=== Necessary Condition === Let $f$ be Darboux integrable. Let $\epsilon \in \R_{>0}$ be given. It is to be proved that a finite subdivision $S$ of $\closedint a b$ exists such that: :$\map U S – \map L S < \epsilon$ As $f$ is Darboux integrable: :$\displaystyle \int_a^b \map f x \rd x$ exists. By the definition of the Darboux integral: :the lower integral $\displaystyle \underline {\int_a^b} \map f x \rd x$ exists. Thus by the definition of lower integral: :$\sup_P \map L P$ exists where: :$\map L P$ denotes the lower sum of $f$ on $\closedint a b$ with respect to the finite subdivision $P$ :$\sup_P \map L P$ denotes the supremum for $\map L P$. Therefore by Supremum of Subset of Real Numbers is Arbitrarily Close: :a finite subdivision $S_1$ of $\closedint a b$ exists, satisfying: ::$\sup_P \map L P - \map L {S_1} < \dfrac \epsilon 2$ In a similar way: By the definition of the Darboux integral: :the upper integral $\displaystyle \overline {\int_a^b} \map f x \rd x$ exists. Thus by the definition of upper integral: :$\inf_P \map U P$ exists where: :$\map U P$ denotes the upper sum of $f$ on $\closedint a b$ with respect to the finite subdivision $P$ :$\inf_P \map U P$ denotes the infimum for $\map U P$. Therefore by Infimum of Subset of Real Numbers is Arbitrarily Close: :a finite subdivision $S_2$ of $\closedint a b$ exists, satisfying: ::$\map U {S_2} - \inf_P \map U P < \dfrac \epsilon 2$ Now let $S := S_1 \cup S_2$ be defined. We observe: :$S$ is either equal to $S_1$ or finer than $S_1$ :$S$ is either equal to $S_2$ or finer than $S_2$ We find: :$\map L S \ge \map L {S_1}$ by the definition of lower sum and $S$ refining $S_1$ :$\map U S \le \map U {S_2}$ by the definition of upper sum and $S$ refining $S_2$ Recall that by the definition of Darboux integrable: :$\displaystyle \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$ Hence we have: {{begin-eqn}} {{eqn | l = \map U S – \map L S       | o = \le       | r = \map U {S_2} – \map L S       | c = as $\map U S \le \map U {S_2}$ }} {{eqn | o = \le       | r = \map U {S_2} – \map L {S_1}       | c = as $\map L S \ge L \map L {S_1}$ }} {{eqn | r = \map U {S_2} - \overline {\int_a^b} \map f x \rd x + \overline {\int_a^b} \map f x \rd x – \map L {S_1} }} {{eqn | r = \map U {S_2} - \overline{\int_a^b} \map f x \rd x + \underline{\int_a^b} \map f x \rd x – \map L {S_1}       | c = as $\displaystyle \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$ }} {{eqn | r = \map U {S_2} - \inf_P \map U P + \sup_P \map L P – \map L {S_1}       | c = {{Defof|Upper Integral}} and {{Defof|Lower Integral}} }} {{eqn | o = <       | r = \frac \epsilon 2 + \sup_P \map L P – \map L {S_1}       | c = as $\map U {S_2} - \inf_P \map U P < \dfrac \epsilon 2$ }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = as $\sup_P \map L P - \map L {S_1} < \dfrac \epsilon 2$ }} {{eqn | r = \epsilon }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $\epsilon \in \R_{>0}$ be given. Let $f$ be such that: :there exists a finite subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$. We need to prove that $f$ is Darboux integrable. First we show that $\inf_P \map U P$ exists. Let $T$ be defined as: :$T := \leftset {\map U P: P}$ is a finite subdivision of $\rightset {\closedint a b}$ By: :$\map U S – \map L S < \epsilon$ we know that $\map U S$ exists. From this we conclude that $T$ is non-empty. Because $f$ is bounded, we know by the definition of upper sum that $T$ is bounded. From the Continuum Property it follows that $\inf_P \map U P$ exists.  Next we show that $\sup_P \map L P$ exists. We do this similarly to how we showed that $\inf_P \map U P$ exists by focusing on lower sums instead of upper sums: We find that $\leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$ is non-empty and bounded. From the Continuum Property it follows that $\sup_P \map L P$ exists. Observe: :$\inf_P \map U P \le \map U S$ by the definition of infimum :$\sup_P \map L P \ge \map L S$ by the definition of supremum We have: {{begin-eqn}} {{eqn | l = \inf_P \map U P - \sup_P \map L P       | o = \le       | r = \map U S - \sup_P \map L P       | c = by $\inf_P \map U P \le \map U S$ }} {{eqn | o = \le       | r = \map U S - \map L S       | c = by $\sup_P \map L P \ge \map L S$ }} {{eqn | o = <       | r = \epsilon       | c = by $\map U S – \map L S < \epsilon$ }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \sup_P \map L P - \inf_P \map U P       | o = \le       | r = \map U S - \inf_P \map U P       | c = Supremum of Lower Sums Never Greater than Upper Sum }} {{eqn | o = \le       | r = \map U S - \map L S       | c = Infimum of Upper Sums Never Smaller than Lower Sum }} {{eqn | o = <       | r = \epsilon       | c = as $\map U S – \map L S < \epsilon$ }} {{end-eqn}} These two results give: :$\size {\inf_P \map U P - \sup_P \map L P} < \epsilon$ Since $\epsilon$ can be chosen arbitrarily small ($>0$), this means that: :$\inf_P \map U P = \sup_P \map L P$ From this it follows by the definitions of upper and lower integrals that: :$\displaystyle \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$ Hence, by the definition of the Darboux integral, $f$ is Darboux integrable. {{qed}}