What is the least crazy/degenerate example of a non-measurable subset of $\mathbb R$ I know that it is possible to construct extremely weird subsets of $\mathbb R$, or of $\mathbb R^3$, that are non-Lebesgue-measurable. For example, the ones used in the Banach-Tarski paradox.
Are there "less crazy" examples of non Lebesgue-measurable subsets of $\mathbb R$?
I am asking in order to get a feeling of how degenerate a set has to be before I have to start worrying about Lebesgue-measurability problems.
 A: The simplest example I know of is a Vitali set (partition $[0,1]$ into equivalence classes of the relation $\sim$, where $x \sim y$ iff $x - y$ is rational, and then pick exactly one member of each equivalence class).
This construction is sort of "characteristic" of non-measurable sets, in my experience - a set that isn't Lebesgue-measurable can't be Borel (or $\mathbf{\Sigma^1_1}$ or $\mathbf{\Pi^1_1}$, which are basically the next level up) which means that they can't have "nice" topological definitions. Another way of thinking about it is that the standard proof that not all subsets of $\mathbb{R}$ are Lebesgue-measurable relies on the Axiom of Choice, so you can expect any examples to be "choicey".
My rule of thumb is that if a set has a natural topological definition, it's probably measurable. If its definition involves "choosing" something infinitely many times, then it's probably not measurable.
A: Not exactly as you want, but non-measurable sets must be “crazy,” because all Borel sets are measurable (in fact, measurable sets are equal to Borel sets up to a set of measure zero).
A: Solovay showed that is consistent with ZF set theory (without the full Axiom of Choice but with Dependent Choice and an inaccessible cardinal) that all sets of reals are measurable.  Thus any example of a nonmeasurable set must be "crazy".
