FALSE: every function of bounded variation is absolutely continuous

The Cantor function $c$ is nondecreasing by The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, so for every partition $0=x_0<\cdots <x_m=1$ the variation sum telescopes to ${\sum }_{j=1}^m\mathrm{\mid }c(x_j)-c(x_{j-1})\mathrm{\mid }={\sum }_{j=1}^m(c(x_j)-c(x_{j-1}))=c(1)-c(0)=1$. Hence $c$ has bounded variation.

Iterating the two affine branches in The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds, the stage-$n$ set consists of $2^n$ pairwise disjoint closed intervals indexed by the $n$-digit words in $\{0,2{\}}^n$, each of length $3^{-n}$. Their total length is therefore $(2\mathrm{/}3{)}^n$, which tends to $0$ by For $\mathrm{\mid }r\mathrm{\mid }<1$ the sequence $r^k$ is null, and for $\mathrm{\mid }r\mathrm{\mid }>1$ the sequence $\mathrm{\mid }r{\mathrm{\mid }}^k$ diverges to $+\mathrm{\infty }$. For the interval indexed by a word $w$, its endpoints have ternary digits $w$ followed respectively by all $0$'s and all $2$'s (The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$). The identity $c=\gamma$ on the Cantor set from claim 1 of The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, together with the binary-digit formula in The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval, therefore gives endpoint increment ${\sum }_{k\ge n}{2}^{-k-1}=2^{-n}$. Thus the sum of the endpoint increments over the stage-$n$ intervals is always $2^n{2}^{-n}=1$. The total input length can be arbitrarily small while this increment sum stays $1$, so $c$ fails the defining $\epsilon$-$\delta$ condition of Absolute continuity on a compact interval. Hence $c$ is not absolutely continuous and the statement is false.