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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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If f is integrable on [a,b] with values in [m,M] and φ is continuous on [m,M], then φ∘f is integrable

Statement

Let a<b and m≤M be reals, let f:[a,b]→R be integrable (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf) with

m  ≤  f(x)  ≤  Mfor every x∈[a,b],

and let φ:[m,M]→R be continuous on [m,M] (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point). Then the composite φ∘f:[a,b]→R is integrable on [a,b].

The order of the hypotheses is the whole content, and it does not reverse. What is assumed is continuous after integrable: the outer function is the continuous one. Weakening the outer function to a merely integrable φ makes the statement false, and the witness is on the companion page. The remaining variant — φ merely integrable with f continuous — is neither proved nor refuted anywhere on this page, and the companion page's witness does not bear on it, its inner function being discontinuous at every rational. Nothing here asserts anything about that variant.

Facts & Assumptions

Given: Reals a<b and m≤M, an integrable f:[a,b]→R with values in [m,M], a continuous φ:[m,M]→R, and a real ε>0. Write h:=φ∘f.

[L4]

A continuous real function on a compact subset of R is bounded there (A continuous real function on a compact subset of R is bounded, Lower bound, bounded below, bounded set).

[L5]

Heine-Cantor: a continuous real function on a compact K⊆R is uniformly continuous on K, so for every real η>0 there is a real δ0>0 with ∣φ(s)−φ(t)∣<η for all s,t∈K with ∣s−t∣<δ0 (Heine-Cantor in R: a continuous real function on a compact subset of R is uniformly continuous, Uniform continuity of f:A→R: one δ serving every pair of points of A).

[L6]

Finite sums: additivity, scaling and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clauses 1, 2 and 4).

[L7]

Ordered-field arithmetic and the absolute value: multiplying an inequality by a nonnegative quantity and adding constants preserve it, the order is total and transitive, a positive real has a positive inverse, and ∣u∣≤c follows from −c≤u≤c (Ordered field, Complete ordered field (least-upper-bound property), Basic properties of the absolute value). The nonstrict forms follow from the strict ones by adjoining the case of equality.

[L8]

For every real η>0 there is a real η′>0 with η′<η, for instance η′=η⋅2−1; and the Archimedean property in reciprocal form (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean).

Proof

technique · direct
1.1

[m,M] is compact, so φ is bounded there: fix a real K≥0 with ∣φ(s)∣≤K for every s∈[m,M]. Hence ∣h(x)∣≤K for every x∈[a,b] and h is bounded.

givenL3L4choose
1.2

By [L5] applied on the compact [m,M] with η:=ε, fix a real δ0>0 with ∣φ(s)−φ(t)∣<ε whenever s,t∈[m,M] and ∣s−t∣<δ0; then put δ:=min⁡{δ0⋅2−1, ε⋅2−1}, a positive real with δ<δ0 and δ<ε.

givenL3L5L7L8choose
2.1

So ∣φ(s)−φ(t)∣≤ε whenever s,t∈[m,M] satisfy ∣s−t∣≤δ, since δ<δ0.

step 1.2L7
2.2

Since δ>0, so is δ2, and [L1] supplies a partition P=(n,t) of [a,b] with U(f,P)−L(f,P)<δ2.

step 1.2givenL1L7choose
3.1

Fix i<n and write Ωi:=Mi(f)−mi(f)≥0. If Ωi≤δ then any x,y∈Ii have ∣f(x)−f(y)∣≤Ωi≤δ with f(x),f(y)∈[m,M], so ∣h(x)−h(y)∣≤ε by step 2.1, whence Mi(h)−mi(h)≤ε by [L2].

step 2.1step 2.2L2L7
3.2

If instead Ωi>δ then Ωi/δ>1, while Mi(h)−mi(h)≤2K always, by [L2] and step 1.1.

step 1.1step 2.2L2L7
4.1

In both cases (Mi(h)−mi(h))Δi≤ε Δi+(2K/δ)ΩiΔi: in the first case the second summand is nonnegative and the first alone dominates, and in the second case (2K/δ)ΩiΔi≥2KΔi dominates by itself.

step 3.1step 3.2L7
5.1

Summing over i<n with [L6] and using ∑i<nΔi=b−a and [L2] gives U(h,P)−L(h,P)≤ε(b−a)+(2K/δ)(U(f,P)−L(f,P)).

step 4.1L2L6L7
6.1

By step 2.2 the second summand is below (2K/δ)δ2=2Kδ, and δ<ε by step 1.2, so U(h,P)−L(h,P)<ε (b−a+2K).

step 2.2step 5.1L7
7.1

Let a real η>0 be given. Running steps 1.2 to 6.1 with ε:=η/(b−a+2K+1), a positive real since b−a+2K+1>0, produces a partition P with U(h,P)−L(h,P)<η (b−a+2K)/(b−a+2K+1)<η.

step 6.1L7L8
8.1

As η>0 was arbitrary and h is bounded by step 1.1, [L1] makes h=φ∘f integrable on [a,b].

step 1.1step 7.1L1∎

Remarks

Depends on

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