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

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A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator

Statement

If $f : [a, b] \to R$ is continuous and $α : [a, b] \to R$ has bounded variation, then $\int_{a}^{b} f d α$ exists.

Facts & Assumptions

Given: A continuous $f : [a, b] \to R$ and a BV function $α : [a, b] \to R$ .

[L1]

The refinement estimate bounds differences of tagged sums by oscillation times total variation (Refinement and tag-change estimates for Stieltjes sums).

[L2]

A Stieltjes mesh limit, when it exists, is unique (The Riemann–Stieltjes integral is unique, Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L3]

Total variation is finite for a BV function (Bounded variation and total variation on an interval).

[L4]

A continuous real function on a compact interval is uniformly continuous (Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L5]

Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

Uniform partitions of arbitrarily small mesh exist, and common refinements exist (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

Proof

technique · direct

1.1L1L3L4L6

Put $V = {Var}_{[a, b]} (α)$ . If $V = 0$ , every integrator increment is $0$ , so every Stieltjes sum is $0$ and the integral exists. Assume $V > 0$ . Given $ε > 0$ , uniform continuity gives $δ > 0$ such that $∣ x - y ∣ < δ$ implies $∣ f (x) - f (y) ∣ < ε / (4 V)$ . Any two tagged partitions of mesh below $δ$ have all local oscillations below $ε / (4 V)$ ; comparing both with their common refinement through [L1] makes their sums differ by less than $ε$ .

2.1step 1.1L5L6choose

Choose one tagged uniform partition with mesh tending to $0$ for each natural index and call its sum $s_{n}$ . Step 1.1 makes $(s_{n})$ Cauchy, so [L5] gives a real limit $I$ .

3.1step 1.1step 2.1L1L2L5L6∎

Given $ε > 0$ , choose the threshold in step 1.1 for error $ε / 2$ and then a uniform sum $s_{n}$ beyond that threshold with $∣ s_{n} - I ∣ < ε / 2$ . Every arbitrary tagged sum with sufficiently small mesh differs from $s_{n}$ by less than $ε / 2$ , hence is within $ε$ of $I$ . This is the mesh-limit definition, and [L2] identifies the unique value.

A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources