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The integral of a product function on a product rectangle is the product of the two integrals

Statement

Let $A\subseteq\mathbb R^p$ and $B\subseteq\mathbb R^q$ be nondegenerate closed rectangles. If $a:A\to\mathbb R$ and $b:B\to\mathbb R$ are continuous and $f(x,y):=a(x)b(y)$ , then $\int_{A\times B}f=\left(\int_Aa\right)\left(\int_Bb\right).$ In particular, if $f(x,y)=a(x)$ is independent of $y$ , then $\int_{A\times B}f=\operatorname{vol}(B)\int_Aa$ .

Facts & Assumptions

Given: Nondegenerate rectangles $A,B$ , continuous functions $a,b$ , and $f(x,y)=a(x)b(y)$ .

[L1]

Riemann--Fubini identifies the integral over a product rectangle with either iterated integral (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

[L2]

A continuous real function on a closed nondegenerate rectangle is Riemann integrable (Every continuous function on a closed nondegenerate rectangle in $\mathbb{R}^m$ is Riemann integrable).

[L3]

The multidimensional Riemann integral is linear (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $\mathbb{R}^m$ ).

Proof

technique · direct

1.1

The product $f$ is continuous and hence integrable by [L2]. For fixed $x$ , linearity [L3] gives $\int_Bf_x=a(x)\int_Bb$ .

L2L3given

2.1

Apply [L1] and [L3] once more: $\int_{A\times B}f=\int_A(a(x)\int_Bb)=(\int_Aa)(\int_Bb)$ .

L1L3step 1.1

3.1

Taking $b$ constantly equal to $1$ gives $\int_Bb=\operatorname{vol}(B)$ and yields the coordinate-independent case, including the case $\int_Aa=0$ .

step 2.1algebra∎

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Sources

J. Lebl, Basic Analysis II, Exercise 10.2.5 (standard reference, not scraped)
A. Leibman, Multidimensional Real Analysis, §5.4 (standard reference, not scraped)