# 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?

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 $$E$$.

• I understood your second paragraph. But I did not understand your first paragraph. What is the difference between the definition of the simple function and its canonical representation, should not they both go to the same thing (with all hypotheses)?
– user965463
Commented Nov 1, 2021 at 2:53
• Why neither of $-1$ or $1$ lies in the range of $\phi(E),$ could you clarify please?
– user965463
Commented Nov 1, 2021 at 2:56
• The range of $\varphi$ is $\varphi(E) = \{\varphi(x)\mid x\in E\} =\{0\}$. Commented Nov 1, 2021 at 3:38
• I mean, simple function has more than one representations. For example, for the constantly zero function $f$, we can write $f = 0\cdot 1_E = (1)1_E + (-1)1_E = (2)1_E + (-2)1_E$. Commented Nov 1, 2021 at 3:40
• Canonical representation is a special representation, where all $\alpha_i$ are elements in the range $\varphi(E)$ and that $A_i = \{x\in E \mid \varphi(x)=\alpha_i \}$. Commented Nov 1, 2021 at 3:42

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.