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Apr 3, 2012 at 8:15 comment added Detelin Apparently, should have been more precise when stating the question. I am assuming ZFC. For the question I have asked, a set is having PSP if it is either countable or contains a perfect subset. So, apparently in ZFC the answer is NO since, as Clinton mentioned, there are universally null sets. Thanks.
Mar 28, 2012 at 16:33 comment added Clinton Conley Well as mentioned above, it a theorem of ZFC that the answer is "no, since there's an uncountable universally null set." If Detelin is not assuming at least ZFC, some clarification is in order.
Mar 28, 2012 at 13:57 comment added Ed Dean Perhaps the question's assertion that projective sets are all universally measurable and have the PSP means that Detelin wants to assume projective determinacy.
Mar 28, 2012 at 13:52 comment added Clinton Conley I assumed perfect set property meant the standard thing, although in retrospect it's somewhat suspicious that projective sets are claimed to possess it. (Also I wish we could edit comments -- I was so thrown by "nullness" vs. "nullity" that I forgot to demote "universally" from adverb to adjective.)
Mar 28, 2012 at 13:31 comment added Gerald Edgar I guess also require $\mu$ to be atomless.
Mar 28, 2012 at 13:29 comment added Gerald Edgar Perhaps the "perfect set property" is: given any nonzero finite measure $\mu$ on the sigma-algebra, there is a perfect set $F$ with $\mu(F)>0$.
Mar 28, 2012 at 13:14 comment added Clinton Conley Are you working in ZFC? If so, you can take any uncountable universally null set as a counterexample. If it contained a continuous injective image of $2^\omega$, you'd be able to push forward the product measure to contradict its universally nullness (nullity?).
Mar 28, 2012 at 12:43 history asked Detelin CC BY-SA 3.0