Let $X$ be a subset of $\mathbb{R}$ and $\lambda$ be the Lebesgue measure on $\mathbb{R}$ and define $$ X_{\epsilon}:=\{x \in \mathbb{R}\,|\,\inf_{y \in X} |x-y|<\epsilon\}. $$ I want to find an open set $X$ that satisfies $$ \inf_{\epsilon >0} \lambda(X_{\epsilon})>\lambda(X). $$
I've found some non-open sets that work. For example $\mathbb{Q}\cap[0,1]$ has Lebesgue measure $0$ since it's countable and other other term is clearly positive. I can't find an open set that works. If I take $X=(a,b)$, then $X_{\epsilon}=(a-\epsilon,b+\epsilon)$. I get $\lambda(X)=b-a$ and I get the same for $\inf_{\epsilon >0} \lambda(X_{\epsilon})$.
Any ideas?
Consider the rational points in $[0,1]$ and enumerate them (in arbitrary fashion) as $\{r_n\}_{n=1}^\infty$, and define $$ X := \bigcup\limits_{n=1}^\infty (r_n - 10^{-n}, r_n + 10^{-n} ). $$ Clearly $X$ is an open set of Lebesgue measure bounded above by $2\sum\limits_{n=1}^\infty 10^{-n} = 2/9$.
Now $X$ contains all rationals of $[0,1]$ which are dense in $[0,1]$. Hence any point of $[0,1]$ can be approximated, with arbitrary precision, by points of $X$. In particular $[0,1] \subset X_\varepsilon$ for any $\varepsilon > 0$. Hence $$ \inf_{\varepsilon>0} \mu(X_\varepsilon ) \geq 1 > \frac {2}{9} > \lambda ( X ). $$