Problem: suppose that $\mu$ is a finite Radon measure in $\Omega \subset \mathbb{R}^N$ such that there exists a finite Radon measure $\nu$ such that: $$\int_{\Omega} \frac{\partial \varphi}{\partial x_i} d\mu = - \int_{\Omega} \varphi d\nu_i$$ for all $\varphi \in C^1_c(\Omega)$. Show that there exists a unique $u \in BV(\Omega)$ such that $\mu=u \mathcal{L}^N$.
Attempt: uniqueness is clear. In fact if the property above holds for $u_1$ and $u_2$ then we obtain that $D(u_1-u_2)=0$ which implies $u_1=u_2$ in every connected component. I am stuck proving existence. I tried using Besicovitch's theorem proving that $\{x : \lim\limits_{\rho \to 0}\frac{|\mu|(B(x,\rho))}{\mathcal{L}^N(B(x,\rho))}= +\infty\}=\emptyset$. In fact this is the support of $\mu^{\perp}$ in the equality $\mu=u \mathcal{L}^N+\mu^{\perp}$.