# Exact statement of the Radon-Nikodym Theorem

I am a bit confused about the exact statement of the Radon-Nikodym Theorem. Suppose that in the usual setup, $v \ll u$. Does it require both $v$ and $u$ to be sigma-finite, or only $u$ to be sigma finite for the theorem to hold?

You need both measures to be $\sigma$-finite. Suppose you have the collection of Lebesgue measurable sets on $[0,1]$ and let $\mu$ be the counting measure. The only set of measure zero with respect to $\mu$ is the empty set, so every measure is absolutely continuous with respect to $\mu$ and you can show that there is no nonnegative Lebesgue measurable function $f$ on $[0,1]$ for which $$m(E)=\int_Ef\,d\mu$$ where $E$ is any Lebesgue measurable set on $[0,1]$.
• You need both measures to be $\sigma$-finite. The counting measure is not $\sigma$-finite and every other measure $\nu$ is absolutely continuous with respect to $\mu$. In particular, the Lebesgue measure is absolutely continuous with respect to $\mu$. So if the Radon-Nikodym result holds in this case, we should be able to find a Lebesgue measurable function such that $m(E)=\int_Ef\,d\mu$, but this is impossible. – Laars Helenius Oct 7 '14 at 15:31
Only $u$ must be $\sigma$-finite. $v$ can be any (signed) measure. The result is proved in Section 2.2 in Ash and Doleans-Dade's book Probability and Measure Theory (2nd edition, 2000).
As a simple illustration, let $\Omega$ be some set and consider the trivial $\sigma$-algebra $\mathcal{F} = \{\emptyset,\Omega\}$. Suppose $u(\Omega) = 1$, $u(\emptyset) = 0$, $v(\Omega) = \infty$ and $v(\emptyset) = 0$. Obviously $u$ is $\sigma$-finite, $v$ is not $\sigma$-finite, and $v$ is absolutely continuous w.r.t. $u$. We can express $v(A) = \int_A g d\mu$, $A \in \mathcal{F}$, where $g = \infty$ is the Radon-Nikodym derivative of $v$ w.r.t. $u$.
The Lebesgue Decomposition Theorem requires both measures to be $\sigma$-finite.