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We know that the algebra $\mathcal{A}$ of finite disjoint unions of intervals of the form $(a,b]$ for $a, b \in \mathbb{R}$ generates the Borel $\sigma$-algebra $\mathcal{B}_{\mathbb{R}}$. Is it true that every Borel set $A \in \mathcal{B}_{\mathbb{R}}$ can be written as a countable disjoint union of elements in $\mathcal{A}$?

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No.

Take the rationals $\Bbb{Q}$

It is a countable union of closed sets(thus Borel),but it cannot be expressed as a countable disjoint union of finite disjoint unions of half-open intervals because then it would have non-empty interior.(which of course cannot be the case)

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  • $\begingroup$ But aren't you able to produce singletons via intersections? Intersections can be represented as unions of complements. Hence, a countable union of singletons will give you $\mathbb{Q}$ $\endgroup$ Dec 3, 2019 at 23:57
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    $\begingroup$ @FredericChopin you ask specifically for unions..an i answred that cannot every Borel set be a countable UNION of elements of $A$.. $\endgroup$ Dec 3, 2019 at 23:59
  • $\begingroup$ Sorry, even though the algebra is closed under finite intersections, this means that there won't be any singletons in $\mathcal{A}$. $\endgroup$ Dec 4, 2019 at 0:10
  • $\begingroup$ So just to add, an example of a uncountable set that cannot be written as a disjoint union of elements of $\mathcal{A}$ is a fat Cantor set because it has empty interior (it also has positive measure). Is this correct? $\endgroup$ Dec 4, 2019 at 0:14
  • $\begingroup$ @FredericChopin i believe it is..because in a union of half open intervals you can find a small open interval inside a half open.. $\endgroup$ Dec 4, 2019 at 0:17

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