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Wednesday, May 29, 2013

1.11

Notice that we can construct \mathcal{A} in a similar fashion to how we construct \mathcal{C}. Proceed as follows. Define: \mathcal{A}_1 = [0, 0.4) \cup [0.5, 1]. \mathcal{A}_2 = \bigcup_{k=0}^{10^1 - 1} 10^{-1}([0 + k, 0.4 + k) \cup [0.5 + k, 1 + k]) \vdots \mathcal{A}_n = \bigcup_{k=0}^{10^{n-1}-1} 10^{-n + 1}([0 + k, 0.4 + k) \cup [0.5 + k, 1+k]).

Now, just define \mathcal{A} = \bigcap_{n=1}^{\infty} \mathcal{A_n}. By corollary 3.3, since \bigcap_{n=1}^{N} \mathcal{A_n} \searrow \mathcal{A} as N \to \infty, and m(\mathcal{A_1}) = 0.9, then: m(A) = \lim_{N \to \infty} m(\bigcap_{n=1}^{N} \mathcal{A_n}) = \lim_{N \to \infty} (0.9)^N = 0

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