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 negligible 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