$\textbf{Exercise 12.10}$ If $E$ and $F$ are sets of locally finite perimeter, then $\mu_E = \mu_F$ on $\mathcal{B}_b(\mathbb{R}^n)$ if and only if $|E \Delta F| = 0$. Characterize the case $\mu_E = - \mu_F$ on $\mathcal{B}_b(\mathbb{R}^n)$.
I know that
Let $E$ a Lebesgue measurable set in $\mathbb{R}^n$. We say that $E$ is a $\textbf{set of locally finite perimeter}$ in $\mathbb{R}^n$ if for every compact set $K \subset \mathbb{R}^n$ we have
$$\sup \left\{ \int_E \text{div} T(x) dx \ ; \ T \in C_c^1(\mathbb{R}^n,\mathbb{R}^n), \text{spt} K, \sup_\limits{\mathbb{R}^n} |T| \leq 1 \right\} < \infty. \ (12.1)$$
If this quantity is bounded independently of $K$, then we say that $E$ is a $\textbf{set of finite perimeter}$ in $\mathbb{R}^n$.
$\textbf{Proposition 12.1}$ If $E$ is a Lebesgue measurable set in $\mathbb{R}^n$, then $E$ is a set of locally finite perimeter if and only if there exists a $\mathbb{R}^n$-valued Radon measure on $\mathbb{R}^n$ such that
$$\int_E \text{div} T = \int_{\mathbb{R}^n} T \cdot d\mu_E, \forall T \in C_c^1(\mathbb{R}^n,\mathbb{R}^n). \ (12.2)$$
Moreover, $E$ is a set of finite perimeter if and only if $|\mu_E|(\mathbb{R}^n) < \infty$.
$\textbf{Remark 12.2}$ Of course, $(12.2)$ is equivalent to
$$\int_E \nabla \varphi = \int_{\mathbb{R}^n} \varphi d\mu_E, \forall \varphi \in C_c^1(\mathbb{R}^n).$$
I would like to know if what I did is right.
$(\Longrightarrow)$
If $\mu_E = \mu_F$ on $\mathcal{B}_b(\mathbb{R}^n)$, then
$$\int_E \nabla \varphi = \int_{\mathbb{R}^n} \varphi d\mu_E = \int_F \nabla \varphi, \forall \varphi \in C_c^1(\mathbb{R}^n).$$
by the remark $12.2$. Thus, $\int_{\mathbb{R}^n} \nabla \varphi (\chi_E - \chi_F) = 0$, which imply that $\chi_E - \chi_F = 0$ at $|\cdot|$-a.e. (I'm denoting by $\mu$ the measures $\mu_E$ and $\mu_F$ because these measures are equal by hypothesis and $|\cdot|$ denotes the Lebesgue measure on $\mathbb{R}^n$), but $\chi_E \neq \chi_F$ on $E \Delta F$, then $|E \Delta F| = 0$.
$(\Longleftarrow)$
If $|E \Delta F| = 0$, then $|E \backslash F| = |F \backslash E| = 0$. I want to use this to state
$$\int_E \nabla \varphi = \int_{E \backslash F} \nabla \varphi + \int_{E \cap F} \nabla \varphi = \int_{E \cap F} \nabla \varphi = \int_{F \backslash E} \nabla \varphi + \int_{E \cap F} \nabla \varphi = \int_F \nabla \varphi,$$
but this imply that $\int_{E \backslash F} \nabla \varphi = \int_{F \backslash E} \nabla \varphi = 0$?
I want use this, the hypothesis that $E$ and $F$ are sets of locally finite perimeter and the remark $12.2$ to ensure that $\mu_E = \mu_F$ on $\mathcal{B}_b(\mathbb{R}^n)$, but why the measures agree only on $\mathcal{B}_b(\mathbb{R}^n)$?
$\textbf{P.S.:}$ I believe $\mathcal{B}_b(\mathbb{R}^n)$ denotes the family of the bounded Borel sets of $\mathbb{R}^n$, because the family of the Borel sets of $\mathbb{R}^n$ is denoted by $\mathcal{B}(\mathbb{R}^n)$.