# A simpler example of a nonmeasurable set

I don’t claim that I’m the first person to have come up with this example; I only claim that yesterday is the first time I thought of it. This is a rather sorry state of affairs, because it’s an obvious and nice example.

First, some background, the example of a nonmeasurable set that I was given when I first learnt measure theory went basically as follows. Start with ${[0,1]}$, and call ${x,y\in[0,1]}$ equivalent if ${x-y\in{\bf Q}}$. This defines an equivalence relation, and thus we can define a set ${Z}$ to consist of exactly one representative from each equivalence class. Then the interval ${[0,1]}$ is just the disjoint union of translates mod ${1}$ of ${Z}$ by rational numbers mod ${1}$ (or something). Because the rationals in ${[0,1]}$ are countably infinite this is incompatible with ${Z}$ being measurable: if ${\mu(Z)>0}$ then then ${\mu([0,1])=\infty}$, and if ${\mu(Z)=0}$ then ${\mu([0,1])=0}$.

This example is fine and beautiful and all that, but I find it rather hard to remember and especially difficult to visualize. Thus, yesterday, when I was explaining to my girlfriend in an intuitive way why not all sets are measurable, the example I produced was, by accident, not the above example.

From an intuitive point of view, if we’re talking about ${[0,1]}$ and we’re generally thinking about mod ${1}$ then we’re probably better off just talking about the circle ${S^1}$ and not confusing the issue. Moreover, now translation mod ${1}$ is just rotation, which is much easier to visualize and, importantly, much easier to believe is measure-preserving. Now we just need to decompose ${S^1}$ into a countably infinite collection of copies of some set ${Z}$. Towards this end, let ${T}$ be any irrational rotation of ${S^1}$. (When I was talking yesterday, I started with the concrete example of rotation by ${1}$ degree, and then hastily changed this to rotation by ${1}$ radian.) Then for any point ${x}$ we can consider the orbit ${\{T^n(x)\}_{n\in\mathbb{Z}}}$, and we thus decompose ${S^1}$ into a union of orbits under the action of ${T}$. Let ${Z}$ be a set consisting of exactly one representative of each of these orbits. Since every orbit is infinite, ${S^1}$ is the disjoint union of $\displaystyle \ldots,T^{-2}(Z), T^{-1}(Z), Z, T(Z), T^2(Z),\ldots.$

Since ${T}$ is measure-preserving, we have a contradiction as before.