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Is it true that all universally measurable sets (say on $[0,1]$ ) have the perfect set property?
I am not an expert in this at all and the answer may be known, but I was not able to find it. I know that all Borel, analytic, and projective sets have the perfect set property and are universally measurable but in such generality the answer to my question may be false.

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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?). –  Clinton Conley Mar 28 '12 at 13:14
    
Perhaps the "perfect set property" is: given any nonzero finite measure $\mu$ on the sigma-algebra, there is a perfect set $F$ with $\mu(F)>0$. –  Gerald Edgar Mar 28 '12 at 13:29
    
I guess also require $\mu$ to be atomless. –  Gerald Edgar Mar 28 '12 at 13:31
    
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.) –  Clinton Conley Mar 28 '12 at 13:52
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Perhaps the question's assertion that projective sets are all universally measurable and have the PSP means that Detelin wants to assume projective determinacy. –  Ed Dean Mar 28 '12 at 13:57
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