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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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