Let $\mu, \nu : \mathcal{F} \to [0, \infty]$ each be an D85: Unsigned basic measure.
Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if \begin{equation} \forall \, E \in \mathcal{F} \left( \mu(E) = 0 \quad \implies \quad \nu(E) = 0 \right) \end{equation}