LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $u,v\in\mathbb{R}$ , and let $f$ and $g$ be integrable between $u$ and $v$ . If $\eta\ge0$ and

$|f(x)-g(x)|\le\eta$

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

$\left|\int_u^v f-\int_u^v g\right|\le \eta\,|v-u|.$

Facts & Assumptions

Given: Reals $u,v$ , functions $f,g$ integrable between them, and a real $\eta\ge0$ with $|f-g|\le\eta$ 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 $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ , 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 $m\le h(x)\le M$ on $[a,b]$ has $m(b-a)\le\int_a^b h\le M(b-a)$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ ).

[L3]

For every real $z$ , $-|z|\le z\le|z|$ 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 $|h(x)|\le\eta$ , while [L3] gives $-|h(x)|\le h(x)\le|h(x)|$ ; hence $-\eta\le h(x)\le\eta$ on $[u,v]$ , and [L2] gives $-\eta(v-u)\le\int_u^v h\le\eta(v-u)$ .

step 1.2L2L3

3.1

Hence $\left|\int_u^v f-\int_u^v g\right|=\left|\int_u^v h\right|\le\eta(v-u)$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 46 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)