GeezerVison

Let $I$ be an interval in the real line $ℝ$. A function $f:I\to ℝ$ is absolutely continuous on $I$ if for every positive number $\epsilon$, there is a positive number $\delta$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k,y_k)$ of $I$ with $x_k<y_k$ satisfies^[1]

$\sum _{k=1}^N(y_k-x_k)<\delta$

then

$\sum _{k=1}^N\left|f\right(y_k)-f(x_k\left)\right|<\epsilon .$

The collection of all absolutely continuous functions on $I$ is denoted $\mathrm{AC}\left(I\right)$.

The following conditions on a real-valued function f on a compact interval [a,b] are equivalent:^[2]

1. f is absolutely continuous;
2. f has a derivative f′ almost everywhere, the derivative is Lebesgue integrable, and $f\left(x\right)=f\left(a\right)+{\int }_a^x{f}^{\prime }\left(t\right)dt$ for all x on [a,b];
3. there exists a Lebesgue integrable function g on [a,b] such that $f\left(x\right)=f\left(a\right)+{\int }_a^{x}g\left(t\right)dt$ for all x in [a,b].

If these equivalent conditions are satisfied, then necessarily any function g as in condition 3. satisfies g = f′ almost everywhere.

Equivalence between (1) and (3) is known as the fundamental theorem of Lebesgue integral calculus, due to Lebesgue.^[3]

For an equivalent definition in terms of measures see the section Relation between the two notions of absolute continuity.

• The sum and difference of two absolutely continuous functions are also absolutely continuous. If the two functions are defined on a (bounded) closed interval, then their product is also absolutely continuous.^[4] On unbounded intervals this may fail: for example on $ℝ$, the function $f\left(x\right)=x$ is absolutely continuous, but $f(x{)}^2$ is not even uniformly continuous.
• If an absolutely continuous function f is defined on a (bounded) closed interval and is nowhere zero then 1/f is absolutely continuous.^[5]
• Every absolutely continuous function (over an interval) is uniformly continuous and, therefore, continuous. Every (globally) Lipschitz-continuous function is absolutely continuous.^[6]
• If f: [a,b] → R is absolutely continuous, then it is weakly differentiable; conversely if f: [a,b] → R is weakly differentiable, then it coincides almost everywhere with an absolutely continuous function;^[7] this provides a characterization of Sobolev spaces on intervals of the real line.
• If f: [a,b] → R is absolutely continuous, then it is of bounded variation on [a,b].^[8]
• If f: [a,b] → R is absolutely continuous, then it can be written as the difference of two monotonic nondecreasing absolutely continuous functions on [a,b].
• If f: [a,b] → R is absolutely continuous, then it has the Luzin N property (that is, for any $N\subseteq [a,b]$ such that $\lambda \left(N\right)=0$, it holds that $\lambda \left(f\right(N\left)\right)=0$, where $\lambda$ stands for the Lebesgue measure on R).
• f: [a,b] → R is absolutely continuous if and only if it is continuous, is of bounded variation and has the Luzin N property. This statement is also known as the Banach-Zareckiǐ theorem.^[9]
• If f: I → R is absolutely continuous and g: R → R is globally Lipschitz-continuous, then the composition g $\circ$ f is absolutely continuous. Conversely, for every function g that is not globally Lipschitz continuous there exists an absolutely continuous function f such that g $\circ$ f is not absolutely continuous.^[10]

The following functions are uniformly continuous but not absolutely continuous:

• The Cantor function on [0, 1] (it is of bounded variation but not absolutely continuous);
• The function:$f\left(x\right)=\{\begin{array}{cc}0,& \text{if }x=0\\ x\sin (1/x),& \text{if }x\ne 0\end{array}$ on a finite interval containing the origin.

The following functions are absolutely continuous but not α-Hölder continuous:

• The function f(x) = x^β on [0, c], for any 0 < β < α < 1

The following functions are absolutely continuous and α-Hölder continuous but not Lipschitz continuous:

• The function f(x) = √x on [0, c], for α ≤ 1/2.

Let (X, d) be a metric space and let I be an interval in the real line R. A function f: I → X is absolutely continuous on I if for every positive number $\epsilon$, there is a positive number $\delta$ such that whenever a finite sequence of pairwise disjoint sub-intervals [x_k, y_k] of I satisfies:

$\sum _k|y_k-x_k|<\delta$

then:

$\sum _{k}d\left(f\right(y_k),f(x_k\left)\right)<\epsilon .$

The collection of all absolutely continuous functions from I into X is denoted AC(I; X).

A further generalization is the space AC^p(I; X) of curves f: I → X such that:^[11]

$d\left(f\right(s),f(t\left)\right)\le {\int }_s^{t}m\left(\tau \right)d\tau \text{ for all }[s,t]\subseteq I$

for some m in the L^p space L^p(I).

• Every absolutely continuous function (over a compact interval) is uniformly continuous and, therefore, continuous. Every Lipschitz-continuous function is absolutely continuous.
• If f: [a,b] → X is absolutely continuous, then it is of bounded variation on [a,b].
• For f ∈ AC^p(I; X), the metric derivative of f exists for λ-almost all times in I, and the metric derivative is the smallest m ∈ L^p(I; R) such that:^[12]$d\left(f\right(s),f(t\left)\right)\le {\int }_s^{t}m\left(\tau \right)d\tau \text{ for all }[s,t]\subseteq I.$

A measure $\mu$ on Borel subsets of the real line is absolutely continuous with respect to the Lebesgue measure $\lambda$ if for every Borel set $A,$ $\lambda \left(A\right)=0$ implies $\mu \left(A\right)=0$. Equivalently, $\mu \left(A\right)>0$ implies $\lambda \left(A\right)>0$. This condition is written as $\mu \ll \lambda .$ We say $\mu$ is dominated by $\lambda .$

In most applications, if a measure on the real line is simply said to be absolutely continuous — without specifying with respect to which other measure it is absolutely continuous — then absolute continuity with respect to the Lebesgue measure is meant.

The same principle holds for measures on Borel subsets of ${ℝ}^n,n\ge 2.$

The following conditions on a finite measure $\mu$ on Borel subsets of the real line are equivalent:^[13]

1. $\mu$ is absolutely continuous;
2. For every positive number $\epsilon$ there is a positive number $\delta >0$ such that $\mu \left(A\right)<\epsilon$ for all Borel sets $A$ of Lebesgue measure less than $\delta ;$
3. There exists a Lebesgue integrable function $g$ on the real line such that: $\mu \left(A\right)={\int }_{A}gd\lambda$ for all Borel subsets $A$ of the real line.

For an equivalent definition in terms of functions see the section Relation between the two notions of absolute continuity.

Any other function satisfying (3) is equal to $g$ almost everywhere. Such a function is called Radon–Nikodym derivative, or density, of the absolutely continuous measure $\mu .$

Equivalence between (1), (2) and (3) holds also in ${ℝ}^n$ for all $n\in \mathbb{N}.$

Thus, the absolutely continuous measures on ${ℝ}^n$ are precisely those that have densities; as a special case, the absolutely continuous probability measures are precisely the ones that have probability density functions.

If $\mu$ and $\nu$ are two measures on the same measurable space $(X,𝒜),$ $\mu$ is said to be absolutely continuous with respect to $\nu$ if $\mu \left(A\right)=0$ for every set $A$ for which $\nu \left(A\right)=0.$^[14] This is written as "$\mu \ll \nu$". That is: $\mu \ll \nu \text{ if and only if }\text{ for all }A\in 𝒜,\left(\nu \right(A)=0\text{ implies }\mu (A)=0).$

When $\mu \ll \nu ,$ then $\nu$ is said to be dominating $\mu .$

Absolute continuity of measures is reflexive and transitive, but is not antisymmetric, so it is a preorder rather than a partial order. Instead, if $\mu \ll \nu$ and $\nu \ll \mu ,$ the measures $\mu$ and $\nu$ are said to be equivalent. Thus absolute continuity induces a partial ordering of such equivalence classes.

If $\mu$ is a signed or complex measure, it is said that $\mu$ is absolutely continuous with respect to $\nu$ if its variation $\left|\mu \right|$ satisfies $\left|\mu \right|\ll \nu ;$ equivalently, if every set $A$ for which $\nu \left(A\right)=0$ is $\mu$-null.

The Radon–Nikodym theorem^[15] states that if $\mu$ is absolutely continuous with respect to $\nu ,$ and both measures are σ-finite, then $\mu$ has a density, or "Radon-Nikodym derivative", with respect to $\nu ,$ which means that there exists a $\nu$-measurable function $f$ taking values in $[0,+\infty ),$ denoted by $f=d\mu /d\nu ,$ such that for any $\nu$-measurable set $A$ we have: $\mu \left(A\right)={\int }_{A}fd\nu .$

Via Lebesgue's decomposition theorem,^[16] every σ-finite measure can be decomposed into the sum of an absolutely continuous measure and a singular measure with respect to another σ-finite measure. See singular measure for examples of measures that are not absolutely continuous.