Let $f: \mathbb R^N \to \mathbb R^N$ be a $BV$ function. Suppose that $\mathrm{div} f$ is absolutely continuous with respect to the Lebesgue measure: $\operatorname{div} f \ll \mathcal L^N$. This implies, as seen in a related question, that $${\rm Tr}\,D_Sf = 0.$$
Does it mean that $D_S f$ is almost absolutely continuous in some sense? What is the correct way to formalize this notion of "almost" absolute continuity?
Here is a more precise question:
- As mentioned in the question Lusin Lipschitz approximation in BV and Sobolev space, $f$ is Lipschitz outside a small set (small with respect to the Lebesgue measure). Does ${\rm Tr} D_S f = 0$ imply that this set is "small" also with respect to the singular measure $D_S f$?
Related question are asked in the posts BV function with absolutely continuous divergence and Role of absolute continuity of divergence of BV function in proof of renormalization property
