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 $\epsilon > 0$, 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 for all subdivisions $P$ of $\left[ {a \,.\,.\, b} \right]$

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

It follows that for all subdivisions $P$ of $\left[{a \,.\,.\, b}\right]$:
 * $L \left({P}\right)$ contains at least one number $L \left({S_1}\right)$ such that $\sup_P L \left({P}\right) - L \left({S_1}\right) < \dfrac \epsilon 2$

where $S_1$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.

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 for all subdivisions $P$ of $\left[{a \,.\,.\, b}\right]$

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

It follows that for all subdivisions $P$ of $\left[{a \,.\,.\, b}\right]$:
 * $U \left({P}\right)$ contains at least one number $U \left({S_2}\right)$ such that $U \left({S_2}\right) - \inf_P U \left({P}\right) < \dfrac \epsilon 2$

where $S_2$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.

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 $S_2$

We have

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

Let an $\epsilon > 0$ 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.

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 which we conclude that {$U\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$} is nonempty.

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

From these two properties of {$U\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$}, we deduce by the Continuum Property that $\inf_P U \left( {P} \right)$, the greatest lower bound for {$U\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$}, exists.

Since $\inf_P U \left( {P} \right)$ is less than or equal to every element of {$U\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$} and $S$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$, we know that $\inf_P U \left( {P} \right) \le U\left( {S} \right)$.

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

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

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 this we deduce by the Continuum Property that $\sup_P L \left( {P} \right)$, the least upper bound of {$L\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$}, exists.

Since $\sup_P L \left( {P} \right)$ is greater than or equal to every element of {$L\left( {P} \right)$: $P$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$} and $S$ is a subdivision of $\left[ {a \,.\,.\, b} \right]$, we know that $\sup_P L \left( {P} \right) \ge L\left( {S} \right)$.

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

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 least upper bound for the same set, we find: $\sup_P L \left( {P} \right) \le U\left( {S} \right)$.

5. We finish by tying all this together.

We have:

Also:

These two results give:


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

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 follows by the definitions of upper and  lower integrals that $\displaystyle \overline{\int_a^b} f \left( {x} \right) \ \mathrm dx = \underline{\int_a^b} f \left( {x} \right) \ \mathrm dx$.

From this we conclude by the definition of the Riemann integral that $f$ is Riemann integrable.