$A \subseteq \mathbb{N}$ for which $\mu(A)=\lim_{n \to\infty}\frac{1}{n} \zeta \big|_\mathbb{N} (A \cap[1,n])$ is not defined

Question

Let $\mu$ be a measure function.

$\mu(A):=\lim_{n \to\infty}\frac{1}{n} \zeta \big|_\mathbb{N} (A \cap[1,n])$ where $\zeta \big|_\mathbb{N}$ denotes the counting measure on $\mathbb{N}$.

I'm looking for a $A \subseteq \mathbb{N}$ for which $\mu(A)$ is not defined.

Answer 1

Jack D'Aurizio's comment above has a very succinct answer.

Here is a more humdrum approach:

Choose $0 < \alpha < \beta < 1$.

Define $A_n$ and choose $d_n$ ('direction') as follows:

$A_1 = \{1\}$, $d_1 =0$.

If $|A_n| \ge \beta n$, let $d_{n+1} = -1$, $A_{n+1} = A_n$.

If $|A_n| \le \alpha n$, let $d_{n+1} = +1$, $A_{n+1} = A_n \cup \{n+1\}$.

If $\alpha n < |A_n| < \beta n$ and $d_n < 0$, let $d_{n+1} = d_n$, $A_{n+1} = A_n$.

If $\alpha n < |A_n| < \beta n$ and $d_n > 0$, let $d_{n+1} = d_n$, $A_{n+1} = A_n \cup \{n+1\}$.

Then let $A=\cup_n A_n$. If we let $m_n = {1 \over n} |A \cap \{1,...,n\}|$ then we see that $\liminf_n m_n \le \alpha$, $ \limsup_n m_n \ge \beta$.