---
tag: 2G173YJC54
kind: definition
name: Absolutely continuous measure
msc: [28A10, 28A15]
defines: P0P21KDV6E
uses: [PR11BZQQPG, N4KDT83V3S, VQQM4Y550T, VG2P7R91Z7, EPBND51CR5]
sources:
  - url: https://en.wikipedia.org/wiki/Absolute_continuity
---

## Definition

Let $(X, \Sigma)$ be a [[measurable space|N4KDT83V3S]] and let $\mu$ and $\nu$ be [[measures|PR11BZQQPG]] on it.
The measure $\nu$ is *absolutely continuous* with respect to $\mu$, written $\nu \ll \mu$, if for every $A \in \Sigma$,
$$\mu(A) = 0 \implies \nu(A) = 0.$$
For a [[signed|VG2P7R91Z7]] or complex measure $\nu$, one says $\nu \ll \mu$ if its [[total variation|EPBND51CR5]] satisfies $\abs{\nu} \ll \mu$.

## Remarks

Example: let $f \colon X \to [0, \infty]$ be measurable and $\nu(A) = \int_A f \, d\mu$.
If $\mu(A) = 0$, then $f \mathbf{1}_A = 0$ $\mu$-almost everywhere, so $\nu(A) = 0$; hence $\nu \ll \mu$.

Non-example: on $\R$ with the Borel sets, the Dirac measure $\delta_0$ is not absolutely continuous with respect to Lebesgue measure $\lambda$.
The set $A = \set{0}$ has $\lambda(A) = 0$ but $\delta_0(A) = 1$.

If $\nu$ is finite, then $\nu \ll \mu$ holds if and only if for every $\varepsilon > 0$ there is $\delta > 0$ with $\nu(A) < \varepsilon$ whenever $\mu(A) < \delta$.
If $\mu$ and $\nu$ are σ-finite, the [[Radon–Nikodym theorem|1ZSZS5G1Q4]] shows that $\nu \ll \mu$ if and only if $\nu$ has a density $f$ with respect to $\mu$.