Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13
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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.

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The arc-length function is continuous and nondecreasing, with increments equal to subpath lengths; it is strictly increasing exactly when no nondegenerate subpath is constant

Statement

Let γ:[a,b]→Rn be rectifiable and let s=sγ. Then s is continuous and nondecreasing. Moreover, s is strictly increasing if and only if γ is constant on no nondegenerate subinterval of [a,b].

On a singleton interval, continuity and nondecrease hold and the strictness equivalence is vacuous on both sides.

Facts & Assumptions

Given: The rectifiable path and its arc-length function.

[L2]

Every coordinate γj has bounded variation, the length of a restriction is at most the sum of its coordinate variations, and variation is additive on adjacent subintervals; hence Var⁡[u,v](γj)=Vj(v)−Vj(u) for Vj(t)=Var⁡[a,t](γj) (A path in Rn is rectifiable exactly when every coordinate has bounded variation, Total variation is additive over adjacent subintervals and decreases under restriction).

[L3]

The variation function of a bounded-variation function is continuous at every point where the function is continuous (The jumps of a variation function equal the absolute jumps of the original function).

Proof

technique · comparison
1.1

From [L1], s(v)≥s(u) whenever u≤v, so s is nondecreasing.

L1
1.2

Let Vj(t)=Var⁡[a,t](γj). Each Vj is continuous by [L2], [L3], and continuity of the path's coordinates.

givenL2L3
1.3

If s(u)=s(v) for some u<v, [L1] says the intervening path has length zero, and [L4] makes it constant on [u,v].

L1L4
2.1

For u≤v, [L1] and [L2] give 0≤s(v)−s(u)≤∑j(Vj(v)−Vj(u)). The finite sum on the right tends to zero as v→u, from either permitted side, so s is continuous.

step 1.2L1L2
3.1

Conversely, if γ is constant on [u,v], every polygonal sum there is zero, so [L1] gives s(u)=s(v). Thus equality at distinct arguments occurs exactly on a constant subinterval, proving the strictness equivalence.

step 1.3L1algebra∎

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