No. Even for a very simple first‑order (indeed, positive existential) definable set, the density need not exist and can oscillate wildly along the chain.

Here is a concrete counterexample.

• Let G be the (locally finite) direct sum of countably many copies of a cyclic group of order 4: G = ⊕m≥1 C4. In this group, the set of squares is exactly 2G (coordinatewise squaring). In particular, an element is a square in G iff each of its C4 coordinates is either 0 or 2. The set of squares is first‑order definable by the formula φ(x): ∃y (y^2 = x).

• Partition the coordinates into disjoint finite blocks I1, I2, … with sizes |Ik| tending to infinity (e.g., |Ik| = k). For each k set: Hk = ⊕j∈Ik C4 and Ak = 2Hk ≤ Hk, so |Hk| = 4|Ik| = 22|Ik| and |Ak| = 2|Ik|.

• Define an increasing chain of finite subgroups whose union is G by G2k−1 = A1 ⊕ ⋯ ⊕ A_{k−1}, G2k = G2k−1 ⊕ Ak = A1 ⊕ ⋯ ⊕ Ak, G2k+1 = G2k−1 ⊕ Hk. Then G = ⋃i Gi, and each Gi is finite.

• Let E ⊆ G be the definable set of squares: E = {x ∈ G : ∃y (y^2 = x)} = 2G. Compute the proportions pi = |E ∩ Gi| / |Gi|: • For even indices, G2k ⊆ 2G, so p2k = 1. • For odd indices, G2k+1 = G2k−1 ⊕ Hk and E ∩ G2k+1 = G2k−1 ⊕ Ak, hence p2k+1 = |G2k−1|·|Ak| / (|G2k−1|·|Hk|) = |Ak|/|Hk| = 2^{−|Ik|}. Since |Ik| → ∞, we have p2k+1 → 0.

Thus the sequence |E ∩ Gi|/|Gi| oscillates between 1 (along the even subsequence) and values tending to 0 (along the odd subsequence), so the density limit does not exist.

This shows that, in general, for a directed union of finite groups, even a very simple first‑order definable subset (here, “being a square”) need not have a well‑defined density with respect to the chain.