Let $(X,\mathcal{S}, \mu)$ be a $\sigma$-finite measure space and let $\nu: \mathcal{S} \to [0,\infty]$ be a measure such that $\nu$ is absolutely continuous with respect to $\mu$. Show that there is a measurable function $f: X \to [0,\infty]$ such that $\nu(A)=\int_{A} f d\mu$ for all $A\in\mathcal{S}$.
It is easy if $\nu$ is $\sigma$-finite, this follows by Radon-Nikodym theorem.
But what if $\nu$ is not $\sigma$-finite? I am interested in the proof of this case.