I'm trying to solve an assignment for my Calc III class.
Let $\Omega$ be an infinite set and $\mathcal{A}= \{ A \subset \Omega : A$ or $A^{c}$ is finite$\}$.
I have shown that $\mathcal{A}$ is a set algebra, which implies:
(i) $\Omega \in \mathcal{A}$,
(ii) $A \in \mathcal{A} => A^{c} \in \mathcal{A} $,
(iii) $A, B \in \mathcal{A} => A \cup B \in \mathcal{A}$.
And I have shown that $\mathcal{A} \subsetneq \sigma(\mathcal{A})$.
Let $\mu : \mathcal{A} \rightarrow [0, \infty]$ with $\mu(A) = 0$ if $A \in \mathcal{A}$ is finite, $\mu(A) = 1$ else. I have shown that $\mu$ is a content, namely:
(i') $\mu(\varnothing) = 0 $,
(ii') $A,B\in \mathcal{A}$ and $A\cap B=\emptyset => \mu (A\cup B)=\mu (A)+\mu (B)$.
I want to show that $\mu$ cannot be extended to a measure on $\sigma(\mathcal{A})$.
In class, we had a generalized version of Carathéodory's extention theorem:
Let $\mathcal{S}$ be a semi-ring, $\mu: \mathcal{S} \rightarrow [0, \infty]$ satisfying $(i'), (ii')$,
(iii'): for $A_{1}, A_{2},\dots \in \mathcal{S}$ with $\cup_{i}^{\infty} A_i \in \mathcal{S}$: $\mu(\cup_{i}^{\infty} A_i) \leq \sum_{i}^{\infty} \mu(A_i)$.
=> There exists a measure $\mu_E$ on $\sigma(\mathcal{S})$ with ${\mu_E}{\vert_\mathcal{S}} = \mu$.
Our definition of a semi-ring:
(i'') $\emptyset \in \mathcal{S}$,
(ii'') $A, B \in \mathcal{S} => A \cap B \in \mathcal{A}$,
(iii'') $A,B \in \mathcal{S}$ with $A \subset B$: there exist $D_{1}, D_{2},\dots \in \mathcal{S}$ with $\cap_{i}^{m} D_i = \emptyset$ such that $B \setminus A = \cup_{i}^{m} D_i$.
I can show that for $\mathcal{A}$:
(i) + (ii) => (i''),
(ii) + (iii) => (ii'').
But is (iii'') too satisfied? Because for $A,B \in \mathcal{A}$ with $A \subset B$:
(ii) => $A^c \in \mathcal{A}$. (ii') => $B \cap A^c \in \mathcal{A}$.
Can I just say $B \cap A^c = B \setminus A =: D_1 $?
If this is out of the way, now my main question:
I was thinking that maybe I could show that (iii') is not satisfied for $\mu$. But there is really no use with the set up we have here. Reversing the implication of the theorem does not help. What am I missing?
Looks to me like countable additivity won't hold. Let $\{\{a_n\}\}_{n \in \mathbb{N}}$ be a countable collection of disjoint singletons in $\Omega$. We have that $\bigcup_n \{a_n\}$ is infinite and thus has measure 1, but each $\{a_n\}$ is finite and thus has measure $0$. Therefore $\mu(\cup_n \{a_n\})=1 \neq 0 =\sum_n \mu(\{a_n\}=\sum 0$(?)