Condition for Darboux Integrability

Theorem
Let $\closedint a b$ be a closed real interval.

Let $f$ be a bounded real function defined on $\closedint a b$.

Then $f$ is Darboux integrable :
 * for every $\epsilon \in \R_{>0}$, there exists a finite subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$

where
 * $\map U S$ is the upper sum of $f$ on $\closedint a b$ with respect to $S$
 * $\map L S$ is the lower sum of $f$ on $\closedint a b$ with respect to $S$

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:

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:

Also:

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.