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The Riemann integral of a compactly supported function is independent of its bounding rectangle

Statement

Let $n \geq 1$ and let $f : R^{n} \to R$ have compact support. If $f$ 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 $f$ and bounding rectangles $Q_{1}, Q_{2}$ 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 $R^{m}$ ).

Proof

technique · common-extension

1.1given

Choose a third rectangle $Q$ whose interior contains $Q_{1} \cup Q_{2}$ . Since $f = 0$ outside its support, extending $f ∣_{Q_{i}}$ by zero to $Q$ recovers exactly $f ∣_{Q}$ .

2.1L1L2step 1.1

If $f ∣_{Q_{1}}$ is integrable, [L1] makes its zero extension integrable on $Q$ with the same integral. Restricting this function to $Q_{2}$ by the coordinate cuts in [L2] gives integrability there, again with zero contribution off the support.

3.1L1step 2.1∎

Applying [L1] to $Q_{1}$ and $Q_{2}$ inside the common rectangle yields $\int_{Q_{1}} f = \int_{Q} f = \int_{Q_{2}} f$ . If the support is empty, all three functions are identically zero, so the same argument gives value $0$ .

Depends on

Dependency tree · 2 levels

Sources

A. Leibman, Multidimensional Real Analysis, §5.5