# Bounded Piecewise Continuous Function is Riemann Integrable

## Theorem

Let $f$ be a real function defined on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $f$ be piecewise continuous and bounded on $\left[{a \,.\,.\, b}\right]$.

Then $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

## Proof

We are given that $f$ is piecewise continuous and bounded on $\left[{a \,.\,.\, b}\right]$.

Therefore, there exists a finite subdivision $\left\{ {x_0, x_1, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \left\{{1, 2, \ldots, n}\right\}$:

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$f$ is bounded on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$.

Note that $n$ is the number of intervals $\left({x_{i - 1} \,.\,.\, x_i}\right)$ defined from the (finite) subdivision $\left\{{x_0, x_1, \ldots, x_n}\right\}$.

We shall use proof by induction on these $n$ intervals.

For all $k \in \left\{{1, 2, \ldots, n}\right\}$, let $P \left({k}\right)$ be the proposition:

$f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_k}\right]$.

### Basis for the Induction

$P \left({1}\right)$ is the case:

$f$ is Riemann integrable on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$

for an arbitrary $i \in \left\{{1, 2, \ldots, k}\right\}$.

$f$ is piecewise continuous and bounded for the case $n = 1$ means that:

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$f$ is bounded on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$.

By Bounded Function Continuous on Open Interval is Riemann Integrable, $f$ is Riemann integrable on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$.

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

$f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_k}\right]$

from which it is to be shown that:

$f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_{k + 1} }\right]$.

### Induction Step

This is the induction step:

By definition of a bounded piecewise continuous function, for every $i \in \left\{ {1, 2, \ldots, k, k + 1}\right\}$:

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$f$ is bounded on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$.

By the induction hypothesis, $f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_k}\right]$.

From the basis for the induction, $f$ is Riemann integrable on $\left[{x_k \,.\,.\, x_{k + 1} }\right]$.

We have that $f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_k}\right]$ and $\left[{x_k \,.\,.\, x_{k + 1} }\right]$.

Therefore, $f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_k}\right] \cup \left[{x_k \,.\,.\, x_{k + 1} }\right]$ by Existence of Integral on Union of Adjacent Intervals.

We have that:

$\left[{x_0 \,.\,.\, x_{k + 1} }\right] = \left[{x_0 \,.\,.\, x_k}\right] \cup \left[{x_k \,.\,.\, x_{k + 1} }\right]$.

Accordingly, $f$ is Riemann integrable on $\left[{x_0 \,.\,.\, x_{k + 1} }\right]$.

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

$\blacksquare$