A singular integrator refutes the unbounded forms of Theorems 6.17, 6.21 and 6.22

In this formalization the upper and lower integrals are the ordinary sSup/sInf of sets of reals, which return the default value $0$ on a set that is unbounded in the relevant direction. Rudin's Chapter 6 assumes throughout that the integrand is bounded, and the three statements of the chapter that omit that clause — Theorem 6.17 (reduction of a Stieltjes integral to a Riemann integral with a density), Theorem 6.21 (the fundamental theorem of calculus) and Theorem 6.22 (integration by parts) — are all refuted at once by a single object.

Call a function $\alpha : \mathbb{R} \to \mathbb{R}$ a singular integrator if it is monotonically increasing, differentiable at every point of $[0,1]$, satisfies $\alpha(0) < \alpha(1)$, has derivative unbounded above on $[0,1]$, and has

This theorem asserts that the existence of such an $\alpha$ makes all three unbounded statements false.

The mechanism is the default value of the suprema and infima. Since $\alpha$ is increasing, $\alpha' \ge 0$, so every infimum occurring in a lower sum is a genuine infimum and every term of an upper sum is nonnegative. The one-interval partition of $[0,1]$ has a single supremum, which is the supremum of an unbounded set and therefore evaluates to $0$, so the upper integral of $\alpha'$ is $0$; the dense small values force every lower sum to vanish, so the lower integral is $0$ as well. Hence $\alpha' \in \mathcal{R}$ with

even though $\alpha$ grows. Taking $f \equiv 1$ (which is bounded, and whose upper and lower sums telescope to $\alpha(1) - \alpha(0)$) contradicts the conclusion of Theorem 6.17; taking $F = \alpha$ contradicts Theorem 6.21; and taking $F = \alpha$, $G \equiv 1$, $g \equiv 0$ contradicts Theorem 6.22, whose right-hand side is then $\alpha(1) - \alpha(0)$ while its left-hand side is $0$.