851lecture04

# Therefore we can write 0 1 as a disjoint countable

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

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