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A continuously differentiable integrator reduces Stieltjes integration to ordinary integration

Statement

Let $f:[a,b]\to\mathbb R$ be Riemann integrable. Suppose $\alpha$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $\alpha'$ extends continuously to $[a,b]$ . Then $f$ is Riemann–Stieltjes integrable with respect to $\alpha$ and

$\int_a^b f\,d\alpha=\int_a^b f(x)\alpha'(x)\,dx.$

Facts & Assumptions

Given: A Riemann-integrable $f$ and an integrator $\alpha$ with continuous derivative on the compact interval.

[L1]

The mean value theorem represents every increment of $\alpha$ as $\alpha'(\eta_i)(t_{i+1}-t_i)$ (The mean value theorem, as the case $g(x) = x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a < b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$ ).

[L2]

The continuous function $\alpha'$ is uniformly continuous (Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness).

[L3]

The continuous function $\alpha'$ is Riemann integrable, so the product $f\alpha'$ is Riemann integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ ).

[L4]

Riemann tagged sums converge to the Darboux integral (The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$ ).

Proof

technique · direct

1.1

A Riemann-integrable function is bounded; choose $M$ with $|f|\le M$ . For each partition interval, [L1] gives $\eta_i\in(t_i,t_{i+1})$ such that $\Delta_i\alpha=\alpha'(\eta_i)\Delta_i t$ . Hence [L1] $S_\alpha(f;P,\xi)-\sum_i f(\xi_i)\alpha'(\xi_i)\Delta_i t=\sum_i f(\xi_i)(\alpha'(\eta_i)-\alpha'(\xi_i))\Delta_i t.$

2.1

By [L2], the absolute value of the right side is at most $M(b-a)\omega_{\alpha'}(\lVert P\rVert)$ , which tends to zero with the mesh. By [L3] and [L4], the second sum in step 1.1 tends to $\int_a^b f\alpha'$ . Thus all Stieltjes sums have the same limit, proving both existence and the formula.

step 1.1L2L3L4∎

Depends on

Used by

∫₀¹ x d(x²)=2/3 Example
A nonlinear reparametrisation leaves a Stieltjes integral unchanged Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 147 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.17 (standard reference, not scraped)