I know this question will get a lot of downvotes but pardon me because after thinking about it in many ways, I can't see why I would end up with disjoint sets after a finite number of steps. Now I state my problem.
Let $(\Omega,\Sigma,\mu)$ be a measured space. Let $A_1,A_2,A_3,...,A_n\in\Sigma$ and let $c_i,c_2,c_3,...,c_n\in(0,\infty)$. Define a quantity $h$ as $$h=\sum_{i=1}^{n}c_i\mu(A_i).$$
$\textbf{Step 1}$
Let's find two sets $A_j,A_k\ (j\neq k)$ such that $A_j\cap A_k\neq\emptyset$. Then $A_j=A_j\setminus A_k\cup (A_j\cap A_k)$, $A_k=A_k\setminus A_j\cup (A_j\cap A_k)$. It is obvious that $A_j\setminus A_k,\ A_k\setminus A_j,\ A_j\cap A_k\in\Sigma$.
$\textbf{Step 2}$
Rewrite the quantity $h$ as $$h=\sum_{i\in [n]\setminus\{j,k\}}c_i\mu(A_i)+ c_j\mu(A_j\setminus A_k)+ c_k\mu(A_k\setminus A_j)+ (c_j+c_k)\mu(A_j\cap A_k).$$
$\textbf{Repeat Step 1 and Step 2}$
Now, my question is, how can I prove that after a finite number of steps, we will end up $h$ in the form $$h=\sum_{i=1}^{m}d_i\mu(B_i),$$ where $d_i\in(0,\infty)$ and $\forall p,q\in[m]\ (p\neq q)$ we have that $B_p\cap B_q=\emptyset.$
Can somebody prove it by contradiction or any other method?
Here is my Venn diagram suggestion from the comments written out more formally
$$\sum_{i=1,\ldots,n}c_i \mu(A_i) = \sum_{I\subseteq\{1,\ldots,n\}}\left(\left(\sum_{i\in I}c_i\right) \cdot \mu\left(\bigcap_{i\in I}A_i\setminus\bigcup_{j\not\in I}A_j\right)\right). $$