Absolutely continuous functions have Luzin's property $(N)$

Given $\epsilon >0$, choose $\delta$ from Absolute continuity on a compact interval. The endpoints have a finite image, so it suffices to consider $E\cap (a,b)$. Outer regularity Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it gives an open $U\subseteq (a,b)$ containing it with $\lambda (U)<\delta$.

Write $U={⨆}_j(a_j,b_j)$. On each $[a_j,b_j]$, continuity gives points $c_j,d_j$ of minimum and maximum value, ordered so that $c_j\le d_j$. Hence $F([a_j,b_j])$ is an interval of length $\mathrm{\mid }F(d_j)-F(c_j)\mathrm{\mid }$. The intervals $[c_j,d_j]$ are disjoint and have total length at most $\lambda (U)$; the AC estimate applies to every finite subfamily, so its nonnegative countable sum is at most $\epsilon$.

Subadditivity gives ${\lambda }^{\ast }(F(E\cap (a,b)))\le {\sum }_j\mathrm{\mid }F(d_j)-F(c_j)\mathrm{\mid }\le \epsilon$. Letting $\epsilon \downarrow 0$ and restoring the finite endpoint image proves property $(N)$ as defined in Luzin's property $(N)$ on a compact interval.