ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-11

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A nonlinear reparametrisation leaves a Stieltjes integral unchanged

Example

For $ϕ (t) = t^{2}$ on $[0, 1]$ , the change-of-variable theorem gives the concrete identity $\int_{0}^{1} x d x = \int_{0}^{1} t^{2} d (t^{2}) = \frac{1}{2} .$

Facts & Assumptions

Given: $f (x) = x$ , $α (x) = x$ , and $ϕ (t) = t^{2}$ on $[0, 1]$ .

[L1]

Increasing continuous reparametrization preserves a Stieltjes integral (Change of variable for the Riemann–Stieltjes integral).

[L2]

The identity integrator gives the ordinary Riemann integral (The identity integrator recovers the Riemann integral).

[L3]

A $C^{1}$ integrator reduces the integral to one against its derivative (A continuously differentiable integrator reduces Stieltjes integration to ordinary integration).

Verification

technique · computation

1.1

The map $ϕ (t) = t^{2}$ is a strictly increasing continuous bijection of $[0, 1]$ onto itself. Applying [L1] gives $\int_{0}^{1} x d x = \int_{0}^{1} t^{2} d (t^{2})$ .

1.2

Direct difference quotients give $(t^{2})^{'} = 2 t$ and $(t^{4} / 2)^{'} = 2 t^{3}$ . Hence [L2], [L3], and the FTC give [L2, L3] $\int_{0}^{1} x d x = \frac{1}{2}, \int_{0}^{1} t^{2} d (t^{2}) = \int_{0}^{1} 2 t^{3} d t = \frac{1}{2} .$ ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.19 (standard reference, not scraped)