= Solution
First assume the stated $\varepsilon$--$\delta$ condition. If $\nu(A)=0$, then $\nu(A)<\delta$ for the $\delta$ belonging to every $\varepsilon>0$, so $\mu(A)<\varepsilon$ for every $\varepsilon$ and $\mu(A)=0$. Hence $\mu\ll\nu$.
Conversely, suppose $\mu\ll\nu$ but the uniform condition fails. Then for some $\varepsilon>0$ there are sets $A_n$ with
$$
\nu(A_n)<2^{-n},\qquad \mu(A_n)\geq\varepsilon.
$$
Put $B_m=\bigcup_{n\geq m}A_n$. Then $\nu(B_m)\leq2^{1-m}$ and $B_m\downarrow B=\limsup A_n$, so $\nu(B)=0$. Absolute continuity gives $\mu(B)=0$. But $\mu(B_m)\geq\varepsilon$, and the finiteness of $\mu(\Omega)$ permits continuity from above:
$$
\mu(B)=\lim_m\mu(B_m)\geq\varepsilon,
$$
a contradiction. This proves <uniform absolute continuity for a finite measure>.
Solved by gpt-5.6-sol high.
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