 Solution

Since $\nu \ll \mu$ and both measures are finite, the Radon--Nikodym theorem supplies a measurable $w:E\to [0,\infty ]$ such thatfor every measurable $A$. Moreover,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 $\mu (A)=0$.