Let $X$ be the universal set, and $E\subset X$ be a measurable set.
In Rudin, suppose the range of $s$ consists of the distinct numbers $c_1, ..., c_n$. Let $E_i =\{x|s(x)=c_i\} ~ (i=1,...,n)$. Then $s = \sum_{n=1}^n c_i K_{E_i}$, that is, every simple function is a finite linear combination of characteristic functions.
Since $c_i,...,c_n$ are distinct, I think that the representation above about $s$ must be unique, and I had a hard time coming up with another different summation of characteristic functions.
However, later in the class, we learned the lemma that "for a simple measurable function $s$, the value of $I_E(s) = \sum_{i=1}^n c_i \mu(E\cap E_i)$ does not depend on the choice of how to write $s$ as a linear combination of characteristic functions." In other words, let $s = \sum_{i=1}^n c_i K_{E_i}$ be simple measurable. and let $\{v_1,...,v_m\}$ be the range of $s$ and set $D_j=s^{-1}(v_j)$, then $\sum_{i=1}^n c_i \mu({E\cap E_i})=\sum_{j=1}^m v_j \mu({E\cap D_j})$.
But how come $\sum_{i=1}^n c_i \mu({E\cap E_i})$ and $\sum_{j=1}^m v_j \mu({E\cap D_j})$ are not same at the first place? How come $\sum_{i=1}^n c_i K_{E_i}$ and $\sum_{j=1}^m v_j K_{ D_j}$ are not the same at the very first place? I think they are same because by definition, $c_1,...,c_n$ are distinct numbers and they form the range of $s$; on the other hand, $v_1,...,v_m$ also for the range of $s$. Doesn't that just imply that $c_1,...,c_n$ is $v_1,...,v_m$?