# 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

 $f(x,y)=\left\{\begin{array}[]{cl}1&\mbox{ if }x=y,\\ 0&\mbox{ otherwise}.\end{array}\right.$

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 CounterexampleToTonellisTheorem 2013-03-22 18:16:36 2013-03-22 18:16:36 rmilson (146) rmilson (146) 4 rmilson (146) Example msc 28A35