# A Comparison Theorem for Integrals of Step Functions on General Intervals

Recall from the Integrals of Step Functions on General Intervals page that if $f$ is a step function on the interval $I$ then 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 $x \in I \setminus [a, b]$ and the integral of $f$ over $I$ is defined as:

(1)Furthermore, since $f$ is a step function on $[a, b]$ we have that there exists a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ such that $f(x) = c_k$ (here $c_k$ is a constant) on each subinterval $(x_{k-1}, x_k)$ for each $k \in \{ 1, 2, ..., n \}$ and $\int_a^b f(x) \: dx$ exists by Lebesgue's criterion and $\displaystyle{\int_a^b f(x) \: dx = \sum_{k=1}^{n} c_k[x_k - x_{k-1}]}$. Therefore:

(2)We will now look at a nice comparison for the integrals of step functions on general intervals.

Theorem 1: Let $f$ and $g$ be step functions on the interval $I$ and suppose that $f(x) \leq g(x)$ for all $x \in I$. Then $\displaystyle{\int_I f(x) \: dx \leq \int_I g(x) \: dx}$. |

**Proof:**Let $f$ and $g$ be step functions on the interval $I$. Then there exists closed and bounded intervals $[a, b]$ and $[c, d]$ such that $f$ is a step function in the usual sense on $[a, b]$ and $f(x) = 0$ for all $x \in I \setminus [a, b]$ as well as $g$ is a step function in the usual sense on $[a, b]$ and $g(x) = 0$ for all $x \in I \setminus [c, d]$.

- Since $f$ is a step function in the usual sense on $[a, b]$ there exists a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ such that $f(x) = a_k$ on each subinterval $(x_{k-1}, x_k)$ for $k \in \{ 1, 2, ..., n \}$. Similarly, since $g$ is a step function in the usual sense on $[c, d]$ there exists a partition $P' = \{ c = y_0, y_1, ..., y_m = d \} \in \mathscr{P}[c, d]$ such that $g(x) = c_k$ on each subinterval $(y_{k-1}, y_k)$ for $k \in \{1, 2, ..., m \}$.

- Let $[e, f]$ be a closed interval such that $[a, b], [c, d] \subseteq [e, f] \subseteq I$. Then:

- Thus $\displaystyle{\int_I f(x) \: dx \leq \int_I g(x) \: dx}$. $\blacksquare$