There is a mistake in the proof of Lemma 4.2(a) in [1], namely there t jn z jn = 1, so that the denominator in K n (z jn , t jn ) is 0. This causes a gap in the proofs of Lemma 4.3(d), and Theorems 1.1 and 1.2 (but not Theorems 1.3 and 1.4). Below we give replacements for Lemmas 4.2(a) and 4.3(d) and revised proofs of Theorems 1.1 and 1.2. Note that the rest of those lemmas, including the hypotheses, remain the same.Revised Lemma 4.2(a) 1. Let ρ > 0 andThere exist C 0 , n 0 such that for n ≥ n 0 and z jn , z kn ∈ L n with j = k, we haveIn particular, all zeros of ϕ n in L n are simple.Proof. If the result is false, we can find a subsequence of integers S and for n ∈ S, z jn ,From the Christoffel-Darboux formula (2.1) and the uniform universality limit (1.1), 0 = K n (z jn , 1 z kn ) K n (τ n , τ n ) = e ιπ(α n −β n (1+o(1))) S(α n − β n (1 + o(1))) + o(1) = 1 + o(1).