# A confusion in understanding Lebesgue measure

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.

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 $$\mathbb R$$.
• @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... – Krishnarjun Apr 15 at 7:37
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.