Absolute continuity on a compact interval

Let $a\le b$ and $f:[a,b]\to \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). The function $f$ is absolutely continuous on $[a,b]$ if for every $\epsilon >0$ there is $\delta >0$ such that every finite family of subintervals $[u_j,v_j]\subseteq [a,b]$, indexed by $j<m$, whose open interiors are pairwise disjoint and which satisfies

${\sum }_{j<m}(v_j-u_j)<\delta$

also satisfies

${\sum }_{j<m}\mathrm{\mid }f(v_j)-f(u_j)\mathrm{\mid }<\epsilon \mathrm{.}$

Finite sums and the empty sum are those of Finite sums and finite products, by recursion and Laws of finite sums and finite products. For $m=0$ both sums are $0$, so the condition is automatic. On $[a,a]$ every permitted interval is a singleton and every endpoint increment is $0$ (Absolute value in an ordered field), so every function on that singleton is absolutely continuous. Absolute continuity implies ordinary continuity (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point); that implication is proved next rather than built into the definition.