Comparing the Lebesgue measure of an open set and its closure

Let $E$ be an open set in $[0,1]^n$ and $m$ be the Lebesgue measure.

Is it possible that $m(E)\neq m(\bar{E})$, where $\bar{E}$ stands for the closure of $E$?

Solution 1:

Yes, this is possible. Already in dimension $1$. If you take a modified Cantor set $C$ in $[0,1]$, that is a nowhere dense compact subset of positive measure $\alpha \gt 0$. Its complement $E = [0,1] \smallsetminus C$ is open, has measure $1 - \alpha$ and its closure is all of $[0,1]$ by density. In higher dimensions simply take the product $E^n$.

Another way of doing it is to enumerate the rationals in $[0,1]$ and taking $E = [0,1] \cap \bigcup_{n=1}^{\infty} (q_{n} - \frac{\varepsilon}{2^{n+1}}, q_{n} + \frac{\varepsilon}{2^{n+1}})$. Then $\mu(E) \leq \sum_{n=1}^{\infty} 2 \cdot \frac{\varepsilon}{2^{n+1}} = \varepsilon$, so for $\varepsilon \lt 1$ the set $E$ will be open and dense in $[0,1]$ but not all of $[0,1]$ and its closure will be all of $[0,1]$ again.

Added: (in view of Davide's comment below). Note that there is a modified Cantor set $C_\alpha \subset [0,1]$ of any measure $0 \lt \alpha \lt 1$. Its complement $E_{\alpha} = [0,1] \smallsetminus C_\alpha$ is open and dense in $[0,1]$ and has measure $1-\alpha$ and by scaling this shows that for every pair of positive numbers $0 \lt a \lt b$ there is an open set $E_a$ of measure $\mu(E_a) = a$ whose closure $\overline{E_a}$ has measure $\mu(\overline{E_a}) = b$. I leave it as an easy exercise to construct an open set of measure $a \gt 0$ whose closure has infinite measure.

Solution 2:

If you remove a fat Cantor set (that is, a nowhere dense positive measure closed subset) from $[0,1]$ you obtain a dense open subset of $[0,1]$ whose measure is $< 1$. So the answer to your question is yes.