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>> No.10345687 [View]
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10345687

Can some one post the Stephen Wolfram quote: "everything has a probability of 1/2 it either happens or it doesn't"

>> No.10136317 [View]
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10136317

1

And that's it

>> No.9719567 [View]
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9719567

>> No.9602668 [View]
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9602668

>> No.9461930 [View]
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9461930

I know the question is silly as the answer should be no, but hear me out.

If [math]E\subset \mathbb{R} [/math] has finite Lebesgue measure, then I can assume it is contained in a bounded open interval by the following argument: Consider the characteristic function [math]\chi_{E\cap I_j} [/math] where [math] I_j = (-j,j) [/math]. Then [math]\chi_{E\cap I_j}\to \chi_E \in L^(\mathbb{R}) [/math]. So by the Dominated Convergence Theorem, [math]|| \chi_{E\cap I_j}-\chi_E \\_{L^1} \to 0[/math]

[math]\mathbb{Q}[/math] has finite measure (0 in fact) so by the above argument, it is contained in some bounded open interval.

>> No.9367639 [View]
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9367639

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