For each n, there is a (lightface) Σ⁰_n set S_n ⊆ ω² that's universal for the Σ⁰_n subsets of ω. Since {n} × S_n is Σ⁰_n, there is a union R of arithmetical sets such that (n, j, k) ∈ R iff (j, k) ∈ S_n. Clearly R is not itself arithmetical, and offhand I don't see why it should be even Δ¹₁.

If we define the sets S_n with care, is there a Δ¹₁ set Q ⊆ ω³ such that (n, j, k) ∈ Q iff (j, k) ∈ S_n?

For each n, there is a (lightface) Σ⁰_n set S_n ⊆ ω² that's universal for the Σ⁰_n subsets of ω. Since {n} × S_n is Σ⁰_n, there is a union R of arithmetical sets such that (n, j, k) ∈ R iff (j, k) ∈ S_n. Clearly R is not itself arithmetical, and offhand I don't see why it should be Δ¹₁.

If we define the sets S_n with care, is there a Δ¹₁ set Q ⊆ ω³ such that (n, j, k) ∈ Q iff (j, k) ∈ S_n?

For each n, there is a (lightface) Σ⁰_n set S_n ⊆ ω² that's universal for the Σ⁰_n subsets of ω. Since {n} × S_n is Σ⁰_n and the Borel sets are closed under countable union, there is a Borel set R ⊆ ω³ such that (n, j, k) ∈ R iff (j, k) ∈ S_n.

If we define the sets S_n with care, is there a Δ¹₁ set Q ⊆ ω³ such that (n, j, k) ∈ Q iff (j, k) ∈ S_n?

For each n, there is a (lightface) Σ⁰_n set S_n ⊆ ω² that's universal for the Σ⁰_n subsets of ω. Since {n} × S_n is Σ⁰_n, there is a union R of arithmetical sets such that (n, j, k) ∈ R iff (j, k) ∈ S_n. Clearly R is not itself arithmetical, and offhand I don't see why it should be even Δ¹₁.

If we define the sets S_n with care, is there a Δ¹₁ set Q ⊆ ω³ such that (n, j, k) ∈ Q iff (j, k) ∈ S_n?