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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 kth 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

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