Bounded Function Continuous on Open Interval is Darboux 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
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:

Define the sum:

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


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


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

From these two inequalities follows:

Hence:


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