The Gauge Integral and Cousin's Lemma: Examples and Counterexamples

Enumerate the rationals in $[0,1]$ as $(q_k)$ using [L1]; for a requested $\epsilon >0$, choose the gauge at $q_k$ below $\epsilon 2^{-k-3}$ and choose any fixed positive gauge at irrational tags.

In a fine partition, a fixed tag occurs on at most two cells, so the total length of rational-tagged cells is below ${\sum }_{k}4\epsilon 2^{-k-3}=\epsilon$; hence $\mathrm{\mid }S(d,P)-1\mathrm{\mid }<\epsilon$, and [L4] gives the stated HK value.

By [L2], every partition cell contains points where $d=0$ and points where $d=1$, so every lower Darboux sum is $0$ and every upper Darboux sum is $1$; [L3] therefore rules out Riemann integrability.

Applying [L3], [L4], and [L5] gives the displayed derivative for $x>0$, while [L6] gives $\mathrm{\mid }F(x)\mathrm{/}x\mathrm{\mid }=\mathrm{\mid }x\sin (1\mathrm{/}{x}^2)\mathrm{\mid }\le x\to 0$ and hence $F^{\mathrm{\prime }}(0)=0$. For every natural $m\ge 1$, at $x_m=1\mathrm{/}\sqrt{2\pi m}$ the values in [L2] give $f(x_m)=-2\mathrm{/}{x}_m$, which is unbounded.

Applying [L1] gives HK integrability and ${\int }_0^{1}f=F(1)-F(0)=\sin 1$.

On $[1,\mathrm{\infty })$ the first factor $\sin x$ is continuous and hence locally Riemann integrable by [L4], while its primitive $-\cos x$ is bounded by [L3] and [L6]. Apply [L1] with second factor $1\mathrm{/}x$, which is nonnegative, decreasing, and tends to zero; the compact truncation integrals therefore have a finite limit.

The derivative clause in [L3] gives the removable limit at zero, so [L4] makes the extension Riemann integrable on each compact truncation and [L2] makes it HK integrable there. By [L7], for every $c>1$ its integral on $[0,c]$ is the fixed integral on $[0,1]$ plus the integral on $[1,c]$; the limit from step 1.1 and [L5] therefore give the noncompact HK integral.

For each $x$, let $r(x)$ be half the supremum of the radii in $(0,1]$ whose centered interval at $x$ lies in some member of $\mathcal{U}$; openness makes this set nonempty, [L2] supplies its positive supremum, and the supremum property ensures the $r(x)$-interval lies in at least one covering member, so $r$ is a gauge.

Apply [L1] to obtain a fine tagged partition; for its finitely many tags, choose covering members containing the corresponding gauge neighborhoods, and these members cover every partition cell and hence $[a,b]$, with a one-point interval covered by one member.

Suppose the claim were true; [L1] supplies a derivative unbounded near zero, while [L3] requires boundedness for Riemann integrability, a contradiction.

The correct conclusion for the same derivative is [L2]: it is HK integrable and [L1] evaluates its integral by endpoint difference.

Suppose the claim were true; the derivative in [L1] is HK integrable and unbounded on the same compact interval.

This contradicts the claimed boundedness, so the claim is false.

By [L1], $f$ is HK integrable on $[0,1]$.

Put $u_k=\pi \mathrm{/}2+k\pi$ and $x_k=u_k^{-1\mathrm{/}2}$. On each compact interval $[x_{k+1},x_k]$, both $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are continuous, so [L6] and [L7] make $\mathrm{\mid }f\mathrm{\mid }$ locally HK integrable. Since $F(x_k)=(-1{)}^k\mathrm{/}{u}_k$, [L4] and [L5] give ${\int }_{x_{k+1}}^{x_k}\mathrm{\mid }f\mathrm{\mid }\ge \mathrm{\mid }F(x_k)-F(x_{k+1})\mathrm{\mid }=1\mathrm{/}{u}_k+1\mathrm{/}{u}_{k+1}$, whose partial sums diverge by comparison with [L3].

Suppose $\mathrm{\mid }f\mathrm{\mid }$ were properly HK integrable on $[0,1]$; [L2] would make its truncation integrals converge as the left endpoint tends to zero, contradicting the unbounded sums of step 2.1.

Suppose the implication were true; [L1] gives a function satisfying its hypothesis but not its conclusion.

This contradiction refutes the implication, so the converse of the Riemann-to-HK theorem is false.

By [L1], $d$ is HK integrable, and [L3] makes it take both values $0$ and $1$ on every nondegenerate subinterval while taking no value strictly between them.

Suppose $d$ were a derivative; step 1.1 contradicts the intermediate-value property [L2], so the false claim is refuted.