Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-11
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A Riemann–Stieltjes integrable integrand need not be bounded

Example

On [0,1], set f(0)=0 and f(x)=1/x for x>0, and let α=H1/2 be the unit step at 1/2. Then f is unbounded but ∫01f dα=2.

Facts & Assumptions

Given: The displayed f and step integrator α.

[L1]

Only the partition interval across the jump of α has a nonzero Stieltjes weight (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

Verification

technique · direct
1.1

For each positive integer n, f(1/n)=n, so the range of f is not bounded above.

given
2.1

Every Stieltjes sum equals f(ξ) for a tag ξ in the interval across 1/2; as the mesh tends to zero, ξ→1/2. By [L2], these sums tend to f(1/2)=2. The unbounded behavior near zero is multiplied only by zero increments of α.

L1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

40 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