Note that the total length of 0 316 1516 1 is 316

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Unformatted text preview: r = 15/16, then [r, 1/4 + r] = [15/16, 19/16] which when “wrapped around” becomes [0, 3/16] ∪ [15/16, 1]. Note that the total length of [0, 3/16] ∪ [15/16, 1] is 3/16 + 1/16 = 1/4. That is, using finite additivity, P {[r, 1/4 + r]} = P {[0, 3/16] ∪ [15/16, 1]} = 4–1 1 = P {[0, 1/4]} . 4 We can write this (allowing for “wrapping around”) using the ⊕ symbol so that ￿ [r, 1/4 + r], if 0 < r ≤ 3/4, [0, 1/4] ⊕ r = [0, 1/4 + r − 1] ∪ [r, 1], if 3/4 < r < 1. Hence, if 0 < r ≤ 3/4, then P {[0, 1/4] ⊕ r} = P {[r, 1/4 + r]} = 1 1 +r−r = 4 4 while if 3/4 < r < 1, then P {[0, 1/4] ⊕ r} = P {[0, 1/4 + r − 1] ∪ [r, 1]} = P {[0, 1/4 + r − 1]} + P {[r, 1]} ￿ ￿ 1 1 = + r − 1 + (1 − r) = . 4 4 In general, if A ⊆ [0, 1] is any subset of [0, 1], then we can define the shift of A by r for any 0 < r < 1 as A ⊕ r = {a + r : a ∈ A, a + r ≤ 1} ∪ {a + r − 1 : a ∈ A, a + r > 1}. 0 1 ⊕r 0 1 A⊕r A And s...
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