# Is $\sigma$-finiteness unnecessary for Radon Nikodym theorem?

Let $(X,\mathfrak{M},\mu)$ be a $\sigma$-finite measure space and $\lambda:\mathfrak{M}\rightarrow [0,\infty]$ be a measure. If $\lambda\ll \mu$, then there exists a measurable $f:(X,\mathfrak{M})\rightarrow [0,\infty)$ such that $d\lambda= fd\mu$.

The above is the Radon Nikodym theorem stated in wikipedia. However, both texts Rudin and Folland prove the Radon Nikodym theorem under assumption that $\lambda$ is $\sigma$-finite. How do I prove the Wikipedia vesion of Radon Nikodym therem? Is there any reference?

The version on wikipedia is wrong (if it's exactly as you say; a link might have been appropriate). This is your chance to do a Good Thing by finding the Edit button and fixing it.

Counterexample with $\mu$ finite but $\nu$ not $\sigma$-finite: Let $X=\{0\}$. Define $\mu(X)=1$, $\nu(X)=\infty$.

That was easy. May as well mention that it's just as easy to give a counterexample with $\nu$ finite but $\mu$ not $\sigma$-finite: $X$ as above, $\mu(X)=\infty$, $\nu(X)=1$.

• Not sure how to fix the link works.. so I'm deleting it. And thank you! – Rubertos Apr 27 '16 at 12:22
• @David I posted a question related to your counterexample here. I didn't have a successful answer, so could you have a look at it? Thank you! – user39756 Oct 23 '16 at 13:33

It is possible to prove Radon Nikodym theorem without assuming $\lambda$ is $\sigma$-finite.