Suppose that there exists such a p dene an

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Unformatted text preview: o if P is to be our candidate for the uniform probability, then it is reasonable to assume that P {A ⊕ r} = P {A} for any 0 < r < 1. To prove that no uniform probability exists for every A ∈ 2[0,1] we will derive a contradiction. Suppose that there exists such a P. Define an equivalence relation on [0, 1] by setting x ∼ y iff y − x ∈ Q. For instance, 1 1 ∼, 2 4 1 1 ∼, 9 27 1 1 ∼, 9 4 1 1 ￿∼ , 3 π 1 1 ￿∼ , e π 11 11 − ∼ +. π4 π2 This equivalence relationship partitions [0, 1] into a disjoint union of equivalence classes (with two elements of the same class differing by a rational, but elements of different classes differing by an irrational). Let Q1 = [0, 1] ∩ Q, and note that there are uncountably many equivalence classes. Formally, we can write this disjoint union as ￿ ￿ [0, 1] = Q1 ∪ {(Q + x) ∩ [0, 1]} = Q1 ∪ {Q1 ⊕ x} . x∈[0,1]\Q1 x∈[0,1]\Q1 4–2 1 4 1 8 15 16 1 π 1 π 1 π...
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