Translation invariance of densities For a set $A\subseteq \mathbb{N}$, let 
$$\overline{d}(A)=\limsup_{n\to\infty}\frac{|A\cap[1,n]|}{n}$$
be the upper asymptotic density of $A$ and 
$$\overline{b}(A)=\lim_{n\to\infty}\max_{k\geq 0}\frac{|A\cap[k+1,k+n]|}{n}$$
be the Banach (or uniform) density of $A$. Being $\delta$ any of these densities, hoe to prove that


*

*$\delta(A+n)=\delta(A)$ and

*$\delta(nA)=\frac{1}{n}\delta(A)$?

 A: (i) Showing that $\overline{d}(A+n) = \overline{d}(A)$:
The following is clear:
$$(A \cap [1,m])+n = (A+n)\cap [1,n+m].$$
Since,
$$\overline{d}(A+n) = \limsup_{m\to\infty}\frac{|(A+n)\cap[1,m]|}{m},$$ 
and the following two facts hold:
$$\limsup_{m\to\infty}\frac{|(A+n)\cap[1,m]|}{m} = \limsup_{m\to\infty}\frac{|(A+n)\cap[1,n+m]|}{n+m},$$
$$|(A+n)\cap [1,n+m]| = |(A \cap [1,m])+n| = |A \cap [1,m]|,$$
then,
$$\overline{d}(A+n) = \limsup_{m\to\infty}\frac{|A\cap[1,m]|}{n+m}.$$
Finally,
$$\limsup_{m\to\infty}\frac{|A\cap[1,m]|}{n+m} = (\limsup_{m\to\infty}\frac{|A\cap[1,m]|}{m})(\lim_{m\to\infty}\frac{m}{n+m}) =\overline{d}(A). $$
(ii) Showing that $\overline{d}(nA) = \frac{1}{n}\overline{d}(A)$:
The following is clear:
$$n(A \cap [1,m]) = (nA)\cap [1,nm].$$
Since,
$$\overline{d}(nA) = \limsup_{m\to\infty}\frac{|(nA)\cap[1,m]|}{m},$$
and the following two facts hold:
$$\limsup_{m\to\infty}\frac{|(nA)\cap[1,m]|}{m} =  \limsup_{m\to\infty}\frac{|(nA)\cap[1,nm]|}{nm},$$
$$|(nA)\cap[1,nm]| = |n(A\cap[1,m])| = |A\cap[1,m]|,$$
then,
$$\overline{d}(nA) = \limsup_{m\to\infty}\frac{|A\cap[1,m]|}{nm}.$$
Finally,
$$\limsup_{m\to\infty}\frac{|A\cap[1,m]|}{nm} = \frac{1}{n}\limsup_{m\to\infty}\frac{|A\cap[1,m]|}{m} = \frac{1}{n}\overline{d}(A).$$
(iii) Showing that $\overline{b}(A+n) = \overline{b}(A):$
It is easy to see that:
$$(A+n)\cap [k+1+n,k+m+n] = (A\cap [k+1,k+m]) + n$$
In order to prove (iii), one needs to show that:
$$\lim_{m\to\infty}max_{k\geq 0}\frac{|A\cap [k+1,k+m]|}{m} = \lim_{m\to\infty}max_{k\geq n}\frac{|A\cap [k+1,k+m]|}{m} \text{, for each } n\in\mathbb{N}.$$
It is sufficient to show that:
$$\lim_{m\to\infty}max_{k\geq 0}\frac{|A\cap [k+1,k+m]|}{m} \leq \liminf_{m\to\infty}max_{k\geq n}\frac{|A\cap [k+1,k+m]|}{m}.$$
One can assume, without loss of generality, that $m>n$ for a fixed $n \in \mathbb{N}.$ It is clear that:
$$(1)\text{ }max_{k\geq 0}\frac{|A\cap [1+k,k+m]|}{m} = max\{\frac{|A\cap [1,m]|}{m}, ...,\frac{|A\cap[n,m+n-1]|}{m}, max_{k\geq n}\frac{|A\cap [1+k,k+m]|}{m}\}.$$
For $k = 0, 1,..., n-1$, then:
$$\frac{|A\cap [1+k,k+m]|}{m} = \frac{|A\cap [1+k,n]|}{m} + \frac{|A\cap [1+n,k+m]|}{m},$$
therefore,
$$\frac{|A\cap [1+k,k+m]|}{m} \leq \frac{|A\cap [1+k,n]|}{m} + \frac{|A\cap [1+n,n+m]|}{m}.$$
From this last inequality one obtains that for each $k = 0, 1,..., n-1$:
$$(2)\text{ }\liminf_{m\to\infty}\frac{|A\cap [1+k,k+m]|}{m} \leq \liminf_{m\to\infty}\frac{|A\cap [1+n,n+m]|}{m}.$$
Since $\frac{|A\cap [1+n,n+m]|}{m} \leq max_{k\geq n}\frac{|A\cap [1+k,k+m]|}{m}$, then:
$$(3)\text{ }\liminf_{m\to\infty}\frac{|A\cap [1+n,n+m]|}{m} \leq \liminf_{m\to\infty}max_{k\geq n}\frac{|A\cap [1+k,k+m]|}{m}.$$
From (1), (2) and (3) one gets the desired inequality.
Now it is clear that:
$$\overline{b}(A+n) = \lim_{m\to\infty}max_{k\geq 0}\frac{|(A+n)\cap [k+1+n,k+m+n]|}{m}.$$
using the fact that:
$$\lim_{m\to\infty}max_{k\geq 0}\frac{|(A+n)\cap [k+1+n,k+m+n]|}{m} = \lim_{m\to\infty}max_{k\geq 0}\frac{|(A\cap [k+1,k+m])+n|}{m},$$
one gets:
$$\lim_{m\to\infty}max_{k\geq 0}\frac{|(A\cap [k+1,k+m])+n|}{m} = \lim_{m\to\infty}max_{k\geq 0}\frac{|A\cap [k+1,k+m]|}{m} = \overline{b}(A).$$
(iv) Showing that  $\overline{b}(nA) = \frac{1}{n}\overline{b}(A):$
It is clear that:
$$\overline{b}(nA) = \lim_{m\to\infty}max_{k\geq 0}\frac{|(nA)\cap [k+1,k+m]|}{m} = \lim_{m\to\infty}max_{k\geq 0}\frac{|(nA)\cap [k+1,k+nm]|}{nm}.$$
To prove the result, it is necessary to show that:
$$(nA)\cap [k+1,k+nm] = n(A\cap [\lfloor \frac{k}{n}\rfloor +1, \lfloor \frac{k}{n}\rfloor +m]).$$
The following is also necessary:
$$(4) \text{ }\frac{k}{n} - \lfloor \frac{k}{n}\rfloor \leq 1 - \frac{1}{n}, \text{ for all } k\in\mathbb{N}.$$
In order to prove it, since $k = qn + r, r = 0 ,1, ..., n-1,$ then:
$$\lfloor \frac{k}{n} \rfloor = \lfloor q + \frac{r}{n}\rfloor = q,$$
therefore,
$$\frac{k}{n} - \lfloor \frac{k}{n}\rfloor = q + \frac{r}{n} - q = \frac{r}{n} \leq 1 - \frac{1}{n}.$$
$\text{Let }a \in A \text{ be such that } \lfloor \frac{k}{n}\rfloor +1 \leq a \leq \lfloor \frac{k}{n}\rfloor +m, \text{ then } n(\lfloor \frac{k}{n}\rfloor +1) \leq na \leq n(\lfloor \frac{k}{n}\rfloor +m).$ Therefore, $n(\lfloor \frac{k}{n}\rfloor +1) \leq na \leq n(\lfloor \frac{k}{n}\rfloor +m).$ From $(4)$, it is clear that $k + 1\leq na \leq k + m.$
$\text{Let }na \in (nA) \text{ be such that } k+1 \leq na \leq k + nm, \text{ then } \frac{k}{n} + \frac{1}{n} \leq a \leq \frac{k}{n} + m.$ Since $a \in \mathbb{N}, $ it is clear that $a \leq \lfloor \frac{k}{n}\rfloor + m.$ It is also clear that $\lfloor \frac{k}{n}\rfloor +1$ is the smallest integer larger or equal to $\frac{k}{n} + \frac{1}{n},$ therefore $\lfloor \frac{k}{n}\rfloor +1 \leq a.$
Applying the findings above,
$$\overline{b}(nA) = \lim_{m\to\infty}max_{k\geq 0}\frac{|n(A\cap [\lfloor \frac{k}{n}\rfloor +1, \lfloor \frac{k}{n}\rfloor +m])|}{nm} = \lim_{m\to\infty}max_{k\geq 0}\frac{|A\cap [\lfloor \frac{k}{n}\rfloor +1, \lfloor \frac{k}{n}\rfloor +m]|}{nm}, $$
$$\overline{b}(nA) = \lim_{m\to\infty}max_{k\geq 0}\frac{|A\cap [\lfloor \frac{k}{n}\rfloor +1, \lfloor \frac{k}{n}\rfloor +m]|}{nm} =\frac{1}{n}\overline{b}(A).$$
