My textbook says the $\int s\, d\mu$ is unaffected by how $s$ is written.
Say a simple function is written in two ways $$s = \sum_{i=1}^{m} a_i\chi_{A_i} =\sum_{j=1}^{n} b_j\chi_{B_j} =t$$ then $\int s\, d\mu =\sum_{i=1}^{m} a_i\mu(A_i)= \sum_{j=1}^{n} b_j\mu(B_j)=\int t\, d\mu$
How is that true?
Here is one standard way to do this, following the presentation in Stein and Shakarchi's book on real analysis. Note that any simple function $\varphi$ can be written in a canonical form $\varphi = \sum_{j=1}^Na_j^*\chi_{E_j^*}$ where the $a_j^*$ are distinct and the $E_j^*$ are disjoint by taking $E_j^* = \varphi^{-1}(\{a_j^*\})$.
Lemma. Given a collection of sets $F_1, F_2, \ldots, F_n$ we can construct another collection $F_1^*, F_2^*, \ldots, F_N^*$ with $N=2^n-1$ so that $\bigcup_{k=1}^n F_k = \bigcup_{j=1}^N F_j^*$; the collection of $\{ F_j^*\}$ is disjoint; also $F_k = \bigcup_{F_j^*\subset F_k} F_j^*$, for every $k$.
Proof. For a fixed $k \in \{1,\dots,n\}$, let $F_k^0 = F_k^c$ and let $F_k^1 = F_k$. Consider the sets $F_1^{\epsilon_1}\cap\dots\cap F_n^{\epsilon_n}$ where $\epsilon_k$ is either $0$ or $1$. If we neglect the set $F_1^0\cap\dots\cap F_n^0= F_1^c\cap\dots\cap F_n^c$ which is not a subset of any of the $F_k$, then we get a collection of $2^n-1$ disjoint sets with the desired properties.
Proposition. $\int \varphi$ is well-defined.
Proof. First suppose that the sets $E_k$ in the definition of $\varphi$ are disjoint. Write $\varphi = \sum_{k=1}^na_k\chi_{E_k}$ in its canonical form $\varphi = \sum_{j=1}^Na_j^*\chi_{E_j^*}$ where the $a_j^*$ are distinct and the $E_j^*$ are disjoint. There are disjoint collections $J_1,\dots,J_N\subset\{1,\dots,n\}$ such that for each fixed $j\in\{1,\dots,N\}$, and every $k\in J_j$ we have that $a_k = a_j^*$ and since the $E_k$ are disjoint, that $\sum_{k\in J_j}\chi_{E_k} = \chi_{E_j^*}$. Thus, \begin{align*} \varphi = \sum_{k=1}^na_k\chi_{E_k} &= \sum_{j=1}^N\sum_{k\in J_j}a_k\chi_{E_k} \\ &= \sum_{j=1}^N a_j^* \sum_{k\in J_j}\chi_{E_k} \\ &= \sum_{j=1}^N a_j^* \chi_{E_j^*}, \end{align*} In this case it is clear that the integral is well defined. Now suppose that the sets $E_k$ in the definition of $\varphi = \sum_{k=1}^na_k\chi_{E_k}$ are not disjoint. By applying part a), we may construct disjoint sets $E_j^*$ such that $E_k = \bigcup_{E_j^*\subset E_k} E_j^*$ for every $k$. Now we apply the same procedure we used in the case where the sets $E_k$ in the definition of $\varphi$ were disjoint.