E$, by Green theorem, we done too...