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?
 A: 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!
A: The sets you are talking about are special cases of what are called $\boldsymbol{\Sigma}^0_3$ sets using the Borel hierarchy notation.
To explain, we start with intervals. Every interval can be written as an intersection of countably many open intervals. Sets obtained this way are also called $G_\delta$ sets, and denoted by $\boldsymbol{\Pi}^0_2$ in the Borel hierarchy. So every interval is a $\boldsymbol{\Pi}^0_2$ set. That means that a set that is a union of countably many intervals is, in particular, a union of countably many $\boldsymbol{\Pi}^0_2$ sets. Such sets are denoted by $\boldsymbol{\Sigma}^0_3$ sets. However, the Borel hierarchy does not stop here, but rather continues up through $\boldsymbol{\Pi}^0_\alpha$ and $\boldsymbol{\Sigma}^0_\alpha$ sets for all countable ordinals $\alpha$. Moreover, it can be shown that every level of this hierarchy is proper. So in conclusion there are Borel sets that are much more complicated than unions of countably many intervals.
