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: {{begin-eqn}} {{eqn | l = U'       | r = K \delta + \map U {S_\delta} + K \delta }} {{eqn | o = \ge       | r = M_a \delta + \map U {S_\delta} + M_b \delta       | c = by $K \ge M_a$ and $K \ge M_b$ }} {{eqn | r = \map U S }} {{end-eqn}} Define the sum: {{begin-eqn}} {{eqn | l = L'       | r = -K \delta + \map L {S_\delta} - K \delta }} {{eqn | o = \le       | r = m_a \delta + \map L {S_\delta} + m_b \delta       | c = by $-K \le m_a$ and $-K \le m_b$ }} {{eqn | r = \map L S }} {{end-eqn}} Therefore, $U'$ and $L'$ satisfy: :$U' \ge \map U S$ :$L' \le \map L S$ From these two inequalities follows: {{begin-eqn}} {{eqn | l = \map U S – \map L S       | o = \le       | r = U' - L' }} {{eqn | r = K \delta + \map U {S_\delta} + K \delta - \paren {-K \delta + \map L {S_\delta} -K \delta}       | c = Definitions of $U'$ and $L'$ }} {{eqn | r = 4 K \delta + \map U {S_\delta} - \map L {S_\delta} }} {{eqn | o = <       | r = 4 K \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} + \map U {S_\delta} - \map L {S_\delta}       | c = by $\delta < \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2}$ }} {{eqn | o = \le       | r = 4 K \dfrac \epsilon {6 K} + \map U {S_\delta} - \map L {S_\delta}       | c = by $\min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} \le \dfrac \epsilon {6 K}$ }} {{eqn | o = <       | r = 4 K \dfrac \epsilon {6 K} + \dfrac \epsilon 3       | c = by $\map U {S_\delta} - \map L {S_\delta} < \dfrac \epsilon 3$ }} {{eqn | r = \frac {2 \epsilon} 3 + \frac \epsilon 3 }} {{eqn | r = \epsilon }} {{end-eqn}} Hence: :$\map U S – \map L S < \epsilon$ {{qed}} Category:Integral Calculus ls000s394m0i1klb4scbrn5l04zhamt