# Lebesgue Measure Inequality (neighbouhood of a set)

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 $$\inf_{\varepsilon>0} \mu(X_\varepsilon ) \geq 1 > \frac {2}{9} > \lambda ( X ).$$