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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 7, 2016 at 5:37 vote accept Marcus Pivato
Jul 6, 2016 at 14:01 answer added Joel David Hamkins timeline score: 7
Jul 6, 2016 at 11:16 comment added Marcus Pivato Hi Joel. I agree ---this was the reason for "Remark (4)" above.
Jul 6, 2016 at 11:07 comment added Joel David Hamkins One should mention that the induced measure is definitely not countably additive, because one can have an open dense set in the unit interval of arbitrarily small Lebesgue measure --- place an interval of size $\epsilon/2^n$ about the $n^{th}$ rational. The join of these intervals is the whole interval, since it is dense, but the measures of the finite joins is bounded by $\epsilon$.
Jul 6, 2016 at 8:51 history asked Marcus Pivato CC BY-SA 3.0