I am trying to show that the measure $\mu_1(E)=\sum_{i=1}^{n}c_i\cdot\mu_2(E\cap E_i)$ is absolutely continuous w.r.t $\mu_2$ and then compute its Radon-Nikodym derivative. Here $E_i$ are measurable sets and $c_i$ are real numbers and both $\mu_1, \mu_2$ are defined on a finite measurable space. At first, i thought that $\mu_1$ is the integral of a simple function but again i realised that $E_i$ are not necessary disjoint and so i am stack. I will appreciate your proof.
[Math] Radon-Nikodym derivative
measure-theoryreal-analysis
Best Answer
First of all, absolute continuity of $\mu_1$ with respect to $\mu_2$ is clear: if $N$ is a $\mu_2$-null set then so is $N \cap E_i$, hence $\mu_1(N) = \sum_{i = 1}^n c_i \mu(N \cap E_i) = 0$ which is the very definition of $\mu_1 \ll \mu_2$. But this is irrelevant anyway, because it follows from what I say below.
To find the Radon–Nikodým derivative, just compute (writing $[E]$ for the characteristic function of $E$ and using $[E \cap F] = [E] \cdot [F]$) that for all $\mu_2$-measurable
$$\mu_1(E) = \sum_{i = 1}^n c_i \mu_2(E \cap E_i) = \sum_{i=1}^n c_i \int [E] \cdot [E_i]\,d\mu_2 = \int [E] \cdot \left(\sum_{i=1}^n c_i [E_i]\right)\,d\mu_2$$
and observe that $f = \sum_{i=1}^n c_i [E_i]$ thus satisfies the defining property of the Radon–Nikodým derivative $\frac{d\mu_1}{d\mu_2}$. If you want to write $f$ as simple function with disjoint supports, decompose the sets $E_i$ into disjoint sets $F_{ij}$, as Thomas suggested, but this is a standard step that is done at the very beginning of every course on measure theory, so I skip it.