# proof of capacity generated by a measure

For a finite measure space $(X,\mathcal{F},\mu )$, define

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

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

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

Note, first of all, that ${\mu}^{*}(S)=\mu (S)$ for any $S\in \mathcal{F}$.
That ${\mu}^{*}$ is increasing follows directly from the definition. If ${A}_{n}\in \mathcal{F}$ is a decreasing sequence of sets then $A={\bigcap}_{n}{A}_{n}$ is also in $\mathcal{F}$ and, by continuity from above (http://planetmath.org/PropertiesForMeasure) for measures^{},

$${\mu}^{*}({A}_{n})=\mu ({A}_{n})\to \mu (A)={\mu}^{*}(A)$$ |

as $n\to \mathrm{\infty}$.

Now suppose that ${S}_{n}$ is an increasing sequence of subsets of $X$ and set $S={\bigcup}_{n}{S}_{n}$. Then, ${\mu}^{*}({S}_{n})\le {\mu}^{*}(S)$ for each $n$ and, hence, ${lim}_{n\to \mathrm{\infty}}{\mu}^{*}({S}_{n})\le {\mu}^{*}(S)$.

To prove the reverse inequality^{}, choose any $\u03f5>0$ and sequence^{} ${A}_{n}\in \mathcal{F}$ with ${S}_{n}\subseteq {A}_{n}$ and $\mu ({A}_{n})\le {\mu}^{*}({S}_{n})+{2}^{-n}\u03f5$.
Then, ${A}_{m}\cap {A}_{n}\supseteq {S}_{n}$ whenever $m\ge n$ and, therefore,

$$\mu ({A}_{n}\setminus {A}_{m})=\mu ({A}_{n})-\mu ({A}_{m}\cap {A}_{n})\le \mu ({A}_{n})-{\mu}^{*}({S}_{n})\le {2}^{-n}\u03f5.$$ |

Additivity of $\mu $ then gives

$$\mu \left(\bigcup _{m\le n}{A}_{m}\right)\le \mu ({A}_{n})+\sum _{m=1}^{n-1}\mu ({A}_{m}\setminus {A}_{n})\le {\mu}^{*}({S}_{n})+\u03f5.$$ |

So, by continuity from below for measures,

$${\mu}^{*}(S)\le \mu \left(\bigcup _{n}{A}_{n}\right)=\underset{n\to \mathrm{\infty}}{lim}\mu \left(\bigcup _{m\le n}{A}_{m}\right)\le \underset{n\to \mathrm{\infty}}{lim}{\mu}^{*}({S}_{n})+\u03f5.$$ |

Choosing $\u03f5$ arbitrarily small shows that ${\mu}^{*}({S}_{n})\to {\mu}^{*}(S)$ and, therefore, ${\mu}^{*}$ is indeed an $\mathcal{F}$-capacity.

Now suppose that $S$ is in the completion of $\mathcal{F}$ with respect to $\mu $, so that there exists $A,B\in \mathcal{F}$ with $A\subseteq S\subseteq B$ and $\mu (B\setminus A)=0$. Then,

$${\mu}^{*}(A)=\mu (A)=\mu (B)\ge {\mu}^{*}(S)$$ |

and $S$ is indeed $(\mathcal{F},{\mu}^{*})$-capacitable. Conversely, let $S$ be $(\mathcal{F},{\mu}^{*})$-capacitable. Then, there exists ${A}_{n},{B}_{n}\in \mathcal{F}$ such that ${A}_{n}\subseteq S\subseteq {B}_{n}$ and

$$\mu ({A}_{n})\ge {\mu}^{*}(S)-1/n,\mu ({B}_{n})\le {\mu}^{*}(S)+1/n.$$ |

Setting $A={\bigcup}_{n}{A}_{n}$ and $B={\bigcap}_{n}{B}_{n}$ gives $A\subseteq S\subseteq B$ and

$$\mu (A)\ge {\mu}^{*}(S)\ge \mu (B).$$ |

So $\mu (B\setminus A)=\mu (B)-\mu (A)=0$, as required.

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

Canonical name | ProofOfCapacityGeneratedByAMeasure |

Date of creation | 2013-03-22 18:47:55 |

Last modified on | 2013-03-22 18:47:55 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 28A12 |

Classification | msc 28A05 |