# Proving Property of Radon-Nikodym Derivative

Wikipedia states in their Radon-Nikodym theorem article (https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem) that if $\nu$ is a signed measure such that $\nu \ll \mu$ for some $\sigma-$finite positive measure, then \begin{align*} \frac{d|\nu|}{d \mu} = \Big|\frac{d \nu}{d \mu}\Big| \end{align*} But I just don't see why this is true.

Does anyone know?

## 1 Answer

If $\mu$ is a signed measure on the measurable space $(X,\mathcal{X})$, then by the Hahn decomposition composition theorem, there is a (up to null sets unique) $P\in\mathcal{X}$ such that the measures $A\mapsto\mu^+(A)=\mu(A\cap P)$ and $A\mapsto\mu^-(A)=-\mu(A\cap X\setminus P)$ are nonnegative. By definition (there are several equivalent ones), $|\mu|=\mu^++\mu^-$.

Now take any Radon-Nikodym derivative $f$ of $\mu$. Show that outside some null set $f$ is nonnegative on $P$ and nonpositive on $X\setminus P$. You can therefore assume without loss of generality that $f$ is nonnegative on $P$ and nonpositive on $X\setminus P$. Let $f_P$ be the funtion that is zero on $X\setminus P$ and agrees with $f$ on $P$, and let $f_{X\setminus P}$ be the funtion that is $0$ on $P$ and agrees with $f$ on $X\setminus P$. Then $f=f_p+f_{X\setminus P}$ and $|f|=|f_P|+|f_{X\setminus P}|$. Also, $|f_P|$ is a Radon-Nikodym derivative of $\mu^+$, $|f_{X\setminus P}|$ is aRadon-Nikodym derivative of $\mu^-$. It follows that $|f|=|f_P|+|f_{X\setminus P}|$ is Radon-Nikodym derivative of $\mu^++\mu^-=|\mu|$. Since Radon-Nikodym derivatives are unique up to null sets, the other direction holds too.

The case of complex measures is analogous.