# completion of a measure space

If the measure space $(X,\mathcal{S},\mu )$ is not complete^{}, then it can be completed in the following way. Let

$$\mathcal{Z}=\bigcup _{E\in \mathcal{S},\mu (E)=0}\mathcal{P}(E),$$ |

i.e. the family of all subsets of sets whose $\mu $-measure is zero. Define

$$\overline{\mathcal{S}}=\{A\cup B:A\in \mathcal{S},B\in \mathcal{Z}\}.$$ |

We assert that $\overline{\mathcal{S}}$ is a $\sigma $-algebra. In fact, it clearly contains the emptyset, and it is closed under countable^{} unions because both $\mathcal{S}$ and $\mathcal{Z}$ are. We thus need to show that it is closed under complements. Let $A\in \mathcal{S}$, $B\in \mathcal{Z}$ and suppose $E\in \mathcal{S}$ is such that $B\subset E$ and $\mu (E)=0$.
Then we have

$${(A\cup B)}^{c}={A}^{c}\cap {B}^{c}={A}^{c}\cap {(E-(E-B))}^{c}={A}^{c}\cap ({E}^{c}\cup (E-B))=({A}^{c}\cap {E}^{c})\cup ({A}^{c}\cap (E-B)),$$ |

where ${A}^{c}\cap {E}^{c}\in \mathcal{S}$ and ${A}^{c}\cap (E-B)\in \mathcal{Z}$. Hence ${(A\cup B)}^{c}\in \overline{\mathcal{S}}$.

Now we define $\overline{\mu}$ on $\overline{\mathcal{S}}$ by $\overline{\mu}(A\cup B)=\mu (A)$, whenever $A\in \mathcal{S}$ and $B\in \mathcal{Z}$. It is easily verified that this defines in fact a measure, and that $(X,\overline{\mathcal{S}},\overline{\mu})$ is the completion of $(X,\mathcal{S},\mu )$.

Title | completion of a measure space |
---|---|

Canonical name | CompletionOfAMeasureSpace |

Date of creation | 2013-03-22 14:06:59 |

Last modified on | 2013-03-22 14:06:59 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 9 |

Author | Koro (127) |

Entry type | Derivation^{} |

Classification | msc 28A12 |