Background/Motivation
First time posting here, so I give the motivation for the question.
Early on in Descriptive Set Theory Sierpinski proved every
${\Sigma}^1_2$ set (PCA set in the older nomenclature) is the union of ${\aleph}_1$ Borel sets. Trivial if we assume the Continuum Hypothesis (use singletons!), in a not-CH context it is essentially a result about "How bad can they be?"
An easy corollary is that such sets can only have cardinality that is countable,
${\aleph}_1$, or that of the continuum.
Around 1970, Solovay sharpened the corollary result to show that if a measurable cardinal exists, such sets enjoy the continuum hypothesis (indeed they have the standard "regularity" properties). Note that Sierpinski's original result stands unimpeached by this.
Then about 1975, D. A. Martin showed every
${\Sigma}^1_3$ set is the union of
${\aleph}_2$ Borel sets, again assuming a measurable cardinal.
—
Yet lately I have been reading that Hugh Woodin has changed has opinion about the truth of the CH (now believing it is true), AND is working toward an "Ultimate L" model which admits large cardinals. Would not such a position undercut Martin's result, or am I missing something?
Best Answer
What is the problem? Large cardinals are consistent with CH. This does not require looking at Ultimate L. But large cardinals are also consistent with failures of CH. And if you are in a model where the continuum is bigger than $\aleph_2$, only then Martin's result gives you the information that every $\Sigma_3^1$ set is the union of fewer than $2^{\aleph_0}$ Borel sets. As long as $2^{\aleph_0}\leq\aleph_2$, every set is the union of (not more than) $\aleph_2$ Borel sets.
Edit: As Andres Caicedo points out in his comment, Martin's result actually say more than just "Assuming a measurable cardinal, every $\Sigma_3^1$-set is the union of $\aleph_2$ Borel sets" and provides nontrivial information even when CH holds.