TITLE: R subset and Borel set
QUESTION [6 upvotes]: Which is the simplest example of a subset of R that is not a Borel set?
Ps: I asked my probability professor how I can see if a subset of R is a Borel set or not? He answered that in the case we will use the set are always Borel set and in fact he said that all subset of R are Borel set. So at this point I was very interested: why I can't imagine a subset of R that is not a Borel set? So I asked him an example of a subset of R and he answered that he will not do that because it will only confuse me and because is not important. So I'm asking here!
Ps2: I don't have any background of topology, geometry and so on. I know analysis (I-II) and linear algebra (I-II) so keep it simple please.

REPLY [6 votes]: I would like to give an answer that is similar to Asaf's but a bit less pessimistic.
One can define the notion of a Borel code, which is a recipe for building up a set of reals from intervals under the operations of countable unions, countable intersections, and complementation.  Every Borel code specifies a particular Borel set.
A bit of elementary set theory shows that there are only as many Borel codes as there are real numbers, so we can list all the Borel codes as $(c_x : x \in \mathbb{R})$.  Let $B_x$ be the Borel set coded by $c_x$.  Define $A = \{x \in \mathbb{R} : x \notin B_x\}$.  Then the set $A$ is not equal to the set $B_x$ for any real $x$ because $x$ itself is in one but not the other.  (This is Cantor's diagonal argument.)
Therefore we have a set $A$ that has no Borel code.  Whether it is simple or not depends on whether the coding system is simple or not.  I would say that conceptually it is simple but the details are rather tedious.
Because $A$ has no Borel code, it has no good reason to be Borel.  It could still be Borel for a silly reason.  For example, if $\mathbb{R}$ were a union of countably many countable sets, then every set would be Borel for a silly reason.  To rule this out, we need the Axiom of Choice (really just the Axiom of Countable Choice.)
Using the Axiom of Choice, we can show that the collection of sets with Borel codes is closed under countable unions and intersections.  Given a countable sequence of sets $A_1$, $A_2$, $A_3$,... with Borel codes, to get a Borel code for their intersection or union, one first has to choose a code $c_i$ for each set $A_i$, which requires the Axiom of Choice because each $A_i$ will have many Borel codes to choose from.  Then the sequence of codes $(c_i : i<\omega)$ gives you a code for the intersection or union.
Because the collection of sets with Borel codes is also closed under complementation and contains every interval, it must contain all the Borel sets.  So the set $A$ we constructed, having no Borel code, is not Borel.