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.

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