absolutely continuousAbsolutelyContinuous2003-02-08 14:29:562008-12-02 23:23:03Definitionabsolute continuity% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $\mu$ and $\nu$ be signed measures or complex measures on the same measurable space
$(\Omega, \mathscr{S})$. We say that $\nu$ is \emph{absolutely continuous}
with respect to $\mu$ if, for each $A\in \mathscr{S}$ such that $|\mu|(A)=0$,
it holds that $\nu(A)=0$. This is usually denoted by $\nu \ll \mu$.
\textbf{Remarks.}
If $\mu$ and $\nu$ are signed measures and $(\nu^+, \nu^-)$ is the Jordan decomposition of $\nu$, the following \PMlinkescapetext{propositions} are equivalent:
\begin{enumerate}
\item $\nu\ll\mu$;
\item $\nu^+\ll\mu$ and $\nu^-\ll\mu$;
\item $|\nu|\ll|\mu|$.
\end{enumerate}
If $\nu$ is a finite signed or complex measure and $\nu\ll\mu$, the following useful property holds: for each $\varepsilon>0$, there is a $\delta>0$ such that
$|\nu|(E)<\varepsilon$ whenever $|\mu|(E)<\delta$.