A finitely additive Probability measure or *is it a measure*? 
Definition 7. Frequency definition of probability: If the sample space $\Omega$ of a random experiment is a countable set, $A \subseteq \Omega$ and $A_n \subseteq \Omega_n$(edit) the probability of $A$ is defined as $P^*(A) := \lim_\limits{n\to \infty} \dfrac{|A_n|}{|\Omega_n|}$ where $\lim_\limits{n\to \infty}  A_n = A$ and $\lim_\limits{n\to \infty}  \Omega_n = \Omega$


It was discussed somewhere that this defines a finitely additive measure but not countably additive and where the measure of any finite set is $0$. 
After this, an example was given to find the probability of picking an even number out of integers using the definition above.
I considered a sequence of sample spaces of the form $[-n,n]$ and found the even number in each of these spaces. The result converged to $0.5$. But I was doubtful that will it be dependent upon what sequence we choose for sample space? For choices like $[-n^2,n^2]$ or $[-e^n,e^n]$ it doesn't seem to be the case. I think even for a general case that $[f(n),g(n)]$ such that $f(n)<g(n)$ for each $n$ and $\lim_\limits{n\to \infty} f(n)=-\infty$
 , $\lim_\limits{n\to \infty} g(n)=\infty$. 
Edit: As shown by gerw, It's not unique. But is it the case that the limit is bounded above by $0.5$ for any sequence $A_n$ such that $\lim_\limits{n\to \infty}  A_n = A$?  Or is there a proper way to write mathematically this "particular kind of sequence of events".
Edit2: We can define a number $0<\varepsilon\leq1$ and consider the sequence $\Omega_{n}=[-n,n]$ and $A_n$ to be all the even numbers between $[-\varepsilon n, \varepsilon n]$, then the limiting conditions for $A_n$ and $\Omega_n$ are satisfied but the limit comes out to be between $0<L\leq \frac{1}{2}$ for various choices of $\varepsilon$. But, I am unable to think of an example which produces $L>0.5$ .
I tried to write it in terms of measure theory notations:
Consider a sequence of countable sample spaces $(\Omega_{n},\mathcal{F}_n,P_n)$ where each $P_n$ is probability based on count measure(division) and define an event $A_n \subset \Omega_n$ in each of these sample spaces. Suppose $A_n \to A$ and $\Omega_{n} \to \Omega$ as $n \to \infty$. Then $$P^*(A):=\lim_{n \to \infty} P_n(A_n)$$ 
This definition won't work obviously if $P_n$ is arbitrary. So, it seems $P_n$ is of certain "class" or "type". Is there a way to define some sufficient or possibly necessary condition for the existence or uniqueness of this limit?
P.S.: Lastly and hopefully not extending the question too much, it possible to generalize it using some "uniformly distributed measures" that doesn't depend upon the location but only the size of the set, so we can include uncountable sample spaces too. Like in the case of countable sample spaces, we can define the uniform probability in terms of the ratio of count measure of the event and sample space.  
 A: In my opinion, this definition is not useful, since it implicitly asserts that the limit $\lim \frac{|A_n|}{|\Omega_n|}$ is independent of the chosen sequences $(A_n)$ and $(\Omega_n)$.
This is, however, not the case: if both $A$ and $\Omega$ are countably infinite sets, this definition can yield any number in $[0,\infty]$! As an example, I use $\Omega = \mathbb N$ and $A = \{ 2 \, n \mid n \in \mathbb N\}.$ Then, the choice
$$\Omega_n = \{1,\ldots, n\}, \qquad A_n = \{2,4,6,\ldots,2\,(n^2-1) , 2 \, n^2\}$$
yields $+\infty$,
whereas
$$\Omega_n = \{1,\ldots, n^2-1, n^2\}, \qquad A_n = \{2,4,\ldots, 2\,n\}$$
yields $0$.
If you add (the reasonable) constraint $A_n \subset \Omega_n$ (or $A_n = A \cap \Omega_n$), you can still reach any number in $[0,1]$
Comment: It should be possible to get higher than $0.5$ by considering odd and even numbers separately: use $\Omega_{n,m} = \{1, 3, \ldots, 2\,m+1\} \cup \{2, 4, \ldots, 2\,n\}$ and $A_{n,m} = \{2,4,\ldots, 2\,n\}$ for various values of $m$ and $n$.
