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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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    $\begingroup$ 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?). $\endgroup$
    – Clinton Conley
    Mar 28, 2012 at 13:14
  • $\begingroup$ Perhaps the "perfect set property" is: given any nonzero finite measure $\mu$ on the sigma-algebra, there is a perfect set $F$ with $\mu(F)>0$. $\endgroup$
    – Gerald Edgar
    Mar 28, 2012 at 13:29
  • $\begingroup$ I guess also require $\mu$ to be atomless. $\endgroup$
    – Gerald Edgar
    Mar 28, 2012 at 13:31
  • $\begingroup$ 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.) $\endgroup$
    – Clinton Conley
    Mar 28, 2012 at 13:52
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    $\begingroup$ Perhaps the question's assertion that projective sets are all universally measurable and have the PSP means that Detelin wants to assume projective determinacy. $\endgroup$
    – Ed Dean
    Mar 28, 2012 at 13:57

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