I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following theorem is Theorem 3.7 on p.76 in Section 3A in this book.
3.7 integral of a simple function
Suppose $(X,\mathcal{S},\mu)$ is a measure space, $E_1,\dots,E_n$ are disjoint sets in $\mathcal{S}$, and $c_1,\dots,c_n\in [0,\infty]$. Then $$\int \left(\sum_{k=1}^n c_k\chi_{E_k}\right) d\mu=\sum_{k=1}^nc_k\mu(E_k).$$
The author wrote as follows on p.79 in this book.
The representation of a simple function $h:X\to [0,\infty]$ in the form $\sum_{k=1}^n c_k\chi_{E_k}$ is not unique. Requiring the numbers $c_1,\dots,c_n$ to be distinct and $E_1,\dots,E_n$ to be nonempty and disjoint with $E_1\cup\dots\cup E_n=X$ produces what is called the standard representation of a simple function [take $E_k=h^{-1}(\{c_k\})$, where $c_1,\dots,c_n$ are the distinct values of $h$]. The following lemma shows that all representations (including representations with sets that are not disjoint) of a simple measurable function give the same sum that we expect from integration.
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 $\sum_{j=1}^m a_j\mu(A_j)$.
The next step is to make the numbers $a_1,\dots,a_m$ distinct. This is done by replacing the sets corresponding to each $a_j$ by the union of those sets, and using finite additivity of the measure $\mu$ to show that the value of the sum $\sum_{j=1}^m a_j\mu(A_j)$ does not change.
Finally, drop any terms for which $A_j=\emptyset$, getting the standard representation for a simple function. We have now shown that the original value of $\sum_{j=1}^m a_j\mu(A_j)$ is equal to the value if we use the standard representation of the simple function $\sum_{j=1}^m a_j\chi_{A_j}$. The same procedure can be used with the representation $\sum_{k=1}^n b_k\chi_{B_k}$ to show that $\sum_{k=1}^n b_k\mu(B_k)$ equals what we would get with the standard representation. Thus the equality of the functions $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}$ implies the equality $\sum_{j=1}^m a_j\mu(A_j)=\sum_{k=1}^n b_k\mu(B_k)$.
The author used the standard representation of a simple function to prove Theorem 3.13.
Why?
Suppose $a'_1,\dots,a'_{m'}\in [0,\infty]$.
Suppose $A'_1,\dots,A'_{m'}$ are disjoint elements of $\mathcal{S}.$
Suppose $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{j=1}^{m'} a'_j\chi_{A'_j}$.
Suppose $\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)$.
Suppose $b'_1,\dots,b'_{n'}\in [0,\infty]$.
Suppose $B'_1,\dots,B'_{n'}$ are disjoint elements of $\mathcal{S}.$
Suppose $\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$.
Suppose $\sum_{k=1}^n b_k\mu(B_k)=\sum_{k=1}^{n'} b'_k\mu(B'_k)$.
Then, by Theorem 3.7, $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)=\int\left(\sum_{j=1}^{m'}a'_j\chi_{A'_j}\right)d\mu=\int\left(\sum_{k=1}^{n'}b'_k\chi_{B'_k}\right)d\mu=\sum_{k=1}^{n'} b'_k\mu(B'_k)=\sum_{k=1}^n b_k\mu(B_k).$$
So, I think we don't necessarily need to use the standard representation of a simple function to prove Theorem 3.13.
Am I wrong?
I modified the proof in this book as follows:
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 without using a standard representation of a simple function:
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 $A'_1,\dots,A'_{m'}$ such that $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{j=1}^{m'} a'_j\chi_{A'_j}$ and $\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)$, where $a'_1,\dots,a'_{m'}\in [0,\infty]$.
Similary, repeating the above process with all pairs of subsets among $B_1,\dots,B_n$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $B_1,\dots,B_n$ into a disjoint list $B'_1,\dots,B'_{n'}$ such that $\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$ and $\sum_{k=1}^n b_k\mu(B_k)=\sum_{k=1}^{n'} b'_k\mu(B'_k)$, where $b'_1,\dots,b'_{n'}\in [0,\infty]$.
By the hypothesis of Theorem 3.13, $\sum_{j=1}^{m'} a'_j\chi_{A'_j}=\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$.
By Theorem 3.7, $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)=\int\left(\sum_{j=1}^{m'}a'_j\chi_{A'_j}\right)d\mu\\=\int\left(\sum_{k=1}^{n'}b'_k\chi_{B'_k}\right)d\mu=\sum_{k=1}^{n'} b'_k\mu(B'_k)=\sum_{k=1}^n b_k\mu(B_k).$$
Best Answer
But you haven't written a proof (of the same theorem).
(EDIT: the OP has since considerably changed the post, so what I say may appear misplaced, out of that context).
You've supposed many things but what if I don't suppose the things you suppose?
Let's agree to suppose we are given a measure space $(X,\mathcal{S},\mu)$ and $n,m\in\Bbb N$ and extended nonnegative reals $a_1,\cdots,a_m,b_1,\cdots,b_n$ and measurable $A_1,\cdots,A_m,B_1,\cdots,B_n$ which satisfy $\sum_{j=1}^ma_j\chi_{A_j}=\sum_{k=1}^nb_j\chi_{B_k}$.
I would now like to conclude $\sum_{j=1}^ma_j\mu(A_j)=\sum_{k=1}^nb_k\mu(B_k)$.
In your approach, you "suppose" also that there are disjoint $A'_1,\cdots,A'_{m'},B'_1,\cdots,B'_{n'}\in\mathcal{S}$ and coefficients $a'_1,\cdots,a'_m,b'_1,\cdots,b'_{n'}$ with $\sum_{j=1}^{m'}a'_j\mu(A'_j)=\sum_{j=1}^{m}a_j\mu(A_j)$, and similarly for $B,b$. But how do you know such $A'_\bullet$ and $a'_\bullet$ exist? It's not a hypothesis of the theorem, so the onus is on you to justify the use of these objects. Otherwise, because you have made extraneous suppositions, you're trying to prove a different theorem!
Similarly, you state $\sum_{j=1}^{m'}a'_j\chi_{A'_j}=\sum_{j=1}^ma_j\chi_{A_j}$... this is essentially the same thing as the "standard representation" because you supposed the $A'_\bullet$ to be disjoint. So your proof is essentially saying: "suppose the standard representation satisfies $\sum_{j=1}^{m'}a'_j\mu(A'_j)=\sum_{j=1}^ma_j\mu(A_j)$ and similarly for $B,b$. Then [conclude the result]". But this is no different to the idea of Axler's proof and makes the mistake of assuming something which you don't yet know - I challenge you to justify these "suppose"s without already proving theorem $3.13$.