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Update. Based on Anthony Quas' comment below, the proof can be made sensibly shorter and the lemma can be slightly generalized by weakening the old assumption (iii).

In a joint paper that I am writing, we need (and prove) the following:

Lemma. Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be sequences of positive real numbers such that:

   (i) $a_n + 1 \le b_n < a_{n+1}$ for all $n \in \mathbf{N}^+$;
   (ii) $a_n/b_n \to \ell$ as $n \to \infty$;
   (iii) $b_n/a_{n+1} \to 0$ as $n \to \infty$.

Then $\mathsf{d}^\ast(X) = 1-\ell$, where $X := \bigcup_{n \ge 1} [\![ a_n+1, b_n ]\!]$.

Here, for a set $X \subseteq \mathbf{Z}$ we let $\mathsf{d}^\ast(X)$ be the upper asymptotic density of $X$, viz. $$\mathsf{d}^\ast(X) := \limsup_{n \to \infty} \frac{|X \cap [\![ 1, n ]\!]|}{n},$$ and for $a,b \in \mathbf{R}$ we set $[\![a,b]\!] := [a,b] \cap \mathbf Z$.

The proof is rather straightforward, but at the same time it takes about half a page to word it in a reasonable amount of details (EDIT: this is no longer the case). Yet, we would prefer to avoid it, so my question is:

Do you know of a reference where the lemma, or a generalization of it, is proved?

E.g., we have tried to look at Halberstam and Roth's Sequences (Springer-Verlag, 1989), but couldn't find anything.

Improve this question
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    $\begingroup$ So I don't see why this would take 1/2 page: If $a_n<N\le a_{n+1}$, then $|X\cap [1,N]|/N \le |X\cap [1,b_n]|/b_n$ so the $\limsup$ is attained along the sequence of $b_n$'s. And $b_n-a_n \le |X\cap [1,b_n]|\le (b_n-a_n)+b_{n-1}$. Dividing by $b_n$ and taking the limit gives the result. $\endgroup$
    – Anthony Quas
    Commented May 25, 2015 at 7:37
  • $\begingroup$ Because "about half a page" is not "half a page" (but an approximation in excess, which takes into account 5 lines for the statement and two more lines for a second point, which I didn't mention in the OP), and because we were working with $b_n \le N < b_{n+1}$ rather than $a_n<N \le a_{n+1}$, which is, indeed, much better (thank you!). Anyway, we would still prefer a reference, if any. $\endgroup$
    – Salvo Tringali
    Commented May 25, 2015 at 8:32
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    $\begingroup$ It seems that you are essentially asking about $d^*(A)=\limsup\frac{(b_1-a_1)+\dots+(b_n-a_n)}{b_n}$. From Stolz-Cesaro Theorem you get that this is in fact equal to $\lim\frac{b_n-a_n}{b_n-b_{n-1}}=1-\ell$. (I know that this does not answer the question, but still it makes the proof a bit shorter, I guess.) $\endgroup$
    – Martin Sleziak
    Commented May 27, 2015 at 15:00
  • $\begingroup$ @MartinSleziak. I'm not so convinced that would make the proof shorter, since you first need to show that ${\sf d}^\ast(X)=\limsup_n\frac{|X \cap [\![1, b_n]\!]|}{b_n}$. AFAICS, this can't avoid a rough estimate of the counting function of $X$, which is possible, e.g., by letting $N$ be as in Anthony Quas' comment and considering that $\frac{x}{y} \le \frac{x+z}{y+z}$ for all $x,y,z \in \mathbf R^+$ with $x \le y$. In any case, thank you for your comment! $\endgroup$
    – Salvo Tringali
    Commented May 27, 2015 at 15:39
  • $\begingroup$ You are probably right. As not to digress from the original question, we can continue this discussion in chat, if needed. $\endgroup$
    – Martin Sleziak
    Commented May 27, 2015 at 15:53

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