851lecture04

# Suppose that there exists such a p dene an

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 04/07/2014.

Ask a homework question - tutors are online