CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)

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A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $a < b$ be reals and let $f : [a, b] \to R$ be bounded, with discontinuity set $D$ . Then $f$ is Riemann integrable on $[a, b]$ if and only if $D$ is Lebesgue measurable with $λ_{1} (D) = 0$ .
Let $m \geq 1$ and let $f$ be a bounded real function on a closed nondegenerate rectangle in $R^{m}$ , with discontinuity set $D$ . Then $f$ is Riemann integrable if and only if $D$ is Lebesgue measurable with $λ_{m} (D) = 0$ .

The choice premise here belongs to measure translation. Both implications of Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero are now proved in ZF. The translation between elementary interval-cover nullity and Lebesgue-measure nullity used here rests on the construction of $λ$ and requires countable choice in the cited suppliers, so the statement above retains that hypothesis throughout.

Facts & Assumptions

Given: The Axiom of Countable Choice, a bounded real function on a closed bounded interval or on a closed nondegenerate rectangle, and its discontinuity set $D$ .

[L1]

Assuming countable choice, $λ_{1}^{*} (A) = 0$ if and only if $A \subseteq R$ has measure zero in the covering sense, and a set of Lebesgue outer measure zero is measurable of measure zero (A subset of $R$ has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[L2]

Assuming countable choice, $λ_{m}^{*} (E) = 0$ if and only if $E \subseteq R^{m}$ is null in the covering sense (A subset of $R^{m}$ has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers).

[F1]

Let $a < b$ be reals, let $f : [a, b] \to R$ be bounded and let $D$ be its set of discontinuities; then $f$ is Riemann integrable on $[a, b]$ if and only if $D$ has measure zero (Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[F2]

A bounded real function on a closed nondegenerate rectangle in $R^{m}$ , $m \geq 1$ , is Riemann integrable if and only if its discontinuity set is null (Lebesgue's criterion in $R^{m}$ : a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null, Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F3]

A measurable set $N \in A$ is $μ$ -null if $μ (N) = 0$ (Measure-null sets and almost-everywhere statements relative to a measure).

Proof

technique · direct

1.1L1F3

On the line, " $D$ has measure zero" in the covering sense of the cited criterion is equivalent to $λ_{1}^{*} (D) = 0$ , and a set of Lebesgue outer measure zero is Lebesgue measurable of measure zero, while conversely $λ_{1} (D) = 0$ for a measurable $D$ says $λ_{1}^{*} (D) = 0$ .

1.2L1L2F3

In $R^{m}$ , "the discontinuity set is null" in the covering sense of the cited criterion is likewise equivalent to $λ_{m}^{*} (D) = 0$ , hence to $D$ being Lebesgue measurable with $λ_{m} (D) = 0$ .

2.1step 1.1step 1.2F1F2∎

Substituting these equivalences into the two published criteria gives claims 1 and 2.

Depends on

Used by

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Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Section 1.2 (standard reference, not scraped)