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Oct 20, 2020 at 9:35 history edited Sofia CC BY-SA 4.0
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Jul 23, 2020 at 16:23 history closed LSpice
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Jul 10, 2020 at 19:34 answer added Sofia timeline score: 1
Jul 10, 2020 at 19:10 comment added Sofia Thank you @NateEldredge for the good example!
Jul 10, 2020 at 18:58 history edited Sofia CC BY-SA 4.0
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Jul 9, 2020 at 14:05 comment added Nate Eldredge I believe that if $\Omega$ is atomless, then (ii) implies (i).
Jul 9, 2020 at 14:05 comment added Nate Eldredge Let $\Omega$ be a single point having probability 1. Then (ii) is trivially satisfied because for any $\delta < 1$, the only $A$ with $P(A) < \delta$ is $A=\emptyset$. So now you can choose any family which violates (i), e.g. $X_n = n$.
Jul 9, 2020 at 12:35 review Close votes
Jul 23, 2020 at 16:23
Jul 9, 2020 at 12:21 comment added LSpice What are you taking the supremum of in (i)? Anyway, any example that satisfies (i) can easily be modified not to do so without breaking (ii). This is not research level.
Jul 9, 2020 at 11:52 review First posts
Jul 9, 2020 at 12:33
Jul 9, 2020 at 11:47 history asked Sofia CC BY-SA 4.0