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 , and call equivalent if . This defines an equivalence relation, and thus we can define a set to consist of exactly one representative from each equivalence class. Then the interval is just the disjoint union of translates mod of by rational numbers mod (or something). Because the rationals in are countably infinite this is incompatible with being measurable: if then then , and if then .

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 and we’re generally thinking about mod then we’re probably better off just talking about the circle and not confusing the issue. Moreover, now translation mod is just rotation, which is much easier to visualize and, importantly, much easier to believe is measure-preserving. Now we just need to decompose into a countably infinite collection of copies of some set . Towards this end, let be any irrational rotation of . (When I was talking yesterday, I started with the concrete example of rotation by degree, and then hastily changed this to rotation by radian.) Then for any point we can consider the orbit , and we thus decompose into a union of orbits under the action of . Let be a set consisting of exactly one representative of each of these orbits. Since every orbit is infinite, is the disjoint union of

Since is measure-preserving, we have a contradiction as before.

Thanks for this blog postt

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