Let $X$ be Polish. It is known that every analytic and coanalytic subset of $X$ is universally measurable. The [Wikipedia article](https://en.wikipedia.org/wiki/Universally_measurable_set) about universally measurable sets notes that (assuming projective determinance) every set in the projective hierarchy is universally measurable. Are there also universally measurable sets that are not related to the projective hierarchy, i.e. can not be constructed in this way?



Relation between projective hierarchy and universally measurable sets
Under the continuum hypothesis, or even just when
$2^{\aleph\_0}<2^{\aleph\_1}$, there must be universally measurable
sets that are not projective. The reason is that Hausdorff proved
that there are always universally null sets of size $\aleph\_1$,
and since any subset of such a set is also universally null and
hence universally measureable, it follows that in ZFC there are at
least $2^{\aleph\_1}$ many universally measurable sets of reals.


But there are only continuum many (boldface) projective sets. So
under CH and more generally if the continuum is less than $2^{\aleph\_1}$, there must be universally
measurable sets that are not projective. 


The following article seems to have a lot of information:


*Paul Larson, Itay Neeman, and Saharon Shelah*, [**Universally measurable sets in generic extensions**](http://dx.doi.org/10.4064/fm208-2-4), *Fund. Math.* **208** (2010), no. 2, 173--192.


Another way to argue is this: if there is a universally null set of size continuum (and this follows from CH by Hausdorff's argument), then all subsets of it will also be universally null and hence universally measurable, which give $2^{\frak c}$ many universally null sets. But there are only continuum many (boldface) projective sets. So in this case, there must be universally null sets that are not projective.


