Let $B \subset \mathbb{N}$ be a set of natural numbers such that $|B \cap [1,N]| \sim N^\gamma$, for some $\gamma > 0$ with the following property:

For any pair of positive integers $k,n$ we define $B_{k,n} = \{b \in B: b \equiv k \pmod{n}\}$. Then $|B_{k,n} \cap [1,N]| \sim \frac{1}{n} |B \cap [1,N]|$ as $N \rightarrow \infty$.

Now suppose that $\gamma > 1/h$. Does it follow that $B$ is necessarily an additive basis of order $h$?

We say a set $A \subset \mathbb{N}$ is an (asymptotic) additive basis of order $h$ if $hA = A + \cdots + A$ is equal to $\mathbb{N} \setminus C$, for some finite set $C$.

The motivation to this question is that $B$ is well-mixed, meaning that it has no bias towards any congruence class with respect to any modulus. A weaker version of this property is held by the primes, where we require $\gcd(k,n) = 1$ and the proportion is $1/\phi(n)$. Note that the primes do not satisfy the desired conclusion, since even though they have density $N/\log N$ they are not an additive basis of order 2, for example (the Goldbach conjecture asserts that they are an additive basis of order 3). In some sense this tries to capture true 'randomness', by which I mean the sets obtained via the probabilistic method of Erdős, constructed by choosing natural numbers at random where the positive integer $x$ has probability $x^{-\gamma}$ of being in the random set $B$. Such a procedure almost surely produces a random set without any bias towards any modulus.