Therefore we can write 0 1 as a disjoint countable

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Q1 − 1 4 + 1 e 1 e 1 e 1 2 − 1 3 + 4 7 (Q + 1 ) ∩ [0, 1] e 1 (Q + π ) ∩ [0, 1] 1 √ 2 1 √ 2 1 √ 2 (Q + − 1 5 + 1 2 1 √) 2 ETC. ∩ [0, 1] Let H be the subset of [0, 1] consisting of precisely one element from each equivalence class. (This step uses the Axiom of Choice.) For definiteness, assume that 0 ￿∈ H . Therefore, we can write (0, 1] as a disjoint, countable union of shifts of H . That is, ￿ (0, 1] = {H ⊕ r } r ∈Q1 ,r ￿=1 with {H ⊕ ri } ∩ {H ⊕ rj } = ∅ for all i ￿= j which implies ￿ ￿ ￿ ￿ P {(0, 1]} = P {H ⊕ r } = P {H ⊕ r} = r ∈Q1 ,r ￿=1 r ∈Q1 ,r ￿=1 In other words, 1= ￿ r ∈Q1 ,r ￿=1 ￿ r ∈Q1 ,r ￿=1 P {H } . P {H } . We have now arrived at our contradiction. Suppose that we wish to assign probability p = P {H } to the set H . The previous line tells us that p satisfies ￿ 1= p. (4.2) r ∈Q1 ,r ￿=1 However, since p is a number between 0 and 1, there are two poss...
View Full Document

This document was uploaded on 04/07/2014.

Ask a homework question - tutors are online