# capacity generated by a measure

Any finite measure^{} (http://planetmath.org/SigmaFinite) can be extended to a set function^{} on the power set^{} of the underlying space. As the following result states, this will be a Choquet capacity.

###### Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{,}\mu \mathrm{)}$ be a finite measure space. Then,

${\mu}^{*}:\mathcal{P}(X)\to {\mathbb{R}}_{+},$ | ||

${\mu}^{*}(S)=inf\{\mu (A):A\in \mathcal{F},A\supseteq S\}$ |

is an $\mathrm{F}$-capacity. Furthermore, a subset $S\mathrm{\subseteq}X$ is $\mathrm{(}\mathrm{F}\mathrm{,}{\mu}^{\mathrm{*}}\mathrm{)}$-capacitable if and only if it is in the completion^{} (http://planetmath.org/CompleteMeasure) of $\mathrm{F}$ with respect to $\mu $.

Note that, as well as being a capacity, ${\mu}^{*}$ is also an outer measure^{} (see here (http://planetmath.org/ConstructionOfOuterMeasures)), which does not require the finiteness of $\mu $.
Clearly, ${\mu}^{*}(A)=\mu (A)$ for all $A\in \mathcal{F}$, so ${\mu}^{*}$ is an extension^{} of $\mu $ to the power set of $X$, and is referred to as the *outer measure generated by $\mu $*.

Recall that a subset $S\subseteq X$ is in the completion of $\mathcal{F}$ with respect to $\mu $ if and only if there are sets $A,B\in \mathcal{F}$ with $A\subseteq S\subseteq B$ and $\mu (B\setminus A)=0$ which, by the above theorem, is equivalent^{} to the capacitability of $S$.

Title | capacity generated by a measure |
---|---|

Canonical name | CapacityGeneratedByAMeasure |

Date of creation | 2013-03-22 18:47:35 |

Last modified on | 2013-03-22 18:47:35 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 7 |

Author | gel (22282) |

Entry type | Theorem |

Classification | msc 28A05 |

Classification | msc 28A12 |

Synonym | outer measure generated by a measure |

Defines | outer measure generated by |