# Can every Borel set be written as a disjoint union of elements in the algebra of half open intervals?

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 $$\mathcal{A}$$?

Take the rationals $$\Bbb{Q}$$
• 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 $\mathbb{Q}$ Dec 3, 2019 at 23:57
• @FredericChopin you ask specifically for unions..an i answred that cannot every Borel set be a countable UNION of elements of $A$.. Dec 3, 2019 at 23:59
• Sorry, even though the algebra is closed under finite intersections, this means that there won't be any singletons in $\mathcal{A}$. Dec 4, 2019 at 0:10
• 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? Dec 4, 2019 at 0:14