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On the upper Banach density of the set of positive integers whose base-$b$ representation misses at least one prescribed digit

Let $b$ be a fixed integer $\ge 2$ and $A$ a proper subset of $\{0, \ldots, b-1\}$. Then define $X$ to be the set of all positive integers whose base-$b$ representation consists only of digits from $A$. It's known that the series $\sum_{x \in X} \frac{1}{x}$ is convergent, see the article on Kempner series on Wiki.en and references therein, which, together with a classical result from additive-theory folklore, implies that the natural density of $X$ is zero. But what is known, say, about the upper Banach density and, for $\alpha \ge -1$, the $\alpha$-density of $X$? I seem to have a proof that these are still zero, yet would like to understand if the result is buried somewhere in the literature (but not too deep, so that someone here around can give a reference) and/or follows from relatively well-known facts and/or standard considerations.

For what it is worth, let me note that the $\alpha$-density is dominated by the natural density for $-1 \le \alpha \le 0$, so you may want to focus on $\alpha > 0$ in the above.

Just in case it may be useful, let me recall that, given $X \subseteq \mathbf N$ and a real exponent $\alpha \ge -1$, we take the natural (or asymptotic) density of $X$ to be the limit: $$\lim_{n \to \infty} \frac{\sum_{0 \ne x \in X} x^\alpha}{\sum_{x \in \mathbf N^+,\, x \le n} x^\alpha},$$ whenever this exists (this gives back the logarithmic density for $\alpha = -1$ and the natural density for $\alpha = 0$), and the upper Banach (or uniform) density of $X$ to be the limit: $$\lim_{n \to \infty} \max_{m \ge 1} \frac{|X \cap [m+1,m+n]|}{n},$$ which always exists (by Fekete's lemma on subadditive real sequences).

Note. I edited the original post to make it clear, I hope, what I mean in the comments below by saying that I'm interested in a kind of scenarios where having more or less precise information about the asymptotic behavior of the counting function of $X$ is unlikely to be useful at all (of course, I may be wrong).

Salvo Tringali
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