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

Statement

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

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 $R^{m}$ 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 $R^{m}$ ).

Proof

technique · direct

1.1L2L3given

The product $f$ is continuous and hence integrable by [L2]. For fixed $x$ , linearity [L3] gives $\int_{B} f_{x} = a (x) \int_{B} b$ .

2.1L1L3step 1.1

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

3.1step 2.1algebra∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources