Let $X$ be Polish. It is known that every analytic and coanalytic subset of $X$ is universally measurable. The Wikipedia article 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?
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, Fund. Math. 208 (2010), no. 2, 173192.
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.

$\begingroup$ Thank you. I haven't thought about the fact that the projective hierarchy defines at most countably many new sets in each stage of the construction (similarly to the Borel hierarchy). Since the induction runs up to $\omega_1$ stages, there are only countably many projective sets. $\endgroup$ – yadaddy Dec 18 '15 at 14:07

1$\begingroup$ I think of the projective sets in terms of an $\omega$ hierarchy, taking projections and complements, rather than an $\omega_1$hierarchy. Alternatively, one may think of the defining formulas, of which there are only countably many. If one allows real parameters, this gives continuum many boldface projective sets. $\endgroup$ – Joel David Hamkins Dec 18 '15 at 14:10

$\begingroup$ Oh yes, of course: replace "countably many" by "continuum many" (which doesn't change your computations). It is really interesting to see the huge gap between the definable sets (projective hierarchy, defined from "inner" by unions) and the universally measurable sets (defined from "outer" by cutting, i.e. intersections of $\sigma$algebras). $\endgroup$ – yadaddy Dec 18 '15 at 14:22
Thanks, Joel, for mentioning our paper.
By MartinSteel, if there exist infinitely many Woodin cardinals, then every uncountable projective set contains a perfect set, so, by the result of Hausdorff mentioned by Joel there are universally null sets which are not projective.
On the other hand, it is apparently an open question whether it is consistent for all universally measurable sets to be $\Delta^{1}_{2}$ (see problem CG on Fremlin's list : https://www.essex.ac.uk/maths/people/fremlin/problems.pdf).
Finally, I should mention two improvements of Hausdorff's theorem :
(Reclaw) Any set of reals which is wellordered by a universally measurable relation is universally null (https://projecteuclid.org/download/pdf_1/euclid.rae/1212763971)
(Grzegorek) There is a universally null set whose cardinality is the smallest cardinality of a nonLebesguenull set. A proof of this can be found in Bukovsky's book (http://www.springer.com/gp/book/9783034800051)