= Solution
The <absolute continuity of measures> condition is
$$
\nu\ll\kappa\quad\Longleftrightarrow\quad \kappa(E)=0\Longrightarrow\nu(E)=0\qquad(E\in\mathfrak M).
$$
The <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 that
$$
\boxed{\nu(E)=\int_Eh\,d\kappa\quad(E\in\mathfrak M).}
$$
One 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.
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