# Absolute continuity

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In mathematics, one may talk about absolute continuity of functions and absolutely continuity of measures, and these two notions are closely connected.

## Absolute continuity of functionsEdit

### DefinitionEdit

Let (X, d) be a metric space and let I be an interval in the real line R. A function f : IX is absolutely continuous on I if for every positive number ε, no matter how small, there is a positive number δ small enough so that whenever a sequence of pairwise disjoint sub-intervals [xk, yk] of I, k = 1, 2, ..., n satisfies

$\sum_{k=1}^{n} \left| y_k - x_k \right| < \delta$

then

$\sum_{k=1}^{n} d \left( f(y_k), f(x_k) \right) < \varepsilon.$

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

A further generalisation is the space ACp(I; X) of curves f : IX such that

$d \left( f(s), f(t) \right) \leq \int_{s}^{t} m(\tau) \, \mathrm{d} \tau \mbox{ for all } [s, t] \subseteq I$

for some m in the Lp space Lp(I; R).

### PropertiesEdit

• The Cantor function is continuous everywhere but not absolutely continuous; as is the function
$f(x) = \begin{cases} 0, & \mbox{if }x =0 \\ x \sin(1/x), & \mbox{if } x \neq 0 \end{cases}$
on a finite interval containing the origin, or the function $f(x)=x^2$ on an infinite interval.
• If f : [a,b] → X is absolutely continuous, then it is of bounded variation on [a,b].
• If f : [a,b] → R is absolutely continuous, then it has the Luzin N property (that is, for any $L \subseteq [a,b]$ that $\lambda(L)=0$, it holds that $\lambda(f(L))=0$, where $\lambda$ stands for the Lebesgue measure on R).
• If f : IR is absolutely continuous, then f has a derivative almost everywhere.
• If f : IR is continuous, is of bounded variation and has the Luzin N property, then it is absolutely continuous.
• For f ∈ ACp(I; X), the metric derivative of f exists for λ-almost all times in I, and the metric derivative is the smallest mLp(I; R) such that
$d \left( f(s), f(t) \right) \leq \int_{s}^{t} m(\tau) \, \mathrm{d} \tau \mbox{ for all } [s, t] \subseteq I.$

## Absolute continuity of measuresEdit

If μ and ν are measures on the same measure space (or, more precisely, on the same sigma-algebra) then μ is absolutely continuous with respect to ν if μ(A) = 0 for every set A for which ν(A) = 0. It is written as "μ << ν". In symbols:

$\mu \ll \nu \iff \left( \nu(A) = 0 \implies \mu (A) = 0 \right).$

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

If μ is a signed or complex measure, it is said that μ is absolutely continuous with respect to ν if its variation |μ| satisfies |μ| << ν; equivalently, if every set A for which ν(A) = 0 is μ-null.

The Radon-Nikodym theorem states that if μ is absolutely continuous with respect to ν, and ν is σ-finite, then μ has a density, or "Radon-Nikodym derivative", with respect to ν, which implies that there exists a ν-measurable function f taking values in [0,∞], denoted by f = dμ/dν, such that for any ν-measurable set A we have

$\mu(A)=\int_A f\,d\nu.$

## The connection between absolute continuity of real functions and absolute continuity of measuresEdit

A measure μ on Borel subsets of the real line is absolutely continuous with respect to Lebesgue measure if and only if the point function

$F(x)=\mu((-\infty,x])$

is locally an absolutely continuous real function. In other words, a function is locally absolutely continuous if and only if its distributional derivative is a measure that is absolutely continuous with respect to the Lebesgue measure.

Example. The Heaviside step function on the real line,

$H(x) \ \stackrel{\mathrm{def}}{=} \ \left\{ \begin{matrix} 0, & x < 0; \\ 1, & x \geq 0; \end{matrix} \right.$

has the Dirac delta distribution $\delta_{0}$ as its distributional derivative. This is a measure on the real line, a "point mass" at 0. However, the Dirac measure $\delta_{0}$ is not absolutely continuous with respect to Lebesgue measure $\lambda$, nor is $\lambda$ absolutely continuous with respect to $\delta_{0}$: $\lambda ( \{ 0 \} ) = 0$ but $\delta_{0} ( \{ 0 \} ) = 1$; if $U$ is any open set not containing 0, then $\lambda (U) > 0$ but $\delta_{0} (U) = 0$.

Example. The Cantor distribution has a continuous cumulative distribution function, but nonetheless the Cantor distribution is not absolutely continuous with respect to Lebesgue measure.