I'm trying to prove that the counting measure on $(X,P(X))$ is $\sigma$-finite if and only if $X$ is countable as an answer to the title.
Intuitively, it feels that $X$ should be countable so I decided to proceed the proof that way:
If the counting measure $\mu$ is $\sigma$-finite, I can write $X=\bigcup_{j=1}^\infty E_j$ where $\mu (E_j)<\infty$ for each $j$ and can make sure that $E_j$'s are disjoint. But I'm not sure how to create an injection to $\mathbb N$.
For the other direction, I have no idea how to go from $X$ being countable to counting measure $\mu$ being $\sigma$-finite.
Any advice or comment would be appreciated, thanks guys.
Best Answer
If $\mu$ is sigma finite then $X=\cup_n A_n$ with $\mu (A_n) <\infty$ for each $n$. This implies that each $A_n$ is a finite set so $X=\cup_n A_n$ is countable. Conversely, if $X$ is countable, say $X=\{x_1,x_2,...\}$ then $A_n=\{x_n\}$ is a set of finite measure and the union of these sets is $X$, so $\mu$ is sigma finite.