Radon–Nikodym Theorem

Definition

Consider sigma-finite measures $\mu ,\lambda$ on a Measurable Space $(S,\mathcal{S})$. If $\mu \ll \lambda$ ($\mu$ is absolutely continuous with respect to $\lambda$), then there exists a $\mathcal{S}$-Measurable Function $f:X\to {\mathbb{R}}_0^{+}$ such that $\forall B\in \mathcal{S},\mu (B)={\int }_{B}fd\lambda$ almost uniquely with respect to $\lambda$. Where the function $f:(S,\mathcal{R})\to ({\mathbb{R}}^{+},{\mathcal{R}}^{+})$ is called a Radon-Nikodym derivative of $\mu$ w.r.t. $\lambda$ $\mu \ll \lambda \Rightarrow \exists !f=\frac{d\mu }{d\lambda }\text{s.t.}\forall B\in \mathcal{S},\mu (B)={\int }_{B}fd\lambda$

Radon–Nikodym Derivative

$f=\frac{d\mu }{d\lambda }$ Consider sigma-finite measures $\mu ,\lambda$ on a Measurable Space $(S,\mathcal{S})$. A Radon–Nikodym derivative $f=\frac{d\mu }{d\lambda }$ is a function satisfying $\forall B\in \mathcal{S},\mu (B)={\int }_{B}fd\lambda$.

Creating New Measure

Given a measure space $(S,\mathcal{S},\lambda )$ and a non-negative Measurable Function $f:S\to {\mathbb{R}}_0^{+}$, we can define a new measure $\mu$ by the relation $\forall B\in S,\mu (B)={\int }_{B}fd\lambda$ The measure $\mu$ is absolutely continuous with respect to $\lambda$ ($\mu \ll \lambda$) and $f$ be a Radon-Nikodym derivative of $\mu$ with respect to $\lambda$.

Examples

Distribution Case

Consider a distribution $F_X:\mathbb{R}\to [0,1]$ and a Lebesgue Measure $\lambda$, and a Measurable Space $(\mathbb{R},\mathcal{R})$ (Borel Sigma-Field on Real Number). If ${\mu }_X\ll \lambda$, then there exists a Density Function $f_X$ of the Distribution Function $F_X$ ${\mu }_X\ll \lambda \Rightarrow \exists !f_X=\frac{d{\mu }_X}{d\lambda }\text{s.t.}{F}_X={\int }_{(-\infty ,x]}{f}_{X}d\lambda$

Facts

If another function $g$ have the same properties, then $f=g$ at Almost Everywhere