Two measurable space - how to check Is it true that


*

*$\left( \mathbb{N}, \mathcal{P}(\mathbb{N}), \mu(M) = \sum_{n \in M} 2^{-n} \right)$ is measurable space?

*$\left( \mathbb{N}, \mathcal{P}(\mathbb{N}), \mu(M) = |M| \right)$ is measurable space?


The first condition is $\mu(\emptyset)=0$. And in these two case it is obviously. But how can I prove the second condition: $$\mu \left( \bigcup_{n=1}^{\infty}A_n \right) = \sum_{n=1}^{\infty} \mu \left( A_n \right)$$
where of course for any $m,n$ such that $n \not= m$ we have $A_n \cap A_m = \emptyset$.
Intuitively it is also obviously but how prove it more formal?
 A: In both cases, the power sets are sigma algebras, so all that remains is to check that the 3rd coordinates of your triples are measures.  To check countable additivity, let $A_i$ be a countable disjoint collection of subsets of $\mathbb{N}$.  Then for the first one, you know that 
$\mu(\bigcup_{i=1}^\infty A_i)=\sum_{k \in \bigcup_{i=1}^\infty A_i} 2^{-k}$ which in particular means that the right side converges. (Which actually should've been checked before even defining $\mu$ but I leave that to you.)  Also, $\mu$ is a finite-valued nonnegative set function, which is also clear by comparison test.  Coming from the other side you wonder if $\sum_{i=1}^\infty \sum_{k \in A_i} 2^{-k}$ converges to the same value.  The inside sum always makes sense as a corollary again of checking that $\mu$ is even well-defined, and the outer sum is still fine because actually its convergence and value is a subsequence of the limit of partial sums that you know to exist when computing $\sum_{k \in \bigcup_{i=1}^\infty A_i} 2^{-k}$.
I hope this has helped, and perhaps you can do the second one now.  I'm having some issues determining exactly what I can use since probably Tonelli's theorem is out of the question.  But Rudin, for example, proves a theorem about interchanging infinite sums without even discussing measure theory.  If you find that relative to your current understanding my proof is slightly circular or lacking detail, you may want to look into that.
