MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that "ZFC+the existence of an inaccessible cardinal" is equconsistent to
"ZFC + every $\mathbf{\Sigma}^1_3$ set is measurable".

Then how about the light face case?

Without large cardinal assumption, can we have a ZFC model in which every analytical set (or lightface $\Sigma^1_3$) is measurable?

Here a set is analytical if it is $\Sigma^1_n$ for some $n$.


This was already answered by Shelah. ''It is known that there is a generic extension of $L$ not collapsing cardinals nor violating CH, in which every definable (with no parameter!) set of reals is measurable..." from the 3rd remark, page 18, Shelah's paper.

share|cite|improve this question
I always thought that analytic sets are $\Sigma^1_1$, then you have co-analytic as $\Pi^1_1$ and the rest are just named "projective sets". – Asaf Karagila Apr 26 '12 at 6:57
You are right. But analytical set is another thing. This follows from Moschovakis's book. – 喻 良 Apr 26 '12 at 8:54
I think Solovay's argument actually shows that "every $\Sigma^1_3(a)$-set is measurable" is equivalent to $\aleph_1^{L[a]} \lt \aleph_1$. – François G. Dorais Apr 26 '12 at 13:21
I agree with you. We just need to collapse the first uncountable cardinal in $L$ to get a zfc model in which every analytical set is measurable. – 喻 良 Apr 26 '12 at 13:29
To answer Asaf's comment/question, "analytic" means boldface $\mathbf{\Sigma^1_1}$ (; and "analytical" (with an "-al" suffix) means lightface $\Sigma^1_n$ for some $n$ ( Hence the analytical sets are the effective projective sets. I was confused myself and had to look it up. – Jason Rute Apr 26 '12 at 15:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.