Measure Theory and Cantor Set (couterexample). > Consider $O_{n} = \\{y \in \mathbb{R}^{d} \mid \exists x \in E \text{ with } |y - x| < \frac{1}{n}\\}$ where $E$ is a mensurable set. > > (a) Prove that if $E$ is compact, then $m(E) = \lim m(O_{n})$. > > (b) Show that there is $E$ closed and unbounded such that $m(E) < \lim m(O_{n})$. > > (c) Show that there is $E$ open and bounded such that $m(E) < \lim m(O_{n})$. I proved the items (a) and (b). But I'm having trouble with other item. For item (c), I'm trying with the following set: Let $\mathcal{C}$ the Cantor Set, then $[0,1]\setminus \mathcal{C}$ is open. But, I don't know how to "work" with $O_{n}$ in this case. I would like help with this example, or some other example would also be helpful

What you need is a fat Cantor set, since the original Cantor set has measure zero and your strict inequality cannot be attained.

Let $C_F \subset [0,1]$ be a fat Cantor set and let $E = [0,1]\setminus C_F$. Then for all $n$ we have $[0,1] \subset O_n$: let $x\in [0,1]$. If $x\in C_F$ then $x\in O_n$ by definition. If $y\in C_F$. Then there is $x\in E$ so that $|y-x|<1/n$, since $E$ is dense in $[0,1]$. Thus $y\in O_n$.

Thus $m(O_n) \ge 1 > m(E)$ for all $n$ and $\lim m(O_n) > m(E)$.

The example by David is probably simpler (and the argument are the same)