The halting set is Sigma_1^0-complete

The universal partial function $(e,x)\mapsto {\phi }_e(x)$ is partial computable. Its halting domain $K$ is computably enumerable, and hence is ${\mathrm{\Sigma }}_1^0$ by Sigma_1^0 sets are exactly the computably enumerable sets.

Write $x\in A⟺\mathrm{\exists }yR(x,y)$ with $R$ primitive recursive. There is a partial computable two-argument procedure which, on parameters $(x,z)$, searches for $y$ and halts exactly when $R(x,y)$; it ignores $z$. The hard-wiring function supplied by acceptability gives, uniformly and totally computably in $x$, an index $e_x$ for the corresponding one-argument procedure.

Therefore $x\in A$ exactly when ${\phi }_{e_x}(0)\downarrow$. After the fixed binary encodings from Completeness at an arithmetical level, the total computable map $x\mapsto ⟨e_x,0⟩$ reduces $A$ to $K$.