Lecture Notes - Department of Mathematics and Statistics - Queen's ...

8 CHAPTER 1. REVIEW OF PROBABILITY
For otherwise there would be a point x which were not in any equivalence class. But x is equivalent with itself
at least. Furthermore, since A contains only one point from each equivalence class, the sets A ⊕ q are disjoint;
for otherwise there would be two sets which could include a common point: A ⊕ q and A ⊕ q ′ would include a
common point, leading to the result that the difference x − q = z and x − q ′ = z are both in A, a contradiction,
since there should be at most one point which is in the same equivalence class as x − q = z. We expect the
uniform distribution to be shift-invariant, therefore P(A) = P(A ⊕ q). But [0, 1] = ∪ q A ⊕ q. Since a countable
sum of identical non-negative elements can either become ∞, or 0, the contradiction follows: We can’t associate
a number with this set.