Generator of the Borel $\sigma$-algebra and measureable sets

I have a question regarding the relationship between measureable Borel sets and the generator of the Borel $$\sigma$$-algebra. I know that $$\sigma$$-algebras are stable under countable unions by definition and the set of intervalls $$\mathcal{E}=\{\, ]r,s]\subset \mathbb{R}: r\leq s\}$$ generates $$\mathcal{B}(\mathbb{R})$$.

So every countable union of elements of $$\mathcal{E}$$ is measureable. But what about the converse? Is every set $$M\in\mathcal{B}(\mathbb{R})$$ representable by a countable union of elements of $$\mathcal{E}$$?

1 Answer

No. Consider the irrational numbers. Since they are totally disconnected, the only way to write them as a union of intervals is $$\bigcup_{x \in \mathbb{R} - \mathbb{Q}} [x,x]$$. This is clearly an uncountable union. Further they are clearly measurable as they are the complement on the rationals, which are a countable set, hence a countable union of length-0 closed intervals (AKA points).