Countably exceptional differentiability and integrable derivative imply absolute continuity

Let $Z$ be the countable exceptional set. It is null by Every at most countable subset of $\mathbb{R}$ has measure zero. If $A\subseteq [a,b]$ is null, Luzin's property $(N)$ gives an integral growth estimate applied to $A\setminus Z$ gives ${\lambda }^{\ast }(F(A\setminus Z))=0$, while $F(A\cap Z)$ is countable and hence null. Thus $F$ has property $(N)$.

For $u<v$, continuity gives the interval between $F(u)$ and $F(v)$ as a subset of $F([u,v])$. The latter image is contained in $F([u,v]\setminus Z)\cup F([u,v]\cap Z)$, whose second part is countable and whose first part has outer measure at most ${\int }_u^v\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ by the growth lemma. Hence $\mathrm{\mid }F(v)-F(u)\mathrm{\mid }\le {\int }_u^v\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$. Summing over any partition bounds its variation by ${\int }_a^b\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\infty }$, so $F$ has bounded variation.

Since $F$ is continuous by hypothesis, step 1.1 gives property $(N)$ and step 2.1 gives bounded variation. The reverse direction of Banach--Zarecki characterisation of absolute continuity therefore makes $F$ AC, and Fundamental theorem of calculus for absolutely continuous functions gives the displayed reconstruction formula at every endpoint.