Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26
How statement and proof provenance work

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.

Tagged sums approximate a contour integral within oscillation times length

Statement

Let γ:[a,b]→C be a rectifiable contour with a<b, let f be continuous on its trace γ∗, let a=t0<t1<⋯<tr=b be a partition of [a,b], and choose a tag ξi∈[ti,ti+1] for each i<r. Write γi for the restriction γ∣[ti,ti+1] and

ωi:=sup⁡{ ∣f(u)−f(v)∣ : u,v∈γ([ti,ti+1]) },

which is a nonnegative real number. Then

∣∫γf(z) dz−∑i<rf(γ(ξi))(γ(ti+1)−γ(ti))∣ ≤ ∑i<rωi L(γi).

In particular, if ω≥0 satisfies ∣f(u)−f(v)∣≤ω for all u,v∈γ∗, then the left-hand side is at most ω L(γ).

The bound is stated with the oscillations themselves and not as a limit, so a modulus of continuity for f on γ∗ converts directly into an error estimate. For a singleton parameter interval [a,a] there is no partition, and both the integral and the empty tagged sum are 0.

Facts & Assumptions

Given: A rectifiable contour γ:[a,b]→C with a<b, a continuous f on γ∗, a partition a=t0<t1<⋯<tr=b, and tags ξi∈[ti,ti+1] for i<r.

[L1]

A complex contour is a rectifiable path γ:[a,b]→C; if α,β:[0,1]→C satisfy α(1)=β(0), their concatenation is (α∗β)(s)=α(2s) for 0≤s≤12 and β(2s−1) for 12≤s≤1 (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L2]

For a rectifiable γ:[a,b]→C and f continuous on its trace, the complex line integral ∫γf dz of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L3]

If ϕ:[c,d]→[a,b] is a strictly increasing continuous bijection, γ:[a,b]→C is rectifiable and f is continuous on the trace of γ, then ∫γ∘ϕf dz=∫γf dz (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L4]

For composable rectifiable contours α,β, ∫α∗βf dz=∫αf dz+∫βf dz (Complex line integrals change sign under reversal and add under concatenation).

[L5]

For continuous f,g on the trace of a rectifiable contour γ and α,β∈C, ∫γ(αf+βg) dz=α∫γf dz+β∫γg dz (Complex line integrals are linear in the integrand).

[L6]

For c∈C and a rectifiable contour γ:[a,b]→C, ∫γc dz=c(γ(b)−γ(a)) (The contour integral of a constant c is c times the endpoint displacement).

[L7]

If ∣f(z)∣≤M on the trace of a rectifiable contour γ, with M≥0, then ∣∫γf(z) dz∣≤M L(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L8]

For a path γ:[a,b]→Rn with n≥1 and c∈[a,b], L[a,b](γ)=L[a,c](γ∣[a,c])+L[c,b](γ∣[c,b]) in the nonnegative extended reals, and γ is rectifiable on [a,b] if and only if both restrictions are rectifiable (Arc length is additive across every subdivision point and decreases under restriction).

[L9]

A partition of [a,b] with a<b consists of a=t0<t1<⋯<tr=b with r≥1, its subintervals [ti,ti+1] being indexed from i=0 (Partition of [a,b] as a finite strictly increasing list a=t0<t1<⋯<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L11]

If ak≤bk for all k<n then ∑k<nak≤∑k<nbk (Laws of finite sums and finite products).

[L12]

If a property holds at 0 and passes from n to n+1, it holds for every n∈N (The principle of mathematical induction).

[L15]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

Proof

technique · direct
1.1givenL8L9L12

By [L8] applied at t1, then to γ∣[t1,b] at t2, and so on, an induction on the number of partition points ([L12]) shows that each γi is rectifiable and that L(γ)=∑i<rL(γi).

1.2givenL13L14L15

Each ωi is a nonnegative real: [ti,ti+1] is a closed bounded interval, hence compact by [L13]; f∘γ is continuous on it, so its image is compact by [L14] and bounded by [L15]; hence {∣f(u)−f(v)∣:u,v∈γ([ti,ti+1])} is a nonempty set of reals bounded above, and it has a supremum, which is ≥0 because u=v is allowed.

1.3givenL1L2L3L4

For a≤u<v<w≤b put α(s)=γ(u+s(v−u)) and β(s)=γ(v+s(w−v)) on [0,1]; then α(1)=γ(v)=β(0), so α∗β is defined by [L1], and α∗β=γ∣[u,w]∘ϕ where ϕ:[0,1]→[u,w] is the strictly increasing continuous bijection that is affine on [0,12] and on [12,1] with ϕ(12)=v. Since α and β are increasing reparametrisations of γ∣[u,v] and γ∣[v,w], [L3] and [L4] give ∫γ∣[u,w]f dz=∫γ∣[u,v]f dz+∫γ∣[v,w]f dz.

1.4L6

For each i<r, [L6] applied to the constant f(γ(ξi)) on γi gives ∫γif(γ(ξi)) dz=f(γ(ξi))(γ(ti+1)−γ(ti)).

2.1step 1.3L12

Applying step 1.3 at t1, then to γ∣[t1,b] at t2, and so on, an induction on the number of partition points ([L12]) gives ∫γf dz=∑i<r∫γif dz.

2.2step 1.2step 1.4L5L7

Fix i<r. The tag value γ(ξi) lies on the trace of γi, so ∣f(z)−f(γ(ξi))∣≤ωi for every z on that trace by the definition of ωi in step 1.2; by [L5] the difference ∫γif dz−∫γif(γ(ξi)) dz equals ∫γi(f(z)−f(γ(ξi)))dz, and [L7] bounds its modulus by ωiL(γi).

3.1step 2.1step 2.2L10L11L12

Subtracting the identity of step 1.4 from that of step 2.1 termwise, the quantity to be estimated is ∑i<r(∫γif dz−f(γ(ξi))(γ(ti+1)−γ(ti))); the finite triangle inequality, obtained from [L10] by induction ([L12]), and then [L11] with the bounds of step 2.2, give the stated estimate ∑i<rωiL(γi).

4.1step 1.1step 3.1L11∎

If ∣f(u)−f(v)∣≤ω for all u,v∈γ∗ then ωi≤ω for every i<r, so step 3.1 and [L11] bound the error by ω∑i<rL(γi), which is ωL(γ) by step 1.1; and on a singleton interval [a,a] the integral is 0 and there is no partition, so the assertion made there is the stated one about the empty sum.

Depends on

Used by

Dependency tree · two levels

85 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