# 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

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

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
$x\ge 0$ the function

$$g(x)={\int}_{0}^{\mathrm{\infty}}f(x,y)\mathit{d}y$$ |

is integrable. Indeed,

$$g(x)=\{\begin{array}{cc}\hfill -x\hfill & \text{for}0\le x\le 1,\hfill \\ \hfill x-2\hfill & \text{for}1\le x\le 2,\hfill \\ \hfill 0\hfill & \text{for}x\ge 2.\hfill \end{array}$$ |

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

$$h(y)={\int}_{0}^{\mathrm{\infty}}f(x,y)\mathit{d}x$$ |

is integrable. Indeed,

$$h(y)=0,y\ge 0.$$ |

Hence, the values of the iterated integrals

$${\int}_{0}^{\mathrm{\infty}}g(x)\mathit{d}x=-1,$$ |

$${\int}_{0}^{\mathrm{\infty}}h(y)\mathit{d}y=0,$$ |

are finite, but do not agree. This does not contradict Fubini’s
theorem because the value of the planar Lebesgue integral^{}

$${\int}_{Q}f(x,y)\mathit{d}\mu (x,y),$$ |

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 |