Bounded Function Continuous on Open Interval is Darboux Integrable

Theorem
Let $f$ be a real function defined on an interval $\closedint a b$ such that $a < b$.

Let $f$ be continuous on $\closedint a b$.

Let $f$ be bounded on $\closedint a b$.

Then $f$ is Darboux integrable on $\closedint a b$.

Proof
By Condition for Darboux Integrability, it suffices to show that, for a given strictly positive $\epsilon$, a subdivision $S$ of $\closedint a b$ exists such that:


 * $\map U S – \map L S < \epsilon$

where $\map U S$ and $\map L S$ are respectively the upper and lower sums of $f$ on $\closedint a b$ with respect to the subdivision $S$.

Since $f$ is bounded, a strictly positive bound $K$ exists for $f$ on $\closedint a b$.

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


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

We have that $f$ is continuous on $\openint a b$.

As $\delta > 0$, $\closedint {a + \delta} {b - \delta}$ is a subset of $\openint a b$.

Thus $f$ is continuous on the interval $\closedint {a + \delta} {b - \delta}$.

By Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint {a + \delta} {b - \delta}$.

Since $f$ is Darboux integrable on $\closedint {a + \delta} {b - \delta}$, there exists a subdivision $S_\delta$ of $\closedint {a + \delta} {b - \delta}$ that satisfies:


 * $\map U {S_\delta} – \map L {S_\delta} < \dfrac \epsilon 3$

where $\map U {S_\delta}$ and $\map L {S_\delta}$ are respectively the upper and lower sums of $f$ on $\closedint {a + \delta} {b - \delta}$ with respect to the subdivision $S_\delta$.

Define the following subdivision of $\closedint a b$:
 * $S = S_\delta \cup \set {a, b}$

The upper sum of $f$ on $\closedint a b$ with respect to $S$ is per definition:


 * $\map U S = M_a \delta + \map U {S_\delta} + M_b \delta$

where:
 * $M_a$ is the supremum of $f$ on $\closedint a {a + \delta}$
 * $M_b$ is the supremum of $f$ on $\closedint {b - \delta} b$.

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

The lower sum of $f$ on $\closedint a b$ with respect to $S$ is per definition:


 * $\map L S = m_a \delta + \map L {S_\delta} + m_b \delta$

where:
 * $m_a$ is the infimum of $f$ on $\closedint a {a + \delta}$
 * $m_b$ is the infimum of $f$ on $\closedint {b - \delta} b$

$m_a$ and $m_b$ exist by the Continuum Property, because $f$ is bounded on $\closedint a {a + \delta}$ and $\closedint {b - \delta} b$.

Define the sum:

Define the sum:

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


 * $U' \ge \map U S$


 * $L' \le \map L S$

From these two inequalities follows:

Hence:


 * $\map U S – \map L S < \epsilon$