# Radon-Nikodym Theorem for two positive measures

Let $$\mu,\nu$$ be two positive measures on $$(X,\mathscr{A})$$ and $$\mu$$ is finite. If $$\nu \ll \mu$$, then does there exist a measurable function $$f: X \to [0,\infty]$$ such that $$\nu(E) = \int_E f d \mu,~\forall E \in \mathscr{A}?$$

I don’t know what happens when $$\nu$$ is not $$\sigma$$-finite.

For the case where $$\nu$$ is not $$\sigma$$-finite the answer is yes provided $$f$$ is allowed to take the value $$\infty$$. See this answer.
• What happens if $\nu$ is not $\sigma$-finite? – Martin Argerami Jun 27 at 3:26
• @Martin Argerami It still holds provided the function $f$ is allowed to take the value $\infty$ (which the OP has indicated is allowed in their setting). Here is another thread where this question is addressed. – user1222 Jun 27 at 3:51
• As it is written, your answer is quoting the Wikipedia article which requires $\nu$ to be $\sigma$-finite. And I don't really see how the other answer contributes to yours; actually, it confuses things, because it says that the Wikipedia article is wrong, but then one would have to check what the article said back in 2016. – Martin Argerami Jun 27 at 4:46
• @Martin Argerami See the second answer in the thread I linked (the one with 1 up vote). The person who wrote it gives a detailed proof that $f$ exists in the case where $\nu$ is not $\sigma$-finite provided $f$ can take the value $\infty$. – user1222 Jun 27 at 4:53