# 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 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 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 $S$

## 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:

- $\displaystyle \int_a^b \map f x \rd x$ 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:

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:

\(\displaystyle \map U S – \map L S\) | \(\le\) | \(\displaystyle \map U {S_2} – \map L S\) | as $\map U S \le \map U {S_2}$ | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \map U {S_2} – \map L {S_1}\) | as $\map L S \ge L \map L {S_1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map U {S_2} - \overline {\int_a^b} \map f x \rd x + \overline {\int_a^b} \map f x \rd x – \map L {S_1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map U {S_2} - \overline{\int_a^b} \map f x \rd x + \underline{\int_a^b} \map f x \rd x – \map L {S_1}\) | as $\displaystyle \overline {\int_a^b} \map f x \rd x = \underline {\int_a^b} \map f x \rd x$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map U {S_2} - \inf_P \map U P + \sup_P \map L P – \map L {S_1}\) | Definition of Upper Integral and Definition of Lower Integral | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \frac \epsilon 2 + \sup_P \map L P – \map L {S_1}\) | as $\map U {S_2} - \inf_P \map U P < \dfrac \epsilon 2$ | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\) | as $\sup_P \map L P - \map L {S_1} < \dfrac \epsilon 2$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \epsilon\) |

$\Box$

### 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 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:

\(\displaystyle \inf_P \map U P - \sup_P \map L P\) | \(\le\) | \(\displaystyle \map U S - \sup_P \map L P\) | by $\inf_P \map U P \le \map U S$ | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \map U S - \map L S\) | by $\sup_P \map L P \ge \map L S$ | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \epsilon\) | by $\map U S – \map L S < \epsilon$ |

Also:

\(\displaystyle \sup_P \map L P - \inf_P \map U P\) | \(\le\) | \(\displaystyle \map U S - \inf_P \map U P\) | Supremum of Lower Sums Never Greater than Upper Sum | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \map U S - \map L S\) | Infimum of Upper Sums Never Smaller than Lower Sum | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \epsilon\) | 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 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.

$\blacksquare$

## 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