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