Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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The Riemann integral of a compactly supported function is independent of its bounding rectangle

Statement

Let n1n\ge1 and let f:RnRf:\mathbb R^n\to\mathbb R have compact support. If ff is Riemann integrable on one closed rectangle whose interior contains its support, then it is integrable on every such rectangle, and all the resulting integrals are equal. This includes the empty-support case.

Facts & Assumptions

Given: Compactly supported ff and bounding rectangles Q1,Q2Q_1,Q_2 whose interiors contain its support.

[L1]

Extending an integrable function on a Jordan set by zero to a bounding rectangle gives a well-defined integral independent of that rectangle (The Riemann integral over a Jordan set is independent of the bounding rectangle).

[L2]

Cutting rectangles along coordinate hyperplanes preserves integrability and adds the integrals of the pieces (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm\mathbb{R}^m).

Proof

technique · common-extension
1.1

Choose a third rectangle QQ whose interior contains Q1Q2Q_1\cup Q_2. Since f=0f=0 outside its support, extending fQif|_{Q_i} by zero to QQ recovers exactly fQf|_Q.

given
2.1

If fQ1f|_{Q_1} is integrable, [L1] makes its zero extension integrable on QQ with the same integral. Restricting this function to Q2Q_2 by the coordinate cuts in [L2] gives integrability there, again with zero contribution off the support.

L1L2step 1.1
3.1

Applying [L1] to Q1Q_1 and Q2Q_2 inside the common rectangle yields Q1f=Qf=Q2f\int_{Q_1}f=\int_Qf=\int_{Q_2}f. If the support is empty, all three functions are identically zero, so the same argument gives value 00.

L1step 2.1

Depends on

Used by

Cited to discharge well-definedness by The support of a function on ℝⁿ and its compactly supported Riemann integral.

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Sources