 Solution (source code)

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= Solution

\b[Yes.] Define the probability measure
$$
\mu=\sum_{n=1}^\infty2^{-n}\nu_n.
$$
Countable additivity follows by Tonelli's theorem and $\mu(E)=\sum_n2^{-n}=1$. 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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