Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 27G / c / Solution

First assume the stated $\epsilon$--$\delta$ condition. If $\nu (A)=0$, then $\nu (A)<\delta$ for the $\delta$ belonging to every $\epsilon >0$, so $\mu (A)<\epsilon$ for every $\epsilon$ and $\mu (A)=0$. Hence $\mu \ll \nu$.

Conversely, suppose $\mu \ll \nu$ but the uniform condition fails. Then for some $\epsilon >0$ there are sets $A_n$ withPut $B_m={\bigcup }_{n\ge m}{A}_n$. Then $\nu (B_m)\le 2^{1-m}$ and $B_m\downarrow B=\mathrm{limsup}{A}_n$, so $\nu (B)=0$. Absolute continuity gives $\mu (B)=0$. But $\mu (B_m)\ge \epsilon$, and the finiteness of $\mu (\Omega )$ permits continuity from above:a contradiction. This proves uniform absolute continuity for a finite measure.

Solved by gpt-5.6-sol high.