For a subset $S\subset[0,1]$ with $0<S<1$ ($S$ is the Lebesgue measure of $S$) we define the multiplicity function of order $n$ $m_{n,S}:[0,1] \rightarrow \{0,1,\ldots,n\}$ in the following way: at point $x$ the function $m_{n,S}(x)$ is the number of elements in the set $S\cap (x+\frac{1}{n}\mathbb{Z})$.
It is clear that if $S$ is dense in some interval then there exists $c>0$ (which may depend on $S$) such that for infinitely many values $n\in\mathbb{N}$ we have $m_{n,S}\geq n\cdot c$ on a set of large measure (could be a set of full measure).
My question is whether it is possible to construct a nowhere dense set $S\subset[0,1]$ (of positive measure) for which there exists a positive constant $c$ and a sequence $n_k\rightarrow\infty$ such that for all $k\in\mathbb{N}$ one has $m_{n_k,S}\geq n_k\cdot c$ on a set of large measure, say at least $\frac{1}{2}$ (for starters I would be happy with having this property on a set of positive measure).
Yes. First fix an increasing sequence of natural numbers $(n_k)$ such that $\frac{k}{n_k} \to 0$ as $k \to \infty$, $\alpha = \sum_{i=1}^\infty \frac{1}{n_i} < 1$, and $\beta_k = n_k\cdot \sum_{i=k+1}^\infty \frac{1}{n_i} \to 0$ as $k \to \infty$. Let $(x_k)$ be a countable dense subset of $[0,1]$ and let $S$ be what remains of $[0,1]$ after an open interval $U_k$ of length $\frac{1}{n_k}$ centered at $x_k$ is removed, for all $k$. The set $S$ is closed and contains no intervals of positive length, so it is nowhere dense. Let $U$ be the complement of $S$, i.e., the union of the removed intervals.
Fix $x \in [0,1]$ and $k \in \mathbb{N}$. For any $i$, the number of elements of $U_i \cap (x + \frac{1}{n_k}\mathbb{Z})$ is at most $\frac{n_k}{n_i} + 1$. So the number of elements of $\bigcup_{i=1}^k U_i \cap (x + \frac{1}{n_k}\mathbb{Z})$ is at most $n_k\alpha + k = n_k\cdot (\alpha + \frac{k}{n_k})$.
Now consider $$A_k = \left\{x \in [0,1]: U_i \cap \left(x + \frac{1}{n_k}\mathbb{Z}\right) \neq \emptyset\mbox{ for some }i > k\right\}.$$ The set of $x$ for which the intersection with $U_i$ is nonempty is precisely $[0,1] \cap (U_i + \frac{1}{n_k}\mathbb{Z})$, which has measure exactly $\frac{n_k}{n_i}$ (for $i > k$), so the entire set $A_k$ has measure at most $\beta_k$. Every point outside of this set will therefore satisfy $$\leftU \cap \left(x + \frac{1}{n_k}\mathbb{Z}\right)\right \leq n_k\cdot \left(\alpha + \frac{k}{n_k}\right)$$ by the previous estimate. That is, for every $x \not\in A_k$, the fraction of $\frac{1}{n_k}$translates of $x$ which lie in $S$ is at least $1  (\alpha + \frac{k}{n_k})$, and the measure of $A_k$ goes to zero as $k \to\infty$.

$\begingroup$ Thanks. It seems that the condition $\beta_k \rightarrow 0$ forces $n_k$ to be extremely fast increasing which should imply the first condition automatically, no? $\endgroup$ – Itay Jan 12 '17 at 14:44

$\begingroup$ Yeah, probably. I guess it's not hard to find a specific sequence with these properties, if you like. $\endgroup$ – Nik Weaver Jan 12 '17 at 14:45