3.15. Necessary Conditions for Existence of Riemann-Stieltjes Integrals

Theorem 3.24

Assume \(\alpha\) is increasing on \([a, b]\). Let \(a \leq c < b\). If \(f\) and \(\alpha\) are discontinuous from the right at \(c\), then \(f \notin \mathfrak{R}(\alpha)\) on \([a, b]\).

Analogously, letting \(a < c \leq b\), if \(f\) and \(\alpha\) are discontinuous from the left at \(c\), then \(f \notin \mathfrak{R}(\alpha)\) on \([a, b]\).

Proof. We only prove the case when \(f\) and \(\alpha\) are discontinuous from the right at \(c\).

Because \(f\) and \(\alpha\) are discontinuous from the right at \(c\), there exists \(\varepsilon > 0\) such that for every \(\delta > 0\), we can find \(x, y \in (c, c+\delta)\) such that

\[\abs{f(x) - f(c)} > \varepsilon\quad\text{and}\quad\abs{\alpha(y) - \alpha(c)} = \alpha(y) - \alpha(c) > \varepsilon\]

We want to show that \(f\) does not satisfy the Riemann’s condition. More precisely, choosing the number \(\varepsilon_0 = \varepsilon^2\), we claim that for every partition \(P_{\varepsilon_0}\) of \([a, b]\), there exists a refinement \(P\) such that \(U(P, f, \alpha) - L(P, f, \alpha) > \varepsilon_0\).

First, exploiting the fact that \(\alpha\) is discontinuous from the right at \(c\) we can find a number \(d > c\) such that \(\alpha(d) - \alpha(c) > \varepsilon\). Let partition \(P = P_{\varepsilon_0} \cup \{c, d\}\). In this way, we guarantee that the subinterval \([c, d]\) is contained in the partition. Then, since \(f\) is discontinuous from the right at \(c\), we can find \(t \in (c, d)\) such that \(\abs{f(t) - f(c)} > \varepsilon\). It then follows that

\[\omega_f([c, d]) = \sup_{x, y \in [c, d]}\abs{f(x) - f(y)}\geq\abs{f(t) - f(c)} > \varepsilon\]

Multiplying by \( [\alpha(d) - \alpha(c)]\) yields

\[\omega_f([c, d]) [\alpha(d) - \alpha(c)] > \varepsilon^2 = \varepsilon_0 \]

Therefore,

\[U(P, f, \alpha) - L(P, f, \alpha) \geq\omega_f([c, d]) [\alpha(d) - \alpha(c)] > \varepsilon_0 \]

This completes the proof.