=== Necessary Condition === We need to prove that $f$ is Darboux integrable on $\closedint a b$. To do this it suffices to show that for all $\epsilon > 0$, there exists a subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$. Here, $\map U S$ and $\map L S$ are, respectively, the upper and lower sums of $f$ on $\closedint a b$ with respect to the subdivision $S$. Let a strictly positive $\epsilon$ be given. Since $f$ is Darboux integrable on $\closedint a c$, we know that a subdivision $S_1$ of $\closedint a c$ exists such that $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$. Since $f$ is Darboux integrable on $\closedint c b$, we know that a subdivision $S_2$ of $\closedint c b$ exists such that $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$. Define the subdivision $S = S_1 \cup S_2$. We observe that $S$ is a subdivision of $\closedint a b$. We get by the definition of upper sum: {{begin-eqn}} {{eqn | l = \map U S       | r = \map U {S_1} + \map U {S_2}       | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent }} {{end-eqn}} Also, by the definition of lower sum: {{begin-eqn}} {{eqn | l = \map L S       | r = \map L {S_1} + \map L {S_2}       | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent }} {{end-eqn}} These two equations give: {{begin-eqn}} {{eqn | l = \map U S – \map L S       | r = \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} } }} {{eqn | r = \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2} }} {{eqn | o = <       | r = \dfrac \epsilon 2 + \dfrac \epsilon 2       | c = since $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$ and $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$ }} {{eqn | r = \epsilon }} {{end-eqn}} This shows that $S$ satisfies $\map U S – \map L S < \epsilon$. We conclude from this that $f$ is Darboux integrable on $\closedint a b$ since $S$ is a subdivision of $\closedint a b$. {{qed|lemma}} === Sufficient Condition === We need to prove that $f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$. Let a strictly positive $\epsilon$ be given. The Darboux integrability of $f$ on $\closedint a b$ implies that a subdivision $P$ of $\closedint a b$ exists such that $\map U P – \map L P < \epsilon$. Here, $\map U P$ and $\map L P$ are, respectively, the upper and lower sums of $f$ on $\closedint a b$ with respect to the subdivision $P$. Define the subdivision $S = P \cup \set c$. We observe that $S$ equals $P$ if $c$ is a point in $P$, otherwise $S$ is a finer subdivision than $P$. We have {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \map U P – \map L P  }} {{eqn | o = \ge       | r = \map U S – \map L P        | c = as $\map U S \le \map U P $ by the definition of upper sum and the fact that $S$ refines $P$ }} {{eqn | o = \ge       | r = \map U S – \map L S        | c = as $\map L S \ge \map L P $ by the definition of lower sum and the fact that $S$ refines $P$ }} {{end-eqn}} This shows that $S$ satisfies $\map U S – \map L S < \epsilon$. Define: : $S_1 = S \cap \set {x: x \le c}$ : $S_2 = S \cap \set {x: x \ge c}$ We observe: :$S_1$ is a subdivision of $\closedint a c$. :$S_2$ is a subdivision of $\closedint c b$. :$S_1$ and $S_2$ are adjacent. :The union of $S_1$ and $S_2$ equals $S$. We get by the definition of upper sum: {{begin-eqn}} {{eqn | l = \map U S       | r = \map U {S_1} + \map U {S_2}       | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent }} {{end-eqn}} Also, by the definition of lower sum: {{begin-eqn}} {{eqn | l = \map L S       | r = \map L {S_1} + \map L {S_2}       | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent }} {{end-eqn}} We have {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \map U S – \map L S }} {{eqn | r = \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} }       | c = as $\map U S = \map U {S_1} + \map U {S_2}$ and $\map L S = \map L {S_1} + \map L {S_2}$ }} {{eqn | r = \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2}  }} {{eqn | o = \ge       | r = \map U {S_1} – \map L {S_1}        | c = as $\map U {S_2} – \map L {S_2} \ge 0$ by the definitions of upper and lower sums }} {{end-eqn}} This shows that $S_1$ satisfies $\map U {S_1} – \map L {S_1} < \epsilon$. A similar deduction focusing on $S_2$ instead of $S_1$ shows that $S_2$ satisfies $\map U {S_2} – \map L {S_2} < \epsilon$. $\map U {S_1} – \map L {S_1} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint a c$ since $S_1$ is a subdivision of $\closedint a c$. $\map U {S_2} – \map L {S_2} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint c b$ since $S_2$ is a subdivision of $\closedint c b$. {{qed}} Category:Integral Calculus 3eh9p5joywrep39apfhisgn7y3kozau