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An integrable function whose sections vanish outside finite sets has multiple integral zero

Statement

Let $A\subseteq\mathbb R^p$ and $B\subseteq\mathbb R^q$ be nondegenerate closed rectangles, and let $f:A\times B\to\mathbb R$ be Riemann integrable. If the set $S:=\{x\in A:f_x\text{ is not identically }0\}$ is finite, then $\int_{A\times B}f=0$ . The analogous assertion holds with the coordinate blocks exchanged.

Facts & Assumptions

Given: An integrable $f:A\times B\to\mathbb R$ whose nonzero $B$ -sections are indexed by a finite set $S$ .

[L1]

Riemann--Fubini permits a content-zero exceptional set of parameters and identifies the multiple integral with the resulting iterated integral (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

[L2]

A set has content zero when it admits finite cube covers of arbitrarily small total volume (Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers).

Proof

technique · direct

1.1

A finite subset of $\mathbb R^p$ has content zero by [L2]: for a given $\varepsilon>0$ , cover its finitely many points by cubes whose total volume is below $\varepsilon$ .

L2given

2.1

Outside $S$ every section is identically zero and has integral zero. Complete the section-integral function by the value $0$ on $S$ and apply [L1]; the resulting outer function is identically zero, so the multiple integral is zero.

L1step 1.1

3.1

If $S$ is empty then $f$ itself is identically zero, and step 2.1 still applies. Exchanging the coordinate blocks proves the symmetric assertion.

step 2.1algebra∎

Depends on

Used by

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Sources

J. Lebl, Basic Analysis II, Exercise 10.2.4 (standard reference, not scraped)