 Solution

The absolute continuity of measures condition isThe Radon-Nikodym theorem says that if both positive measures are sigma-finite measures, then this condition is equivalent to the existence of a nonnegative measurable function $h$, unique up to equality $\kappa$-almost everywhere, such thatOne writes $h=d\nu /d\kappa$, the Radon-Nikodym derivative. Under these hypotheses $h$ is finite $\kappa$-almost everywhere. If $\nu$ is finite, then $h\in L^1(\kappa )$. A finite signed measure absolutely continuous with respect to a sigma-finite positive measure likewise has an integrable signed Radon-Nikodym derivative.

The sigma-finiteness hypothesis on the dominating measure cannot simply be omitted. On the Borel subsets of $[0,1]$, let $\kappa$ be counting measure and $\nu$ Lebesgue measure. Absolute continuity holds because the only $\kappa$-null set is empty. A Radon-Nikodym derivative would satisfy $h(x)=\nu (\{x\})=0$ at every point, contradicting $\nu ([0,1])=1$. Here counting measure is not sigma-finite.