# The $E_k$ in the definition of the simple function on pg. 61 in Royden real analysis “4th edition”.

analysislebesgue-measuremeasurable-functionsmeasure-theoryreal-analysis

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?

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 $$E$$.