Is every Borel set a countable union of intervals? Since the Borel $\sigma$ algebra is generated by all open subsets of $\mathbb{R}$ and all open sets are the countable union of disjoint open intervals, I figured that any Borel set is the countable union of intervals. I also used the fact that if we only use '$\sigma $ algebra operations' on intervals then we get an interval or countable union of intervals.
But I ask if this is actually true?

The Cantor set $C$ is an uncountable Borel set and it does not contain any non-trivial interval (i.e. a singleton).
Hence $C$ can not be written an as countable union of intervals.
However, its complement is indeed a countable union of disjoint open intervals!
