Show mean of a sequence of sets is zero but mean of countably infinite set is infinity if increasing Let $\Omega$ be a countably infinite set, and let $F$ consist of all subsets of $\Omega$. Define $\mu(A) = 0$ if $A$ is finite and $\mu(A) = \infty$ if $A$ is infinite. Show that $\Omega$ is the limit of an increasing sequence of sets $A_n$ with $\mu(A_n) = 0$ for all $n$ but $\mu(\Omega) = \infty$.

Does anyone know which theorem might help me start to show this? I was already able to show that μ is finitely additive but not countable additive using finite subadditivity. Any help would be greatly appreciated. I haven't had a proof course in 10 years and am trying to study for the GRE.
$\Omega$ is infinit thus $\mu(\Omega) = \infty$

Now $\Omega$ is countable, thus let $f:\Bbb N \rightarrow \Omega$ be a bijection.

Let $A_n = \{f(k), k\in [0,n]\}$, what follows is that for any $n$, $A_n$ is finite thus $\mu(A_n) = 0$ but $\Omega = \lim_{n\to+\infty}A_n$
You have to show that for any countably infinite set $\Omega$, there exists some sequence of sets

$A_1, A_2, ... \in \mathscr F := 2^{\Omega}$ (I believe this is equivalent to: $A_1, A_2, ... \subseteq \Omega$)

s.t.


$$A_n \subseteq A_{n+1}$$
$$\bigcup_{n=1}^{\infty} A_n = \Omega$$




Relevant Theorem 1 Continuity of measure:

If we have a measure space $(\Omega, \mathscr F, \mu)$ and $A, A_1, A_2, ... \in \mathscr F$ s.t.


$$A_n \subseteq A_{n+1}$$
$$\bigcup_{n=1}^{\infty} A_n = A$$


then

$$\mu(A) = \lim_{n \to \infty} \mu(A_n)$$



Relevant Theorem 2

If we have a measurable space $(\Omega, \mathscr F)$ and a set function $\mu: \mathscr F \to [0,\infty]$ s.t. 


$\mu(\emptyset) = 0$
If $A, A_1, A_2, ... \in \mathscr F$ s.t.


2.1. $$A_n \subseteq A_{n+1}$$
2.2. $$\bigcup_{n=1}^{\infty} A_n = A$$

then

$$\mu(A) = \lim_{n \to \infty} \mu(A_n)$$

then $\mu$ is a measure on $(\Omega, \mathscr F)$



It follows that $\mu$ defined as

$$\mu(\text{finite}) = 0$$

$$\mu(\text{infinite}) = \infty$$

is not a measure on $(\Omega, \mathscr F)$ for any countably infinite set $\Omega$ with $\mathscr F = 2^{\Omega}$



If $\Omega = \mathbb N$, then consider $$A_n:= \{1,2,\cdots,n\}$$



As for any other countably infinite set $\Omega = A$, we want to choose $A_n$ as follows:

$$A_n = \{\text{first n elements of A, assuming A has some...}\}$$

$$ \{\text{...order by which one element precedes another}\}$$

With $\Omega = A = \mathbb N$, we had the elements in ascending order. But what if we had $A = \mathbb Z$? What if we had some countably infinite set of names for instance such as $A = \{Elsa, Anna, ElsaA, AnnaA, ElsaB, AnnaB, \cdots\}$ ?



Solution 1:

A countably infinite set can be enumerated: $A=\{a_n\}_{n=1}^{\infty}$

So let $$A_n := \{a_1, ..., a_n\}$$



Solution 2:

Note that by definition (or a property, depending on the textbook) of a countably infinite set, there exists a bijection $f: A \to \mathbb N$.

This bijection is what allows us to give order to the elements of $A$:

$$A_n := \{f(1), ..., f(n)\}$$



Observe that $f(n)$ from Solution 2 equals $a_n$ from Solution 1.