Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A common jump can destroy Riemann–Stieltjes integrability

Example

Let a<c<ba<c<b and put f=α=Hcf=\alpha=H_c, the unit step that is zero left of cc and one at and right of cc. Both functions are BV, but ∫abf dα\int_a^b f\,d\alpha does not exist.

Facts & Assumptions

Given: The two identical unit-step functions.

[L1]

A Stieltjes integral must be the same limit for every sufficiently fine choice of partition and tags (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, The Riemann–Stieltjes integral is unique).

[L2]

For a nondecreasing function every partition increment is nonnegative, so the absolute values in the variation sum may be removed and the finite sum telescopes to the endpoint increment (Bounded variation and total variation on an interval, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

Verification

technique · counterexample
1.1

By [L2], both ff and α\alpha have total variation one. For every h>0h>0 small enough, choose a partition containing c−hc-h and cc. The only nonzero integrator increment occurs on [c−h,c][c-h,c].

L2
2.1

Tag that interval first at c−hc-h and then at cc. The corresponding sums are respectively f(c−h)=0f(c-h)=0 and f(c)=1f(c)=1, although both partitions have mesh tending to zero after the other intervals are refined. Thus no common mesh limit exists, contradicting the necessary condition [L1].

L1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources