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 $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$

Let $a<b$ be reals, let $f:[a,b]\to \mathbb{R}$ be bounded (Lower bound, bounded below, bounded set), so that there is a real $M\ge 0$ with $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ for every $x\in [a,b]$, and let $P=(n,t)$ be a partition of $[a,b]$ with subintervals $I_i=[t_i,t_{i+1}]$ and lengths ${\mathrm{\Delta }}_i=t_{i+1}-t_i$ for $i<n$ (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).

The two extreme values on a subinterval

For $i<n$ put

$m_i:=\inf f[I_i],M_i:=\sup f[I_i],f[I_i]=\{f(x):x\in I_i\}\mathrm{.}$

Both exist. The set $f[I_i]$ is nonempty, because $t_i<t_{i+1}$ makes $I_i$ nonempty (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), and it is bounded, because $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ for every $x$ (Lower bound, bounded below, bounded set). 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 is unique, so the notations $\inf f[I_i]$ and $\sup f[I_i]$ name single real numbers (Suprema and infima are unique).

They bracket the values, and each other. For $x\in I_i$,

$-M\le m_i\le f(x)\le M_i\le M,$

the outer inequalities because $-M$ is a lower bound and $M$ an upper bound of $f[I_i]$, and the middle ones by the definitions of infimum and supremum. In particular $m_i\le M_i$ and $M_i-m_i\le 2M$.

The dependence of $m_i$ and $M_i$ on $f$ and on $P$ is suppressed in the notation, as is customary; where two partitions are in play the sums below carry the partition and the extreme values are written out.

The two Darboux sums

$L(f,P):={\sum }_{i<n}{m}_i{\mathrm{\Delta }}_i,U(f,P):={\sum }_{i<n}{M}_i{\mathrm{\Delta }}_i,$

the finite sums of Finite sums and finite products, by recursion, indexed by $i\in \mathbb{N}$ with $i<n$. Both are real numbers, being finite sums of reals, and

$L(f,P)\le U(f,P),$

by monotonicity of finite sums, clause 4 of Laws of finite sums and finite products, since $m_i{\mathrm{\Delta }}_i\le M_i{\mathrm{\Delta }}_i$ for every $i<n$: multiplying $m_i\le M_i$ by ${\mathrm{\Delta }}_i>0$ preserves the inequality (Ordered field).

The gap on a subinterval is the oscillation there

For every $i<n$,

$M_i-m_i={\omega }_f(I_i)=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in I_i\},$

the oscillation of $f$ on the set $I_i$ (The oscillation ${\omega }_f(S)=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in S\}$ of $f$ on a set and the oscillation ${\omega }_f(c)={\inf }_{\delta >0}{\omega }_f(A\cap N_{\delta }(c))$ at a point, both taken in the extended reals). The supremum is a real number here rather than an extended one, because $f$ is bounded (The oscillation ${\omega }_f(S)=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in S\}$ of $f$ on a set and the oscillation ${\omega }_f(c)={\inf }_{\delta >0}{\omega }_f(A\cap N_{\delta }(c))$ at a point, both taken in the extended reals, The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined). The identity is proved in two inequalities.

The oscillation is at most the gap. For $x,y\in I_i$ both $f(x)$ and $f(y)$ lie in $[m_i,M_i]$, so $f(x)-f(y)\le M_i-m_i$ and $f(y)-f(x)\le M_i-m_i$, whence $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M_i-m_i$ (Basic properties of the absolute value). So $M_i-m_i$ is an upper bound of the set whose supremum is ${\omega }_f(I_i)$.

The gap is at most the oscillation. Let $\epsilon >0$ be real. By the $\epsilon$-characterisations of the supremum and the infimum (Epsilon characterisation of the supremum, Epsilon characterisation of the infimum) there are $x,y\in I_i$ with $f(x)>M_i-\epsilon \mathrm{/}2$ and $f(y)<m_i+\epsilon \mathrm{/}2$; then

$\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\ge f(x)-f(y)>(M_i-m_i)-\epsilon ,$

so ${\omega }_f(I_i)>(M_i-m_i)-\epsilon$. As $\epsilon >0$ was arbitrary, ${\omega }_f(I_i)\ge M_i-m_i$: otherwise $\epsilon :=(M_i-m_i)-{\omega }_f(I_i)$ would be positive and give ${\omega }_f(I_i)>{\omega }_f(I_i)$.

This identity is what connects the Darboux machinery to the pointwise oscillation of The oscillation ${\omega }_f(S)=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in S\}$ of $f$ on a set and the oscillation ${\omega }_f(c)={\inf }_{\delta >0}{\omega }_f(A\cap N_{\delta }(c))$ at a point, both taken in the extended reals, and it is the hinge of 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.

• Boundedness is a hypothesis of the definition, not of the theorems. Without it $f[I_i]$ may fail to have a supremum in $\mathbb{R}$ and $U(f,P)$ is not defined at all. Every statement on this page that mentions $L$ or $U$ therefore carries "bounded $f$" in its hypotheses, and none of them is a restriction that could be lifted: an unbounded function has no Darboux sums to compare.

• Why the infimum and the supremum, and not a value of $f$. Replacing $m_i$ by $f({\xi }_i)$ for a point ${\xi }_i\in I_i$ gives the Riemann sums of Tagged partitions of $[a,b]$, with a tag ${\xi }_i$ in each subinterval, and the Riemann sum $S(f,P,\xi )={\sum }_{i}f({\xi }_i){\mathrm{\Delta }}_i$, which depend on a choice of points and are not extremal. The Darboux sums are canonical functions of $f$ and $P$ alone, which is what makes the supremum and infimum over all partitions in 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 ${\int }_a^{b}f$ well posed without any selection.

• The empty sum does not occur. A partition has $n\ge 1$ (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), so $L(f,P)$ and $U(f,P)$ are sums of at least one term. The first term is $m_0{\mathrm{\Delta }}_0$, over the subinterval $[a,t_1]$; the indexing starts at $i=0$ throughout this page.