Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 23H / b / Solution

Yes. Define the probability measureCountable additivity follows by Tonelli's theorem and $\mu (E)={\sum }_n{2}^{-n}=1$. If $\mu (A)=0$, every nonnegative term $2^{-n}{\nu }_n(A)$ is zero, so ${\nu }_n(A)=0$ for every $n$. Hence every ${\nu }_n\ll \mu$.

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