Condition for Darboux Integrability

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

Let $U \left({S}\right)$ be the upper sum of $f$ on $\left[{a \,.\,.\, b}\right]$.

Let $L \left({S}\right)$ be the lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$.

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

Necessary Condition
Let $f$ be Riemann integrable.

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

It is to be proved that a subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ exists such that:
 * $U \left({S}\right) – L \left({S}\right) < \epsilon$

As $f$ is Riemann integrable:
 * $\displaystyle \int_a^b f \left({x}\right) \ \mathrm d x$ exists.

By the definition of the Riemann 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 subdivision $P$
 * $\sup_P L \left({P}\right)$ denotes the supremum for $L \left({P}\right)$.

It follows that a subdivision $S_1$ of $\left[{a \,.\,.\, b}\right]$ exists, satisfying:
 * $\sup_P L \left({P}\right) - L \left({S_1}\right) < \dfrac \epsilon 2$

In a similar way:

By the definition of the Riemann 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 subdivision $P$
 * $\inf_P U \left({P}\right)$ denotes the infimum for $U \left({P}\right)$.

It follows that a subdivision $S_2$ of $\left[{a \,.\,.\, b}\right]$ exists, satisfying:
 * $U \left({S_2}\right) - \inf_P U \left({P}\right) < \dfrac \epsilon 2$

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:


 * $L \left({S}\right) \ge L \left({S_1}\right)$ 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 definition of Riemann 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:

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

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

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

We need to prove that $f$ is Riemann integrable.

Lemma
Let $P$ and $Q$ be subdivisions of $\left[{a \,.\,.\, b}\right]$.

Then $L \left({P}\right) \le U \left({Q}\right)$.

Proof
Let $P' = P \cup Q$.

We observe:


 * $P'$ is either equal to $P$ or finer than $P$


 * $P'$ is either equal to $Q$ or finer than $Q$

We find:


 * $L \left({P}\right) \le L \left({P'}\right)$ by the definition of lower sum and $P'$ refining $P$


 * $L \left({P'}\right) \le U \left({P'}\right)$ by the definitions of upper and  lower sums


 * $U \left({P'}\right) \le U \left({Q}\right)$ by the definition of upper sum and $P'$ refining $Q$

By combining these inequalities, we conclude:
 * $L \left({P}\right) \le U \left({Q}\right)$


 * Step $1$: First we show that $\inf_P U \left({P}\right) \le U \left({S}\right)$.

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 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 subdivision of $\left.{\left[{a \,.\,.\, b}\right]}\right\}$ is bounded.

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

Hence:
 * $\inf_P U \left({P}\right) \le U \left({S}\right)$


 * Step $2$: Next we show that $\inf_P U \left({P}\right) \ge L \left({S}\right)$.

The lemma gives that $L \left({S}\right)$ is a lower bound for {$U \left({P}\right)$: $P$ is a subdivision of $\left[{a \,.\,.\, b}\right]$}.

Since $\inf_P U \left({P}\right)$ is the greatest lower bound for the same set, we find:
 * $\inf_P U \left({P}\right) \ge L \left({S}\right)$.


 * Step $3$: Next we show that $\sup_P L \left({P}\right) \ge L \left({S}\right)$.

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

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

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

Hence:
 * $\sup_P L \left({P}\right) \ge L \left({S}\right)$.


 * Step $4$: Next we show that $\sup_P L \left({P}\right) \le U \left({S}\right)$.

The lemma gives that $U \left({S}\right)$ is an upper bound for {$L \left({P}\right)$: $P$ is a subdivision of $\left[{a \,.\,.\, b}\right]$}.

Since $\sup_P L \left({P}\right)$ is a supremum for the same set, we find:
 * $\sup_P L \left({P}\right) \le U \left({S}\right)$


 * Step $5$: We finish by tying all this together.

We have:

Also:

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 Riemann integral, $f$ is Riemann integrable.