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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.

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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 $\mathbb R$.

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  • $\begingroup$ Yes I guessed where the problem may come from but still doesn't solve the paradox: number of added points still is countable so will add zero measure. $\endgroup$ – L.G. Apr 15 at 7:33
  • $\begingroup$ I think the closure formula is true here because open intervals are considered $\endgroup$ – Krishnarjun Apr 15 at 7:33
  • $\begingroup$ @Krishnarjun Do you have a proof? $\endgroup$ – Kavi Rama Murthy Apr 15 at 7:34
  • $\begingroup$ @KaviRamaMurthy I am thinking of one now... Say $A$ is the union of the open sets and $B$ is the union of respective closures. Then clearly $B\subset \bar{A}$. Suppose you have a convergent sequence in $A$, then considering possibly a subsequence, we can consider the sequence is inside one of the open intervals of $A$. Then the limit is inside $B$. Thus $\bar{A}\subset B$ I am trying to see if there is any mistake here... $\endgroup$ – Krishnarjun Apr 15 at 7:37
  • $\begingroup$ What OP has written shows that the union of the closures cannot be the closure of the union in this case. @Krishnarjun $\endgroup$ – Kavi Rama Murthy Apr 15 at 7:39
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That the closure of $E$ equals $[0,1]$ is clear because it contains $\Bbb Q \cap[0,1]$ which is dense. This actually shows very clearly that $\operatorname{cl}(\bigcup_n E_n ) \neq \bigcup_n \operatorname{cl}(E_n)$ in general (the fomer set has measure $1$, the sum of the measures on the right is at most $\varepsilon$). The equality holds for finite and locally finite unions.

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