Absolute continuity

2020 Mathematics Subject Classification: Primary: 28A33 [MSN][ZBL]

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 \[ \left|\int_E f (x) \rd\lambda (x)\right| < \varepsilon \qquad \text{for every measurable set '"`UNIQ-MathJax6-QINU`"' with '"`UNIQ-MathJax7-QINU`"'}. \] This property can be generalized to measures $\mu$ on a $\sigma$-algebra $\mathcal{B}$ of subsets of a space $X$ and to functions $f\in L^1 (X, \mu)$.

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 $\mu (A) =0$. This definition can be generalized to signed measures $\nu$ and even to vector-valued measure $\nu$. 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 Signed measure for the relevant definition).

The Radon-Nikodym theorem 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 \[ \nu (A) = \int_A f\rd\mu \qquad \text{for every '"`UNIQ-MathJax36-QINU`"'.} \] A corollary of the Radon-Nikodym, the Hahn decomposition theorem, characterizes signed measures as differences of nonnegative measures. We refer to Signed measure for more on this topic.

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 any $a_1<b_1<a_2<b_2<\ldots < a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have \[ \sum_i |f(a_i)-f (b_i)| <\varepsilon. \] This notion can be easily generalized when the target of the function is a metric space.

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 (if $I$ is bounded, this is equivalent to require $f\in W^{1,1} (I)$). Vice versa, for any function with $L^1$ distributional derivatie there is an absolutely continuous representative, i.e. an absolutely continuous $\tilde{f}$ such that $\tilde{f} = f$ a.e.. 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.

The differentiability almost everywhere does not imply the absolute continuity: a notable example is the Cantor ternary function or devil staircase. Though such function is differentiable almost everywhere, it fails to satisfy \ref{e:fundamental} (indeed the generalized derivative of the Cantor ternary function is a measure which is not absolutely continuous with respect to the Lebesgue measure).

An absolutely continuous function maps a set of measure zero into a set of measure zero, and a measurable set into a measurable set. Any continuous function of finite variation which maps each set of measure zero into a set of measure zero is absolutely continuous. Any absolutely continuous function can be represented as the difference of two absolutely continuous non-decreasing functions.

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