Which subset of $\mathbb{R}$ is not on the Borel $\sigma$-algebra of $\mathbb{R}$?

Question

When I learned this $\sigma$-algebra, which is the smallest $\sigma$-algebra that contains all the open intervals of $\mathbb{R}$ (let's denote it by $\mathcal{B}$), I've got the feeling that no subset of $\mathbb{R}$ is not in it. Because:

1. Of course $\emptyset \in \mathcal{B}$ and $\mathbb{R} \in \mathcal{B}$
2. Also obvious that $(a,b) \in \mathcal{B}$, with $a<b$.
3. Since $[a,b] = \displaystyle\bigcap_{n=1}^{\infty} \left(a - \frac{1}{n}, b + \frac{1}{n}\right)$, then $[a,b] \in \mathcal{B}$
4. $(-\infty,x], (x, \infty), (-\infty,x), [x, \infty) \in \mathcal{B}$
5. In the particular case of (3), if $x=y$, then $\{x\} \in \mathcal{B}$, hence all $\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{Q}^c \in \mathcal{B}$.

So, which subset $X$ of $\mathbb{R}$ is not on $\mathcal{B}$?

I appreciate the attention.

Answer 1

Without the axiom of choice, it is consistent that all sets of real numbers are Borel (and besides, many definitions which usually coincide and are taken to define Borel sets may in fact describe different families).

This means that you cannot describe any Borel set without (maybe implicitly) invoking some choice principle.

Obviously, any non-measurable set is necessarily non-Borel (but the converse is certainly not true: any uncountable set contains a non-Borel subset). This includes e.g. the Vitali sets mentioned in the comments, as well as even more exotic objects like the Bernstein sets (a set which intersects, but does not contain any uncountable closed set).

As mentioned in the comments, there are only $2^{\aleph_0}$ Borel subsets of the reals, compared to $2^{2^{\aleph_0}}>2^{\aleph_0}$ subsets in total. This shows that in some sense, "most" sets are not Borel. Furthermore, one can show that Borel sets fail some desirable closure properties: for example, the projection of a Borel set need not be Borel. This is the motivation for the introduction of analytic and projective sets (the smallest class of sets containing the open sets, closed under countable unions, continuous images and complements) --- there are still only $2^{\aleph_0}$ many of those, but I think it is not a very controversial thing to say that "reasonable" sets are projective.

Furthermore, one can show that every Borel set is either countable or of cardinality $2^{\aleph_0}$ (in fact, contains a subset homeomorphic to the Cantor set). So if the CH fails, you can just take any set of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$, and this set will be non-Borel.