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 $\mathcal{A}$?
Can every Borel set be written as a disjoint union of elements in the algebra of half open intervals
measure-theoryreal-analysis
Best Answer
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)