Step Functions on General Intervals

When we looked at Riemann-Stieltjes integration we looked at a special type of integrator function known as Step Functions. We will soon look into another type of integral known as a Lebesgue integral but we will first need to give a more general definition of a step function - this time on any interval $I$.

Definition: A function $f$ is said to be a Step Function on the interval $I$ if for some interval $[a, b] \subseteq I$ we have that $f$ is a step function on $[a, b]$ in the usual sense and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$.

It is important to note that the interval $I$ described above can be any type of interval. It can itself be a closed and bounded interval; an unbounded interval; and open interval, etc… We just require that there does indeed exist a closed and bounded interval $[a, b]$ contained in $I$ for which $f$ is a step function on $[a, b]$ in the usual sense.

For example, consider the interval $I = [0, \infty)$ and the following function $f$ defined for all $x \in [0, \infty)$ by:

(1)

\begin{align} \quad f(x) = \left\{\begin{matrix} 1 & \mathrm{if} \: 0 \leq x < 1\\ 2 & \mathrm{if} \: x = 1\\ -1 & \mathrm{if} \: 1 < x \leq 3\\ 0 & \mathrm{if} \: 3 < x < \infty \end{matrix}\right. \end{align}

Note that on the closed and bounded interval $[0, 3] \subset I = [0, \infty)$ we have that $f$ is a step function in the usual sense, and that for $x \in I \setminus [0, 3] = (3, \infty)$ we have that $f(x) = 0$. Therefore $f$ is a step function on $I$. The graph of $f$ is given below: