An absolutely continuous function with zero derivative almost everywhere is constant

Fix $s<t$ and $\epsilon >0$. Absolute continuity gives $\delta >0$ such that every permitted finite interval family of total length below $\delta$ has total $F$-increment below $\epsilon \mathrm{/}2$. Put $D=\{x\in (s,t):F^{\mathrm{\prime }}(x)=0\}$ and $\eta =\epsilon \mathrm{/}(2(t-s+1))$. At every $x\in D$, differentiability supplies arbitrarily short closed intervals $I=[u,v]\subseteq [s,t]$ centred at $x$ such that $\mathrm{\mid }F(v)-F(u)\mathrm{\mid }\le \mathrm{\mid }F(v)-F(x)\mathrm{\mid }+\mathrm{\mid }F(x)-F(u)\mathrm{\mid }<\eta (v-u)\mathrm{.}$ These intervals form a fine cover of $D$.

Apply The Vitali covering theorem for fine covers on the real line with residual outer measure below $\delta$. It gives pairwise disjoint $I_1,\dots ,I_N$ from the fine cover such that ${\lambda }^{\ast }\text{\hspace{0.17em}⁣}(D\setminus {\bigcup }_{j=1}^N{I}_j)<\delta \mathrm{.}$ If $N_0=(s,t)\setminus D$, then ${\lambda }^{\ast }(N_0)=0$, so $R=[s,t]\setminus {\bigcup }_{j=1}^N{I}_j$ has outer measure below $\delta$ (the endpoints add no outer measure).

Order the selected intervals from left to right. The finitely many closed gaps between them, including the end gaps from $s$ and to $t$, have pairwise disjoint open interiors and total length $\lambda (R)<\delta$. Absolute continuity therefore bounds the sum of the $F$-increments over the gaps by $\epsilon \mathrm{/}2$. The estimate in step 1.1 bounds the corresponding sum over the selected intervals by $\eta {\sum }_{j=1}^N\mathrm{\mid }{I}_j\mathrm{\mid }\le \eta (t-s)<\epsilon \mathrm{/}2.$

Telescoping across the alternating selected intervals and gaps gives $\mathrm{\mid }F(t)-F(s)\mathrm{\mid }<\epsilon$. Since $\epsilon$ is arbitrary, $F(t)=F(s)$. This includes $a=b$, and arbitrary $s,t$ prove constancy.