Radon-Nikodym theoremRadonNikodymTheorem2003-02-08 15:20:572004-08-04 23:01:47TheoremRadon-Nikodym derivative% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $\mu$ and $\nu$ be two $\sigma$-finite measures on the same measurable space $(\Omega, \mathscr{S})$, such that $\nu\ll \mu$
(i.e. $\nu$ is absolutely continuous with respect to $\mu$.)
Then there exists a measurable function $f$, which is nonnegative
and finite, such that for each $A\in \mathscr{S}$,
\[\nu(A)=\int_A fd\mu.\]
This function is unique (any other function satisfying these
conditions is equal to $f$ $\mu$-almost everywhere,) and it is called
the \emph{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$,
denoted by $f = \frac{d\nu}{d\mu}$.
\textbf{Remark.} The theorem also holds if $\nu$ is a signed measure. Even if $\nu$ is not $\sigma$-finite the theorem holds, with the exception that $f$ is not necessarely finite.
\textbf{Some properties of the Radon-Nikodym derivative}
Let $\nu$, $\mu$, and $\lambda$ be $\sigma$-finite measures in
$(\Omega,\mathscr{S})$.
\begin{enumerate}
\item If $\nu \ll \lambda$ and $\mu\ll\lambda$, then
\[\frac{d(\nu+\mu)}{d\lambda} =
\frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}\;\;
\mu\mbox{-almost everywhere};\]
\item If $\nu\ll\mu\ll\lambda$, then
\[\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}
\;\; \mu\mbox{-almost everywhere};\]
\item If $\mu\ll\lambda$ and $g$ is a $\mu$-integrable function, then
\[\int_\Omega gd\mu = \int_\Omega g\frac{d\mu}{d\lambda}d\lambda;\]
\item If $\mu\ll\nu$ and $\nu \ll\mu$, then
\[\frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1}.\]
\end{enumerate}