counting measure on naturals not equal to limit 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
 A: You're right to say that in any example we must have $\mu(A_n) = \infty$ for each $n$. The first thing that comes to mind is that you could start with an infinite set $A_1$ and pick away elements one by one such that there is nothing in the intersection, so that $\mu(\bigcap_{n = 1}^\infty A_n) = 0$. Can you write down something like this?
A: I do not know if following solution is correct or not. However I am also attempting to understand this problem and did not want to replicate this question. Please guide me towards correct solution.
We know that the set of integers has same cardinality as set of rational numbers. Specifically according to Cantor's scheme:
We list all rational numbers for which sum of numerator and denominator is 2:1/1. Next we we list rational for which the sum is 3: 1/2, 2/1. similarly 4:1/3, 2/2, 3/1 and so on. Now if we list these numbers together omitting the ones already listed:
1,1/2,2,1/3,3,1/4,2/3,3/2,4,1/5,5.....
then there is one to one correspondence between integers and rational numbers. Let us call the integers the index of these rational numbers. 
Now if we remove the indexes of rational numbers greater than or equal to 1/2 at first tep, indexes of rational numbers greater or equal to 1/4 at next step and so on for  $1/8....1/2^n...$ from the set of integers we still have a set of infinite cardinality which is same as cardinality of set of integers as the number of rationals in every open interval is same. So the limit of counting measure is $\infty$. However the sequence of sets converges to the set of integers indexing rationals in interval $(0,0)$, which is an empty set.
