Every step function is a linear combination of characteristic functions

We define a characteristic function, $\chi_S$ , on a set $S$ of points on $\mathbb{R}$ by

$\chi_S(x) = \begin{cases} 1 & \text{if } x \in S \\ 0 & \text{if } x \notin S. \end{cases}$

Let $f$ be a step functions taking the (constant) value $c_k$ on the $k$ th open subinterval $I_k$ of a partition of $[a,b]$ . Prove that for each $x \in I_1 \cup \cdots \cup I_n$ , we have

$f(x) = \sum_{k=1}^n c_k \chi_{I_k} (x).$

Proof. First, we note that the open subintervals of some partition of $[a,b]$ are necessarily disjoint since $P = \{ x_0, x_1, \ldots, x_n \} \ \implies \ x_0 < x_1 < \cdots < x_n$ . Hence, if $x \in I_1 \cup \cdots \cup I_n$ then $x \in I_j$ for exactly one $j, \ 1 \leq j \leq n$ .
So, we have

$\chi_{I_k}(x) = \begin{cases} 1 & \text{if } j = k \\ 0 & \text{if } j \neq k \end{cases}$

for all $1 \leq k \leq n$ , and for any $x$ . Further, by definition of $f$ , we know $f(x) = c_k$ if $x \in I_k$ . So,

$\begin{align*} \sum_{k=1}^n c_k \chi_{I_k}(x) &= c_1 \cdot 0 + c_2 \cdot 0 + \cdots + c_k \cdot 1 + \cdots + c_n \cdot 0 \\ &= c_k \\ &= f(x) \end{align*}$

for each $x \in I_1 \cup \cdots \cup I_n. \qquad \blacksquare$

One comment

September 3, 2020 at 8:43 am

Anonymous says:

I think the summation result should be c(j) instead of c(k) since you have already let x is an element of I(j)

Stumbling Robot

Every step function is a linear combination of characteristic functions

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