absolutely continuous measure

One measure is absolutely continuous with respect to another when it agrees about what is negligible: anything the reference measure considers to have zero size, the first measure also considers to have zero size. Intuitively, the new measure has no mass hiding on sets the old measure cannot see; it is ‘carried along’ by the reference measure and never lights up where the reference is dark.

Formally, for measures nu and mu on the same sigma-algebra, nu is absolutely continuous with respect to mu — written nu << mu — if mu(E) = 0 implies nu(E) = 0 for every measurable set E. For finite nu there is an equivalent epsilon–delta form that explains the name: for every epsilon > 0 there is a delta > 0 such that mu(E) < delta forces |nu(E)| < epsilon. This mirrors the absolute continuity of a function, where small total length forces small total variation.

The opposite extreme is mutual singularity, where the two measures live on disjoint sets. The Lebesgue decomposition theorem says any sigma-finite nu splits uniquely as an absolutely continuous part plus a singular part relative to mu. The absolutely continuous part is exactly the piece the Radon–Nikodym theorem represents as an integral of a density. A classic non-example: the measure that places mass 1 at the point 0 is not absolutely continuous with respect to Lebesgue measure, since {0} has Lebesgue measure 0 but point-mass 1.