Absolute continuity

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2020 Mathematics Subject Classification: Primary: 28A33 [MSN][ZBL] (Absolute continuity of measures)

2020 Mathematics Subject Classification: Primary: 26A46 [MSN][ZBL] (Absolute continuity of functions)

Absolute continuity of the Lebesgue integral

Describes a property of absolutely Lebesgue integrable functions. Consider the Lebesgue measure $\lambda$ on the $n$-dimensional euclidean space and let $f\in L^1 (\mathbb R^n, \lambda)$. Then for every $\varepsilon>0$ there is a $\delta>0$ such that \[ \left|\int_A f (x) \rd\lambda (x)\right| < \varepsilon \qquad \mbox{for every measurable set}\, A \mbox{ with } \lambda (A)< \delta\, . \] This property can be generalized to measures $\mu$ on a $\sigma$-algebra $\mathcal{B}$ of subsets of a space $X$ and to functions $f\in L^1 (X, \mu)$ (cp. with Theorem 12.34 of [HS]).

Absolute continuity of measures

A concept in measure theory (see also Absolutely continuous measures). If $\mu$ and $\nu$ are two measures on a σ-algebra $\mathcal{B}$ of subsets of $X$, we say that $\nu$ is absolutely continuous with respect to $\mu$ if $\nu (A) =0$ for any $A\in\mathcal{B}$ such that $\mu (A) =0$ (cp. with Defininition 2.11 of [Ma]). This definition can be generalized to signed measures $\nu$ and even to vector-valued measures $\nu$. Some authors generalize it further to vector-valued $\mu$'s: in that case the absolute continuity of $\nu$ with respect to $\mu$ amounts to the requirement that $\nu (A) = 0$ for any $A\in\mathcal{B}$ such that $|\mu| (A)=0$, where $|\mu|$ is the total variation of $\mu$ (see for instance Section 30 of [Ha]).

The Radon-Nikodym theorem (see Theorem B, Section 31 of [Ha]) characterizes the absolute continuity of $\nu$ with respect to $\mu$ with the existence of a function $f\in L^1 (\mu)$ such that $\nu = f \mu$, i.e. such that \[ \nu (A) = \int_A f\, \rd\mu \qquad \text{for every } A\in\mathcal{B}. \] A corollary of the Radon-Nikodym, the Jordan decomposition Theorem, characterizes signed measures as differences of nonnegative measures. We refer to Signed measure for more on this topic (see also Hahn decomposition).

Absolute continuity of a function

A function $f:I\to \mathbb R$, where $I$ is an interval of the real line, is said absolutely continuous if for every $\varepsilon> 0$ there is $\delta> 0$ such that, for every finite collection of pairwise disjoint intervals $(a_1,b_1), (a_2,b_2), \ldots , (a_n,b_n) \subset I$ with $\sum_i (b_i-a_i) <\delta$, we have \[ \sum_i |f(b_i)-f (a_i)| <\varepsilon \] (see Section 4 in Chapter 5 of [Ro]).

An absolutely continuous function is always continuous. Indeed, if the interval of definition is open, then the absolutely continuous function has a continuous extension to its closure, which is itself absolutely continuous. A continuous function might not be absolutely continuous, even if the interval $I$ is compact. Take for instance the function $f:[0,1]\to \mathbb R$ such that $f(0)=0$ and $f(x) = x \sin x^{-1}$ for $x>0$. The space of absolutely continuous (real-valued) functions is a vector space.

A characterization of absolutely continuous functions on an interval might be given in terms of Sobolev spaces: a continuous function $f:I\to \mathbb R$ is absolutely continuous if and only its distributional derivative is an $L^1$ function, cp. with Theorem 1 in Section 4.9 of [EG] (if $I$ is bounded, this is equivalent to require $f\in W^{1,1} (I)$). Vice versa, for any function with $L^1$ distributional derivative there is an absolutely continuous representative, i.e. an absolutely continuous $\tilde{f}$ such that $\tilde{f} = f$ a.e. (cp. again with [EG]). The latter statement can be proved using the absolute continuity of the Lebesgue integral.

An absolutely continuous function is differentiable almost everywhere and its pointwise derivative coincides with the generalized one. The fundamental theorem of calculus holds for absolutely continuous functions, i.e. if we denote by $f'$ its pointwise derivative, we then have \begin{equation}\label{e:fundamental} f (b)-f(a) = \int_a^b f' (x)\rd x \qquad \forall a<b\in I. \end{equation} In fact this is yet another characterization of absolutely continuous functions (see Theorem 13 and Corollary 11 of Section 4 in Chapter 5 of [Ro]).

The differentiability almost everywhere does not imply the absolute continuity: a notable example is the Cantor ternary function or "Devil's staircase" (see Problem 46 in Chapter 2 of [Ro]). Though such function is differentiable almost everywhere, it fails to satisfy \ref{e:fundamental} since the derivative vanishes almost everywhere but the function is not constant, cp. with Problems 11 and 12 of Chapter 5 in [Ro] (indeed the generalized derivative of the Cantor ternary function is a measure which is not absolutely continuous with respect to the Lebesgue measure, see [AFP]).

It follows from \ref{e:fundamental} that an absolutely continuous function maps a set of (Lebesgue) measure zero into a set of measure zero (i.e. it has the Luzin-N-property), and a (Lebesgue) measurable set into a measurable set. Any continuous function of bounded variation which maps each set of measure zero into a set of measure zero is absolutely continuous (this follows, for instance, from the Radon-Nikodym theorem). Any absolutely continuous function can be represented as the difference of two absolutely continuous non-decreasing functions.

Metric setting

This notion can be easily generalized when the target of the function is a metric space $(X,d)$. In that case the function $f:I\to X$ is absolutely continuous if for every positive $\varepsilon$ there is a positive $\delta$ such that for any $a_1<b_1\leq a_2<b_2 \leq \ldots \leq a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have \[ \sum_i d (f (b_i), f(a_i)) <\varepsilon\, . \] The absolute continuity guarantees the uniform continuity. As for real valued functions, there is a characterization through an appropriate notion of derivative.

Theorem 1 A continuous function $f$ is absolutely continuous if and only if there is a function $g\in L^1_{loc} (I, \mathbb R)$ such that \begin{equation}\label{e:metric} d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\, \end{equation} (cp. with [AGS]). This theorem motivates the following

Definition 2 If $f:I\to X$ is absolutely continuous and $I$ is a closed interval, the metric derivative of $f$ is the function $g\in L^1$ with the smallest $L^1$ norm such that \ref{e:metric} holds (cp. with [AGS]).

The definition can be easily generalized to more general domains of definition. Observe also that, if $X$ is the standard Euclidean space $\mathbb R^k$, then the metric derivative of $f$ is the norm of the classical derivative.


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How to Cite This Entry:
Absolute continuity. Encyclopedia of Mathematics. URL:
This article was adapted from an original article by A.P. Terekhin, V.F. Emel'yanov, L.D. Kudryavtsev, V.V. Sazonov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article