Let $(X, \mathfrak{M}, \mu)$ be a finite measure space, let $\{E_{k}\}_{k=1}^{n}$ a collection of measurable sets, and $\{ c_{k}\}_{k=1}^n$ a collection of real numbers. For $E \in \mathfrak{M}$ define $$\nu(E) = \sum_{k=1}^{n} c_{k} \mu(E \cap E_{k}).$$ Show that $\nu$ is a measure on $(X, \mathfrak{M})$ that is absolutely continuous with respect to $\mu$ and find its Radon-Nikodym derivative $d\nu / d\mu$.
Why is the measure $\nu$ even nonnegative here?
If you do not assume that the $c_k$ are nonnegative, the statement is false. For $n=1$, $E_1=X$ and $c_1=-1$, you'd have $\nu=-\mu$, which is only nonnegative for $\mu(X)=0$.
However, even without the nonnegativity of the $c_k$, $\nu$ would still be a signed measure and it would indeed be absolutely continuous with respect to $\mu$.