[proofplan] Choose a Lipschitz constant $L$ for $f$. If $L=0$, then all values of $f$ on $I$ are equal, so every absolute-continuity variation sum is zero. If $L>0$, choose $\delta=\varepsilon/L$ and apply the Lipschitz inequality separately to each interval in a finite pairwise disjoint family. Summing the resulting inequalities gives the required $\varepsilon$-bound, which is exactly the finite-family definition of [absolute continuity](/page/Absolute%20Continuity) on $I$. [/proofplan] [step:Fix a Lipschitz constant and separate the constant case] Since $f:I\to\mathbb C$ is Lipschitz, there exists a real number $L\ge 0$ such that, for all $x,y\in I$, \begin{align*} |f(x)-f(y)|\le L|x-y|. \end{align*} If $L=0$, then $|f(x)-f(y)|=0$ for all $x,y\in I$, so $f(x)=f(y)$ for all $x,y\in I$. Hence for every $m\in\mathbb N$ and every family of subintervals $[a_k,b_k]\subset I$, $1\le k\le m$, \begin{align*} \sum_{k=1}^m |f(b_k)-f(a_k)|=0. \end{align*} Thus the absolute-continuity condition holds for any choice of $\delta>0$. [guided] The definition of a [Lipschitz function](/page/Lipschitz%20Function) gives a single constant controlling all oscillations of $f$ on $I$. Concretely, there is a real number $L\ge 0$ such that, whenever $x,y\in I$, \begin{align*} |f(x)-f(y)|\le L|x-y|. \end{align*} We first handle the degenerate case $L=0$. In that case the inequality becomes \begin{align*} |f(x)-f(y)|\le 0 \end{align*} for all $x,y\in I$. Since a complex modulus is always nonnegative, this forces $|f(x)-f(y)|=0$, and hence $f(x)=f(y)$ for all $x,y\in I$. Therefore $f$ is constant on $I$. Now let $m\in\mathbb N$ and let $[a_k,b_k]\subset I$, $1\le k\le m$, be any finite family of subintervals. Because $f(b_k)=f(a_k)$ for every $k$, each summand in the variation sum vanishes: \begin{align*} |f(b_k)-f(a_k)|=0. \end{align*} Summing over $1\le k\le m$ gives \begin{align*} \sum_{k=1}^m |f(b_k)-f(a_k)|=0. \end{align*} Thus, for every $\varepsilon>0$, the desired inequality is satisfied for any positive $\delta$. This proves absolute continuity in the case $L=0$. [/guided] [/step] [step:Choose $\delta$ from the positive Lipschitz constant] Assume now that $L>0$. Let $\varepsilon>0$ be given, and define \begin{align*} \delta:=\frac{\varepsilon}{L}. \end{align*} Then $\delta>0$. [/step] [step:Sum the Lipschitz bounds over a finite disjoint interval family] Let $m\in\mathbb N$, and let $[a_k,b_k]\subset I$, $1\le k\le m$, be a finite pairwise disjoint family of subintervals with $a_k\le b_k$ and \begin{align*} \sum_{k=1}^m (b_k-a_k)<\delta. \end{align*} For each $k$, the endpoints $a_k,b_k$ belong to $I$, so the Lipschitz inequality gives \begin{align*} |f(b_k)-f(a_k)|\le L|b_k-a_k|=L(b_k-a_k). \end{align*} Summing this inequality over $1\le k\le m$ yields \begin{align*} \sum_{k=1}^m |f(b_k)-f(a_k)|\le L\sum_{k=1}^m (b_k-a_k). \end{align*} Using the assumed bound on the total length and the definition of $\delta$, we obtain \begin{align*} \sum_{k=1}^m |f(b_k)-f(a_k)|<L\delta=\varepsilon. \end{align*} Therefore the finite-family absolute-continuity condition holds for this $\varepsilon$. Since $\varepsilon>0$ was arbitrary, $f$ is absolutely continuous on $I$. [/step]