This is a [crosspost](https://math.stackexchange.com/questions/4445539/is-there-an-open-subset-a-of-0-12-with-measure-frac1100-that-sati) from MSE.


Can we find for any given $\varepsilon>0$ an open subset $A\subseteq[0,1]^2$ with measure $>\frac{1}{100}$ such that, for any smooth curve $\gamma:[0,1]\to\mathbb{R}^2$ of length $1$, the set $\gamma+A=\{\gamma(t)+a;t\in[0,1],a\in A\}$ does not contain any balls of radius $\varepsilon$?


I wouldn't mind changing the $\frac{1}{100}$ for any other positive constant. Also, I ask about smooth curves but it may make more sense to consider in general $1$-Lipschitz functions $\gamma:[0,1]\to[0,1]^2$.


I came across (something similar to) this question while thinking about [this one](https://mathoverflow.net/questions/420897/what-is-the-smallest-size-of-a-shape-in-which-all-fixed-n-polyominos-can-fit/421175#421175).



Is there an open subset $A$ of $[0,1]^2$ with measure $>\frac{1}{100}$ that satisfies this property?
The answer is no: if $\varepsilon$ is small enough, then for every open $A \subset [0,1]^2$ of measure at least $1/100$, there exists a smooth curve $\gamma$ of length $\leq 1$ such that $\gamma+A$ contains a $\varepsilon$-ball. The idea is to randomly shift by translates first at extremely small scales to cover a lot of balls (or squares) at that scale, then at less small scales to cover a lot of balls (or squares) at this larger scale, and so forth until one is able to cover a macroscopic ball or square.


First some reductions. By enlarging $\varepsilon$ a little bit, we can assume it is a negative power of two, and replace "$\varepsilon$-ball" by "dyadic $\varepsilon$-square". By inner regularity, $A$ contains a finite union of balls of measure at least $1/200$, hence contains a finite union $A'$ of closed dyadic squares of measure at least $1/400$. If $N$ is large enough, we can view $A'$ as a union of dyadic $2^{-N}$-squares, and assume that the $2^{-N}$-neighbourhood of $A'$ lies in $A$. It will then suffice to find a piecewise polygonal path $\gamma'$ of length at most $1/2$ such that $\gamma'+A'$ contains a dyadic $\varepsilon$-square, since one can smooth $\gamma'$ out at a scale much less than $2^{-N}$ to find a smooth curve $\gamma$ of length at most $1$ such that $\gamma+A \subset \gamma'+A'$.


Now introduce the hyperdyadic scales $\varepsilon\_n := \varepsilon^{2^n}$ and integers $J\_n := \lfloor \log^{100} \frac{1}{\varepsilon\_n} \rfloor$. Clearly there is $n\_0$ such that $\varepsilon\_{n\_0+1} \leq 2^{-N}$, hence $A'$ is now a union of dyadic $\varepsilon\_{n\_0+1}$-squares. We will choose, for each $n=0,\dots,n\_0$, a set $H\_n$ of $J\_n$ shifts in the ball $B(0, 10 \varepsilon\_n)$. By an easy induction on $n\_0$, it is possible to find a piecewise polygonal path $\gamma'$ passing through the sumset
$$ H\_0 + H\_1 + \dots + H\_{n\_0}$$
(which is a finite set of cardinality at most $J\_0 \dots J\_{n\_0}$) of length at most
$$ \sum\_{n=0}^{n\_0} 20 \varepsilon\_n J\_0 \dots J\_n$$
which one can calculate to be less than $1/2$ if $\varepsilon$ is small enough (the double exponential decay of the $\varepsilon\_n$ beats the quadratic exponential growth of the $J\_0 \dots J\_n$). Hence it will suffice to locate $H\_0,\dots,H\_{n\_0}$ such that
$$ H\_0 + H\_1 + \dots + H\_{n\_0} + A'$$
contains a dyadic $\varepsilon$-square.


We will establish the more general claim that for any $0 \leq n \leq n\_0+1$, we can select $H\_n,\dots,H\_{n\_0}$ such that
$$ H\_n + \dots + H\_{n\_0} + A'$$
contains a union of dyadic $\varepsilon\_{n}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-1}$; setting $n=0$ will give the claim.


We prove this claim by downward induction on $n$. When $n=n\_0+1$ the claim follows since $A'$ itself is a union of dyadic $\varepsilon\_{n\_0+1}$ squares in $[0,1]^2$ of measure at least $\frac{1}{400}$. Now assume inductively that $0 \leq n \leq n\_0$ and the claim has already been proven for $n+1$, thus we have already located $H\_{n+1},\dots,H\_{n\_0}$ such that
$$ H\_{n+1} + \dots + H\_{n\_0} + A'$$
contains a union $B$ of dyadic $\varepsilon\_{n+1}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-2}$.


We cover $B$ by dyadic $\varepsilon\_n$-squares. If we let $B'$ be the union of all the dyadic $\varepsilon\_n$-squares in which $B$ has relative density at least $1000^{-n-2}$, then $B \backslash B'$ has measure at most $1000^{-n-2}$, hence $B'$ (which contains $B \cap B'$) has measure at least $\frac{1}{400} - 2 \times 1000^{-n-2}$. We now use the probabilistic method, picking $H\_n$ to be $J\_n$ random elements of $B(0,10\varepsilon\_n)$. A standard union bound calculation (discretising each $\varepsilon\_n$-square at scale $\frac{1}{10} \varepsilon\_{n+1}$, say into a lattice of cardinality $O( (\varepsilon\_n/\varepsilon\_{n+1})^2 )$) shows that each $\varepsilon\_n$-square in $B'$ will lie in $H\_n+B$ with probability at least
$$ 1 - O( (\varepsilon\_n/\varepsilon\_{n+1})^2 ( 1 - 1000^{-n-3} )^{J\_n} )$$
which by the choice of parameters can be seen to be at least $1 - 1000^{-n-2}$. Thus, by linearity of expectation (first moment method), one can choose $H\_n$ so that $H\_n+B$ covers a subcollection of $\varepsilon\_n$-squares in $B'$ of measure at least $\frac{1}{400} - 1000^{-n-1}$, closing the induction.


