# proof of Carathéodory’s extension theorem

The first step is to extend the set function^{} ${\mu}_{0}$ to the power set^{} $P(X)$. For any subset $S\subseteq X$ the value of ${\mu}^{*}(S)$ is defined by taking sequences ${S}_{i}$ in $A$ which cover $S$,

$${\mu}^{*}(S)\equiv inf\{\sum _{i=1}^{\mathrm{\infty}}{\mu}_{0}({S}_{i}):{S}_{i}\in A,S\subseteq \bigcup _{i=1}^{\mathrm{\infty}}{S}_{i}\}.$$ | (1) |

We show that this is an outer measure^{} (http://planetmath.org/OuterMeasure2). First, it is clearly non-negative. Secondly, if $S=\mathrm{\varnothing}$ then we can take ${S}_{i}=\mathrm{\varnothing}$ in (1) to obtain ${\mu}^{*}(S)\le {\sum}_{i}{\mu}_{0}(\mathrm{\varnothing})=0$, giving ${\mu}^{*}(\mathrm{\varnothing})$=0. It is also clear that ${\mu}^{*}$ is increasing, so that if $S\subseteq T$ then ${\mu}^{*}(S)\le {\mu}^{*}(T)$. The only remaining property to be proven is subadditivity. That is, if ${S}_{i}$ is a sequence in $P(X)$ then

$${\mu}^{*}\left(\bigcup _{i}{S}_{i}\right)\le \sum _{i}{\mu}^{*}({S}_{i}).$$ | (2) |

To prove this inequality^{}, choose any $\u03f5>0$ and, by the definition (1) of ${\mu}^{*}$, for each $i$ there exists a sequence ${S}_{i,j}\in A$ such that ${S}_{i}\subseteq {\bigcup}_{j}{S}_{i,j}$ and,

$$\sum _{j=1}^{\mathrm{\infty}}{\mu}_{0}({S}_{i,j})\le {\mu}^{*}({S}_{i})+{2}^{-i}\u03f5.$$ |

As ${\bigcup}_{i}{S}_{i}\subseteq {\bigcup}_{i,j}{S}_{i,j}$, equation (1) defining ${\mu}^{*}$ gives

$${\mu}^{*}\left(\bigcup _{i}{S}_{i}\right)\le \sum _{i,j}{\mu}_{0}({S}_{i,j})=\sum _{i}\sum _{j}{\mu}_{0}({S}_{i,j})\le \sum _{i}({\mu}^{*}({S}_{i})+{2}^{-i}\u03f5)=\sum _{i}{\mu}^{*}({S}_{i})+\u03f5.$$ |

As $\u03f5>0$ is arbitrary, this proves subadditivity (2). So, ${\mu}^{*}$ is indeed an outer measure.

The next step is to show that ${\mu}^{*}$ agrees with ${\mu}_{0}$ on $A$. So, choose any $S\in A$. The inequality ${\mu}^{*}(S)\le {\mu}_{0}(S)$ follows from taking ${S}_{1}=S$ and ${S}_{i}=\mathrm{\varnothing}$ in (1), and it remains to prove the reverse inequality. So, let ${S}_{i}$ be a sequence in $A$ covering $S$, and set

$${S}_{i}^{\prime}=(S\cap {S}_{i})\setminus \bigcup _{j=1}^{i-1}{S}_{j}\in A.$$ |

Then, ${S}_{i}^{\prime}$ are disjoint sets satisfying ${\bigcup}_{j=1}^{i}{S}_{j}^{\prime}=S\cap {\bigcup}_{j=1}^{i}{S}_{j}$ and, therefore, ${\bigcup}_{i}{S}_{i}^{\prime}=S$. By the countable additivity^{} of ${\mu}_{0}$,

$$\sum _{i}{\mu}_{0}({S}_{i})=\sum _{i}({\mu}_{0}({S}_{i}^{\prime})+{\mu}_{0}({S}_{i}\setminus {S}_{i}^{\prime}))\ge \sum _{i}{\mu}_{0}({S}_{i}^{\prime})={\mu}_{0}(S).$$ |

As this inequality hold for any sequence ${S}_{i}\in A$ covering $S$, equation (1) gives ${\mu}^{*}(S)\ge {\mu}_{0}(S)$ and, by combining with the reverse inequality, shows that ${\mu}^{*}$ does indeed agree with ${\mu}_{0}$ on $A$.

We have shown that ${\mu}_{0}$ extends to an outer measure ${\mu}^{*}$ on the power set of $X$. The final step is to apply Carathéodory’s lemma on the restriction^{} of outer measures. A set $S\subseteq X$ is said to be ${\mu}^{*}$-measurable if the inequality

$${\mu}^{*}(E)\ge {\mu}^{*}(E\cap S)+{\mu}^{*}(E\cap {S}^{c})$$ | (3) |

is satisfied for all subsets $E$ of $X$. Carathéodory’s lemma then states that the collection^{} $\mathcal{F}$ of ${\mu}^{*}$-measurable sets^{} is a $\sigma $-algebra^{} (http://planetmath.org/SigmaAlgebra) and that the restriction of ${\mu}^{*}$ to $\mathcal{F}$ is a measure^{}.
To complete^{} the proof of the theorem it only remains to be shown that every set in $A$ is ${\mu}^{*}$-measurable, as it will then follow that $\mathcal{F}$ contains $\mathcal{A}=\sigma (A)$ and the restriction of ${\mu}^{*}$ to $\mathcal{A}$ is a measure.

So, choosing any $S\in A$ and $E\subseteq X$, the proof will be complete once it is shown that (3) is satisfied. Given any $\u03f5>0$, equation (1) says that there is a sequence ${E}_{i}$ in $A$ such that $E\subseteq {\bigcup}_{i}{E}_{i}$ and

$$\sum _{i}{\mu}_{0}({E}_{i})\le {\mu}^{*}(E)+\u03f5.$$ |

As $E\cap S\subseteq {\bigcup}_{i}({E}_{i}\cap S)$ and $E\cap {S}^{c}\subseteq {\bigcup}_{i}({E}_{i}\cap {S}^{c})$,

$${\mu}^{*}(E\cap S)+{\mu}^{*}(E\cap {S}^{c})\le \sum _{i}{\mu}_{0}({E}_{i}\cap S)+\sum _{i}{\mu}_{0}({E}_{i}\cap {S}^{c})=\sum _{i}{\mu}_{0}({E}_{i})\le {\mu}^{*}(E)+\u03f5.$$ |

Since $\u03f5$ is arbitrary, this shows that (3) is satisfied and $S$ is ${\mu}^{*}$-measurable.

Title | proof of Carathéodory’s extension theorem |
---|---|

Canonical name | ProofOfCaratheodorysExtensionTheorem |

Date of creation | 2013-03-22 18:33:28 |

Last modified on | 2013-03-22 18:33:28 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 4 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 28A12 |

Related topic | CaratheodorysLemma |

Related topic | Measure |

Related topic | OuterMeasure2 |