A question about the natural density from number theory Take a set $ S \subseteq N $, define the sequence $ x_{n}=\#(A\cap [1,n])/n $, and then if $ \lim x_{n} $ exists, call it $ D(A) $ , the (natural) density of A on N. 
Prop: If for any natural number $n$, $ D(A_n)=0 $,  then we can find a common $ A , D(A)=0 $  with the property for any natural number $n$, $ A_n-A$  is finite. Here $ A, A_n $ are all subsets of the natural number system N.
How to prove this prop? It's  really difficult for me. Thanks ahead.
 A: We can easily prove (for example by induction) that for any fixed natural number $N$ the union
$$
A^{[N]}:=\bigcup_{n=1}^NA_n
$$
also has density $D(A^{[N]})=0$.
Using this Lemma we can construct the set $A$. I define a nested sequence of subsets of density zero
$$
B_1\subseteq B_2\subseteq B_3\subseteq \cdots
$$
recursively as follows. 


*

*Set $B_0:=\emptyset$, $M_0=0$.

*Assume that we have already constructed the set $B_k$ and selected an integer $M_k$ for some natural number $k$. Then define
$$
B_{k+1}:=B_k\cup \{n\in A_{k+1}\mid n> M_k\}.
$$

*Select a large integer $M_{k+1}$, and go back to the previous step.


Then consider the union $A=\bigcup_{n=1}^\infty B_n$. For all $n$ we have that $A$ contains all the elements of $A_{n+1}$ that are larger than $M_n$. In other words all but finitely many of them. Therefore the set $A\setminus A_n$ is finite for all $n$ irrespective of how we selected the sequence of numbers $(M_n)$.
The trick is to show that the sequence $M_n$ can be selected in such a way that $D(A)=0$.
We rely on the Lemma to do that. More precisely, we select the integers $M_k$ in such a way that
$$
\frac{\#(A^{[k]}\cap [1,M])}{M}<\frac1k
$$
holds for all $M\ge M_k$. The Lemma guarantees that such an integer $M_k$ exists. Because $B_k\subseteq A^{[k]}$ and $B_k\cap [1,M]=A\cap [1,M]$ for all $M<M_{k+1}$, we also have, for all $k$,
$$
\frac{\#(A\cap [1,M])}{M}<1/k
$$
for all $M$ such that $M_k\le M<M_{k+1}$. Thus $D(A)=0$.
