([Apo74, 7.2]) If \(f\in\mathcal R(\alpha)\) on \([a,b]\) and \(\int_a^b f\,d\alpha=0\) for every \(f\) which is monotonic on \([a,b]\), prove that \(\alpha\) must be constant on \([a,b]\).
Solution
1. Apply the hypothesis to \(f \equiv 1\) (constant, hence monotone).Proof: \(\int_a^b 1\,d\alpha = \alpha(b) - \alpha(a) = 0\), so \(\alpha(a) = \alpha(b)\).
2. Apply the hypothesis to \(f(x) = x\) (monotone increasing).Proof: \(\int_a^b x\,d\alpha = 0\).By the Stieltjes integration-by-parts formula (valid since \(f\) is continuous and \(\alpha\) is monotone, both of bounded variation), \[\int_a^b x\,d\alpha = b\alpha(b) - a\alpha(a) - \int_a^b \alpha(x)\,dx.\] Using \(\alpha(a) = \alpha(b) =: c\) from 1: \[0 = b c - a c - \int_a^b \alpha(x)\,dx = c(b-a) - \int_a^b \alpha(x)\,dx,\] so \(\int_a^b (\alpha(x) - c)\,dx = 0\).
3. \(\alpha \equiv c\) on \([a,b]\).Proof: \(\alpha\) is increasing, so \(\alpha(x) - c = \alpha(x) - \alpha(a) \ge 0\) for all \(x\).A non-negative Riemann-integrable function with zero integral is \(0\) a.e., so \(\alpha(x) = c\) for a.e. \(x \in [a,b]\).Since \(\alpha\) is monotone, if \(\alpha(x_0) > c\) at some point then \(\alpha \ge \alpha(x_0) > c\) on \([x_0, b]\), a set of positive measure — contradiction.Hence \(\alpha(x) = c\) for all \(x \in [a,b]\); \(\alpha\) is constant.
4. Q.E.D.