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).



Nowhere dense set with high multiplicity
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 $$\left|U \cap \left(x + \frac{1}{n\_k}\mathbb{Z}\right)\right| \leq n\_k\cdot \left(\alpha + \frac{k}{n\_k}\right)$$ 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 $k \to\infty$.


