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 $\openint a b$.
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 Darboux sum and lower Darboux sum 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 Darboux sum and lower Darboux sum 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 Darboux 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 Darboux 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:
\(\ds U'\) | \(=\) | \(\ds K \delta + \map U {S_\delta} + K \delta\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds M_a \delta + \map U {S_\delta} + M_b \delta\) | by $K \ge M_a$ and $K \ge M_b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map U S\) |
Define the sum:
\(\ds L'\) | \(=\) | \(\ds -K \delta + \map L {S_\delta} - K \delta\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds m_a \delta + \map L {S_\delta} + m_b \delta\) | by $-K \le m_a$ and $-K \le m_b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map L S\) |
Therefore, $U'$ and $L'$ satisfy:
- $U' \ge \map U S$
- $L' \le \map L S$
From these two inequalities follows:
\(\ds \map U S – \map L S\) | \(\le\) | \(\ds U' - L'\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds K \delta + \map U {S_\delta} + K \delta - \paren {-K \delta + \map L {S_\delta} -K \delta}\) | Definitions of $U'$ and $L'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 K \delta + \map U {S_\delta} - \map L {S_\delta}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 4 K \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} + \map U {S_\delta} - \map L {S_\delta}\) | by $\delta < \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2}$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 4 K \dfrac \epsilon {6 K} + \map U {S_\delta} - \map L {S_\delta}\) | by $\min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} \le \dfrac \epsilon {6 K}$ | |||||||||||
\(\ds \) | \(<\) | \(\ds 4 K \dfrac \epsilon {6 K} + \dfrac \epsilon 3\) | by $\map U {S_\delta} - \map L {S_\delta} < \dfrac \epsilon 3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \epsilon} 3 + \frac \epsilon 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Hence:
- $\map U S – \map L S < \epsilon$
$\blacksquare$