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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 $E$?

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How about $\mathbb{Q} \cap [0,1]$? –  user17762 Mar 20 '11 at 3:49
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@Sivaram: This subset is not open. –  Matt E Mar 20 '11 at 3:57
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Oh Right. I didn't notice the open set part... But won't an epsilon covering of $\mathbb{Q} \cap [0,1] $ i.e. $\displaystyle \cup_{i=1}^{\infty} \left(q_i-\epsilon/2^i,q_i+\epsilon/2^i \right)$ work? –  user17762 Mar 20 '11 at 4:00
    
@Sivaram: sure, this works. Sorry I hadn't seen your comment until now. –  t.b. Mar 20 '11 at 4:12
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@Matt, @Sivaram: On a related note: If you write $E_{\varepsilon}$ for Sivaram's set and take $E = \bigcap E_{1/n}$ then you'll get a dense null set of second Baire category (of course no longer open) and the complement will be a meager set of full measure. This construction shows that genericity in the sense of Baire and genericity in the sense of "with probability $1$" can be rather 'orthogonal' notions. There are two famous papers in ergodic theory, one by Rohlin and one by Halmos, with the respective titles "In general, a transformation is [is not] mixing" playing with exactly this issue. –  t.b. Mar 20 '11 at 13:00
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2 Answers

up vote 14 down vote accepted

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.

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In fact, given $a<b$ two real numbers (or $b=+\infty$) we can find an open set $E$ such that $m(E)=a$ and $m(\bar E)=b$. –  Davide Giraudo Dec 29 '11 at 17:44
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Yes, that's not too hard to achieve from what I say here if $b$ is finite. Just scale the interval $[0,1]$ to $[0,b]$ and take $\alpha = b-a$ (and of course you want $a \gt 0$). If $b$ is infinite, partitioning $[0,\infty)$ into $[n,n+1]$ and choosing fat Cantor sets of appropriate measure will do. Anyway: thanks for pointing it out! –  t.b. Dec 29 '11 at 18:24
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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.

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Dear Matt, We're very simultaneous today... –  t.b. Mar 20 '11 at 3:57
    
@Theo: Indeed! Best wishes, –  Matt E Mar 20 '11 at 3:59
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