My question is regarding the Radon–Nikodym derivative $\frac{d\nu}{d\mu}$, when $\nu \ll \mu$ and both measures are $\sigma$-finite. So Wikipedia says in the article about the Radon–Nikodym theorem that the derivative is a function that maps to $[0, \infty)$, i.e., the positive reals including zero. In all the books I have looked in, however, the theorem states that the density is a function mapping to either
1.) $[0, \infty]$,
or to
2.) $\mathbb{R}$
or to
3.) $[0, \infty)$ almost everywhere,
or to
4.) $[0, \infty)$ but there exists a measureable set $F$ such that $$\nu = \int f d\mu + \int(\infty \cdot \chi_F)d\mu.$$
But nowhere could I find that the density in fact maps to $[0, \infty)$ (without any other side conditions).
Can someone tell me, if what Wikipedia says is correct? And maybe comment on the tfour different codomains, why they are different, and which assumptions for the theorem may be different for each of these. Also can we even say anything that holds for all measurable sets instead of just making a statement that holds almost everywhere?
The derivative is finite a.e. for nonnegative, $\sigma$-finite measures.
In the wikipedia page https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem you have the proof of R-N and $\mu$ a.e. finitness of the derivative for finite measures.
Then from this they explicitly write down the implication that R-N holds for all $\sigma$-finite measures; that's the point at which they write down the construction of the derivative as a countable sum of derivatives of finite measures with pairwise disjoint supports; it's easy to see that thus the derivative is $\mu$ a.e. finite for infinite measures as well.
The point 4. in your question is equivalent to saying that the derivative is infinite on the set $F$, which by the finiteness and uniqueness of R-N derivative is only true if $\mu(F) = 0$ (for $\sigma$-finite measures!), thus it's kind of a superfluous way of writing this equation.