Let $n\ge 2$.
Let $u \in W^{1,1}(\mathbb{R}^n)$, i.e. $u \in L^1(\mathbb{R}^n)$ and its distributional gradient is represented by an element of $L^1(\mathbb{R}^n;\mathbb{R}^n)$. By the Sobolev embedding theorem, if $\Pi$ is a hyperplane then $u$ has trace values in $L^1(\Pi, \mathcal{H}^{n-1})$.
On the other hand, if $E \subset \mathbb{R}^n$ is a measurable set of finite perimeter, i.e. $\chi_E \in BV(\mathbb{R}^n)$, then its measure-theoretic boundary $\partial^*E$ can be decomposed into a denumerable union of $C^1$-hypersurfaces (plus a "negligible set") and its measure-theoretic unit outward normal to $\partial^* E$, say $\nu_E$, coincides with the "classic" normal of these hypersurfaces.
Now, I suspect that an element of $W^{1,1}(\mathbb{R}^n)$ should also admits trace values on these hypersurfaces (after all, they are locally hyperplanes) and that some kind of Green-Gauss theorem should hold, in the following sense: if $ u \in W^{1,1}(\mathbb{R}^n;\mathbb{R}^n)$ then $$\int_E \operatorname{div}(u) \operatorname{d}x = \int_{\partial^*E} u \cdot \nu_E \operatorname{d}\mathcal{H}^{n-1},$$ where in the RHS $u$ should be interpreted as the trace of $u$ on the various hypersurfaces. This formula is known to hold under the further assumption that $u \in C^1_c(\mathbb{R}^n;\mathbb{R}^n)$. However, what if we just assume that $ u \in W^{1,1}(\mathbb{R}^n;\mathbb{R}^n)$? What if we work in the bigger space of functions $u \in L^1(\mathbb{R}^n;\mathbb{R}^n)$ with their distributional divergence that is represented by $L^1(\mathbb{R}^n)$ functions?