Reduction of a Stieltjes integral to a Riemann integral

1. Set up: \(\alpha\) increasing, \(\alpha' \in \mathcal{R}\) on \([a,b]\), \(f\) bounded.Proof: since \(\alpha' \in \mathcal{R}\), \(\alpha'\) is bounded, so \(\alpha\) is Lipschitz (by the fundamental theorem of calculus for Riemann integrals, \(\alpha(x) = \alpha(a) + \int_a^x \alpha'(t)\,dt\)). In particular \(\alpha\) is continuous and of bounded variation.We prove the two implications.

2. (⟹) If \(f \in \mathcal{R}(\alpha)\) then \(f\alpha' \in \mathcal{R}\) and \(\int f\,d\alpha = \int f\alpha'\).Proof: let \(M = \sup|f|\).For any partition \(P = \{x_0 < \cdots < x_n\}\) and any choice of tags \(t_i \in [x_{i-1}, x_i]\), write \(m_i' = \inf_{[x_{i-1},x_i]}\alpha'\) and \(M_i' = \sup_{[x_{i-1},x_i]}\alpha'\).Since \(\Delta\alpha_i = \int_{x_{i-1}}^{x_i}\alpha'\) (fundamental theorem for monotone functions) and \(\int_{x_{i-1}}^{x_i}\alpha' \in [m_i'\Delta x_i,\ M_i'\Delta x_i]\), we have \(|\Delta\alpha_i - \alpha'(t_i)\Delta x_i| \le (M_i' - m_i')\Delta x_i\).Hence the Stieltjes sum and the Riemann sum of \(f\alpha'\) with the same tags differ by \[\left|\sum_i f(t_i)\Delta\alpha_i - \sum_i f(t_i)\alpha'(t_i)\Delta x_i\right| \le M\sum_i (M_i' - m_i')\Delta x_i = M\big(U(\alpha',P) - L(\alpha',P)\big).\] Since \(f \in \mathcal{R}(\alpha)\), Stieltjes sums converge to \(\int f\,d\alpha\) along partitions with mesh \(\to 0\); since \(\alpha' \in \mathcal{R}\), \(U(\alpha',P) - L(\alpha',P) \to 0\) along the same refinement.So the Riemann sums \(\sum f(t_i)\alpha'(t_i)\Delta x_i\) also converge, to \(\int f\,d\alpha\).Riemann sums converging to a common limit along partitions with mesh \(\to 0\) is exactly Riemann integrability, so \(f\alpha' \in \mathcal{R}\) and \(\int f\alpha' = \int f\,d\alpha\).

3. (⟸) If \(f\alpha' \in \mathcal{R}\) then \(f \in \mathcal{R}(\alpha)\) with the same value.Proof: the same inequality of 2 shows that along any partition sequence with mesh \(\to 0\), the Stieltjes sums \(\sum f(t_i)\Delta\alpha_i\) are within \(M\big(U(\alpha',P) - L(\alpha',P)\big) \to 0\) of the convergent Riemann sums of \(f\alpha'\); hence the Stieltjes sums converge, to \(\int_a^b f\alpha'\).So \(f \in \mathcal{R}(\alpha)\) and \(\int f\,d\alpha = \int_a^b f\alpha'\).

4. Q.E.D. Proof: both directions established: \(f \in \mathcal{R}(\alpha)\) iff \(f\alpha' \in \mathcal{R}\), and in that case the integrals agree.