Exercise on BV functions and Radon measures I don't how to solve the following exercise (3.2 from "Ambrosio, Fusco, Pallara - Functions of bounded variations and free discontinuity problems").
Let $\mu$ be a finite Radon measure in $\Omega \subset \mathbb{R}^n$ whose first order derivative $\nu$ is a finite Radon measure, i.e.
$$ \int_{\Omega} \frac{\partial\phi}{\partial x_i} d\mu = -\int_{\Omega} \phi d \nu_i \quad \forall \phi \in C^1_c(\Omega), \quad i=1, ..., n $$
Show that there exists a unique $u \in BV(\Omega)$ such that $\mu = u\mathcal{L}^n $.
I have no idea of how to proceed
 A: I am not very knowledgeable on measure theory, I just recently started to learn about it, but I want to give it a try. This means of course that the following might not be perfectly correct. From the assumptions and the definition of a BV function, it is clear that it suffices to find a unique $u$ in $L^1$ such that $\mu = u L^n$, so this is what I will prove.
Let us start with $\mathbf{n = 1}$. Since $\mu$ and $\nu$ are finite measures it is no restriction to assume $\Omega$ to be bounded. Define $f_\nu(x) = \nu((-\infty, x) \cap \Omega)$. We have that $f_\nu$ is bounded and hence in $L^1(\Omega, \mathcal{L}^1)$. 
Now let $(x_i, x_{i+1})$ a partition of $\Omega$ with $x_{i+1}-x_i = \delta$, then
\begin{align*}
\int \phi' f d \mathcal{L}^1 = \lim_{\delta \to 0} \sum_{i} \phi'(x_{i}) f(x_{i}) \delta = \lim_{\delta \to 0} \sum_{i} (\phi(x_{i+1})-\phi(x_i)) f(x_i) \\
= -\lim_{\delta \to 0} \sum_{i} (\phi(x_{i+1}) \nu([x_i, x_{i+1})) = -\int\phi d\nu,
\end{align*}
where the second equality is by definition of the derivative and the second to last equality is by a telescoping sum argument. We conclude that 
$$
\int \phi' f d \mathcal{L}^1 = \int \phi' d \mu \text{ for all } \phi \in C^1_c(\Omega).
$$
It is a standard result that the above inequality implies that $u \mathcal{L}^1:= (f + c) \mathcal{L}^1 = \mu$ for a constant $c\in \mathbb R$. It is clear from the assumptions that $f + c \in BV(\Omega)$. Moreover, since $\mu$ is a finite measure the constant $c$ needs to be unique. This completes the case $n = 1$. 
I am not sure if it is possible to prove the case $\mathbf{n\geq 1}$ without deeper theorems and I would be very interested if there was a simpler solution. I could only do it via mollification and the compactness theorem for BV functions. Let $(\rho_\epsilon)_{\epsilon>0}$ be a family of mollifiers, 
then, by Theorem 2.2 in Ambrosio, Fusco, Pallara, we have that $\rho_\epsilon * \mu \in BV(\Omega)$ for all $\epsilon>0$. By the finiteness of $\mu$ and Theorem 2.2 b) we have that
$$
\sup_{\epsilon>0}\int_{\Omega} |\rho_\epsilon * \mu|dx < \infty.
$$
Moreover, using standard arguments for interchanging convolution and derivatives 
$$
\int_{\Omega} \frac{\partial\phi}{\partial x_i} d \rho_\epsilon * \mu = \int_{\Omega} \rho_\epsilon* \frac{\partial\phi}{\partial x_i} d \mu = \int_{\Omega} \frac{\partial (\rho_\epsilon* \phi)}{\partial x_i}d \mu = -\int_{\Omega} \rho_\epsilon* \phi d \nu_i = -\int_{\Omega} \phi d \rho_\epsilon* \nu_i.
$$
Hence $(\rho_\epsilon* \nu)_i = D \rho_\epsilon* \mu$ and again by Theorem 2.2 b) $\rho_\epsilon* \nu_i$ is a bounded measure such that 
$$
\sup_\epsilon | \rho_\epsilon*\nu_i|(\Omega) < \infty.
$$
By Theorem 3.23 in Ambrosio, Fusco, Pallara (Compactness in BV) we get that a subsequence of $\rho_\epsilon * \mu$ converges weakly$^*$ to a function $u$ in $BV$. By the weak* convergence, we get that 
$$
\int_\Omega \frac{\partial \phi}{\partial x_i} u d \mathcal{L}^n = \lim_{\epsilon \to 0} \int_\Omega \frac{\partial \phi}{\partial x_i} d \rho_\epsilon *\mu = -\lim_{\epsilon \to 0} \int_\Omega \phi d \rho_\epsilon *\nu_i = -\lim_{\epsilon \to 0} \int_\Omega \phi *\rho_\epsilon d \nu_i =  -\int_\Omega \phi d \nu_i.
$$
How this implies the result was already shown in the 1d case.
