# proof of Carathéodory’s lemma

A set $S\subseteq X$ is $\mu $-measurable if and only if

$$\mu (E)\ge \mu (E\cap S)+\mu (E\cap {S}^{c})$$ | (1) |

for every $E\subseteq X$.
As this inequality^{} is clearly satisfied if $S=\mathrm{\varnothing}$ and is unchanged when $S$ is replaced by ${S}^{c}$, then $\mathcal{A}$ contains the empty set^{} and is closed under taking complements^{} of sets.
To show that $\mathcal{A}$ is a $\sigma $-algebra, it only remains to show that it is closed under taking countable^{} unions of sets. Choose any sets $A,B\in \mathcal{A}$ and $E\subseteq X$. Then,

$$\begin{array}{cc}\hfill \mu (E)& \ge \mu (E\cap A)+\mu (E\cap {A}^{c})\hfill \\ & \ge \mu (E\cap A)+\mu (E\cap {A}^{c}\cap B)+\mu (E\cap {A}^{c}\cap {B}^{c})\hfill \\ & \ge \mu (E\cap (A\cup B))+\mu (E\cap {A}^{c}\cap {B}^{c})\hfill \end{array}$$ |

The first two inequalities here follow from applying (1) with $A$ and then $B$ in place of $S$, and the third uses the subadditivity of $\mu $ together with $A\cup ({A}^{c}\cap B)=A\cup B$. So (1) is satisfied with $A\cup B$ in place of $S$, showing that $\mathcal{A}$ is closed under taking pairwise unions and is therefore an algebra of sets^{} on $X$. If $A,B$ are disjoint sets in $\mathcal{A}$ then replacing $E$ by $E\cap (A\cup B)$ and $S$ by $A$ in (1) gives $\mu (E\cap (A\cup B))\ge \mu (E\cap A)+\mu (E\cap B)$. As the reverse inequality follows from subadditivity of $\mu $, this implies that

$$\mu (E\cap (A\cup B))=\mu (E\cap A)+\mu (E\cap B).$$ |

So, the map $A\mapsto \mu (E\cap A)$ is an additive set function on $\mathcal{A}$. In particular, taking $E=X$ shows that $\mu $ is additive on $\mathcal{A}$.

Now choose a sequence ${A}_{i}\in \mathcal{A}$, and set ${B}_{i}\equiv {\bigcup}_{j=1}^{i}{A}_{j}$ which are in the algebra $\mathcal{A}$. To prove that $\mathcal{A}$ is a $\sigma $-algebra it needs to be shown that $A\equiv {\bigcup}_{i}{A}_{i}={\bigcup}_{i}{B}_{i}$ is itself in $\mathcal{A}$. First, as ${B}_{i}\in \mathcal{A}$ and ${A}^{c}\subseteq {B}_{i}^{c}$,

$$\mu (E)\ge \mu (E\cap {B}_{i})+\mu (E\cap {B}_{i}^{c})\ge \mu (E\cap {B}_{i})+\mu (E\cap {A}^{c}).$$ |

As ${C}_{i}\equiv {B}_{i}\setminus {B}_{i-1}$ are pairwise disjoint sets in $\mathcal{A}$ satisfying ${\bigcup}_{j=1}^{i}{C}_{j}={B}_{i}$ the additivity of $C\mapsto \mu (E\cap C)$ on $\mathcal{A}$ gives

$$\mu (E)\ge \sum _{j=1}^{i}\mu (E\cap {C}_{j})+\mu (E\cap {A}^{c}).$$ |

So, letting $i$ increase to infinity^{}, the subadditivity of $\mu $ applied to ${\bigcup}_{j}(E\cap {C}_{j})=E\cap A$ gives

$$\mu (E)\ge \sum _{j}\mu (E\cap {C}_{j})+\mu (E\cap {A}^{c})\ge \mu (E\cap A)+\mu (E\cap {A}^{c}).$$ |

This shows that $A$ is $\mu $-measurable and so $\mathcal{A}$ is a $\sigma $-algebra.

It only remains to show that the restriction^{} of $\mu $ to $\mathcal{A}$ is a measure^{}, for which it needs to be shown that $\mu $ is countably additive on $\mathcal{A}$. So, choose any pairwise disjoint sequence ${A}_{i}\in \mathcal{A}$ and set $A={\bigcup}_{i}{A}_{i}$. The following inequality

$$\sum _{j=1}^{i}\mu ({A}_{j})=\mu \left(\bigcup _{j=1}^{i}{A}_{j}\right)\le \mu (A)\le \sum _{j}\mu ({A}_{j})$$ |

follows from the additivity of $\mu $ on $\mathcal{A}$, the requirement that $\mu $ is increasing and from the countable subadditivity of $\mu $. Letting $i$ increase to infinity gives $\mu (A)={\sum}_{j}\mu ({A}_{j})$ and $\mu $ is indeed countably additive on $\mathcal{A}$.

Title | proof of Carathéodory’s lemma |
---|---|

Canonical name | ProofOfCaratheodorysLemma |

Date of creation | 2013-03-22 18:33:25 |

Last modified on | 2013-03-22 18:33:25 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 28A12 |

Related topic | CaratheodorysLemma |

Related topic | OuterMeasure |