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Definition: If $f$ is a simple function with range $R(f)=\{a_1,a_2,\dots,a_n\}$ and $E_k=\{x\in D(f):f(x)=a_k\}$ for $k\in \{1,2,...,n\}$ then the Canonical Representation of $f$ is $$\begin{align} \quad f(x) &= \sum_{k=1}^{n} a_k \chi_{E_k}(x) \\ &= a_1\chi_{E_1}(x) + a_2\chi_{E_2}(x) + ... + a_n\chi_{E_n}(x) \end{align}.$$

If each term in the sum is a constant $\times$ a characteristic function $\{0,1\},$ is the sum a constant? The confusion comes from picturing each term as a separate step in things like this:

enter image description here

If the function is composed of all the different steps in red segments in the plot, then adding arithmetically the different constant values doesn't seem to represent the function well, because you end up with just a single constant - not multiple constant values, one for each step. On the other hand, if each step is a separate simple function, then the sum doesn't make sense.

My problem may lie in understanding what is involved in the operation $a_k\chi_{E_k}(x).$ Is this sum an arithmetic sum, or is the $\sum$ symbol simply a representation of a sequence of simple functions in this context?

Source for definition and images.

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  • $\begingroup$ Each term in the series is a constant times a characteristic function. A simple function will be constant on the sets in the canonical representation. It needn't be constant in general (for example, $\chi_{\mathbb{Q}}$, the function that is one on the rationals and zero elsewhere, is not constant on any set containing an interval). $\endgroup$
    – Xander Henderson ♦
    Commented Mar 14, 2018 at 13:50

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No. The function is not constant, since there are different points in which it takes different values. At most one term in the sum is not zero, since the $E_k$ are disjoint. The sum is equivalent to the following: $$ f(x)=\begin{cases} a_1 & \text{if }x\in E_1\\ a_2 & \text{if }x\in E_2\\ \dots &\\ a_n & \text{if }x\in E_n\\ 0 & \text{if }x\notin\bigcup_{k=1}^nE_k \end{cases} $$

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  • $\begingroup$ I suspect a basic misunderstanding on my part, so let me be concrete... The function would include all the different red lines on the left plot, each corresponding to one of the lines in your $\LaTeX$ formula, correct? But multiplying a constant $\times$ a characteristic function (either $1$ or $0$) you get a constant number, and summing leaves you with a single number... $\endgroup$
    – Antoni Parellada
    Commented Mar 14, 2018 at 13:57
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    $\begingroup$ No. If $x\in E_1$, then $\chi_{E_1}(x)=1$ and $\chi_{E_k}(x)=0$ if $k\ne1$. The sum is then $a_1\times1+0+\dots+0=a_1$. Similarly, if $x\in E_2$, then $\chi_{E_2}(x)=1$ and $\chi_{E_k}(x)=0$ if $k\ne2$. The sum is then $0+a_2\times1+0+\dots+0=a_2$. And so on. $\endgroup$
    – Julián Aguirre
    Commented Mar 14, 2018 at 15:00
  • $\begingroup$ eskerrik asko!! $\endgroup$
    – Antoni Parellada
    Commented Mar 14, 2018 at 15:02

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