Any open subset $G$ of the real line can be written as a countable disjoint union of open intervals say $G=\bigcup_{i=1}^{\infty} (a_i,b_i)$ which its closure is $\operatorname{cl}(G)=\bigcup_{i=1}^{\infty} [a_i,b_i]$. My thought is that $\bigcup_{i=1}^{\infty} (a_i,b_i)$ and $\bigcup_1^{\infty} [a_i,b_i]$ differ by a countable set and since every countable set has Lebesgue measure zero, it follows that the following example must give the same Lebesgue measure $1> \epsilon =1 $ which is not possible :
There is a contradiction comparing Lebesgue measure of an open set and its closure and I can't solve the following dilemma from this answer:
Another way of doing it is to enumerate the rationals in $[0,1]$ and taking
$$E = [0,1] \cap \bigcup_{n=1}^{\infty} \left(q_{n} – \frac{\varepsilon}{2^{n+1}}, q_{n} + \frac{\varepsilon}{2^{n+1}}\right).$$
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.
Best Answer
It is not true that $cl(G)=\cup_i [a_i,b_i]$. The formula $Cl(\cup_i A_i)=\cup_i (Cl(A_i)$ holds for finite families of sets but not in general. For example if you consider the collection of all singleton sets $\{r\}$ where $r$ is rational you can see that the union of the closures is countble but the closure of the union is $\mathbb R$.