Having some degree of trouble trying to solve this question. Firstly, let us call the power set of the integers $A$. I want to find a function $p: A \to [0,1]$ where the function $p$ satisfies:
1) $p(\mathbb{Z}) = 1$
2) for $a, b \in A$, if $a\cap b = \emptyset \hspace{1mm}$ then $\hspace{1mm}p(a\cup b) = p(a) + p(b)$
I.e. it is a probability measure but only finitely additive. The additional condition on this function is given below. Let $a \in A$ and define:
$f_n(a) = \frac{|a \cap\{0,1,2,\cdots,n\}|}{n+1}$ (intepreted as the proportion of the first $n+1$ integers in $a$).
Let $\overline{f}(a) = \lim \sup_n f_n(a)\hspace{4mm}$ and $\hspace{4mm}\underline{f}(a) = \lim \inf_n f_n(a)$
I have to ensure that $\overline{f}(a) \geq p(a) \geq \underline{f}(a)$.
The textbook had me check whether $p(a) = \frac{\overline{f}(a) + \underline{f}(a)}{2}$ was a valid function and I produced a counter-example to that already. I however, do not have much intuition on how to proceed from there. Does such a function exist? If so could you give me a hint on how to tackle it? Thank you!