Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 ERmE\subseteq\mathbb R^m is Riemann integrable over EE.

Facts & Assumptions

Given: Compact Jordan measurable EE and continuous f:ERf:E\to\mathbb R.

Proof

technique · direct
1.1

If E=E=\varnothing, its zero extension is identically zero and the conclusion is immediate. Otherwise [L1] makes ff 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 EE, by continuity of ff, and at every point outside the closure of EE, because it is locally zero. Its discontinuities are therefore contained in E\partial 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 140 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources