CorollaryStatement: AI-generatedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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A Riemann integrable function on a closed bounded interval is almost everywhere equal to a Borel function

Statement

Assume the Axiom of Countable Choice. Let $a < b$ and let $f : [a, b] \to R$ be Riemann integrable. Then there is a Borel function $g : [a, b] \to R$ such that $f = g$ almost everywhere on $[a, b]$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, reals $a < b$ , and a Riemann integrable function $f : [a, b] \to R$ .

[L1]

The envelope lemma gives bounded Borel functions $φ, ψ : [a, b] \to R$ with $φ \leq f \leq ψ$ and $\int_{[a, b]} (ψ - φ) d λ_{1} = 0.$ (A bounded Riemann integrable function admits Borel Darboux envelopes with the same Lebesgue integral)

[L2]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

Proof

technique · direct

1.1

By [L1], choose bounded Borel functions $φ, ψ$ with [L1, L2] $φ \leq f \leq ψ$ and $\int_{[a, b]} (ψ - φ) d λ_{1} = 0.$ Since $ψ - φ \geq 0$ , [L2] gives $ψ = φ$ almost everywhere on $[a, b]$ .

2.1

On the same full-measure set one has [step 1.1, L1] $φ \leq f \leq ψ = φ$ , so $f = φ$ almost everywhere. Taking $g : = φ$ proves the claim, and $g$ is Borel by step 1.1. ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Richard F. Bass, Real Analysis for Graduate Students, Version 5.0, Section 9.1 (standard reference, not scraped)