I'm wondering if there is some reference you may know that gives an explicit set which is not analyticallymeasurable (i.e., not in the sigmaalgebra generated by $\Sigma^1_1$), but which is in $\Delta^1_2$? I can prove that such a set exists but just wondering if there's a (fairly) concrete example.

1$\begingroup$ Are you sure you wrote the question correctly? There is some awkward double negation in there, and that question seems different from what is mentioned in the title of the question. $\endgroup$– WojowuJun 18 at 20:33

$\begingroup$ @Wojowu Sorry, corrected! $\endgroup$– John LevyJun 18 at 21:42

$\begingroup$ Not a genuine example but a conditional one: If $V=L$ then there is a $\Delta^1_2$ wellordering of the reals with ordertype $\omega_1$. Such a wellordering, as a subset of $\mathbb R^2$, is not Lebesgue measurable and therefore not in the $\sigma$algebra generated by analytic sets. $\endgroup$– Andreas BlassJun 18 at 22:22
1 Answer
You write:
I can prove that such a set exists but just wondering if there's a (fairly) concrete example.
The following might just be the proof you allude to, but I think it is fairly concrete:
A set $X$ is analyticallymeasurable iff it has a code of a particular form  namely, say that a $\sigma\Sigma^1_1$code is a labelled tree $T$ such that
$T$ is a wellfounded subtree of $\omega^{<\omega}$,
each terminal node of $T$ is labelled with either a $\Sigma^1_1$ or $\Pi^1_1$ set, and
each nonterminal node is labelled with either $\bigwedge$ or $\bigvee$.
There is a natural way to evaluate a $\sigma\Sigma^1_1$code $T$, that is, a function $\mathsf{eval}$ from $\sigma\Sigma^1_1$codes to analyticallymeasurable sets; crucially, the set of $\sigma\Sigma^1_1$codes and the relation "$x\in\mathsf{eval}(T)$" are each $\Delta^1_2$ (indeed much better than that). This lets us directly diagonalize; fixing some (relatively)lowcomplexity bijection $b$ between $\sigma\Sigma^1_1$codes and reals, the set $$\mathscr{D}=\{x:x\not\in \mathsf{eval}(b(x))\}$$ is $\Delta^1_2$ and not analyticallymeasurable.
(Note that the above assumes a small amount of choice, specifically in assuming that all analytically measurable sets actually have $\sigma\Sigma^1_1$codes; this issue is analogous to the need for choice to make the various notions of "Borel" coincide. See here for more on this.)