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.

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 $D(A)=0$.
