# Uniqueness in the decomposition of a simple function in the canonical form.

I know that every simple function $\varphi$ is a finite sum

$$\varphi (x) = \sum_{k=1}^N a_k \chi_{E_k} (x).$$

where the $E_k$ are measurable sets of finite measure and $a_k$ are constants.This representation of a simple function is not unique.For instance $0= \chi_E - \chi_E$ for any measurable set $E$ of finite measure.In order to make this representation unique we have to decompose $\varphi$ in a unique way known as canonical form of $\varphi$.It is the similar decomposition as mentioned above where $a_k$ are distinct and non-zero constants and $E_k$ are disjoint.

In this connection let me quote Stein and Shakarchi's statement - "Finding the canonical form of $\varphi$ is straightforward $:$ since $\varphi$ can take only finitely many distinct and non-zero values, say $c_1 , c_2 , \cdots ,c_M$, we may set $F_k=\{x: \varphi (x) = c_k \}$, and note that the sets $F_k$ are disjoint. Therefore $\varphi = \sum_{k=1}^M c_k \chi_{F_k}$ is the desired canonical form of $\varphi$."

Now my question is : "Is there any other form of $\varphi$ which can also be treated as a canonical form of $\varphi$."I am not much sure of that.Please help me in this regard.

• Why do you want another "canonical form"? There are countless ways you could define a canonical form, most of which are rather artificial and not useful. – Eric Wofsey Mar 24 '18 at 16:54
• Which are they? Would you please tell me one such? – Arnab Chatterjee. Mar 24 '18 at 17:00

No, there can't be any other canonical form (I assume that $\varphi=\sum_{k=1}^M b_k 1_{E_k}$ is a cannonical form if and only if $E_k$ and $b_k$ are distinct).
If $\varphi = \sum_{k=1}^M b_k 1_{E_k}$ is another canonical form then for every $x\in F_k$ you have that
$a_k = \varphi(x) = \sum_{k=1}^M b_k 1_{E_k}(x)$
By definition, $E_k$ are distinct, so there exists a unique $i$ such that $1_{E_i}(x)=1$. You then have $a_k = b_i$.
Therefore the only way to produce another canonical form is to rearrange the sets $F_k$.