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The Riemann integral of a compactly supported function is independent of its bounding rectangle
Statement
Let and let have compact support. If 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 and bounding rectangles whose interiors contain its support.
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).
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 ).
Proof
Choose a third rectangle whose interior contains . Since outside its support, extending by zero to recovers exactly .
If is integrable, [L1] makes its zero extension integrable on with the same integral. Restricting this function to by the coordinate cuts in [L2] gives integrability there, again with zero contribution off the support.
Applying [L1] to and inside the common rectangle yields . If the support is empty, all three functions are identically zero, so the same argument gives value .
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
- A. Leibman, Multidimensional Real Analysis, §5.5 (standard reference, not scraped)