LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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Tagged sums approximate a contour integral within oscillation times length

Statement

Let $γ : [a, b] \to C$ be a rectifiable contour with $a < b$ , let $f$ be continuous on its trace $γ^{*}$ , let $a = t_{0} < t_{1} < \dots < t_{r} = b$ be a partition of $[a, b]$ , and choose a tag $ξ_{i} \in [t_{i}, t_{i + 1}]$ for each $i < r$ . Write $γ_{i}$ for the restriction $γ ∣_{[t_{i}, t_{i + 1}]}$ and

$ω_{i} : = \sup {∣ f (u) - f (v) ∣ : u, v \in γ ([t_{i}, t_{i + 1}])},$

which is a nonnegative real number. Then

$∣ \int_{γ} f (z) d z - \sum_{i < r} f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i})) ∣ \leq \sum_{i < r} ω_{i} L (γ_{i}) .$

In particular, if $ω \geq 0$ satisfies $∣ f (u) - f (v) ∣ \leq ω$ for all $u, v \in γ^{*}$ , 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 $γ : [a, b] \to C$ with $a < b$ , a continuous $f$ on $γ^{*}$ , a partition $a = t_{0} < t_{1} < \dots < t_{r} = b$ , and tags $ξ_{i} \in [t_{i}, t_{i + 1}]$ for $i < r$ .

[L1]

A complex contour is a rectifiable path $γ : [a, b] \to C$ ; if $α, β : [0, 1] \to C$ satisfy $α (1) = β (0)$ , their concatenation is $(α * β) (s) = α (2 s)$ for $0 \leq s \leq \frac{1}{2}$ and $β (2 s - 1)$ for $\frac{1}{2} \leq s \leq 1$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L2]

For a rectifiable $γ : [a, b] \to C$ and $f$ continuous on its trace, the complex line integral $\int_{γ} f d z$ 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 $ϕ : [c, d] \to [a, b]$ is a strictly increasing continuous bijection, $γ : [a, b] \to C$ is rectifiable and $f$ is continuous on the trace of $γ$ , then $\int_{γ \circ ϕ} f d z = \int_{γ} f d z$ (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L4]

For composable rectifiable contours $α, β$ , $\int_{α * β} f d z = \int_{α} f d z + \int_{β} f d z$ (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 $α, β \in C$ , $\int_{γ} (α f + β g) d z = α \int_{γ} f d z + β \int_{γ} g d z$ (Complex line integrals are linear in the integrand).

[L6]

For $c \in C$ and a rectifiable contour $γ : [a, b] \to C$ , $\int_{γ} c d z = c (γ (b) - γ (a))$ (The contour integral of a constant c is c times the endpoint displacement).

[L7]

If $∣ f (z) ∣ \leq M$ on the trace of a rectifiable contour $γ$ , with $M \geq 0$ , then $∣ \int_{γ} f (z) d z ∣ \leq M L (γ)$ (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L8]

For a path $γ : [a, b] \to R^{n}$ with $n \geq 1$ and $c \in [a, b]$ , $L_{[a, b]} (γ) = L_{[a, c]} (γ ∣_{[a, c]}) + L_{[c, b]} (γ ∣_{[c, b]})$ 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 $a = t_{0} < t_{1} < \dots < t_{r} = b$ with $r \geq 1$ , its subintervals $[t_{i}, t_{i + 1}]$ being indexed from $i = 0$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L10]

$∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for all $z, w \in C$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L11]

If $a_{k} \leq b_{k}$ for all $k < n$ then $\sum_{k < n} a_{k} \leq \sum_{k < n} b_{k}$ (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 $n \in N$ (The principle of mathematical induction).

[L13]

A closed box in $R^{n}$ is compact, and a subset of $R^{n}$ is compact exactly when it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L14]

The continuous image of a compact subset is a compact subset (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[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 $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) shows that each $γ_{i}$ is rectifiable and that $L (γ) = \sum_{i < r} L (γ_{i})$ .

1.2givenL13L14L15

Each $ω_{i}$ is a nonnegative real: $[t_{i}, t_{i + 1}]$ is a closed bounded interval, hence compact by [L13]; $f \circ γ$ is continuous on it, so its image is compact by [L14] and bounded by [L15]; hence ${∣ f (u) - f (v) ∣ : u, v \in γ ([t_{i}, t_{i + 1}])}$ is a nonempty set of reals bounded above, and it has a supremum, which is $\geq 0$ because $u = v$ is allowed.

1.3givenL1L2L3L4

For $a \leq u < v < w \leq b$ 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 $α * β = γ ∣_{[u, w]} \circ ϕ$ where $ϕ : [0, 1] \to [u, w]$ is the strictly increasing continuous bijection that is affine on $[0, \frac{1}{2}]$ and on $[\frac{1}{2}, 1]$ with $ϕ (\frac{1}{2}) = v$ . Since $α$ and $β$ are increasing reparametrisations of $γ ∣_{[u, v]}$ and $γ ∣_{[v, w]}$ , [L3] and [L4] give $\int_{γ ∣_{[u, w]}} f d z = \int_{γ ∣_{[u, v]}} f d z + \int_{γ ∣_{[v, w]}} f d z$ .

1.4L6

For each $i < r$ , [L6] applied to the constant $f (γ (ξ_{i}))$ on $γ_{i}$ gives $\int_{γ_{i}} f (γ (ξ_{i})) d z = f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i}))$ .

2.1step 1.3L12

Applying step 1.3 at $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) gives $\int_{γ} f d z = \sum_{i < r} \int_{γ_{i}} f d z$ .

2.2step 1.2step 1.4L5L7

Fix $i < r$ . The tag value $γ (ξ_{i})$ lies on the trace of $γ_{i}$ , so $∣ f (z) - f (γ (ξ_{i})) ∣ \leq ω_{i}$ for every $z$ on that trace by the definition of $ω_{i}$ in step 1.2; by [L5] the difference $\int_{γ_{i}} f d z - \int_{γ_{i}} f (γ (ξ_{i})) d z$ equals $\int_{γ_{i}} (f (z) - f (γ (ξ_{i}))) d z$ , and [L7] bounds its modulus by $ω_{i} L (γ_{i})$ .

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 $\sum_{i < r} (\int_{γ_{i}} f d z - f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i})))$ ; the finite triangle inequality, obtained from [L10] by induction ([L12]), and then [L11] with the bounds of step 2.2, give the stated estimate $\sum_{i < r} ω_{i} L (γ_{i})$ .

4.1step 1.1step 3.1L11∎

If $∣ f (u) - f (v) ∣ \leq ω$ for all $u, v \in γ^{*}$ then $ω_{i} \leq ω$ for every $i < r$ , so step 3.1 and [L11] bound the error by $ω \sum_{i < r} L (γ_{i})$ , which is $ω L (γ)$ 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.

Tagged sums approximate a contour integral within oscillation times length

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources