Condition for Riemann Integrability

Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f$ be a bounded real function defined on $\left[{a \,.\,.\, b}\right]$.

for every $\epsilon \in \R_{>0}$, there exists a finite subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ such that $U \left({S}\right) – L \left({S}\right) < \epsilon$

where

$U \left({S}\right)$ is the upper sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$
$L \left({S}\right)$ is the lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$

Proof

Necessary Condition

Let $f$ be Riemann (Darboux) integrable.

Let $\epsilon \in \R_{>0}$ be given.

It is to be proved that a finite subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ exists such that:

$U \left({S}\right) – L \left({S}\right) < \epsilon$

As $f$ is Riemann (Darboux) integrable:

$\displaystyle \int_a^b f \left({x}\right) \ \mathrm d x$ exists.

By the definition of the Darboux integral:

the lower integral $\displaystyle \underline {\int_a^b} f \left({x}\right) \ \mathrm d x$ exists.

Thus by the definition of lower integral:

$\sup_P L \left({P}\right)$ exists

where:

$L \left({P}\right)$ denotes the lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the finite subdivision $P$
$\sup_P L \left({P}\right)$ denotes the supremum for $L \left({P}\right)$.

Therefore a finite subdivision $S_1$ of $\left[{a \,.\,.\, b}\right]$ exists, satisfying:

$\sup_P L \left({P}\right) - L \left({S_1}\right) < \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close

In a similar way:

By the definition of the Darboux integral:

the upper integral $\displaystyle \overline {\int_a^b} f \left({x}\right) \ \mathrm d x$ exists.

Thus by the definition of upper integral:

$\inf_P U \left({P}\right)$ exists

where:

$U \left({P}\right)$ denotes the upper sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the finite subdivision $P$
$\inf_P U \left({P}\right)$ denotes the infimum for $U \left({P}\right)$.

Therefore a finite subdivision $S_2$ of $\left[{a \,.\,.\, b}\right]$ exists, satisfying:

$U \left({S_2}\right) - \inf_P U \left({P}\right) < \dfrac \epsilon 2$ by Infimum of Subset of Real Numbers is Arbitrarily Close

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:

$L \left({S}\right) \ge L \left({S_1}\right)$ by the definition of lower sum and $S$ refining $S_1$
$U \left({S}\right) \le U \left({S_2}\right)$ by the definition of upper sum and $S$ refining $S_2$

Recall that by the definition of Riemann (Darboux) integrable:

$\displaystyle \overline {\int_a^b} f \left({x}\right) \ \mathrm d x = \underline {\int_a^b} f \left({x}\right) \ \mathrm d x$

Hence we have:

 $\displaystyle U \left({S}\right) – L \left({S}\right)$ $\le$ $\displaystyle U \left({S_2}\right) – L \left({S}\right)$ $\quad$ as $U \left({S}\right) \le U \left({S_2}\right)$ $\quad$ $\displaystyle$ $\le$ $\displaystyle U \left({S_2}\right) – L \left({S_1}\right)$ $\quad$ as $L \left({S}\right) \ge L \left({S_1}\right)$ $\quad$ $\displaystyle$ $=$ $\displaystyle U \left({S_2}\right) - \overline {\int_a^b} f \left({x}\right) \ \mathrm d x + \overline {\int_a^b} f \left({x}\right) \ \mathrm d x – L \left({S_1}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle U \left({S_2}\right) - \overline{\int_a^b} f \left({x}\right) \ \mathrm d x + \underline{\int_a^b} f \left({x}\right) \ \mathrm d x – L \left({S_1}\right)$ $\quad$ as $\displaystyle \overline {\int_a^b} f \left({x}\right) \ \mathrm d x = \underline {\int_a^b} f \left({x}\right) \ \mathrm d x$ $\quad$ $\displaystyle$ $=$ $\displaystyle U \left({S_2}\right) - \inf_P U \left({P}\right) + \sup_P L \left({P}\right) – L \left({S_1}\right)$ $\quad$ Definitions of Upper Integral and Lower Integral $\quad$ $\displaystyle$ $\lt$ $\displaystyle \frac \epsilon 2 + \sup_P L \left({P}\right) – L \left({S_1}\right)$ $\quad$ as $U \left({S_2}\right) - \inf_P U \left({P}\right) < \dfrac \epsilon 2$ $\quad$ $\displaystyle$ $\lt$ $\displaystyle \frac \epsilon 2 + \frac \epsilon 2$ $\quad$ as $\sup_P L \left({P}\right) - L \left({S_1}\right) < \dfrac \epsilon 2$ $\quad$ $\displaystyle$ $=$ $\displaystyle \epsilon$ $\quad$ $\quad$

$\Box$

Sufficient Condition

Let $\epsilon \in \R_{>0}$ be given.

Let $f$ be such that:

there exists a finite subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ such that $U \left({S}\right) – L \left({S}\right) < \epsilon$.

We need to prove that $f$ is Riemann (Darboux) integrable.

First we show that $\inf_P U \left({P}\right)$ exists.

By:

$U \left({S}\right) – L \left({S}\right) < \epsilon$ we know that $U \left({S}\right)$ exists.

From this we conclude that:

$\left\{ {U \left({P}\right): P}\right.$ is a finite subdivision of $\left.{\left[{a \,.\,.\, b}\right]}\right\}$

is non-empty.

Since $f$ is bounded, we know by the definition of upper sum that $\left\{ {U \left({P}\right): P}\right.$ is a finite subdivision of $\left.{\left[{a \,.\,.\, b}\right]}\right\}$ is bounded.

From the Continuum Property it follows that $\inf_P U \left({P}\right)$ exists.

Next we show that $\sup_P L \left({P}\right)$ exists.

We do this similarly to how we showed that $\inf_P U \left({P}\right)$ exists by focusing on lower sums instead of upper sums:

We find that {$L \left({P}\right)$: $P$ is a finite subdivision of $\left[{a \,.\,.\, b}\right]$} is nonempty and bounded.

From the Continuum Property it follows that $\sup_P L \left({P}\right)$ exists.

Observe:

$\inf_P U \left({P}\right) \le U \left({S}\right)$ by the definition of infimum
$\sup_P L \left({P}\right) \ge L \left({S}\right)$ by the definition of supremum

We have:

 $\displaystyle \inf_P U \left({P}\right) - \sup_P L \left({P}\right)$ $\le$ $\displaystyle U \left({S}\right) - \sup_P L \left({P}\right)$ $\quad$ by $\inf_P U \left({P}\right) \le U \left({S}\right)$ $\quad$ $\displaystyle$ $\le$ $\displaystyle U \left({S}\right) - L \left({S}\right)$ $\quad$ by $\sup_P L \left({P}\right) \ge L \left({S}\right)$ $\quad$ $\displaystyle$ $<$ $\displaystyle \epsilon$ $\quad$ by $U \left({S}\right) – L \left({S}\right) < \epsilon$ $\quad$

Also:

 $\displaystyle \sup_P L \left({P}\right) - \inf_P U \left({P}\right)$ $\le$ $\displaystyle U \left({S}\right) - \inf_P U \left({P}\right)$ $\quad$ by Supremum of Lower Sums Never Greater than Upper Sum $\quad$ $\displaystyle$ $\le$ $\displaystyle U \left({S}\right) - L \left({S}\right)$ $\quad$ by Infimum of Upper Sums Never Smaller than Lower Sum $\quad$ $\displaystyle$ $<$ $\displaystyle \epsilon$ $\quad$ by $U \left({S}\right) – L \left({S}\right) < \epsilon$ $\quad$

These two results give:

$\left\lvert{\inf_P U \left({P}\right) - \sup_P L \left({P}\right)}\right\rvert < \epsilon$

Since $\epsilon$ can be chosen arbitrarily small ($>0$), this means that:

$\inf_P U \left({P}\right) = \sup_P L \left({P}\right)$

From this it follows by the definitions of upper and lower integrals that:

$\displaystyle \overline{\int_a^b} f \left({x}\right) \ \mathrm d x = \underline{\int_a^b} f \left({x}\right) \ \mathrm d x$

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