Why is $\mathcal{A}$ not a $\sigma$-algebra?

Question

Let $\Omega = \mathbb N$ and define $\mathcal{A}:=\{A\subseteq \Omega: \lim_{n \to \infty}\frac{1}{n} \mid A \cap \{1,...,n\}|$ exists $\}$

By definition a $\sigma$-algebra fulfills:

i) $\Omega \in \mathcal{A}$

ii) $\mathcal{A}$ is $^{c}$-stable

iii) $\mathcal{A}$ is stable under countable unions

For all points, I am not able to construct an argument that would render $\mathcal{A}$ not a $\sigma$-algebra. particularly as $|A \cap \{1,...,n\}| \leq n$ and then $\lim_{n\to \infty }\frac{1}{n}|A \cap \{1,...,n\}|$ would always exist.

Answer 1

This is in response to your comment on user10354138's answer. I suggested constructing a sequence by adding numbers until the "partial density" is at least $1/2,$ omitting numbers until it falls to $1/4$ or less, and iterating. Let $$d_n= \frac{1}{n}|A \cap \{1,...,n\}|,\ n=1,2,3,\dots$$ Then $d_n$ is what I intended by the phrase "partial density".

So we would get $$A=\{1,5,6,13,14,15,16,17,18,37,38,\dots\}$$

$d_1=1,$ so we omit the following numbers

$d_4=1/4,$ so we include the following numbers

$d_6=1/2,$ so we omit the following numbers

$d_{12}=1/4,$ so we include the following numbers

$d_{18}=1/2,$ so we omit the following numbers

et cetera.

Answer 2

$\lvert A\cap\lbrace1,\dots,n\rbrace\rvert\leqslant n$ only gives $\limsup_{n\to\infty}\frac1n\lvert A\cap\lbrace1,\dots,n\rbrace\rvert\leqslant 1$, not the equality of limsup and liminf (which is what was needed for $\lim_{n\to\infty}\frac1n\lvert A\cap\lbrace1,\dots,n\rbrace\rvert$ to exist).

You can check explicitly that every singleton is in $\mathcal{A}$, so if $\mathcal{A}$ is a $\sigma$-algebra it must be $2^\Omega$ since $\Omega$ is countable. However, there are subsets of $\mathbb{N}$ which does not have an asymptotic density (i.e., (iii) fails).