# Is there a weaker form of the Divergence Theorem?

I know that the divergence theorem holds in $W^{1,1}(\mathbb{R}^n)$. I'm wondering what can be proven for functions less smooth than this. The biggest challenge is obviously the fact that functions without a certain level of smoothness can't be restricted to a lower-dimensional surface, and that integrability of the divergence is not guaranteed for a general distribution less smooth than $W^{1,1}(\mathbb{R}^n)$.

For example, suppose that $\nabla\cdot\mathbf{v}=0$ (in the distributional sense), but that $\mathbf{v}$ is only known to be $L^{1}$. Further let $\Omega_R$ be a nested continuum of domains satisfying some regularity condition. For now, let's say $\Omega_R=B_R(\vec{0})$ for $R>0$.

Can we conclude that for every subset $I\subset \mathbb{R}$, $\int_{\left\{\mathbf{x}\,|\exists R\in I\,\textrm{s.t. }\mathbf{x}\in\partial\Omega_R\right\}} \mathbf{v}\cdot\mathbf{n}\,\mathbf{dx}=0$?

Similarly, if $\nabla\cdot\mathbf{v}=0$ and $\mathbf{v}\in H^{s}(\Omega)$ for some $s>\frac{1}{2}$, does the standard divergence theorem hold ($\int_{\partial\Omega}\mathbf{v}\cdot\mathbf{n}\,dS=0$)?