I came across this problem in Probability and Measure Theory by Robert Ash and Catherine Doléans-Dade. I think my attempt at the proof lacks rigour. How can I improve it?
Problem: Let $\Omega$ be a countably infinite set, and let $\mathscr{F}$ be the field consisting of all finite subsets of $\Omega$ and their complements. If $A$ is finite, set $\mu(A) = 0$ and if $A^c$ is finite then set $\mu(A) = 1$.
Show that $\Omega$ is the limit of an increasing sequence of sets $A_n \in \mathscr{F}$ with $\mu(A_n) = 0 \space \forall \space n$, but $\mu(\Omega) = 1$.
My attempt: Any countably infinite set can be broken down into a sequence of subsets by the following construction. Start with $\Omega$ and exclude some elements from it to form the non-empty sets $A$ and $A^c$. This can always be done as $\Omega$ is infinite. Discard $A^c$ and breakdown $A$ recursively into smaller sets by the same construction. You are then left with sets $A_1 \subset A_2 \subset \dots \subset A_n \subset \Omega$. In the limit $n \rightarrow \infty$ we have obtained a limiting sequence.
Question: To me this argument feels a bit circular and can be strengthened if I can formally express an infinitesimal portion of $\Omega$. How can I do that?
Regardless, once I have proved the construction the actual problem that $\mu(A_n) = 0 \space \forall \space n$, but $\mu(\Omega) = 1$ shall follow by the definition of $\mu$. Is that argument also correct?
One issue with the way your proof reads (which could very well be my misunderstanding, but this still suggests an area of improvement) is that you seem to start your construction with the "last" element of an infinite sequence. You've actually constructed a sequence of sets that shrink into the empty set, which isn't quite what the problem needs you to do. This is why it's not really clear what $A_1$ is since it seems to be the element at the end of your construction.
I believe your approach is salvageable, but let me offer another, perhaps more standard one.
Let $\{\omega_1,\omega_2,\ldots\}$ be an enumeration of $\Omega$. This is possible because $\Omega$ is countable. Now set $A_1 = \{ \omega_1 \}$, and $A_n = A_{n-1} \cup \{ \omega_n \}$ for $n > 1$. You should be able to see that $\{ A_n \}$ satisfies our requirements.