Can you implicitly construct a non-Borel subset $X \subset \mathbb{R}$ such that $\overline{X} = \mathbb{R}$? I'm just now reading into measure theory and the idea of a $\sigma$ algebra. My textbook says implicit constructions of non-Borel subset of $\mathbb{R}$ with AOC are possible.  Can you construct such a non-Borel subset $X \subset \mathbb{R}$ such that $\overline{X} = \mathbb{R}$?
If so, how would you go about such a construction?
 A: It is possible to construct such a set even without the axiom of choice, but the axiom of choice certainly makes it easier:
First construct any non-Borel set $A\subseteq\mathbb{R}$. For example, the Vitali set is a construction by first partitioning $\mathbb{R}$ into the subsets $\{a+\mathbb{Q}:a\in\mathbb{R}\}$, then from each set $a+\mathbb{Q}$ choosing a distinct member to form the set $A$. This $A$ is not even in the Lebesgue $\sigma$-algebra, so it's not in the Borel $\sigma$-algebra. There are other wats to make such $A$ (and even ways without invoking the axiom of choice), so I'll assume that you've seen at least one.
Then the key observation is, we can make $A$ into a dense set without losing its non-Borel-ness! Consider the set $B=A\cup\mathbb{Q}$, which is certainly dense in $\mathbb{R}$. If $B$ were Borel, then note that $A\cap\mathbb{Q}$ is a countable set and hence Borel, and $A=(B\setminus\mathbb{Q})\cup(A\cap\mathbb{Q})$ is the union of two Borel sets and hence Borel, a contradiction.
This $B\subseteq\mathbb{R}$ is what you wanted.
