absolutely continuous measure

A measure $\mu$ is absolutely continuous with respect to $\nu$ if we can think of $\mu$ as a weighted variation on $\nu$.

Fix a measurable space $X$ and let $\mu$ and $\nu$ be two measures on $X$.

The measure $\mu$ is **absolutely continuous** with respect to $\nu$ if every $\nu$-full set is also $\mu$-full.

Because null sets are more familiar than full sets, we may equivalently express things as follows (but this is not correct in constructive mathematics):

The measure $\mu$ is **absolutely continuous** with respect to $\nu$ if every $\nu$-null set is also $\mu$-null.

If $\mu$ and $\nu$ are positive measures, then we may also express this as follows:

The positive measure $\mu$ is **absolutely continuous** with respect to $\nu$ if, for every measurable set $A$, $\mu(A) = 0$ if $\nu(A) = 0$.

Since the absolute value of a measure is a positive measure, we can also express the general definition as follows:

The measure $\mu$ is **absolutely continuous** with respect to $\nu$ if, for every measurable set $A$, ${|\mu|}(A) = 0$ if ${|\nu|}(A) = 0$.

Since only the full sets (or, classically, the null sets) matter, we do not need to have the full structure of a measure. Sometimes we equip a measurable space with a $\delta$-filter $\mathcal{F}$ of full sets (or a $\sigma$-ideal $\mathcal{N}$ of null sets) without specifying a measure that produces these. (For example, a smooth manifold is so equipped, effectively the full/null sets under Lebesgue measure, even though there is no canonical such measure, since these sets are the same regardless of coordinate chart.) Then we say:

The measure $\mu$ is **absolutely continuous** with respect to $\mathcal{F}$ (or $\mathcal{N}$) if every element of $\mathcal{F}$ is $\mu$-full (or every element of $\mathcal{N}$ is $\mu$-null).

In fact, only the full/null sets of $\mu$ matter either, but until somebody has use for the notion of one $\delta$-filter (or $\sigma$-ideal) being absolutely continuous with respect to another, I will refrain from writing it down. (See centipede mathematics.)

If one calls a measure on the real line ‘absolutely continuous’, this means with respect to Lebesgue measure.

This generalises to any cartesian space (with Lebesgue measure) or indeed to any smooth manifold of finite dimension (where there is no canonical Lebesgue measure but a family of local ones and so still a notion of Lebesgue-full and Lebesgue-null sets).

Or, this generalises to any compact group (with Haar measure) or indeed to any locally compact group (where there is no canonical Haar measure but a proportional family of them and so still a canonical notion of Haar-full and Haar-null sets).

Let $\nu$ be a measure, and let $f \in L^1(\nu)$ be an absolutely integrable function with respect to $\nu$; then integration defines a measure $f \nu$:

$(f \nu)(E) = \int_E f \nu = \int_E f(x) \nu(\mathrm{d}x) .$

This measure $f \nu$ is absolutely continuous with respect to $\nu$. Conversely, given any absolutely continuous measure $\mu$, there is (at most) a unique (up to almost equality) absolutely integrable function $f$ such that $\mu = f \nu$; and this function must exist if $\mu$ and $\nu$ are localizable. This converse is the subject of the Radon–Nikodym theorem.

A function $f\colon \mathbb{R} \to \mathbb{R}$ is absolutely continuous? iff the Lebesgue–Stieltjes measure? $\mathrm{d}f$ is absolutely continuous with respect to Lebesgue measure. (All such functions are continuous, and this example is actually the origin of the term.)

Revised on October 1, 2014 07:51:27
by Toby Bartels
(98.19.41.214)