CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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The identity integrator recovers the Riemann integral

Statement

Let $a \leq b$ . For the identity function $id (x) = x$ , a bounded $f : [a, b] \to R$ is Riemann–Stieltjes integrable with respect to $id$ exactly when it is Riemann integrable, and then

$\int_{a}^{b} f d id = \int_{a}^{b} f (x) d x .$

For $a = b$ both sides are $0$ by the singleton conventions. For reversed endpoints the statement is about a function defined on the sorted interval: if $a > b$ and $f$ is bounded on $[b, a]$ , then both oriented conventions negate the corresponding sorted integral, so the equality is inherited from the case just proved. The hypothesis $a \leq b$ is needed for the displayed clause itself, because $[a, b]$ is empty when $a > b$ and a function typed on it supplies no values to integrate.

Facts & Assumptions

Given: A bounded function $f$ on the compact interval with endpoints $a, b$ .

[L1]

A Stieltjes sum is $\sum_{i} f (ξ_{i}) (α (t_{i + 1}) - α (t_{i}))$ (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

A Riemann sum is $\sum_{i} f (ξ_{i}) (t_{i + 1} - t_{i})$ (Tagged partitions of $[a, b]$ , with a tag $ξ_{i}$ in each subinterval, and the Riemann sum $S (f, P, ξ) = \sum_{i} f (ξ_{i}) Δ_{i}$ ).

[L3]

Darboux and tagged-sum Riemann integrability agree (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 $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ ).

[L4]

Oriented integrals reverse sign when endpoints are interchanged (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

Proof

technique · direct

1.1

When $a < b$ and $α = id$ , every increment $α (t_{i + 1}) - α (t_{i})$ equals $t_{i + 1} - t_{i}$ . Thus [L1] and [L2] are termwise identical for every tagged partition.

L1L2

2.1

Consequently the two mesh limits exist simultaneously and have the same value; [L3] identifies that tagged limit with the Darboux integral. For $a = b$ both conventions give zero, and [L4] handles $a > b$ .

step 1.1L3L4∎

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Sources

W. Rudin, Principles of Mathematical Analysis, Ch. 6, Definition 6.1 (standard reference, not scraped)