$\mathcal{A} = \left\{ A \subset \mathbb{N} ; \lim_{n \to \infty} \frac{1}{n} \#(A \cap \{1,\ldots,n\}) \text{ exists} \right\}$ is not a algebra I must show that the set $$\mathcal{A} = \left\{ A \subset \mathbb{N} ; \lim_{n \to \infty} \frac{1}{n} \#(A \cap \{1,\ldots,n\}) \text{ exists} \right\}$$
is not a algebra.
It's easy to see that $\emptyset,\mathbb{N}\in \mathcal{A}$ and that if $A \in \mathcal{A}$ then $\mathbb{N}-A \in \mathcal{A}$. So I must find $A, B \in \mathcal{A}$ with $A \cup B \notin \mathcal{A}$.   I'm having problems simply finding a set that is not contained in $\mathcal{A}$. From my understanding the limit is, in some way, representing the "presence of A" in $\mathbb{N}$ as I look upon arbitrarily large natural numbers.
Any help is appreciated.
 A: Let $A$ be the set of even integers and define $J: \mathbb{N} \rightarrow \mathbb{N}$ at an $n$ by letting $k$ be the unique integer such that $2^k\leq n<2^{k+1}$ and let $J(n)$ be $2n$ if $k$ is even and $2n-1$ if $k$ is odd. Putting $B:=J(\mathbb{N})$ we see that $|(A\cup B)\cap\{1,\cdots ,n\}|/n$ will in fact oscillate between $2/3$ and $5/6$.
A: The idea is to "just do it". We will define $A$ and $B$ by hand so that $A$ and $B$ both have density $1/2$ but $A \cup B$ has undefined density. Let $N_1 < N_2 < N_3 < \cdots$ be a sequence of larger and larger scales such that $N_{k+1} / N_k \to \infty$, say $N_k = k!$. On $[N_{k-1}, N_k)$ with $k$ even let $A$ and $B$ both consist of all even numbers. On $[N_{k-1}, N_k)$ with $k$ odd let $A$ be all even numbers and $B$ all odd numbers. Now if you measure the density of $A \cup B$ in $\{1, \dots, N\}$, the question is how much do $A$ and $B$ overlap, and that massively depends on what interval $[N_{k-1}, N_k)$ contains $N$. I'll leave the remaining details to you.
