Absolute continuity

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In mathematics, a real-valued function f of a real variable is absolutely continuous on a specified finite or infinite interval 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 [x_k, y_k], k = 1, ..., n satisfies

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

then

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

Every absolutely continuous function is uniformly continuous and, therefore, continuous. Every Lipschitz-continuous function is absolutely continuous.

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 is absolutely continuous on a finite interval [a,b], then it is of bounded variation on [a,b].
If f is absolutely continuous on the interval [a,b], then it has the Luzin N property (that is, for any $L\subseteq [a,b]$ that $λ(L) = 0$ , it holds that $λ(f (L)) = 0$ , where $λ$ stands for the Lebesgue measure).
If f is absolutely continuous, then f has a derivative almost everywhere.
If f is continuous, is of bounded variation and has the Luzin N property, then it is absolutely continuous.

1 Absolute continuity of measures
2 The connection between absolute continuity of real functions and absolute continuity of measures
3 See also
4 Reference

[edit] Absolute continuity of measures

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

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

[edit] The connection between absolute continuity of real functions and absolute continuity of measures

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

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