A measure is absolutely continuous with respect to if we can think of as a weighted variation on .
The measure is absolutely continuous with respect to if every -full set is also -full.
The measure is absolutely continuous with respect to if every -null set is also -null.
If and are positive measures, then we may also express this as follows:
The positive measure is absolutely continuous with respect to if, for every measurable set , if .
Since the absolute value of a measure is a positive measure, we can also express the general definition as follows:
The measure is absolutely continuous with respect to if, for every measurable set , if .
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 -filter of full sets (or a -ideal 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:
In fact, only the full/null sets of matter either, but until somebody has use for the notion of one -filter (or -ideal) being absolutely continuous with respect to another, I will refrain from writing it down. (See centipede mathematics.)
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).
This measure is absolutely continuous with respect to . Conversely, given any absolutely continuous measure , there is (at most) a unique (up to almost equality) absolutely integrable function such that ; and this function must exist if and are localizable. This converse is the subject of the Radon–Nikodym theorem.
A function is absolutely continuous? iff the Lebesgue–Stieltjes measure? is absolutely continuous with respect to Lebesgue measure. (All such functions are continuous, and this example is actually the origin of the term.)