Let $\mu$ 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 Radon-Nikodym derivative  of $\nu$ with respect to $\mu$, denoted by $f=\frac{d\nu}{d\mu}$.

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.

Some properties of the Radon-Nikodym derivative

Let $\nu$, $\mu$, and $\lambda$ be $\sigma$-finite measures in $(\Omega,\mathscr{S})$.

1. 1.

If $\nu\ll\lambda$ and $\mu\ll\lambda$, then

 $\frac{d(\nu+\mu)}{d\lambda}=\frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}\;\;\mu% \mbox{-almost everywhere};$
2. 2.

If $\nu\ll\mu\ll\lambda$, then

 $\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\;\;\mu\mbox{-% almost everywhere};$
3. 3.

If $\mu\ll\lambda$ and $g$ is a $\mu$-integrable function, then

 $\int_{\Omega}gd\mu=\int_{\Omega}g\frac{d\mu}{d\lambda}d\lambda;$
4. 4.

If $\mu\ll\nu$ and $\nu\ll\mu$, then

 $\frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1}.$
Title Radon-Nikodym theorem RadonNikodymTheorem 2013-03-22 13:26:15 2013-03-22 13:26:15 Koro (127) Koro (127) 9 Koro (127) Theorem msc 28A15 AbsolutelyContinuous BoundedLinearFunctionalsOnLpmu MartingaleProofOfTheRadonNikodymTheorem BoundedLinearFunctionalsOnLinftymu Radon-Nikodym derivative