The Gauge Integral and Cousin's Lemma

If $a=b$, the declared degenerate partition is fine; otherwise suppose, for contradiction, that $[a,b]$ has no fine partition, bisect it, and at each stage retain the left half if it has no fine partition and otherwise the right half, which must have none because two fine half-partitions concatenate; the retained closed intervals are nested and have length $(b-a)2^{-k}$, so [L2] and [L1] give one common point $c$.

Since $\delta (c)>0$ and the retained lengths tend to zero, a sufficiently late retained interval lies inside $(c-\delta (c),c+\delta (c))$; tagged by $c$, [L3] makes that one cell a fine partition, contradicting its construction.

Suppose, for contradiction, that $I\mathrm{\ne }J$; choose gauges controlling errors below $\mathrm{\mid }I-J\mathrm{\mid }\mathrm{/}3$, take their pointwise minimum, and use [L1] to obtain one tagged partition $P$ fine for both.

The triangle inequality gives $\mathrm{\mid }I-J\mathrm{\mid }\le \mathrm{\mid }I-S(f,P)\mathrm{\mid }+\mathrm{\mid }S(f,P)-J\mathrm{\mid }<2\mathrm{\mid }I-J\mathrm{\mid }\mathrm{/}3$, a contradiction, so $I=J$.

For $f+g$, take the pointwise minimum of gauges from [L1] with half the requested error; by [L2], $S(f+g,P)=S(f,P)+S(g,P)$, and the triangle inequality gives the required estimate.

For a scalar multiple, the case of scalar $0$ is immediate, and otherwise [L1] with tolerance $\epsilon \mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ and [L2] gives $\int cf=c\int f$; combining the sum and scaling conclusions proves the formula.

If a nonnegative integrable function $h$ had integral $H<0$, choose a gauge making every fine sum differ from $H$ by less than $-H\mathrm{/}2$ and use [L2]; every such sum is nonnegative, contradicting $S(h,P)<H\mathrm{/}2<0$.

Apply step 1.1 to $h=g-f$ and use [L1] to obtain $0\le \int (g-f)=\int g-\int f$. Every constant $k$ is HK integrable with integral $k(b-a)$ because every tagged sum equals that value; applying the first conclusion to $f-m$ and $M-f$ therefore gives the constant bounds, with equality on a degenerate interval.

For the forward direction, apply [L5] with error $\epsilon \mathrm{/}2$; any two sums fine for the resulting gauge differ by less than $\epsilon$.

For the reverse direction, use [L3] to choose a diameter-controlling gauge ${\gamma }_n$ for tolerance $2^{-n}$, set ${\delta }_n={\min }_{k\le n}{\gamma }_k$, and let $H_n$ be the closed interval hull of the nonempty set of ${\delta }_n$-fine sums supplied by [L1]; the $H_n$ are nested, have length at most $2^{-n}$, and [L4] and [L2] give a common point $I$.

Given $\epsilon >0$, [L4] gives $n$ with $2^{-n}<\epsilon$; every ${\delta }_n$-fine sum lies in $H_n$ with $I$, hence within $\epsilon$ of $I$, which is precisely HK integrability and selects no partition.

For the forward direction, fix a whole-interval gauge whose fine sums are within $\epsilon \mathrm{/}2$ of the integral. Given two fine partitions $P,Q$ of $[a,c]$, use [L1] to choose one fine partition $R$ of $[c,b]$ for the restricted gauge. Then $P\cup R$ and $Q\cup R$ are whole-interval fine partitions, so $\mathrm{\mid }S(f,P)-S(f,Q)\mathrm{\mid }<\epsilon$; [L2] proves integrability on $[a,c]$, and the symmetric completion proves it on $[c,b]$.

For the reverse direction, choose side gauges for error $\epsilon \mathrm{/}2$. For $x<c$ shrink the left gauge below $(c-x)\mathrm{/}2$, for $x>c$ shrink the right gauge below $(x-c)\mathrm{/}2$, and at $c$ take the minimum of the two gauges. Thus a fine cell can cross $c$ only when tagged at $c$, in which case splitting it at $c$ produces one fine cell for each side. The two side estimates then add, proving whole-interval integrability and the displayed additivity, including $c=a$ or $c=b$.

Order $u,v,w$, apply step 1.2 on the two adjacent compact subintervals, and reverse any necessary limits with the orientation convention in the Statement. The resulting signed equality is the oriented three-point identity in every ordering.

The empty family has error $0$. Otherwise fix a whole-interval gauge whose full-partition error is below $\epsilon \mathrm{/}4$. After a partial partition fine for that fixed gauge is given, [L2] makes $f$ integrable on each of its finitely many complementary compact intervals. For any prescribed complement error, [L3] supplies a local accuracy gauge there; [L1] supplies a partition fine for the minimum of that local gauge and the already fixed whole-interval gauge. Thus the resulting completions are both arbitrarily accurate and fine for the original gauge.

For the cells with nonnegative local error, complete their complement with fine partitions whose total local error is below $\epsilon \mathrm{/}4$. Additivity [L2] identifies the resulting full-partition error with the selected positive errors plus those complement errors, so the positive total is below $\epsilon \mathrm{/}2$. Repeating the construction for the negative cells bounds the absolute value of their total by $\epsilon \mathrm{/}2$; adding the two bounds gives the displayed strict estimate.

If $a=b$, both integrals are $0$; otherwise take the constant gauge $\gamma (x)=\delta \mathrm{/}2$ from [L1], so [L2] makes every $\gamma$-fine cell shorter than $\delta$ and the whole partition has mesh below $\delta$.

The universal estimate in [L1] therefore applies to every $\gamma$-fine tagged partition, which is exactly the HK definition with the same value $I$.

Given $\epsilon >0$, put $\eta =\epsilon \mathrm{/}(2(b-a))$. For each $\xi \in [a,b]$, let $N(\xi )$ be the least natural $n\ge 1$ such that the derivative estimate with error $\eta$ holds whenever $0<\mathrm{\mid }y-\xi \mathrm{\mid }<1\mathrm{/}n$ in the domain. Differentiability and [L2] make this set nonempty, and [L3] makes its least element unique; hence $\delta (\xi )=1\mathrm{/}N(\xi )$ is a gauge defined without an uncountable choice.

For a fine tagged cell, split $F(v_i)-F(u_i)$ at its tag, apply the two estimates from step 1.1, sum over all cells, and telescope by [L4]; the total error is below $2\eta (b-a)=\epsilon$, proving the displayed HK value.

For $x>a$, restriction to $[a,x]$ leaves the domain-relative difference quotients of $F$ unchanged at every limit point of that interval, so [L4] makes the restriction differentiable with derivative $f$. Apply [L1] to obtain $G(x)=F(x)-F(a)$; at $x=a$ the same identity follows from [L2].

The constant $F(a)$ cancels from every domain-relative difference quotient in step 1.1, so [L4] gives $G^{\mathrm{\prime }}=F^{\mathrm{\prime }}=f$ throughout $[a,b]$, endpoints included.

By [L2] and [L1], $F^{\mathrm{\prime }}G+F{G}^{\mathrm{\prime }}$ is HK integrable and its integral equals $F(b)G(b)-F(a)G(a)$.

If either summand is integrable, [L3] applied to its difference from the integrable sum in step 1.1 makes the other integrable; rearranging gives the formula, and the same argument in the other order proves the reverse implication.

Applying [L1] throughout $[a,b]$ identifies the derivative of $F\circ \varphi$ as $(f\circ \varphi ){\varphi }^{\mathrm{\prime }}$, including a constant $\varphi$.

Applying [L2] to the composite gives its HK integrability and the displayed endpoint formula, which also covers reversed endpoint values of $\varphi$.

For the forward direction, the truncation primitive $A(c)={\int }_a^{c}f$ has a finite limit, so late values $A(c),A(d)$ are close; [L1] identifies their difference with ${\int }_c^{d}f$.

For the reverse direction, take the explicit cofinal sequence $c_n=b-(b-a)\mathrm{/}(n+1)$ at a finite endpoint, or $c_n=a+n$ at $+\mathrm{\infty }$. The tail condition makes $A(c_n)$ Cauchy, so [L2] gives a finite limit $I$. For an arbitrary sufficiently late $c$, choose $n$ with $c_n>c$; then [L1] gives $\mathrm{\mid }A(c)-I\mathrm{\mid }\le \mathrm{\mid }{\int }_c^{c_n}f\mathrm{\mid }+\mathrm{\mid }A(c_n)-I\mathrm{\mid }$, and the two terms are small. This is exactly the limit in [L3] or [L4]; reflection handles a missing left endpoint.

On every sufficiently late compact tail, $-g\le f\le g$. By [L3], $-g$ is integrable with integral $-\int g$, so two applications of [L1] give $-\int g\le \int f\le \int g$ and hence $\mathrm{\mid }\int f\mathrm{\mid }\le \int g$; the tail criterion [L2] proves claim 1, and taking $g=\mathrm{\mid }f\mathrm{\mid }$ proves claim 2.

If $f\mathrm{/}g\to c\in (0,\mathrm{\infty })$, it lies between two positive constants near the endpoint, so two applications of step 1.1 give equivalence; for limit $0$ or $+\mathrm{\infty }$, the corresponding one-sided eventual bound gives exactly the stated implication.

For the forward direction, [L3] restricts a proper HK integral to every compact prefix; apply [L2] to the one-cell partial partition $([c,b],b)$ after taking $c$ sufficiently close to $b$, and also make $\mathrm{\mid }f(b)\mathrm{\mid }(b-c)$ small, to obtain uniformly small tail integrals, so [L1] and [L3] make the truncation values converge to the proper integral.

For the reverse direction, let $A={\lim }_{c\uparrow b}{\int }_a^{c}f$ and set $c_i=b-(b-a)\mathrm{/}(i+1)$, so $c_0=a$ and $c_i\uparrow b$. By [L4], choose for each band $[c_{i-1},c_i]$ a gauge whose Saks–Henstock partial-partition error is below $\epsilon 2^{-i-4}$.

On each open band, take the minimum of its local gauge and half the distances to the two band endpoints; at each $c_i$, take the minimum of the adjacent gauges and half the adjacent band lengths. At $b$, choose a radius $\gamma$ so that the truncation tail error is below $\epsilon \mathrm{/}4$ and $\mathrm{\mid }f(b)\mathrm{\mid }\gamma <\epsilon \mathrm{/}4$. A partition fine for this global gauge can cross a band boundary only when tagged there, so it splits into finitely many complete band partitions, one final fine partial band partition, and a possible last cell tagged at $b$.

Additivity [L3], the summable local error budget from step 1.2, the Saks–Henstock estimate on the final partial band, the truncation bound, and the endpoint-cell bound from step 2.1 show that every fine sum differs from $A$ by less than $\epsilon$. Thus the extension is properly HK integrable with integral $A$. Changing the assigned value at $b$ alters only the last endpoint-tagged term, whose length the gauge can make arbitrarily small, so the integral is independent of that value; reflection gives the left-endpoint form.