# counter-example of Fubini’s theorem for the Lebesgue integral

The following observation demonstrates the necessity of the integrability assumption  in Fubini’s theorem. Let

 $Q=\{(x,y)\in\mathbb{R}^{2}:x\geq 0,y\geq 0\}$

denote the upper, right quadrant. Let $R\subset Q$ be the region in the quadrant bounded by the lines $y=x,y=x-1$, and let let $S\subset Q$ be a similar region, but this time bounded by the lines $y=x-1,\;y=x-2$. Let

 $f=\chi_{S}-\chi_{R},$

Observe that the Lebesgue measure  of $R$ and of $S$ is infinite   . Hence, $f$ is not a Lebesgue-integrable function. However for every $x\geq 0$ the function

 $g(x)=\int_{0}^{\infty}f(x,y)\,dy$

is integrable. Indeed,

 $g(x)=\left\{\begin{array}[]{cl}-x&\mbox{ for }0\leq x\leq 1,\\ x-2&\mbox{ for }1\leq x\leq 2,\\ 0&\mbox{ for }x\geq 2.\end{array}\right.$

Similarly, for $y\geq 0$, the function

 $h(y)=\int_{0}^{\infty}f(x,y)\,dx$

is integrable. Indeed,

 $h(y)=0,\quad y\geq 0.$

Hence, the values of the iterated integrals

 $\int_{0}^{\infty}g(x)\,dx=-1,$
 $\int_{0}^{\infty}h(y)\,dy=0,$

are finite, but do not agree. This does not contradict Fubini’s theorem because the value of the planar Lebesgue integral  $\int_{Q}f(x,y)\,d\mu(x,y),$

where $\mu(x,y)$ is the planar Lebesgue measure, is not defined.

Title counter-example of Fubini’s theorem for the Lebesgue integral CounterexampleOfFubinisTheoremForTheLebesgueIntegral 2013-03-22 18:18:15 2013-03-22 18:18:15 rmilson (146) rmilson (146) 6 rmilson (146) Example msc 28A35