TITLE: What sets are obtained by adding $\aleph_1$ unions and intersections to the Borel algebra?
QUESTION [9 upvotes]: The Borel sigma algebra on $\mathbb{R}$ is obtained by starting with open sets in $\mathbb{R}$ and repeatedly applying the operations of complement, countable union, and countable intersections.  My question is, what would happen if you don't just allow countable unions and intersections, but also allow unions and intersections of $\aleph_1$ many sets?
Now obviously if we assume the Continuum Hypothesis, the answer is "we obtain all subsets of $\mathbb{R}$". But we just assume $ZFC$, then what sets do we know we can obtain?  Can we at least obtain all Lebesgue measurable sets?  Can we still obtain all subsets of $\mathbb{R}$?

REPLY [6 votes]: This is an excellent question! Note first that, just from cardinal arithmetic considerations, whether we obtain all sets is independent (and therefore so is whether we obtain all Lebesgue measurable sets). 
The right context to study this question is descriptive set theory, and indeed the problem was considered early on. For example, Sierpiński proved in Sur une classe d’ensembles, Fundamenta Mathematicae 7 (1925), 237–243, that all $\mathbf\Sigma^1_2$ sets are unions of $\aleph_1$ many Borel sets. Recall that a $\mathbf\Sigma^1_2$ set is one which is the continuous image of the complement of the continuous image of a Borel set.
[Note that it is consistent that there are $\mathbf\Sigma^1_2$ sets that are not Lebesgue measurable.]
Around 1970, Martin and Solovay proved that the conjunction of Martin's axiom, $\lnot\mathrm{CH}$, and the statement that $\omega_1$ is accessible to reals implies the converse: Every union of $\aleph_1$ many Borel sets is $\mathbf\Sigma^1_2$. The statement that $\omega_1$ is accessible to reals means that there is a real $r$ such that $\omega_1^{L[r]}=\omega_1$. On the other hand, Solovay showed that not much more can be said: It is consistent with the continuum being arbitrarily large that there is a $\mathbf\Pi^1_2$ set (that is, a set which is the complement of a $\mathbf\Sigma^1_2$ set) that is not the union of fewer than $2^{\aleph_0}$ many Borel sets.

One can nevertheless soldier on and wonder what can be said if we allow longer unions. The right tool to study the problem is now the notion of $\kappa$-Suslin set. It is simpler to work on Baire space rather than the Euclidean line, but this makes no difference since Baire space $\omega^\omega$ is homeomorphic to the irrationals. A subset $A$ of  $\omega^\omega$ is $\kappa$-Suslin if and only if there is a tree $T$ on $\omega\times\kappa$ such that $A=p[T]$. Recall that a (descriptive set-theoretic) tree on a set $B$ is a set of finite sequences of elements of $B$ closed under initial segments. For $B=\omega\times\kappa$, we can identify any such sequence with two sequences of the same length, one consisting of finite numbers, and one of members of $\kappa$. A branch through $T$ is an infinite sequence all of whose finite initial segments are in $T$ or, thinking of $T$ as consisting of pairs of sequences, a branch would be a pair $(x,t)$ where $x\in\omega^\omega$ (a "real") and $t\in\kappa^\omega$; the set of all branches through $T$ is denoted $[T]$. The projection $p[T]$ is defined as the set of reals $x$ for which we can find a $t\in\kappa^\omega$ with $(x,t)\in[T]$.
Martin proved that for any (finite) $n>0$, the $\omega_n$-Suslin sets are precisely the ones that are the union of $\aleph_n$ many Borel sets. In the context of determinacy, the extent of the $\kappa$-Suslin sets has been identified precisely for various $\kappa$ (significantly beyond the $\omega_n$). This implies results about the extent of these sets under appropriate large cardinal assumptions, and has also been studied from forcing axioms.