TITLE: Can any subset of $\mathbb{R}$ be generated from open intervals?
QUESTION [8 upvotes]: Can any subset of $\mathbb{R}$ be generated by taking countable unions, countable intersections and complements of open intervals?

Clearly, singletons can be generated from the complement of the union of half-rays, e.g.: $$a=((-\infty, a) \cup (a, \infty))^C .$$ 
Closed intervals can also be generated by a countable intersection of open sets of the form $(a- \frac1n, b+\frac1n)$ and similarly for half-open intervals.  
From this, it seems obvious that any countable union/intersection of intervals can be generated. 
How about for example uncountable unions/intersections of intervals?  It is unclear to me whether this is enough to generate all subsets of  $\mathbb{R}$. Can this be done?

REPLY [2 votes]: ______Editing my comment to extend the answer_______________
Hint: There is an example for Lebesgue but not Borel measurable sets.
Additionally:-
If you're familiar with measure theory, then there's a couple results that are somewhat related to this matter.

Every non-empty open subset $U \subseteq \mathbb{R}$ is a disjoint union of open intervals.
Littlewood's first principle of Analysis

Every measurable set $E\subset \mathbb{R}$ with $m(E)<$ $\infty$ is almost a finite union of intervals.