absolutely continuous
Let $\mu $ and $\nu $ be signed measures or complex measures^{} on the same measurable space^{} $(\mathrm{\Omega},\mathcal{S})$. We say that $\nu $ is absolutely continuous^{} with respect to $\mu $ if, for each $A\in \mathcal{S}$ such that $\mu (A)=0$, it holds that $\nu (A)=0$. This is usually denoted by $\nu \ll \mu $.
Remarks.
If $\mu $ and $\nu $ are signed measures and $({\nu}^{+},{\nu}^{})$ is the Jordan decomposition of $\nu $, the following are equivalent^{}:

1.
$\nu \ll \mu $;

2.
${\nu}^{+}\ll \mu $ and ${\nu}^{}\ll \mu $;

3.
$\nu \ll \mu $.
If $\nu $ is a finite signed or complex measure and $\nu \ll \mu $, the following useful property holds: for each $\epsilon >0$, there is a $\delta >0$ such that $$ whenever $$.
Title  absolutely continuous 

Canonical name  AbsolutelyContinuous 
Date of creation  20130322 13:26:12 
Last modified on  20130322 13:26:12 
Owner  Koro (127) 
Last modified by  Koro (127) 
Numerical id  10 
Author  Koro (127) 
Entry type  Definition 
Classification  msc 28A12 
Related topic  RadonNikodymTheorem 
Related topic  AbsolutelyContinuousFunction2 
Defines  absolute continuity 