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}

▾ Set of symbols

▾ Alphabet

▾ Deduction system

▾ Theory

▾ Zermelo-Fraenkel set theory

▾ Set

▾ Binary cartesian set product

▾ Binary relation

▾ Map

▾ Function

▾ Measure

▾ Real measure

▾ Euclidean real measure

▾ Complex measure

▾ Basic measure

▾ Unsigned basic measure

▾ Alphabet

▾ Deduction system

▾ Theory

▾ Zermelo-Fraenkel set theory

▾ Set

▾ Binary cartesian set product

▾ Binary relation

▾ Map

▾ Function

▾ Measure

▾ Real measure

▾ Euclidean real measure

▾ Complex measure

▾ Basic measure

▾ Unsigned basic measure

Formulation 2

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}

Then $\nu$ is

Child definitions