If two sets have a natural (asymptotic) density, does their union? 
Let $\Omega=\mathbb{N}$. For each $E\subset\Omega$ let $N_n(E)$ be the cardinality of the set $E\cap \{1,2,\ldots,n\}$. Define $C=\left\{E: \lim_{n\rightarrow \infty} \frac{N_n(E)}{n} \text{ exists}\right\}$.
Show that $C$ isn't a field.

I already know that it isn't closed to finite union of non-disjoint set but I can't see why. I saw a post that somebody said "the set of natural numbers whose first digit is 1 doesn't have this limit".
I'm in trouble to see why this limit wouldn't exists in finite unions.
Thanks in advance.
 A: We show that the intersection of two sets that have "natural density" does not necessarily have natural density.  Then union can be dealt with by taking complements.  It is basically an example involving decimal expansions, though for smoothness we use base $4$.
Let $A$ be the set of even integers. The natural density exists and is equal to $1/2$.  
Let $B=B_0 \cup B_1$, where $B_0$ is the set of all even numbers in intervals of the shape $[2^{2m}, 2^{2m+1}]$ and $B_1$ is the set of all odd numbers in intervals of the shape $[2^{2m+1},2^{2m+2}]$.  Again, $B$ has natural density $1/2$.
Now look at $A\cap B=E$.  These are all the even numbers in intervals of the shape $[2^{2m},2^{2m+1}]$. There is very considerable fluctuation in $\frac{N_n(E)}{n}$.  For large $m$, if $n$ has shape $2^{2m+1}$, then $\frac{N_n(E)}{n}$ is very close to $1/3$, while if $n$ has the shape $2^{2m+2}$, then $\frac{N_n(E)}{n}$ is very close to $1/6$. 
Somewhat less prettily, we can play the same game with decimal representations. The set $A$ could be all numbers that end in a $0$. For $B$, between $10^{2m}$ and $10^{2m+1}$,  use the numbers that end in a $0$, and between $10^{2m+1}$ and $10^{2m+2}$ use the numbers that end in a $1$. Each of $A$ and $B$ has natural density $1/10$. But $A\cap B$ has huge relatively thick regions, followed by huge empty gaps, and the natural density does not exist. 
