Show that $\mu^\#$ (Counting Measure) is $\sigma$-finite iff $X$ is countable

Question

For the '$\Rightarrow$' is easy because if we assume $\mu^\#$ sigma finite then

$$X = \bigcup_{j=1}^{\infty} A_j$$

We have that $\mu^\#(A_j) \lt \infty \hspace{3mm}\forall j$ so this means that $A_j$ is finite for all j and countable union of finite sets is still countable. I'm having troubles showing the '$\Leftarrow$' because if X is countable it could be union of countable sets, so if exists a $j_0$ such that $A_{j_0}$ is countable as well then how is it possible to have $\mu^\#(A_{j_0}) \lt \infty$?

Answer 1

It is true a countable set can be written as a countable union of countable sets. But it can also be written as a countable union of singletons. For example, $\Bbb N$ is $\bigcup_{n\in\Bbb N}\{n\}$.

Answer 2

To be explicit about something implicit in the existing answer: When you say that $A_{j_0}$ could be countable it seems that you're confused about the definition of "$\sigma$-finite".

Say $\mu$ is a measure on $X$. Consider the following four conditions:

(i) $X$ is a countable union of sets of finite measure.

(ii) $X=\bigcup_{j=1}^\infty A_j$ where $\mu(A_j)<\infty$.

(iii) There exist $A_1,A_2,\dots$ such that $X=\bigcup_{j=1}^\infty A_j$ and $\mu(A_j)<\infty$.

(iv) If $X=\bigcup_{j=1}^\infty A_j$ then $\mu(A_j)<\infty$.

The definition of "$\mu$ is $\sigma$-finite" is often stated as (i). It's clear that (ii) is the same as (i), with some of the English replaced by mathematical notation. The point it seems you may be missing is this:

(iii) is the same as (ii). Not for any deep reason; (iii) is simply what one means when one writes (ii).

The reason I conjecture this is what you're missing is that your concern "but $A_j$ could be countable" makes no sense if you're thinking in terms of (iii). Because in (iii) we're not given the $A_j$, we're saying that there exists a sequence $(A_j)$ with a certain property.

Otoh "but $A_j$ could be countable" makes perfect sense if you think the definition is (iv). But (iv) is not the definition, (iv) is something entirely different.

(In fact, easy exercise (iv) holds if and only if $\mu(X)<\infty$.)