tag:blogger.com,1999:blog-3891434218564545511.post1063938273696004752..comments2024-03-28T13:23:50.623-05:00Comments on Alexander Pruss's Blog: Popper probability measureAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-3891434218564545511.post-74087746080817307702013-07-02T13:01:16.144-05:002013-07-02T13:01:16.144-05:00It's also not hard to prove that an isometrica...It's also not hard to prove that an isometrically invariant Popper measure on all Borel subsets of [0,1]^2 makes negligible every line segment. <br /><br />So, roughly speaking, we am here: Invariant Popper functions on [0,1]^3 make line segments negligible (see comment on an earlier post) while invariant Popper measures on [0,1]^2 make line segments negligible. Wonder how far one could improve on these results (are rectangles negligible for invariant Popper functions on [0,1]^3?).Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.com