**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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