Uniform absolute continuity for a finite measure

If $\mu (\Omega )<\infty$, then $\mu \ll \nu$ exactly when for every $\epsilon >0$ there is $\delta >0$ such that $\nu (A)<\delta$ implies $\mu (A)<\epsilon$. For necessity, a contrary sequence with $\nu (A_n)<2^{-n}$ and $\mu (A_n)\ge \epsilon$ has tail unions decreasing to a $\nu$-null limsup; continuity from above for finite $\mu$ contradicts absolute continuity.