Radon-Nikodym theorem

Let $\mu$ and $\nu$ be two $\sigma$-finite measures on the same measurable space $(\mathrm{\Omega },𝒮)$, 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 $A\in 𝒮$,

$$\nu (A)={\int }_{A}f𝑑\mu .$$

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 $(\mathrm{\Omega },𝒮)$.

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

   $${\int }_{\mathrm{\Omega }}g𝑑\mu ={\int }_{\mathrm{\Omega }}g\frac{d\mu }{d\lambda }𝑑\lambda ;$$

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

   $$\frac{d\mu }{d\nu }={\left(\frac{d\nu }{d\mu }\right)}^{-1}.$$

Title	Radon-Nikodym theorem
Canonical name	RadonNikodymTheorem
Date of creation	2013-03-22 13:26:15
Last modified on	2013-03-22 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	Radon-Nikodym derivative