Proof of existence of nonmeasurable sets

[mathmonkey]

Hi,

I'm reading through a proof of the existence of a nonmeasurable set. I've copied down the proof below more or less verbatim:

Let $C$ be a circumference of length 1, and let $\alpha$ be an irrational number. Partition the points of $C$ into classes by the following rule: Two points of $C$ belong to the same class if and only if one can be carried into the other by a rotation of $C$ through an angle $n\alpha$ where $n \in \mathbb{Z}$. Each class is clearly countable. We now select a point from each class. We show that the resulting set $\Phi$ is nomeasurable. Denote by $\Phi _n$ the set obtained by rotating $\Phi$through the angle $n\alpha$. It is easily seen that all the sets $\Phi _n$ are pairwise disjoint and that their union is $C$. If the set $\Phi$ were measurable, the sets $\Phi _n$ congruent to it would also be measurable. Since $C = \bigcup _{n= -\infty}^\infty \Phi _n$ and $\Phi _n \cap \Phi _m = \emptyset$, the additivity of the measure would imply that $\sum _{n=-\infty}^\infty m(\Phi _n) = 1$. But $m(\Phi _n) = m(\Phi)$ for all $n$. Hence, $\Phi$ cannot be measurable.

In particular, I am trying to understand the significance of why $\alpha$ has to be an irrational number. Would the proof not hold if we used any other number $\alpha \in \mathbb{R}$? For your reference, this proof was given in Kolmogorov's "Elements of the Theory of Functions and Functional Analysis". I'd really appreciate if someone could explain this for me. Thanks!

 

[HallsofIvy]

If \alpha were a rational number, say \alpha= p/q for integers p and q, then n\apha would be an integer for some n. The sets would not be "pairwise disjoint".

 

[mathmonkey]

If \alpha were a rational number, say \alpha= p/q for integers p and q, then n\apha would be an integer for some n. The sets would not be "pairwise disjoint".

Ah okay I see. In the case that $\alpha \in \mathbb{Q}$, each $\Phi _n$ is either pairwise disjoint or exactly equal to $\Phi _m$ (for example $\Phi _0 = \Phi _{360}$ if $\alpha = 1$, assuming we are using degree angles), so that consequently we wouldn't have the the infinite union $C = \bigcup _{n=-\infty}^\infty \Phi _n$ but rather a finite union $C = \bigcup _{n=1}^k \Phi _n$ after removing redundant sets, where the contradiction would no longer hold. Is this more or less what you are hinting at? Thanks!