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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 $n$.

Edited:

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.

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    $\begingroup$ 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". $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 26, 2012 at 6:57
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    $\begingroup$ You are right. But analytical set is another thing. This follows from Moschovakis's book. $\endgroup$
    – 喻 良
    Commented Apr 26, 2012 at 8:54
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    $\begingroup$ I think Solovay's argument actually shows that "every $\Sigma^1_3(a)$-set is measurable" is equivalent to $\aleph_1^{L[a]} \lt \aleph_1$. $\endgroup$
    – François G. Dorais
    Commented Apr 26, 2012 at 13:21
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    $\begingroup$ 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. $\endgroup$
    – 喻 良
    Commented Apr 26, 2012 at 13:29
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    $\begingroup$ To answer Asaf's comment/question, "analytic" means boldface $\mathbf{\Sigma^1_1}$ (en.wikipedia.org/wiki/Analytic_set); and "analytical" (with an "-al" suffix) means lightface $\Sigma^1_n$ for some $n$ (en.wikipedia.org/wiki/Analytical_hierarchy). Hence the analytical sets are the effective projective sets. I was confused myself and had to look it up. $\endgroup$
    – Jason Rute
    Commented Apr 26, 2012 at 15:23

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