Double and Iterated Integrals

Let $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ be two $\sigma$ -finite measure spaces, and let $\lambda=\mu\times\nu$ be the product measure on $X\times Y$ .

If $h$ is a function on $X\times Y$ so that its integral is defined — either $h$ is integrable, or its integral diverges definitely — then we write it as any of

$\displaystyle\int h\,d\lambda=\int h\,d(\mu\times\nu)=\int h(x,y)\,d\lambda(x,y)=\int h(x,y)\,d(\mu\times\nu)(x,y)$

and call it the “double integral” of $h$ over $X\times Y$ . We can also consider the sections $h_x(y)=h(x,y)$ and $h^y(x)=h(x,y)$ . For any given $x\in X$ , we set

$\displaystyle\int h_x(y)\,d\nu(y)=f(x)$

if the integral exists. If the resulting function $f$ is integrable, then we write

$\displaystyle\int f\,d\mu=\iint h(x,y)\,d\nu(y)\,d\mu(x)=\iint h\,d\nu\,d\mu=\int\,d\mu(x)\int h(x,y)\,d\nu(y)$

The latter notation, with the measure $\mu$ before the integrand is less common, but it can be seen in older texts. I’ll usually stick to the other order.

Similarly, we can define the function $g(y)$ as the integral of the $Y$ -section $h^y$ if it exists. If $g$ is integrable, we write

$\displaystyle\int g\,d\nu=\iint h(x,y)\,d\mu(x)\,d\nu(y)=\iint h\,d\mu\,d\nu=\int\,d\nu(y)\int h(x,y)\,d\mu(x)$

where, again, the latter notation is deprecated. These integrals are called the “iterated integrals” of $h$ . We can also define double and iterated integrals over a measurable subset $E\subseteq X\times Y$ , as usual, and write

$\displaystyle\begin{aligned}\int\limits_Eh&\,d\lambda\\\int\limits_Eh&\,d(\mu\times\nu)\\\iint\limits_Eh&\,d\mu\,d\nu\\\iint\limits_Eh&\,d\nu\,d\mu\end{aligned}$

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July 27, 2010 - Posted by John Armstrong | Analysis, Measure Theory

2 Comments »

[...] if is a non-negative measurable function on , then we have the following equalities between the double integral and the two iterated [...]

Pingback by Fubini’s Theorem « The Unapologetic Mathematician | July 28, 2010 | Reply
[...] can turn this into an iterated integral, which Fubini’s theorem tells us we can evaluate in any order we [...]

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