The $E_k$ in the definition of the simple function on pg. 61 in Royden real analysis "4th edition". I want to prove that the sum of 2 simple functions is a simple function and to do so I want to use the following facts:
$$\chi_{A_i}=\sum_{j}\chi_{A_i\cap B_j} \text{ and } \chi_{B_j}=\sum_{i}\chi_{A_i\cap B_j}.$$
But say if I want to prove the first fact, upon fixing $i,$ I want $A_i$ to be subset of $E$ and $E = \cup_{j}B_j$
Here is the definition of the simple function stated in Royden:
If $\varphi$ is simple, has domain $E$ and takes the distinct values $c_1, \dots, c_n,$ then $$\varphi = \sum_{i=1}^{n_1} c_k \chi_{E_k} \text{ where } E_k = \{x \in E| \varphi(x) = c_k \}.$$
My question is:
Are we considering the $E_k's$ in the definition of the simple function to be a partition of $E$? Could someone clarify this to me please?
 A: We say that $\varphi$ is a simple function if it is a linear combination
of indicator functions of measurable sets. That is, $\varphi=\sum_{i=1}^{n}\alpha_{i}1_{A_{i}}$,
where $\alpha_{i}\in\mathbb{R}$ and $A_{i}$ is a measurable set.
We do not require that $A_{1},A_{2},\ldots,A_{n}$ are pairwisely
disjoint. Moreover, it is also false that $\varphi$ will take values
$\alpha_{1},\ldots\alpha_{n}$. (For example, for the constant function
$0$, we can write it as $0=(1)1_{E}+(-1)1_{E}$ but neither $1$
nor $-1$ lies in the range $\varphi(E)$.)
Your representation is called "canonical" representation of $\varphi$.
In that case, the range $\varphi(E)$ is precisely $\{\alpha_{1},\ldots,\alpha_{n}\}$
(Here, $\alpha_{1},\ldots,\alpha_{n}$ are pairwisely distinct) and
$A_{i}:=\{x\in E\mid\varphi(x)=\alpha_{i}\}$ will form a partition
of $E$.
A: Simple functions on a measurable space $(E, \mathcal{F})$ are measurable functions $\phi \colon E \to \mathbb{C}$ with finite range. Royden writes the simple function in the canonical way as a linear combination of characteristic functions of disjoint measurable sets. Even if the sets in the linear combination are not disjoint, the function is still simple since it has finite range.
