I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following theorem is Theorem 3.13 on p.79 in Section 3A in this book.
3.13 integral-type sums for simple functions
Suppose $(X,\mathcal{S},\mu)$ is a measure space. Suppose $a_1,\dots,a_m,b_1,\dots,b_n\in [0,\infty]$ and $A_1,\dots,A_m,B_1,\dots,B_n\in\mathcal{S}$ are such that $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}.$ Then $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{k=1}^n b_k\mu(B_k).$$
Proof We assume $A_1\cup\dots\cup A_m=X$ (otherwise add the term $0\chi_{X\setminus (A_1\cup\dots\cup A_m)}$).
Suppose $A_1$ and $A_2$ are not disjoint. Then we can write $$a_1\chi_{A_1}+a_2\chi_{A_2}=a_1\chi_{A_1\setminus A_2}+a_2\chi_{A_2\setminus A_1}+(a_1+a_2)\chi_{A_1\cap A_2},\tag{3.14}\label{3.14}$$ where the three sets appearing on the right side of the equation above are disjoint.
Now $A_1=(A_1\setminus A_2)\cup (A_1\cap A_2)$ and $A_2=(A_2\setminus A_1)\cup (A_1\cap A_2);$ each of there unions is a disjoint union. Thus $\mu(A_1)=\mu(A_1\setminus A_2)+\mu(A_1\cap A_2)$ and $\mu(A_2)=\mu(A_2\setminus A_1)+\mu(A_1\cap A_2).$ Hence $$a_1\mu(A_1)+a_2\mu(A_2)=a_1\mu(A_1\setminus A_2)+a_2\mu(A_2\setminus A_1)+(a_1+a_2)\mu(A_1\cap A_2).$$ The equation above, in conjunction with $\ref{3.14}$, shows that if we replace the two sets $A_1,A_2$ by the three disjoint sets $A_1\setminus A_2,A_2\setminus A_1,A_1\cap A_2$ and make the appropriate adjustments to the coefficients $a_1,\dots,a_m$, then the value of the sum $\sum_{j=1}^m a_j\mu(A_j)$ is unchanged (although $m$ has increased by $1$).
Repeating this process with all pairs of subsets among $A_1,\dots,A_m$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $A_1,\dots,A_m$ into a disjoint list of subsets without changing the value of $\sum_{j=1}^m a_j\mu(A_j)$.
$\dots$
I executed the above process for $m=3$ as follows:
But for $m$ which is larger than $3$, I could not execute the above process because of the complexity.
How to prove the following fact?
Repeating this process with all pairs of subsets among $A_1,\dots,A_m$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $A_1,\dots,A_m$ into a disjoint list of subsets without changing the value of $\sum_{j=1}^m a_j\mu(A_j)$.