Prove that if \(f \in \mathcal{R}\) on \([a,b]\) and \(\alpha \in C^1[a,b]\), then the Riemann integral \(\int_a^b f(x)\alpha'(x)dx\) exists and \[\int_a^b f(x) d\alpha(x)= \int_a^b f(x)\alpha'(x)dx.\]
Solution
1. \(\alpha'\) is continuous, hence uniformly continuous and bounded on \([a,b]\); \(\alpha\) is of bounded variation.
\(\alpha \in C^1[a,b]\), so \(\alpha'\) is continuous on the compact set \([a,b]\), hence uniformly continuous and bounded.
2. \(f\alpha'\) is Riemann integrable on \([a,b]\).
\(f\) is Riemann integrable, hence bounded and continuous a.e.; \(\alpha'\) is continuous, hence Riemann integrable.
3. Compare Riemann–Stieltjes sums to Riemann sums for \(f\alpha'\).
let \(P = \{a = x_0 < \cdots < x_n = b\}\) be a partition and pick tags \(\xi_k \in [x_{k-1}, x_k]\).
4. Such sums converge to \(\int_a^b f\alpha'\).
since \(f\) is bounded (\(|f| \le B\)) and \(\alpha'\) is uniformly continuous, the difference between the sums in 3 and the ordinary Riemann sums \(\sum_k f(\xi_k)\alpha'(\xi_k)(x_k-x_{k-1})\) tends to \(0\) as the mesh goes to \(0\): \[ \Big|\sum_k f(\xi_k)\big(\alpha'(\eta_k) - \alpha'(\xi_k)\big)\Delta x_k\Big| \le B \sum_k |\alpha'(\eta_k)-\alpha'(\xi_k)|\Delta x_k \le B\,(b-a)\,\omega_\alpha'(\text{mesh}) \to 0. \] Both sums converge to \(\int_a^b f(x)\alpha'(x)\,dx\) (<1>2), so the Stieltjes sums converge to the same value.
5. Conclude.
the Riemann–Stieltjes integral \(\int_a^b f\,d\alpha\) exists (as \(f\) is Riemann integrable and \(\alpha\) is of bounded variation, 1) and equals the limit of the Stieltjes sums, which by 4 is \(\int_a^b f(x)\alpha'(x)\,dx\).
6. Q.E.D.