# 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

$$f(x,y)=\{\begin{array}{cc}\hfill 1\hfill & \text{if}x=y,\hfill \\ \hfill 0\hfill & \text{otherwise}.\hfill \end{array}$$ |

Observe that

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

while

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

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 |