Banach--Zarecki characterisation of absolute continuity

If $F$ is AC, its defining interval estimate directly gives continuity and bounded variation; it has $(N)$ by Absolutely continuous functions have Luzin's property $(N)$.

Conversely assume continuity, BV, and $(N)$. The normalized Jordan decomposition Jordan decomposition for functions of bounded variation gives $F=F(a)+P-N$, and For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase gives $F^{\mathrm{\prime }}=P^{\mathrm{\prime }}-N^{\mathrm{\prime }}\in L^1$ wherever both derivatives exist. The BV differentiability theorem Every function of bounded variation is differentiable almost everywhere makes that a full-measure set.

Apply the consequence of Luzin's property $(N)$ gives an integral growth estimate to each interval $[u_j,v_j]$ in a finite disjoint family. Since continuity makes $F([u_j,v_j])$ an interval, $\mathrm{\mid }F(v_j)-F(u_j)\mathrm{\mid }\le {\int }_{u_j}^{v_j}\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$. The absolute continuity of the integral Absolute continuity of the integral now gives the finite-family AC condition.

Thus the reverse implication holds; both directions include the singleton interval.