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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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Arzela's bounded convergence theorem for Riemann integrals

Statement

Assume the Axiom of Countable Choice. Let a<b, let fn:[a,b]→R be Riemann integrable for every n, and let f:[a,b]→R be Riemann integrable. Suppose that fn(x)→f(x) for every x∈[a,b] and that there is a real M≥0 with ∣fn(x)∣≤M for all n and all x∈[a,b]. Then ∫abfn(x) dx⟶∫abf(x) dx.

Facts & Assumptions

Given: The Axiom of Countable Choice, reals a<b, Riemann integrable functions f,fn:[a,b]→R with fn(x)→f(x) for every x∈[a,b], and a real M≥0 with ∣fn(x)∣≤M for all n and x.

[L1]

A bounded Riemann integrable function on [a,b] is Lebesgue measurable, Lebesgue integrable, and has the same Lebesgue and Riemann integrals. (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral)

[L2]

On a finite measure space, almost-everywhere pointwise convergence of a uniformly bounded measurable sequence implies convergence of the integrals. (Bounded convergence on a finite measure space)

Proof

technique · direct
1.1L1L3

By [L1], each fn and f is Lebesgue measurable and integrable on [a,b], and ∫[a,b]fn dλ1=∫abfn(x) dx,∫[a,b]f dλ1=∫abf(x) dx. Also [L3] makes ([a,b],L([a,b]),λ1) a finite measure space.

2.1step 1.1L2∎

The convergence hypothesis is pointwise, hence almost everywhere, and the uniform bound ∣fn∣≤M holds everywhere. So [L2] applies on [a,b] and gives ∫[a,b]fn dλ1⟶∫[a,b]f dλ1. Translating the two sides with step 1.1 yields the stated convergence of the Riemann integrals.

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