Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
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A continuous real function on a compact Jordan measurable set is Riemann integrable over that set

Statement

Every continuous real function on a compact Jordan measurable set E⊆Rm is Riemann integrable over E.

Facts & Assumptions

Given: Compact Jordan measurable E and continuous f:E→R.

[L3]

A bounded function on a rectangle is integrable exactly when its discontinuity set is null (Lebesgue's criterion in Rm: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null).

Proof

technique · direct
1.1

If E=∅, its zero extension is identically zero and the conclusion is immediate. Otherwise [L1] makes f bounded. Choose a bounding rectangle and form its zero extension as in The Riemann integral of a bounded function over a bounded Jordan measurable set.

L1choose
2.1

The extension is continuous at every point of the interior of E, by continuity of f, and at every point outside the closure of E, because it is locally zero. Its discontinuities are therefore contained in ∂E.

step 1.1given
3.1

The containing boundary is null by [L2], so subset closure and [L3] make the extension integrable. Bounding-rectangle independence is The Riemann integral over a Jordan set is independent of the bounding rectangle.

step 2.1L2L3∎

Depends on

Used by

Dependency tree · two levels

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Sources