Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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  • 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.
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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
1.1L1L2L3choose

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.

2.1step 1.1L1L4∎

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.

Depends on

Dependency tree · 2 levels

Sources