Show that $\Omega$ is the limit of an increasing sequence of sets $A_n$; I am (self) studying probability theory and measure using the book from Ash, R. et al.. I am trying to solve one of the basic problems (Section 1.2, problem 1.b ) but with no avail...here is the link to the problem:

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*Let $\Omega$ be countably infinite set, and let $\mathcal{F}$ consist of all subsets of $\Omega$. Defina $\mu (A)=0$ if $A$ if finite,  $\mu (A)=\infty$ if A if infinite.

a)....
b) Show that $\Omega$ is the limit of an increasing sequence of sets $A_n$, with $\mu (A_n)=0$, but $\mu (\Omega)=\infty$.
I understan that if $A_n$ is finite $\forall$ $\{1,2,3,...,n\}$, then I have $\mu (A_n)=0$; but than why would $\bigcup_{n=1}^{n\rightarrow \infty}A_n$ converge to $\Omega$? Is there somewhere assumed that $\Omega$ consists of only finite elements?
 A: Since $\Omega$ is countable, we can find the enumeration $\Omega=\{\omega_1,\omega_2,\omega_3,...\}$. Define
$$A_n=\bigcup_{k\leq n}\{\omega_k\}$$
we have that $A_n \subset A_{n+1}$ and $A_n\uparrow \Omega$. We have $\mu(A_n)=0,\,\forall n \in \mathbb{N}$ but $\mu(\Omega)=\infty$. To see that $A_n\uparrow \Omega$, notice that for any $\omega_k \in \Omega$ we have $\omega_k\in \cup_{n \in \mathbb{N}}A_n$ but also the converse, so $\Omega = \cup_{n \in \mathbb{N}}A_n$.
A: To complement on the previous answer let me mention the following useful concepts from set theory, which may be convenient to recall since they play a role in your question.
Let $\{A_n\}_{n=1}^{+\infty}$ be a sequence of subsets of a given set $\Omega$ (for now, arbitrary).

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*Define $\limsup_{n\to+\infty} A_n:=\cap_{n=1}^{+\infty}\cup_{m=n}^{+\infty}A_m$. In other terms, $x\in \limsup_{n\to+\infty} A_n$ if and only if $x$ belongs to infinitely many $A_n$'s.


*Define $\liminf_{n\to+\infty} A_n:=\cup_{n=1}^{+\infty}\cap_{m=n}^{+\infty}A_m$. In other terms, $x\in \liminf_{n\to+\infty} A_n$ if and only there is $n_*$ such that $x$ belongs to all $A_n$'s with $n\geq n_*$.


*We say that the sequence $A_n$ converges (set-theoretically) to a subset $\lim_{n\to+\infty}A_n\subseteq \Omega$ if and only $\limsup_{n\to+\infty} A_n =\liminf_{n\to+\infty} A_n =: \lim_{n\to+\infty}A_n$.
That being said, if the sequence $A_n$ is increasing then it always converges (according to this definition) to $\lim_{n\to+\infty} A_n = \cup_{n=1}^{+\infty} A_n$.
Now, in your case $\Omega$ is countably infinite so that $\Omega=\{\omega_1,\omega_2,\dots\}$ and you can set $A_n:=\{\omega_1,\dots,\omega_n\}$, which is increasing and converges to $\Omega$.
