I am reading the definition of Lebesgue integration that turns into an exercise$^{(1)}$:
Definition: Let $(X, \mathcal A, \mu)$ be a measure space. If $$s = \sum_{i=1}^{n}a_i \chi_{E_i}$$ is a non-negative measurable simple function, define the Lebesgue integral $s$ to be $$\int s d\mu = \sum_{i=1}^{n}a_i \mu (E_i).$$ A few remarks are in order. A function $s$ might be written as a simple function in more than one way. For example, $s = \chi_{A \cup B} = \chi_A + \chi_B$ if $A$and $B$ are disjoint. It is not hard to check that the definition of $\int s d \mu$ is unaffected by how $s$ is written. If
$$\begin{align} s = \sum_{i=1}^{m} a_i \chi_{A_i} &= \sum_{j=1}^{n} b_j \chi_{B_j}, \text{ then} \tag{a}\\ \sum_{i=1}^{m} a_i \mu(A_i) &= \sum_{j=1}^{n} b_j \mu(B_j). \tag{b} \end{align}$$ We leave the proof of this (b) to the reader as an exercise.
Here is my folksy homespun attempt to prove the (b):
(1) I am visualizing Lebesgue integration as the area under the curve partitioned horizontally, with $a_i, b_i$ as partition of $y$-axis and $A_i, B_i$ as the intervals on the $x$-axis.
(2) Since the characteristic function is defined as $$\chi_{A_i}= \begin{cases} 1, &\text{for } A_i \neq 0\\ 0, &\text{for } A_i = 0 \end{cases} ,$$ may I then define the measure $\mu$ as follow, $$\mu (A_i)= \begin{cases} \mu(A_i), &\text{for } \chi_{A_i} = 1 \\ 0, &\text{for } \chi_{A_i} = 0 \end{cases} ,$$ such that this following represents the area under the curve? $$\sum_{i=1}^{m} a_i \mu(A_i)$$
(3) Similarly, following the same argument, I can claim that for the $B_i$ this following also represents the area under the same curve: $$\sum_{j=1}^{n} a_i \mu(B_j).$$
(4) Since (a) is true and both represent the same area under the same curve, therefore (b) is true.
Please let me know if the above is legit, and if it isn't, the elegant way of proving it. Thanks for your time and help.
(1) Richard F. Bass' Real Analysis, 2nd. edition, chapter 6: The Lebesgue Integration, Definition 6.1, page 47-48.