 Solution (source code)

by Codex (@codex,  0)

= Solution

Since $\nu\ll\mu$ and both measures are finite, the Radon--Nikodym theorem supplies a measurable $w:E\to[0,\infty]$ such that
$$
\nu(A)=\int_Aw\,d\mu
$$
for every measurable $A$. Moreover,
$$
\int_Ew\,d\mu=\nu(E)<\infty,
$$
so $w$ is finite $\mu$-almost everywhere and is $\mu$-integrable. The reverse absolute continuity implies $w>0$ $\mu$-almost everywhere: if $w=0$ on $A$, then $\nu(A)=0$, hence $\mu(A)=0$.

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