If \(\mu_1\) and \(\mu_2\) are finite measures on a measurable space \((\Omega, \mathcal{F})\) with \(\mu_1(\Omega)=\mu_2(\Omega)\), then the class \(\mathcal{L}:= \{A\in\mathcal{F}: \mu_1(A)=\mu_2(A)\}\) is a \(\lambda\)-system.
증명은 간단하다.
주어진 조건 \(\mu_1(\Omega)=\mu_2(\Omega)\)로 인하여 \(\Omega \in \mathcal{L}\)은 trivial하다.
또한 \(A\in \mathcal{F}\)에 대하여 \(\mu_1(A^c)=\mu_1(\Omega)-\mu_1(A)=\mu_2(\Omega)-\mu_2(A)=\mu_2(A^c)\)를 얻을 수 있다.
그리고 \(A_1,A_2,\ldots\in\mathcal{F}\), \(A_m\cap A_n =\phi, m\ne n\)에 대해
\(\mu_1(\uplus_nA_n)=\sum_n \mu_1(A_n)=\sum_n \mu_2(A_n)=\mu_2(\uplus_nA_n)\)이다.
때문에 \(\uplus_nA_n \in \mathcal{F}\) 이다.
Suppose \(\mu_1\) and \(\mu_2\) are finite measures on \(\sigma(\mathcal{P})\), where \(\mathcal{P}\) is a \(\pi\)-system with \(\Omega\in \mathcal{P}\). If \(\mu_1\) and \(\mu_2\) agree on \(\mathcal{P}\), then they agree on \(\sigma(\mathcal{P})\).
(만약 두 measure가 finite measure일 때 주어진 field에서 agree하다면, \(\sigma\)-field에서도 agree한다. 즉 Unique함을 의미한다.)
Suppose \(\mu_1\) and \(\mu_2\) are finite measures on \(\sigma(\mathcal{P})\), where \(\mathcal{P}\) is a \(\pi\)-system and \(\Omega\) can be written as a finite or countably infinite union of sets in \(\mathcal{P}\). If \(\mu_1\) and \(\mu_2\) agree on \(\mathcal{P}\), then they agree on \(\sigma(\mathcal{P})\).
An outer measure is a set function \(\mu^*\) defined on \(2^\Omega\) and satisfying
\(0\le \mu^*(A)\le \infty\) for all \(A\subset\Omega\);
\(\mu^*(\phi)=0\) ;
Monotonicity: \(A\subset B\) implies \(\mu^*(A)\le \mu^*(B)\).
Countable subadditivity: for all \(A_1, A_2, \ldots \subset \Omega\), \(\mu^*(\cup_n A_n)\le \sum_n \mu^*(A_n)\).
Let \(\mu\) be a set function on a class \(\mathcal{A}\) of subsets of \(\Omega\), where
\(\phi\in \mathcal{A}\); (field의 조건.)
\(\mu(\phi)=0\); (measure의 조건)
\(0 \le \mu(A) \le \infty\) for all \(A\in \mathcal{A}\). (measure의 조건)
If \(\mu^*\) is defined by \[ \mu^*(A)=\inf\left\{ \sum_n \mu(A_n): A_1,A_2,\ldots\in \mathcal{A}, A\subset \cup_n A_n \right\}, A\subset \Omega, \] then \(\mu^*\) is an outer measure.
If \(\mu^*\) is an outer measure, then a set \(A\subset \Omega\) is \(\mu^*\)-measurable if
\[ \mu^*(E)= \mu^*(A\cap E)+ \mu^*(A^c \cap E) \mbox{ for all }E\subset \Omega, \] and let \(M(\mu^*)\) denote the class of \(\mu^*\)-measurable sets.
\(M(\mu^*)\) is a \(\sigma\)-field, and \(\mu^*\) restricted to \(M(\mu^*)\) is a measure.
첫번째로, \(M(\mu^*)\)는 field임을 간단히 증명할 수 있다. \(\mu^*\)는 symmetric하기 때문에 당연히 \(\Omega\)를 포함하고, closed under complementation이다. Closedness under finite intersection을 보이려면 \(A, B\in M(\mu^*)\)에 대하여 \[ \mu^*(E)=\mu^*((A\cap B ) \cap E)+ \mu^*((A\cap B )^c \cap E) \] 를 보이면 된다. \(\mu^*(E)\le \mu^*((A\cap B ) \cap E)+ \mu^*((A\cap B )^c \cap E)\)는 countable subadditivity에 의해 trivial 하고 때문에 \(\mu^*(E)\ge \mu^*((A\cap B ) \cap E)+ \mu^*((A\cap B )^c \cap E)\)만 보이면 된다. (Exercise) 이미 \(M(\mu^*)\)가 field임을 보였기 때문에 \(M(\mu^*)\)가 countable disjoint union에 의해 닫혀있다면 \(M(\mu^*)\)는 \(\sigma\)-field임을 증명할 수 있다. (Exercise)
\(\mu^*\)는 outer measure이기 때문에 공집합을 포함하고 nonnegative하다. 때문에 countable additivity만 증명하면 \(\mu^*\) restricted to \(M(\mu^*)\) 는 measure이다. (Exercise)
A class \(\mathcal{A}\) of subsets of \(\Omega\) is a semiring if
\(\phi\in \mathcal{A}\);
\(A,B \in \mathcal{A}\) implies \(A\cap B \in \mathcal{A}\); (closed under finite union)
If \(A,B \in \mathcal{A}\) and \(A\subset B\), then there exists disjoint \(\mathcal{A}\)-sets \(C_1,\ldots, C_n\) such that \(B-A = \uplus_{k=1}^n C_k\).
매우 중요
Suppose that \(\mu\) is a set function on a semiring \(\mathcal{A}\) satisfying
\(\mu\) is nonnegative: \(0\le \mu\le \infty\) for all \(A\in \mathcal{A}\);
\(\mu(\phi)=0\);
\(\mu\) is finitely additive on \(\mathcal{A}\): if \(A_1,A_2,\ldots,A_n\in \mathcal{A}\) are disjoint, and \(\uplus_{k=1}^n A_k\in \mathcal{A}\), then \(\mu(\uplus_{k=1}^n A_k)= \sum_{k=1}^n \mu (A_k)\),
\(\mu\) is countably subadditive on \(\mathcal{A}\): if \(A_n\in \mathcal{A}, n\ge 1\) and \(\uplus_n A_n\in \mathcal{A}\), then \(\mu(\cup_{n} A_n)\le \sum_n \mu (A_n)\).
Then, \(\mu\) extends to a measure on \(\sigma(\mathcal{A})\). (일반적인 measure \(\mu\)는 위의 4가지 properties를 갖고있다.)
증명을 위해 semiring에서의 \(\mu^*\) 를 다음과 같이 정의하자; \[ \mu^*(A)=\inf\left\{ \sum_n \mu(A_n): A_1,A_2,\ldots\in \mathcal{A}, A\subset \cup_n A_n \right\}, A\subset \Omega, \] 위의 4가지 properties로 인해 \(\mu^*\)는 위의 outer measure를 설명한 lemma에서 1~3번을 만족시킨다. 때문에 \(\mu^*\)는 outer measure on \(\mathcal{A}\)이다. 또한 \(M(\mu^*)\)는 \(\sigma\)-field이고, \(\mu^*\) restricted to \(M(\mu^*)\)은 measure이다.
여기서 아래 두가지를 증명한다면 Caratheodory Extension Theorem을 증명해낼 수 있다.
결론
A measure \(\mu\) on a field \(\mathcal{F}\) has an extension to \(\sigma(\mathcal{F})\). If \(\mu\) is \(\sigma\)-finite on \(\mathcal{F}\), then the extension is unique.