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CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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The total-variation bound for a Riemann–Stieltjes integral

Statement

Suppose ∫abf dα exists, α has bounded variation, and ∣f(x)∣≤M on [a,b]. Then

∣∫abf dα∣≤MVar⁡[a,b](α).

Facts & Assumptions

Given: An existing Stieltjes integral, a BV integrator α, and a bound ∣f∣≤M.

[L1]

Stieltjes sums converge to the integral in the mesh sense (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, The Riemann–Stieltjes integral is unique).

[L2]

Every sum of absolute integrator increments is bounded by total variation (Bounded variation and total variation on an interval).

[L3]

Finite sums and the triangle inequality give ∣∑ui∣≤∑∣ui∣ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

[L4]

Non-strict inequalities pass to limits (Limits preserve non-strict inequalities).

Proof

technique · direct
1.1L2L3

Every tagged sum satisfies ∣S(f,α;P,ξ)∣≤∑i<n∣f(ξi)∣∣Δiα∣≤M∑i<n∣Δiα∣≤MVar⁡[a,b](α).

2.1step 1.1L1L4∎

Take a sequence of tagged partitions with mesh tending to 0. Their sums converge to the integral by [L1], and [L4] passes the bound in step 1.1 to the limit. Orientation and the singleton case preserve the same absolute-value inequality.

Depends on

Dependency tree · 2 levels

Sources