# Extension of Radon-Nikodym Theorem

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.

## 2 Answers

This is still true, see for instance exercise 3 page 113 in the book "A course in abstract analysis" by J.B. Conway.

There is a related exercise problem in Folland's Real Analysis. It is problem 3.14, which is stated a little more generally, i.e. $\nu$ is a signed measure. It has the following parts:

(a) It suffices to assume that $\mu$ is finite and $\nu$ is positive.

(b) With these assumptions, there exists $E\in\mathcal{S}$ that is $\sigma$-finite for $\nu$ such that $\mu(E)\geq\mu(F)$ for all sets $F$ that are $\sigma$-finite for $\nu$.

(c) The Radon-Nikodym theorem applies on $E$. If $F\cap E=\emptyset$, then either $\nu(F)=\mu(F)=0$ or $\mu(F)>0$ and $|\nu(F)|=\infty$.