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 $\mu^\#(A_{j_0}) \lt \infty$?
Best Answer
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 $\bigcup_{n\in\Bbb N}\{n\}$.