# Proof of existence of nonmeasurable sets

1. Dec 22, 2012

### 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:

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!

2. Dec 22, 2012

### HallsofIvy

Staff Emeritus
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".

3. Dec 22, 2012

### mathmonkey

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!