Luzin's property $(N)$ gives an integral growth estimate

Fix $\eta >0$. On the differentiability set, split $E$ into the levels $(k-1)\eta \le \mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }<k\eta$ and then into sets on which the differentiability estimate has one common radius. A cover of each latter set by intervals shorter than that radius shows that its image has outer measure at most $k\eta$ times the outer measure of the set: two points in one covering interval have image distance at most $k\eta$ times its length.

Sum the level estimates. Since $(k-1)\eta \le \mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ on level $k$, this gives ${\lambda }^{\ast }(F(E))\le {\int }_E\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }+\eta \lambda (E)$. Letting $\eta \downarrow 0$ proves the first assertion; the integrability convention is that of Integrable real and complex functions, and their integrals.

If $[a,b]\setminus D$ is null, property $(N)$ Luzin's property $(N)$ on a compact interval makes $F(E\setminus D)$ null. Subadditivity and step 2.1 applied to $E\cap D$ give the stated consequence. The singleton interval is immediate.