- #1
mathmonkey
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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!
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!