# Integrals of Step Functions on General Intervals

Recall from the Step Functions on General Intervals page that a function $f$ is said to be a step function on the general interval $I$ if there exists a closed and bounded interval $[a, b] \subseteq I$ such that $f$ is a step function in the usual sense on $[a, b]$ and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$.

Now let $f$ be a step function on the interval $I$. Then for some closed and bounded interval $[a, b]$ we have that $f$ is a usual step function on this interval, and so there exists a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ such that $f$ is constant on each open subinterval $(x_{k-1}, x_k)$. The set of discontinuities of $f$ occur at the endpoints of these intervals, but nevertheless, this set of discontinuities must be a finite set since the partition $P$ breaks $[a, b]$ into at most finitely many subintervals. Regardless, the measure of the set of discontinuities is $0$ and so by Lebesgue's criterion for the Riemann integrability of a function we have that $\int_a^b f(x) \: dx$ exists. Furthermore, if $f(x) = c_k$ for $x \in (x_{k-1}, x_k)$ for each $k \in \{1, 2, ..., n \}$ we have that:

(1)Now, since $[a, b] \subseteq I$ and $f(x) = 0$ for all $x \in I \setminus [a, b]$ we have that the integral of $f$ on $I$ will equal to that of above. We formally define this integral below.

Definition: Let $f$ be a step function on the interval $I$. Then there exists an $[a, b] \subseteq I$ such that $f$ is a step function in the usual sense on $[a, b]$ and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$. The Integral of $f$ over $I$ is defined to be $\displaystyle{\int_I f(x) \: dx = \int_a^b f(x) \: dx}$. |

*For brevity, the notation $\int_I f$ can be used in place of $\int_I f(x) \: dx$ when no ambiguity arises.*

For example, consider the following step function $f$ on $[0, \infty)$ given by:

(2)Then the integral of $f$ over $I$ is:

(3)