Does recent work of Woodin clash with an older result in Descriptive Set Theory?

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.

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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?

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People voting to close: for the sake of courtesy, please leave at least one comment with a brief explanation. – S. Carnahan May 26 '11 at 5:53
"Undercut" is pretty vague, but Tom's comment on Stefan's answer suggests that he's worried about the significance, not the correctness, of Martin's result when CH holds. I would answer that the statement of Martin's result indeed becomes trivial in the presence of CH (but see Andres's comment on Stefan's answer), as do lots of other theorems (for example, inequalities between cardinal characteristics of the continuum). But so what? I'll vote to close as "not a real question". – Andreas Blass May 26 '11 at 14:06

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.
Stefan: Martin's result is stronger. It is not simply that you have a union of a certain length (which would of course be trivial with a small continuum), but that you have an effective representation of $\Sigma^1_3$ sets. The result gives nontrivial information, regardless of the size of the continuum. – Andrés E. Caicedo May 26 '11 at 6:06
@ Andres: Thanks for the expansion on Stefan's point. I knew that Ultimate L is not necessary to have large cardinals + CH, but that is "cutting edge" research these days it seems. And it is still not clear to me how the ${\Sigma}^1_3$ sets are in some meaningful sense representable in Martin's way in a CH context (since there are only continnum many Borel sets), and obviously something like "just tack on" ${\aleph}_2$ more copies of the empty set, for instance, is not a meaningful thing to do. I would appreciate any additional elucidation, thanks in advance. -- Tom Dunion – Tom Dunion May 26 '11 at 12:38
@Tom: Effective representations are meaningful and provide you with information even without choice. For example: it is consistent with ZF that the reals are a countable union of countable sets. However, this is not possible in an effective way. You can develop measure theory working with Borel codes, and it is meaningful even in these pathological models. You are giving a very powerful effective result (Martin's) and all you are taking from it is an obvious consequence (under CH, any $\Sigma^1_3$ set is a union of a few singletons) which in no way is what the result is about. – Andrés E. Caicedo May 26 '11 at 15:14