Condition for Darboux Integrability

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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 if and only if:

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 Darboux sum of $f$ on $\closedint a b$ with respect to $S$
$\map L S$ is the lower Darboux sum of $f$ on $\closedint a b$ with respect to $S$


Proof

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:

$\ds \int_a^b \map f x \rd x$ exists.

By the definition of the Darboux integral:

the lower Darboux integral $\ds \underline {\int_a^b} \map f x \rd x$ exists.

Thus by the definition of lower Darboux integral:

$\sup_P \map L P$ exists

where:

$\map L P$ denotes the lower Darboux 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 Darboux integral $\ds \overline {\int_a^b} \map f x \rd x$ exists.

Thus by the definition of upper Darboux integral:

$\inf_P \map U P$ exists

where:

$\map U P$ denotes the upper Darboux 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 Lower Sum of Refinement
$\map U S \le \map U {S_2}$ by Upper Sum of Refinement


Recall that by the definition of Darboux integrable:

$\ds \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$


Hence we have:

\(\ds \map U S – \map L S\) \(\le\) \(\ds \map U {S_2} – \map L S\) as $\map U S \le \map U {S_2}$
\(\ds \) \(\le\) \(\ds \map U {S_2} – \map L {S_1}\) as $\map L S \ge L \map L {S_1}$
\(\ds \) \(=\) \(\ds \map U {S_2} - \overline {\int_a^b} \map f x \rd x + \overline {\int_a^b} \map f x \rd x – \map L {S_1}\)
\(\ds \) \(=\) \(\ds \map U {S_2} - \overline{\int_a^b} \map f x \rd x + \underline{\int_a^b} \map f x \rd x – \map L {S_1}\) as $\ds \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$
\(\ds \) \(=\) \(\ds \map U {S_2} - \inf_P \map U P + \sup_P \map L P – \map L {S_1}\) Definition of Upper Darboux Integral and Definition of Lower Darboux Integral
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \sup_P \map L P – \map L {S_1}\) as $\map U {S_2} - \inf_P \map U P < \dfrac \epsilon 2$
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \frac \epsilon 2\) as $\sup_P \map L P - \map L {S_1} < \dfrac \epsilon 2$
\(\ds \) \(=\) \(\ds \epsilon\)

$\Box$


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 Darboux 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 Darboux sums instead of upper Darboux 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:

\(\ds \inf_P \map U P - \sup_P \map L P\) \(\le\) \(\ds \map U S - \sup_P \map L P\) by $\inf_P \map U P \le \map U S$
\(\ds \) \(\le\) \(\ds \map U S - \map L S\) by $\sup_P \map L P \ge \map L S$
\(\ds \) \(<\) \(\ds \epsilon\) by $\map U S – \map L S < \epsilon$

Also:

\(\ds \sup_P \map L P - \inf_P \map U P\) \(\le\) \(\ds \map U S - \inf_P \map U P\) Supremum of Lower Sums Never Greater than Upper Sum
\(\ds \) \(\le\) \(\ds \map U S - \map L S\) Infimum of Upper Sums Never Smaller than Lower Sum
\(\ds \) \(<\) \(\ds \epsilon\) as $\map U S – \map L S < \epsilon$

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 Darboux integrals that:

$\ds \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$

Hence, by the definition of the Darboux integral, $f$ is Darboux integrable.

$\blacksquare$


Historical Note

The necessary and sufficient conditions for the existence of a Riemann integral were determined by Riemann in his paper Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe of $1854$, on the subject of Fourier series.

It is noted that the proof given on this page refers specifically to the Darboux integral, which is a simpler but equivalent concept to that introduced by Riemann.

That they are equivalent is given in Equivalence of Definitions of Riemann and Darboux Integrals


Sources

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