Here is the problem:
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 $$\nu(E) = \sum_{k=1}^{n} c_{k} \mu(E \bigcap E_{k}).$$ Show that $\nu$ is a measure on $(X, \mathfrak{M})$ that is absolutely continuous with respect to $\mu$ and find its Radon-Nikodym derivative $d\nu / d\mu.$
I want to find the Radon-Nikodym derivative of $\nu$.
My trial is
Since the definition of the integral of a simple function $\psi$ is given by: let $c_{1}, c_{2}, ... , c_{n}$ be positive values taken by $\psi$ on X and, for $1 \leq k \leq n,$ define $E_{k} = \{ x \in X | \psi(x) = c_{k}\}.$ Define $$\int_{X} \psi d\mu = \sum_{k=1}^{n} c_{k}. \mu(E_{k}).$$
Then the given definition of $\nu$ is the integral of the simple function $\psi$, then the Radon-Nikodym derivative of $\nu$ w.r.t $\mu$ is the simple function $\psi$,
but I have a question about my trial: the given definition of $\nu$ in the question contains $E \bigcap E_{k}$ but the definition of the integral of a simple function contains $E_{k}$ only, so how can I deal with this?
Note the Radon Nikodym derivative of the sum is the sum of the Radon Nikodym derivatives. It also respects scaling so you need only find it for $\nu_1(E)=\mu(E\cap E_1)=\int 1_E 1_{E_1}d\mu$. Thus $\frac{d\nu}{d\mu}=\sum_{i=1}^n c_i1_{E_i}$.
It is perfectly fine that the Radon Nikodym derivative doesn't reference $E$. You only need E when caluclating $\nu(E)$ but you can write down $\nu$ only referencing your preferred measurable sets.