Absolutely continuous measure

Remarks

Example: let $f \colon X \to [0, \infty]$ be measurable and $\nu(A) = \int_A f \, d\mu$. If $\mu(A) = 0$, then $f \mathbf{1}_A = 0$ $\mu$-almost everywhere, so $\nu(A) = 0$; hence $\nu \ll \mu$.

Non-example: on $\R$ with the Borel sets, the Dirac measure $\delta_0$ is not absolutely continuous with respect to Lebesgue measure $\lambda$. The set $A = \set{0}$ has $\lambda(A) = 0$ but $\delta_0(A) = 1$.

If $\nu$ is finite, then $\nu \ll \mu$ holds if and only if for every $\varepsilon > 0$ there is $\delta > 0$ with $\nu(A) < \varepsilon$ whenever $\mu(A) < \delta$. If $\mu$ and $\nu$ are σ-finite, the Radon–Nikodym theorem shows that $\nu \ll \mu$ if and only if $\nu$ has a density $f$ with respect to $\mu$.