If $A \subset B \subset \mathbb{R}$, with $A$, $B$ intervals, how do I show that
$$\limsup_{n \to \infty} \, \frac{1}{n} \#(A \cap \frac{\mathbb{Z}}{n}) \leq \liminf_{n \to \infty} \,\frac{1}{n} \#(B \cap \frac{\mathbb{Z}}{n})?$$
here $\#$ denotes cardinality, and $\frac{\mathbb{Z}}{n} : = \{\frac{z}{n} : z \in \mathbb{Z} \}$.
In particular, the proof I've given here seems to require this fact, but I just wanted to confirm it's true. Namely, I want to verify this without using the fact that both of these limits actually exist, because that would make the proof I've given useless. I only need this weaker fact.
Intuitively it seems rather easy since these are intervals. Here's a sketch of what I mean:
let $\{a_n\}$ be a subsequence of the values $\#(A \cap \frac{\mathbb{Z}}{n})$ converging to the limsup and $\{b_n\}$ the liminf of the expression with $B$. Then, for each $n$ that gives $a_n$, we can take a much larger $N$ for $b_N$. If $N$ is taken large enough (say, so that $\frac{1}{N}$ is smaller than the leftover space in $A \setminus \frac{\mathbb{Z}}{n}$, and smaller than the space between $A$ and $B$), then $a_n \leq b_N$. Since we can always find an element of the latter subsequence bigger than the former, we have the inequality.
Does this sketch work?
I've added an answer below, please comment on it if you have a chance
Here's what I believe to be an answer:
Let $n, a_n$ be as in the question. Let $\#(A \cap \frac{\mathbb{Z}}{n}) = k+1$, which means there are $k$ intervals of length $\frac{1}{n}$. Now let $\epsilon_1,\epsilon_2$ be the left and right gaps between $A$ and $B$, i.e., if $A := [a,b]$ and $B = [c,d]$ (or any type of interval), $\epsilon_1 = c-a$, $\epsilon_2 = b-d$.
Let $\frac{1}{N} < \frac{1}{k}\min\{\epsilon_1,\epsilon_2\}$, or smaller than the non-zero $\epsilon$ times $\frac{1}{k}$. Then let $b_N$ be an element of the liminf sequence. We see first that there are $k *\operatorname{floor}(\frac{N}{n})$ intervals of length $\frac{1}{N}$ in $A$. By choice of $N$, we have the following inequalities:
$$\begin{eqnarray*} a_n = \frac{k+1}{n} = \frac{(k+1)\frac{N}{n}}{N} &<& \frac{(k+1)*\operatorname{floor}(\frac{N}{n}) + k+1}{n} \\ &<& \frac{1}{N}\#(B \cap \frac{\mathbb{Z}}{N}) \\ &=& b_N \end{eqnarray*}$$
where the second inequality comes from the choice of $N$, since we may fit enough intervals into the gap between $A$ and $B$ to get at least $k+1$ more endpoints.
Thus, the limit of $a_n$ which is the limsup is less than the limit of $b_n$, the lim inf.