# Piecewise Continuous Function with One-Sided Limits is Riemann Integrable

## Theorem

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

Let $f$ be piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$.

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

## Proof 1

We are given that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$.

The result follows from Bounded Piecewise Continuous Function is Riemann Integrable.

$\blacksquare$

## Proof 2

We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.

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

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} f \left({x}\right)$ exist.

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

We shall use proof by induction on the intervals.

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

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

### Basis for the Induction

$\map P 1$ is the case:

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

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

Piecewise continuity with one-sided limits of $f$ for the case $n = 1$ means that:

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} f \left({x}\right)$ exist.

By Integrability Theorem for Functions Continuous on Open Intervals, $f$ is Riemann integrable on $\left[{x_{i - 1} \,.\,.\, x_i}\right]$.

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ 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 piecewise continuous function with one-sided limits, for every $i \in \left\{ {1, 2, \ldots, k, k + 1}\right\}$:

$f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$
the one-sided limits $\displaystyle \lim_{x \mathop \to x_{i - 1}^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to x_i^-} f \left({x}\right)$ exist.

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, by Existence of Integral on Union of Adjacent Intervals:

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

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 $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$f$ is Riemann integrable on $\closedint a b$.

$\blacksquare$