# The proof of theorem is wrong

Let’s create a very simple measurable space^{}: $X=\{a,b\}$, $\mathcal{A}=\{\mathrm{\varnothing},\{a\},\{b\},X\}$.

Let’s take the $\pi $-system $P=\{\{a\}\}$ containing only one subset of $X$.

Let’s create two measures^{} $\mu ={\delta}_{a}+{\delta}_{b}$ and $\nu ={\delta}_{a}+2{\delta}_{b}$. Then obviously $\mu $ and $\nu $ agree on $P$ and are finite, but they obviously are not equal on $\mathcal{A}$.

The proof, however, claims that it is sufficient if $\mu $ and $\nu $ are finite. I believe that $\mu (X)=\nu (X)$ is a necessary condition.

Title | The proof of theorem is wrong |
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Canonical name | TheProofOfTheoremIsWrong |

Date of creation | 2013-03-22 19:16:05 |

Last modified on | 2013-03-22 19:16:05 |

Owner | tomprimozic (26284) |

Last modified by | tomprimozic (26284) |

Numerical id | 4 |

Author | tomprimozic (26284) |

Entry type | Example |

Classification | msc 28A12 |