The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$

Let $a<b$ be reals and let $f:[a,b]\to \mathbb{R}$ be bounded (Lower bound, bounded below, bounded set). Write $\mathcal{P}$ for the set of all partitions of $[a,b]$ (Partition of $[a,b]$ as a finite strictly increasing list $a=t_0<t_1<\cdots <t_n=b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions) and put

$\mathcal{L}:=\{L(f,P):P\in \mathcal{P}\},\mathcal{U}:=\{U(f,P):P\in \mathcal{P}\}$

for the sets of lower and of upper Darboux sums (For bounded $f$ on $[a,b]$ and a partition $P$: the infimum $m_i$ and supremum $M_i$ of $f$ on the $i$-th subinterval, and the lower and upper Darboux sums $L(f,P)={\sum }_i{m}_i{\mathrm{\Delta }}_i$ and $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$).

Both extrema exist

$\mathcal{P}$ is nonempty: the pair $(1,t)$ with $t_0:=a$ and $t_k:=b$ for $k\ge 1$ is a partition of $[a,b]$, since $a<b$. So $\mathcal{L}$ and $\mathcal{U}$ are nonempty.

$\mathcal{L}$ is bounded above and $\mathcal{U}$ is bounded below. Fix any $Q\in \mathcal{P}$. By claim 2 of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L(f,P)\le L(f,P^{\mathrm{\prime }})\le U(f,P^{\mathrm{\prime }})\le U(f,P)$ when $P^{\mathrm{\prime }}$ refines $P$, and $L(f,P)\le U(f,Q)$ for arbitrary partitions $P$ and $Q$; moreover the two changes are at most $2M(n^{\mathrm{\prime }}-n)\mathrm{\parallel }P\mathrm{\parallel }$, $L(f,P)\le U(f,Q)$ for every $P\in \mathcal{P}$, so $U(f,Q)$ is an upper bound of $\mathcal{L}$; and $L(f,Q)\le U(f,P)$ for every $P$, so $L(f,Q)$ is a lower bound of $\mathcal{U}$.

Hence a nonempty set bounded above has a supremum (Complete ordered field (least-upper-bound property)) and a nonempty set bounded below has an infimum (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)), each unique (Suprema and infima are unique). The lower and upper Darboux integrals of $f$ over $[a,b]$ are the real numbers

$\underset{‾}{{\int }_a^b}f:=\sup \mathcal{L}={\sup }_{P}L(f,P),\overline{{\int }_a^b}f:=\inf \mathcal{U}={\inf }_{P}U(f,P)\mathrm{.}$

The lower integral never exceeds the upper one

$\underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\mathrm{.}$

Indeed, for each fixed $Q\in \mathcal{P}$ the number $U(f,Q)$ is an upper bound of $\mathcal{L}$, so the least upper bound satisfies $\underset{‾}{{\int }_a^b}f\le U(f,Q)$. As $Q$ was arbitrary, $\underset{‾}{{\int }_a^b}f$ is a lower bound of $\mathcal{U}$, and the greatest lower bound satisfies $\underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f$ (Greatest lower bound (infimum)).

Moreover, for every partition $P$,

$L(f,P)\le \underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P),$

the outer inequalities because a member of a set is at most its supremum and at least its infimum.

Integrability

$f$ is Darboux integrable on $[a,b]$, and on this page simply integrable, when

$\underset{‾}{{\int }_a^b}f=\overline{{\int }_a^b}f,$

and then the common value is written

${\int }_a^{b}f\text{or}{\int }_a^{b}f(x)\mathrm{d}x,$

the integral of $f$ over $[a,b]$. It is a single well-determined real number, being the common value of two numbers each of which is unique (Suprema and infima are unique). Without the displayed equality the symbol ${\int }_a^{b}f$ is not defined and is never written.

The inequality above is the whole difficulty. By the previous paragraph integrability is never a question of one integral exceeding the other, only of the gap $\overline{{\int }_a^b}f-\underset{‾}{{\int }_a^b}f\ge 0$ being $0$; and by Riemann's criterion: a bounded $f$ on $[a,b]$ is Darboux integrable if and only if for every real $\epsilon >0$ there is a partition $P$ with $U(f,P)-L(f,P)<\epsilon$ that gap is $0$ exactly when a single partition can be found making $U(f,P)-L(f,P)$ small. Whether that is possible is settled completely, in terms of the discontinuities of $f$, by Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero.

"Riemann integrable" means the same thing here. The definition above is Darboux's. Riemann's own definition, in terms of tagged partitions of small mesh, is Tagged partitions of $[a,b]$, with a tag ${\xi }_i$ in each subinterval, and the Riemann sum $S(f,P,\xi )={\sum }_{i}f({\xi }_i){\mathrm{\Delta }}_i$, and the two define the same class of functions with the same integral by The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\epsilon >0$ there is a real $\delta >0$ such that $\mathrm{\mid }S(f,P,\xi )-I\mathrm{\mid }<\epsilon$ for every tagged partition of mesh below $\delta$. Until that theorem is proved the two phrases are kept apart; after it they are used interchangeably, as they are throughout the literature.

• The supremum is over all partitions, and nothing is selected. Both $\mathcal{L}$ and $\mathcal{U}$ are sets determined by $f$ and $[a,b]$ alone, and $\sup$ and $\inf$ are canonical, so no choice principle is involved in forming either integral. Where a choice does enter on this page is recorded in What this page costs in choice: Riemann's criterion, the Darboux-Riemann equivalence and integrability of a monotone function are theorems of ZF; integrability of a continuous function inherits the single use of countable choice inside Heine-Cantor; and only the forward half of the Lebesgue criterion spends countable choice, once, at the countable union of null sets.

• Why the lower integral is a supremum and not an infimum. Refining a partition can only increase a lower sum and decrease an upper sum (Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L(f,P)\le L(f,P^{\mathrm{\prime }})\le U(f,P^{\mathrm{\prime }})\le U(f,P)$ when $P^{\mathrm{\prime }}$ refines $P$, and $L(f,P)\le U(f,Q)$ for arbitrary partitions $P$ and $Q$; moreover the two changes are at most $2M(n^{\mathrm{\prime }}-n)\mathrm{\parallel }P\mathrm{\parallel }$), so the lower sums push up towards the integral and the upper sums push down towards it. Taking $\inf \mathcal{L}$ would return the sum over the coarsest partition and would carry no information about $f$ beyond its infimum on $[a,b]$.

• A bounded $f$ always has both integrals; only their equality can fail. The Dirichlet function on $[0,1]$ has lower integral $0$ and upper integral $1$ (FALSE: every bounded function on $[a,b]$ is Riemann integrable), which is the standard witness that the definition above is not vacuous in either direction.