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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 $\mathcal{E}$?

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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).

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