Uniform Probability Measure on $\mathbb{N}$? Hello this problem I am working on has four parts, and I have figured out the first three, but I am stuck on the fourth.
For some clarification, the title of the problem I am working on is, Uniform Probability Measure on $\mathbb{N}$? That is why I titled my question so. In searching for a solution to this problem, I found out that such a thing does not exist so I am not asserting that I may have found one or anything like that.

For each $A\subset \mathbb{N}$ define $$\rho_n(A)= \frac{1}{n}|A\cap[1,n]|,$$ and say that A has density $\rho(A)=\lim\limits_{n\rightarrow \infty}\rho_n(A)$ if that limit exists. Let $D=\{ A\subset\mathbb{N}:\rho(A) \text{exists}\}.$
1) Show that $D$ is closed under complements.
2) Show that $D$ is closed under finite disjoint unions.
3) Show that $D$ is not closed under countable disjoint unions.
4) Show that $D$ is not closed under finite non-disjoint unions.

For 1) I found that for any $A\in D,$ $\lim\limits_{n\rightarrow\infty} \rho_n(A^c)=1-\lim\limits_{n\rightarrow\infty} \rho_n(A)$ using the $\varepsilon$ definition of limit of a sequence.
For 2) I get the answer to be the sum of the limits, again using the $\varepsilon$ definition and this time the triangle inequality.
For 3) I defined a countable disjoint sequence so that the $\rho_n \text{'s}$ of their union oscillates between $3/4$ and $1/2$.
I want to do something similar as 3) for 4), but I am a little stuck as to how to find a finite number of sets that will give me a nice oscillation like that (I would prefer to find 2 such sets). I do know that the only way to get the sequence to diverge is to make it oscillate in some way. Otherwise, we would get convergence because the sequence is bounded.
Any help would be greatly appreciated.
 A: You're right about the approach for (4). You want to construct $A, B \in D$ so that $\rho_n(A\cup B)$ has two limit points. Let $A$ be the even integers and define $B$ by its characteristic sequence $\chi_B: \mathbb{N} \to \{0, 1\}$ as follows: 
Consider the $2^n$-truncations of $\chi_A$:
$$0,
01,
0101,
01010101,
0101010101010101, \dots$$
Define $\chi_B$ by taking the $2^{2n}$-truncation of $\chi_A$, reversing it, and appending to the $2^{2n}$-truncation to the $2^{2n+1}$-truncations of $\chi_B$ and reversing:
$$0,
01,
01{\bf 10},
01100101,
01{\bf 10} 0101{\bf 10101010},\dots$$
Then $\rho(A) = \rho(B) = \frac12$, so $A, B \in D$, and $\chi_{A \cup B}$ is given by
$$0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,\dots$$
Note that
$$
\begin{align*}
\rho_{2^{2n+1}} (A\cup B) &= \frac{1}{2^{2n+1}} \left(1+ \sum_{k=0}^{n-1} 2^{2k+1} +\sum_{k=0}^{n-1} 2^{2k+1}\right)
= \sum_{k=0}^n \frac{1}{2^{2k+1}} \to \frac23
\end{align*}
$$
as $n \to \infty$. On the other hand,
$$
\rho_{2^{2n}}(A\cup B) = \frac12 \rho_{2^{2n-1}}(A\cup B) + \frac12 \to \frac56.
$$
Thus the sequence $(\rho_n)$ has two limits points, so $A \cup B \not \in D$. 
