# Bounded Function Continuous on Open Interval is Riemann Integrable

## Theorem

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

Let $f$ be continuous on $\left({a \,.\,.\, b}\right)$.

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

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

## Proof

By Condition for Riemann Integrability, it suffices to show that, for a given strictly positive $\epsilon$, a subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ exists such that:

$U \left({S}\right) – L \left({S}\right) < \epsilon$

where $U \left({S}\right)$ and $L \left({S}\right)$ are respectively the upper and lower sums of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $S$.

Since $f$ is bounded, a strictly positive bound $K$ exists for $f$ on $\left[{a \,.\,.\, b}\right]$.

Let a strictly positive $\epsilon$ be given, and choose a $\delta$ that satisfies:

$0 < \delta < \min \left({\dfrac \epsilon {6K}, \dfrac {b-a} 2}\right)$

We have that $f$ is continuous on $\left({a \,.\,.\, b}\right)$.

As $\delta > 0$, $\left[{a + \delta \,.\,.\, b - \delta}\right]$ is a subset of $\left({a \,.\,.\, b}\right)$.

Thus $f$ is continuous on the interval $\left[{a + \delta \,.\,.\, b - \delta}\right]$.

By Continuous Function is Riemann Integrable, $f$ is Riemann integrable on $\left[{a + \delta \,.\,.\, b - \delta}\right]$.

Since $f$ is Riemann integrable on $\left[{a + \delta \,.\,.\, b - \delta}\right]$, there exists a subdivision $S_\delta$ of $\left[{a + \delta \,.\,.\, b - \delta}\right]$ that satisfies:

$U \left({S_\delta}\right) – L \left({S_\delta}\right) < \dfrac \epsilon 3$

where $U \left({S_\delta}\right)$ and $L \left({S_\delta}\right)$ are respectively the upper and lower sums of $f$ on $\left[{a + \delta \,.\,.\, b - \delta}\right]$ with respect to the subdivision $S_\delta$.

Define the following subdivision of $\left[{a \,.\,.\, b}\right]$:

$S = S_\delta \cup \left\{{a, b}\right\}$.

The upper sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$ is per definition:

$U \left({S}\right) = M_a \delta + U \left({S_\delta}\right) + M_b \delta$

where:

$M_a$ is the supremum of $f$ on $\left[{a \,.\,.\, a + \delta}\right]$
$M_b$ is the supremum of $f$ on $\left[{b - \delta \,.\,.\, b}\right]$.

$M_a$ and $M_b$ exist by the least upper bound property of the real numbers because $f$ is bounded on $\left[{a \,.\,.\, a + \delta}\right]$ and $\left[{b - \delta \,.\,.\, b}\right]$.

The lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$ is per definition:

$L \left({S}\right) = m_a \delta + L \left({S_\delta}\right) + m_b \delta$

where:

$m_a$ is the infimum of $f$ on $\left[{a \,.\,.\, a + \delta}\right]$
$m_b$ is the infimum of $f$ on $\left[{b - \delta \,.\,.\, b}\right]$

$m_a$ and $m_b$ exist by the greatest lower bound property of the real numbers because $f$ is bounded on $\left[{a \,.\,.\, a + \delta}\right]$ and $\left[{b - \delta \,.\,.\, b}\right]$.

Define the sum:

 $\displaystyle U'$ $=$ $\displaystyle K \delta + U \left({S_\delta}\right) + K \delta$ $\displaystyle$ $\ge$ $\displaystyle M_a \delta + U \left({S_\delta}\right) + M_b \delta$ by $K \ge M_a$ and $K \ge M_b$ $\displaystyle$ $=$ $\displaystyle U \left({S}\right)$

Define the sum:

 $\displaystyle L'$ $=$ $\displaystyle -K \delta + L \left({S_\delta}\right) -K \delta$ $\displaystyle$ $\le$ $\displaystyle m_a \delta + L \left({S_\delta}\right) + m_b \delta$ by $-K \le m_a$ and $-K \le m_b$ $\displaystyle$ $=$ $\displaystyle L \left({S}\right)$

Therefore, $U'$ and $L'$ satisfy:

$U' \ge U \left({S}\right)$
$L' \le L \left({S}\right)$

From these two inequalities follows:

 $\displaystyle U \left({S}\right) – L \left({S}\right)$ $\le$ $\displaystyle U' - L'$ $\displaystyle$ $=$ $\displaystyle K \delta + U \left({S_\delta}\right) + K \delta - (-K \delta + L \left({S_\delta}\right) -K \delta)$ by the definitions of $U'$ and $L'$ $\displaystyle$ $=$ $\displaystyle 4 K \delta + U \left({S_\delta}\right) - L \left({S_\delta}\right)$ $\displaystyle$ $<$ $\displaystyle 4 K \min \left({\dfrac \epsilon {6 K}, \dfrac {b - a} 2}\right) + U \left({S_\delta}\right) - L \left({S_\delta}\right)$ by $\delta < \min \left({\dfrac \epsilon {6 K}, \dfrac {b - a} 2}\right)$ $\displaystyle$ $\le$ $\displaystyle 4 K \dfrac \epsilon {6 K} + U \left({S_\delta}\right) - L \left({S_\delta}\right)$ by $\min \left({\dfrac \epsilon {6 K}, \dfrac {b - a} 2}\right) \le \dfrac \epsilon {6 K}$ $\displaystyle$ $<$ $\displaystyle 4 K \dfrac \epsilon {6 K} + \dfrac \epsilon 3$ by $U \left({S_\delta}\right) - L \left({S_\delta}\right) < \dfrac \epsilon 3$ $\displaystyle$ $=$ $\displaystyle \frac {2 \epsilon} 3 + \frac \epsilon 3$ $\displaystyle$ $=$ $\displaystyle \epsilon$

Hence:

$U \left({S}\right) – L \left({S}\right) < \epsilon$

$\blacksquare$