Given a $\sigma$-algebra, $\mathcal{A}$, of all subsets of $\mathcal{N}$ with a counting measure $\mu$. Give a decreasing sequence $\{A_n\}$ of sets in the $\sigma$-algebra such that $\mu(\cap_k A_k) \neq \lim_k \mu(A_k)$?
Somehow this has something to do with an assumption about finiteness involved in the theorem stating:
For a measure space $(X,\mathcal{A},\mu)$, if $\{A_k\}$ is a decreasing sequence of sets that belong to $\mathcal{A}$, and if $\mu(A_n) < +\infty$ holds for some $n$, then $\mu(\cap_k A_k) = \lim_k \mu(A_k)$.
Now I know that the counting measure is: $\mu(A) = +\infty$, if $A$ is infinite, and $\mu(A) = n$ if set $A$ has $n$ elements.
I guess this means that $\mu(A_n) < +\infty$ does NOT hold for all $n$....?
The other part of the theorem that is used is that: if $\{A_k\}$ is an increasing sequence of sets that belong to $\mathcal{A}$, then $\mu(\cup_k A_k) = \lim_k \mu(A_k)$.
Any direction and/or explanations would be appreciated!
Nate