counter-example to Tonelli’s theorem

The following observation demonstrates the necessity of the $\sigma$-finite assumption in Tonelli’s and Fubini’s theorem. Let $X$ denote the closed unit interval $[0,1]$ equipped with Lebesgue measure and $Y$ the same set, but this time equipped with counting measure $\nu$. Let

Observe that

$${\int }_Y\left({\int }_{X}f(x,y)𝑑\mu (x)\right)𝑑\nu (y)=0,$$

while

$${\int }_X\left({\int }_{Y}f(x,y)𝑑\nu (y)\right)𝑑\mu (x)=1.$$

The iterated integrals do not give the same value, this despite the fact that the integrand is a non-negative function.

Also observe that there does not exist a simple function on $X\times Y$ that is dominated by $f$. Hence,

$${\int }_{X\times Y}f(x,y)d(\mu (x)\times \nu (y)=0.$$

Therefore, the integrand is $L^1$ integrable relative to the product measure. However, as we observed above, the iterated integrals do not agree. This observation illustrates the need for the $\sigma$-finite assumption for Fubini’s theorem.

Title	counter-example to Tonelli’s theorem
Canonical name	CounterexampleToTonellisTheorem
Date of creation	2013-03-22 18:16:36
Last modified on	2013-03-22 18:16:36
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	4
Author	rmilson (146)
Entry type	Example
Classification	msc 28A35