LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error

Statement

Let $u, v \in R$ , and let $f$ and $g$ be integrable between $u$ and $v$ . If $η \geq 0$ and

$∣ f (x) - g (x) ∣ \leq η$

throughout the closed interval with endpoints $u$ and $v$ , then

$∣ \int_{u}^{v} f - \int_{u}^{v} g ∣ \leq η ∣ v - u ∣ .$

Facts & Assumptions

Given: Reals $u, v$ , functions $f, g$ integrable between them, and a real $η \geq 0$ with $∣ f - g ∣ \leq η$ on the interval between them.

[L1]

Linear combinations of integrable functions are integrable and their integrals are the corresponding linear combinations, including for oriented limits (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L2]

If $a < b$ , an integrable function $h$ satisfying $m \leq h (x) \leq M$ on $[a, b]$ has $m (b - a) \leq \int_{a}^{b} h \leq M (b - a)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L3]

For every real $z$ , $- ∣ z ∣ \leq z \leq ∣ z ∣$ and $∣ - z ∣ = ∣ z ∣$ ; for $c > 0$ , $∣ z ∣ < c$ exactly when $- c < z < c$ (Basic properties of the absolute value).

Proof

technique · direct

1.1L1algebra

If $u = v$ , both oriented integrals are $0$ and the asserted inequality holds.

1.2L1

Suppose $u < v$ and put $h : = f - g$ . Then $h$ is integrable and $\int_{u}^{v} h = \int_{u}^{v} f - \int_{u}^{v} g$ .

2.1step 1.2L2L3

The hypothesis gives $∣ h (x) ∣ \leq η$ , while [L3] gives $- ∣ h (x) ∣ \leq h (x) \leq ∣ h (x) ∣$ ; hence $- η \leq h (x) \leq η$ on $[u, v]$ , and [L2] gives $- η (v - u) \leq \int_{u}^{v} h \leq η (v - u)$ .

3.1step 1.2step 2.1L3

Hence $∣ \int_{u}^{v} f - \int_{u}^{v} g ∣ = ∣ \int_{u}^{v} h ∣ \leq η (v - u)$ when $u < v$ .

4.1step 3.1L1L3

If $u > v$ , apply step 3.1 to the ordered pair $(v, u)$ and use antisymmetry of oriented integrals; the same bound results because $∣ u - v ∣ = ∣ v - u ∣$ .

5.1step 1.1step 3.1step 4.1∎

The alternatives $u = v$ , $u < v$ , and $u > v$ are exhaustive, and steps 1.1, 3.1, and 4.1 give the claimed inequality.

Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources