---
tag: 5NB63N9W3P
kind: definition
name: Absolute continuity
msc: [26A46]
defines: ZT6F4C2YHK
uses: [0TK8Y3J5RW]
sources:
  - url: https://en.wikipedia.org/wiki/Absolute_continuity
---

## Definition

Let $a < b$ be real numbers and let $f \colon [a, b] \to \R$ be a function.
The function $f$ is *absolutely continuous* on $[a, b]$ if for every $\varepsilon > 0$ there is $\delta > 0$ with the following property.
For every finite family of pairwise disjoint open [[intervals|0TK8Y3J5RW]] $(a_1, b_1), \dots, (a_n, b_n)$ contained in $[a, b]$,
$$\sum_{k=1}^{n} (b_k - a_k) < \delta \quad \text{implies} \quad \sum_{k=1}^{n} \abs{f(b_k) - f(a_k)} < \varepsilon.$$

## Remarks

Example: a [[Lipschitz function|0QNVQ4VD9T]] with constant $L > 0$ is absolutely continuous.
Given $\varepsilon$, take $\delta = \varepsilon / L$; then $\sum_k \abs{f(b_k) - f(a_k)} \le L \sum_k (b_k - a_k) < \varepsilon$.

Non-example: the [[Cantor function|NPBYYK63FE]] $c$ on $[0, 1]$ is continuous but not absolutely continuous.
For each $\delta > 0$, the Cantor set is covered by finitely many disjoint open intervals of total length less than $\delta$.
The changes of $c$ across these intervals add up to $c(1) - c(0) = 1$.

Equivalent form: $f$ is absolutely continuous on $[a, b]$ if and only if there is a Lebesgue integrable $g$ on $[a, b]$ with $f(x) = f(a) + \int_a^x g(t)\, dt$ for all $x$.
Then $f' = g$ almost everywhere.
With $n = 1$ only, the definition reduces to uniform continuity.