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Assume $V=L(\mathbb{R})$ and the Axiom of Determinacy. Is every set of reals generated by ordinal-definable sets of reals under the operations of countable union and intersection?

The class of sets generated in this way is Wadge-cofinal and not wellorderable (it contains $\lbrace x\rbrace$ for every $x \in \mathbb{R}$) so there don't seem to be obvious limitations on its extent.

This question came up when I was trying to answer Asaf Karagila's "bonus question" here: Generating family for the Lebesgue $\sigma$-algebra

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Thanks for taking interest in my question, by the way! – Asaf Karagila Jul 26 '12 at 17:50

This does not answer the question, but for every set $A\subset\mathbb{R}$ in $L(\mathbb{R})$, you can get $A\times\{r\}$ in your algebra, for some real $r$, and this does not use AD.

The reason is that if $V=L(\mathbb{R})$, then every set is definable from an ordinal and a real. So if $A\subset\mathbb{R}$, then $A=\{ x\in\mathbb{R}\mid \varphi(x,\alpha,r)\}$ for some definition $\varphi$ and parameters $\alpha\in\text{Ord}$, $r\in\mathbb{R}$. For any finite binary sequence $t$, let $B_t=\{(x,s)\mid t\subset s,\varphi(x,\alpha,s)\}$, which is an ordinal definable subset of the plane. Finally, observe that $A\times\{r\}=\{(x,r)\mid \varphi(x,\alpha,r)\}$ is precisely $\bigcap_{t\subset r} B_t$, a countable intersection of OD sets.

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Thanks, yes, I noticed this too. So one way to get an affirmative answer would be to show that the $\sigma$-algebra is closed under $\exists^{\mathbb{R}}$. – Trevor Wilson Jul 26 '12 at 19:58

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