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 measureand $Y$ 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 thefact 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 $L^1$ integrablerelative to the product measure. However, as we observed above, theiterated integrals do not agree. This observation illustrates the need for the$\sigma$-finite assumption for Fubini’s theorem.