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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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

1.1

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

L1algebra

1.2

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

2.1

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 $- η (v - u) \leq \int_{u}^{v} h \leq η (v - u)$ .

step 1.2L2L3

3.1

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

step 1.2step 2.1L3

4.1

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

step 3.1L1L3

5.1

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

step 1.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

22 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

MIT OpenCourseWare 18.100B, Real Analysis, Lectures 20–21 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)