$A$ so that $\lim\limits_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n])$ is not defined. ($\xi$ = counting measure)

Question

$$\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n]),$$

where $\xi_{\big| \Bbb{N}}$ is the counting measure on $\Bbb{N}$.

I am looking for $A \subset \Bbb{N}$ for which $\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n])$ is not defined. So I need to find $A$ so that the limit above does not converge but I don't come to any solution.

Answer 1

Let $A=\bigcup\limits_{k=0}^\infty \{2^{2k},\ldots,2^{2k+1}-1\}$ (be a set of integers whose binary expansion contains an odd number of digits).

\begin{align} n &= 2^{2m+1} - 1, \frac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n]) = \frac{1+2^2+\cdots +2^{2m}}{2^{2m+1}-1} = \frac{2^{2m+2}-1}{3(2^{2m+1}-1)} \xrightarrow[m \to \infty]{} \frac 23; \\ n &= 2^{2m+2} - 1, \frac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n]) = \frac{1+2^2+\cdots +2^{2m}}{2^{2m+2}-1} = \frac{2^{2m+2}-1}{3(2^{2m+2}-1)} \xrightarrow[m \to \infty]{} \frac 13. \end{align}

Therefore, the limit of $\dfrac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n])$ when $m \to \infty$ doesn't exist.

Answer 2

First of all let's put $p_n = \frac{1}{n} \xi (A \cap [1, n])$ to sinplify notations.

Then we define $A$ in such a way: $1 \in A$, so $p_1 = 1$, $[2, 10] \subset A^c$ (so that $A \cap [1, 10]$ only contains $1$) and we have $p_{10} = \frac{1}{10}$, next $[11, 100] \subset A$ so we have $p_{100}> \frac{9}{10}$... We can continue like this (in general $[10^k + 1, 10^{k + 1}] \subset A$ if $k$ is odd and $[10^k + 1, 10^{k + 1}] \subset A^c$ if $k$ is even).

In this way we have $p_{10^k} \geq \frac{9}{10}$ if $k$ is odd and $p_{10^{k + 1}} \leq \frac{1}{10}$ if $k$ is even and we see that $p_n$ can't converge.