The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$

Why this item is first. The published definition of the integral does not cover this page. 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 ${\int }_a^{b}f$ is stated for reals $a<b$, because the partitions it quantifies over are those of 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, whose standing hypothesis is $a<b$: with $a=b$ the chain $a=t_0<\cdots <t_n=b$ is unsatisfiable. So ${\int }_a^{b}f$ is an undefined symbol whenever $a\ge b$, and every additivity statement below would be ill-formed as it is usually written. This item extends the notation, and nothing else: the object it names is still the Darboux integral of 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 $u,v\in \mathbb{R}$ and write

$[u\wedge v,u\vee v]:=\text{the closed interval with endpoints }u\text{ and }v$

(Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Let $f$ be a real-valued function whose domain contains that interval. Say that $f$ is integrable between $u$ and $v$ when either $u=v$, or $u\mathrm{\ne }v$ and the restriction of $f$ to $[u\wedge v,u\vee v]$ is bounded (Lower bound, bounded below, bounded set) and Darboux integrable there (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 ${\int }_a^{b}f$, 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 $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$). For such $f$ define

${\int }_u^{v}f:=\{\begin{array}{ll}\text{the Darboux integral of }f\text{ over }[u,v]& \text{if }u<v,\\ 0& \text{if }u=v,\\ -{\displaystyle {\int }_v^{u}f}& \text{if }u>v\mathrm{.}\end{array}$

There is nothing to check for consistency. The three clauses are indexed by the three cases of trichotomy, $u<v$, $u=v$ and $u>v$, which are mutually exclusive and exhaustive; no pair of them ever applies to the same $(u,v)$. In particular the first clause is untouched, so on $u<v$ this is the published integral verbatim and every published theorem about it applies unchanged.

The middle clause is a stipulation, not a computation. It is not claimed that $0$ is a value forced by the $u<v$ definition in any limiting sense; that definition simply says nothing at $u=v$, and ${\int }_u^{u}f:=0$ is what is written there. It is also unconditional: no hypothesis on $f$ beyond being defined at $u$ is asked for, since the case $u=v$ never refers to a partition.

The two consequences used throughout the page

Antisymmetry, for every pair. For all reals $u,v$ with $f$ integrable between them,

${\int }_u^{v}f=-{\int }_v^{u}f\mathrm{.}$

Indeed if $u<v$ then $v>u$ and the third clause reads ${\int }_v^{u}f=-{\int }_u^{v}f$, which rearranges to the display; if $u=v$ both sides are $0$; and if $u>v$ the third clause is the display itself.

Absolute values agree. Consequently $\mid {\int }_u^{v}f\mid =\mid {\int }_v^{u}f\mid$ for every such pair.

An obligation recorded here and discharged elsewhere. With this convention the additivity identity

${\int }_u^{v}f+{\int }_v^{w}f={\int }_u^{w}f$

holds for every arrangement of $u,v,w$ in an interval on which $f$ is integrable, not only for $u<v<w$. That is a theorem and not part of this definition; it is proved as the last clause of For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$; with the oriented form for arbitrary $a,b,c$, and nothing on this page uses it before it is proved there.

• This is notation, and it is a real notation. Without it the substitution theorem could not be stated with the limits $\phi (c)$ and $\phi (d)$ in the order the map produces them, since a differentiable $\phi$ need be neither injective nor monotone; and the integral function $x\mapsto {\int }_a^{x}f$ would be undefined at $x=a$.

• One published inequality is not orientation-invariant, and that is a trap. The estimate $\mid {\int }_u^{v}f\mid \le {\int }_u^v\mathrm{\mid }f\mathrm{\mid }$ is guaranteed only for $u\le v$: at $u>v$ the right-hand side is $-{\int }_v^u\mathrm{\mid }f\mathrm{\mid }\le 0$ while the left-hand side is $\ge 0$, so the inequality fails whenever ${\int }_v^u\mathrm{\mid }f\mathrm{\mid }>0$. The form valid for every pair is $\mid {\int }_u^{v}f\mid \le \mid {\int }_u^v\mathrm{\mid }f\mathrm{\mid }\mid$, and this is stated where it is proved (If $f,g$ are integrable on $[a,b]$ then so are $\mid f\mid$, $f^2$, $fg$, $\max (f,g)$ and $\min (f,g)$, and $\mid {\int }_a^{b}f\mid \le {\int }_a^b\mid f\mid$).

• Integrability is a property of the unordered pair. By construction, $f$ is integrable between $u$ and $v$ if and only if it is integrable between $v$ and $u$, since both refer to the same closed interval; only the sign of the value remembers the order.