RadonNikodym theorem
Let $\mu $ and $\nu $ be two $\sigma $finite measures^{} on the same measurable space^{} $(\mathrm{\Omega},\mathcal{S})$, 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 $A\in \mathcal{S}$,
$$\nu (A)={\int}_{A}f\mathit{d}\mu .$$ 
This function is unique (any other function satisfying these conditions is equal to $f$ $\mu $almost everywhere,) and it is called the RadonNikodym 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 RadonNikodym derivative
Let $\nu $, $\mu $, and $\lambda $ be $\sigma $finite measures in $(\mathrm{\Omega},\mathcal{S})$.

1.
If $\nu \ll \lambda $ and $\mu \ll \lambda $, then
$$\frac{d(\nu +\mu )}{d\lambda}=\frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}\mu \text{almost everywhere};$$ 
2.
If $\nu \ll \mu \ll \lambda $, then
$$\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\mu \text{almost everywhere};$$ 
3.
If $\mu \ll \lambda $ and $g$ is a $\mu $integrable function, then
$${\int}_{\mathrm{\Omega}}g\mathit{d}\mu ={\int}_{\mathrm{\Omega}}g\frac{d\mu}{d\lambda}\mathit{d}\lambda ;$$ 
4.
If $\mu \ll \nu $ and $\nu \ll \mu $, then
$$\frac{d\mu}{d\nu}={\left(\frac{d\nu}{d\mu}\right)}^{1}.$$
Title  RadonNikodym theorem 

Canonical name  RadonNikodymTheorem 
Date of creation  20130322 13:26:15 
Last modified on  20130322 13:26:15 
Owner  Koro (127) 
Last modified by  Koro (127) 
Numerical id  9 
Author  Koro (127) 
Entry type  Theorem 
Classification  msc 28A15 
Related topic  AbsolutelyContinuous 
Related topic  BoundedLinearFunctionalsOnLpmu 
Related topic  MartingaleProofOfTheRadonNikodymTheorem 
Related topic  BoundedLinearFunctionalsOnLinftymu 
Defines  RadonNikodym derivative 