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},

where $\chi$ denotes a characteristic function.

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
Canonical name	CounterexampleOfFubinisTheoremForTheLebesgueIntegral
Date of creation	2013-03-22 18:18:15
Last modified on	2013-03-22 18:18:15
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	6
Author	rmilson (146)
Entry type	Example
Classification	msc 28A35