Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The Riemann–Stieltjes integral is unique

Statement

For fixed f,α:[a,b]→R, at most one real number satisfies the mesh-limit condition defining ∫abf dα.

Facts & Assumptions

Given: Two reals I,J satisfying the defining mesh condition for the same functions f,α.

[L1]

The mesh-limit condition quantifies over every sufficiently fine tagged partition (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L4]

∣u+v∣≤∣u∣+∣v∣ and ∣u∣=0 exactly when u=0 (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct
1.1

Given ε>0, choose positive thresholds δI,δJ for error ε/2 in the two mesh conditions. By [L3] choose a natural N≥1 whose uniform partition has mesh smaller than both thresholds, and give it arbitrary tags.

L1L2L3choose
2.1

For its sum S, ∣I−J∣≤∣I−S∣+∣S−J∣<ε. Since this holds for every ε>0, ∣I−J∣=0 and I=J. The singleton interval has only the prescribed value 0.

step 1.1L1L4∎

Depends on

Used by

Dependency tree · two levels

33 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