When is a Radon Nikodym derivative integrable?

Question

Let $\nu \ll \mu$ be two measures on the same measure space. Consider the Radon Nikodym derivative $$\nu(A) = \int_A f d\mu$$ Which requirements must $\nu$ and $\mu$ meet for $f$ to be integrable, i.e., $\int |f|d\mu <\infty$, or quasi-integrable, i.e., at least one of the following is true $\int f^+d\mu <\infty$ or $\int f^- d\mu <\infty$

Answer 1

The existence of the Radon Nikodym derivative $f$ is guaranteed by the Radon Nikodym theorem, which requires that $\mu$ is $\sigma$-finite. See e.g. Donald Cohn's Measure Theory, page 132. It may be good to note that the theorem as stated there for positive $\nu$ also requires $\nu$ to be $\sigma$-finite, but on page 138 in Exercise 6 it says that this assumption can be dropped if $f$ is allowed to take values in $[0,\infty]$.

If $\nu$ is a positive measure then $f$ will be nonnegative almost everywhere. Then (trivially) the integrability of $f$ is equivalent to $\nu$ being finite.

If $\nu$ is a signed measure then the existence of $f$ is guaranteed by (another version of) the Radon Nikodym theorem, which requires $\nu$ to be finite. See e.g. page 135 in Cohn's Measure Theory. This would then imply integrability of $f$. Here in fact integrability of $f$ is also equivalent to $\nu$ being finite: $f$ is integrable iff $f^+$ and $f^-$ are, i.e., iff the positive and negative parts of $\nu$ are finite, i.e., iff $\nu$ is finite. I am not sure whether the condition of $\nu$ being finite can be dropped, to then only have integrability of $f^+$ of $f^-$.