To 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 shown in a related question,

The $$ { rm Tr} , D_Sf = 0. $$

Does that mean that? $ D_S f $ is *nearly* absolutely uninterrupted in a sense? How can one properly formulate this notion of "almost" absolute continuity?

Here is a more specific question:

- As mentioned in the question Lusin Lipschitz approximation in BV and Sobolev room, $ f $ Lipschitz is outside a small amount (small in relation to the Lebesgue measure). does $ { rm Tr} D_S f = 0 $ imply that this quantity is also "small" in relation to the singular measure $ D_S f $?

Related questions are placed in the BV function with absolutely continuous divergence and role of the absolute continuity of the divergence of the BV function to demonstrate renormalization properties