= Solution
First apply <Lebesgue decomposition> to the <derivative> <measure> relative to $\mathcal L^n$:
$$
Du=D^au+D^su,\qquad D^au=\nabla u\,\mathcal L^n,\qquad D^su\perp\mathcal L^n.
$$
Here $\nabla u$ is the almost-everywhere <approximate gradient>, rather than an assertion that $u$ belongs to $W^{1,1}$. Split the singular part into its <jump part of a BV derivative> and <Cantor part of a BV derivative>:
$$
\boxed{Du=\nabla u\,\mathcal L^n+(u^+-u^-)\nu_u\,\mathcal H^{n-1}\!\lfloor J_u+D^cu.}
$$
An <approximate jump point> has a unit normal $\nu_u$ and distinct finite <BV traces on a hypersurface> $u^+,u^-$, obtained as mean limits on the corresponding two half-balls. Their set $J_u$ is the <jump set of a BV function>, countably $(n-1)$-rectifiable. Reversing the normal swaps the <BV traces on a hypersurface> and leaves the displayed <measure> unchanged. The <approximate discontinuity set> $S_u$ differs from $J_u$ only by an $\mathcal H^{n-1}$-null set.
The remaining $D^cu=D^su-D^ju$ is singular to <Lebesgue measure> and gives zero mass to every set with sigma-finite $\mathcal H^{n-1}$ <measure>. It is diffuse rather than a second jump contribution. In one dimension the three parts are illustrated by an <affine function>, a <step function> and the <Cantor function>, respectively. Countably many jumps therefore do not imply that the singular <derivative> has no <Cantor part of a BV derivative>.
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