$\mu$ is a $\sigma-$finite measure on and $\{E_n\}$ measurable sets. When $\nu(E)=\sum \mu(E\cap E_n)$, is $\nu$ is $\sigma$-finite? My question stems from the following problem.
Suppose $\mu$ is a $\sigma-$finite measure on $(X,M)$ and $\{E_n\}$ a sequence of measurable sets. Define $\nu$ on $M$ by $\nu(E)=\sum \mu(E\cap E_n)$. Find the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$.
For all measurable set $E$, $\nu(E)=\sum \mu(E\cap E_n)=\sum \int \chi_{E\cap E_n} d\mu = \sum \int_E \chi_{E_n}d\mu=\int (\sum \chi_{E_n}) d\mu$, so that $\frac{d\nu}{d\mu}=\sum \chi_{E_n}$. 
However, for the Radon-Nikodym derivative to exist, I need the fact that $\nu$ is $\sigma-$finite, but I can't show how this is possible. Is $\nu$ sigma finite even when the $E_n$'s are not disjoint? And how can I show this in that case? I would greatly appreciate any help.

For the Radon–Nikodym theorem you don't need the measure $\nu$ to be $\sigma$-finite, only $\mu$.
