From Folland, the theorem is as follows:
The Lebsgue–Radon–Nikodym Theorem Let $\nu$ be a $\sigma$-finite signed measure and $\mu$ a $\sigma$-finite positive measure on $(X,\mathcal{M})$. There exists unique $\sigma$-finite signed measure $\lambda,\rho$ on $(X,\mathcal{M})$ such that $\lambda\perp \mu$, $\rho\ll\mu$, and $\nu=\lambda+\rho$. Moreover, there is an extended $\mu$-integrable function $f: X\to\mathbb{R}$ such that $d\rho=f\,d\mu$, and any two functions are equal $\mu$-a.e.
The statement of the theorem itself seems straightforward. However, I am confused as to the later claim in Folland:
In the case where $\nu\ll\mu$, the theorem says that $d\nu = f \, d\nu$ for some $f$
I would really appreciate a simple and intuitive explanation of this.
Best Answer
The LRN theorem lets us decompose one measure with respect to another -- one part singular and one part absolutely continuous. If $\nu \ll \mu$, then there can be no singular part.
This is because $\nu \ll \mu$ says that if $\mu(E) = 0$, then $\nu(E) = 0$ too. But if $\nu = \lambda + \rho$ with $\lambda \perp \mu$, then $X = E \sqcup F$ where $F$ is $\mu$-null and $E$ is $\lambda$-null, but then $F$ is $\mu$-null, $F$ must also be $\nu$-null because in this case, $\nu \ll \mu$, so $F$ must also be $\lambda$-null. But then $X$ is $\lambda$-null, so that $\lambda \equiv 0$.