An unbounded integrable derivative of a monotone function has integral zero and dense small values

Let $\alpha : \mathbb{R} \to \mathbb{R}$ be monotonically increasing and differentiable at every point of $[0,1]$, and suppose its derivative is unbounded above on $[0,1]$. This theorem says that if $\alpha' \in \mathcal{R}$ on $[0,1]$ then

The upper and lower integrals here are the ordinary supremum and infimum of sets of reals, which take the default value $0$ on a set unbounded in the relevant direction. Since $\alpha$ increases, $\alpha' \ge 0$, so every term of an upper sum is nonnegative, while the one-interval partition of $[0,1]$ produces the supremum of an unbounded set and hence the value $0$: the upper integral is exactly $0$, and integrability transports this to the lower integral.

For the second assertion, no default value intervenes in the lower sums, because $\alpha' \ge 0$ makes every infimum genuine. If the derivative were bounded below by some $\varepsilon > 0$ on a subinterval $[u,v]$, then the partition with division points $0, u, v, 1$ would have lower sum at least $\varepsilon (v-u) > 0$; the mean value theorem bounds every lower sum by the total increment $\alpha(1) - \alpha(0)$, so the lower integral is a genuine supremum and would be positive, contradicting its vanishing.

The statement is the converse half of the criterion recorded in Rudin.ch06_singular_integrator_refutes: a monotone integrator whose derivative is unbounded and Riemann integrable in this sense is necessarily a singular integrator in the sense used there. Together the two results say that the versions of Rudin's Theorems 6.17, 6.21 and 6.22 stated without the boundedness hypothesis of Definition 6.2 fail exactly when such an integrator exists.