I have two questions about universally measurable sets:

(1) Is there a universally measurable set of reals which does not have the Baire property?

(2) Is there a universally measurable set of reals which is not universally Baire?

Note that a Luzin set is universally measurable (null) but it does not have the Baire property so these are consistent. But are there ZFC examples?

Recall that a set of reals $A$ is universally measurable if for every Borel (probability) measure $m$ on reals, $A$ equals some Borel set modulo an $m$-null set. $A$ is universally Baire if for every topological space $X$ and continuous function $f: X \rightarrow \mathbb{R}$ the preimage $f^{-1}[A]$ has the Baire property in $X$ (equals an open set modulo meager). Also, universally Baire set are universally measurable.


  • $\begingroup$ Regarding the first question, according to section 4 in the paper "Universally measurable sets in generic extensions" by Larson, Neeman and Shelah, the answer to (1) is positive in most natural models. It's still unknown whether it's consistent that every universally measurable set has the Baire property. $\endgroup$ – Haim Nov 30 '14 at 22:24
  • $\begingroup$ Reclaw ("On a construction of universally null sets") has shown that every set of reals which is wellordered by a Borel relation is universally null. There are many ways to construct such sets which are uncountable (e.g., a set of functions from $\omega$ to $\omega$ which has ordertype $\omega_{1}$ under mod-finite domination) and no such set can contain a perfect set. If there exists a Woodin cardinal, however, every uncountable universally Baire set contains a perfect set. Without large cardinals, universally Baire sets can behave very differently, so I don't know what happens in general. $\endgroup$ – Paul Larson Feb 14 '15 at 5:25
  • $\begingroup$ Dear Haim and Paul, thank you for your interesting remarks and references. $\endgroup$ – Ashutosh Feb 16 '15 at 23:11
  • $\begingroup$ Dear @PaulLarson, Do you know the answer to (2) if we replace univ. meas. by univ. null? Thanks! $\endgroup$ – Ashutosh Feb 28 '15 at 0:45
  • $\begingroup$ I don't. I would expect the easiest solution to (2) to be showing that there is a universally null set which is not universally Baire. $\endgroup$ – Paul Larson Apr 16 '15 at 2:35

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