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Suppose $t_{n}$ is a sequence of positive real numbers such that $c_{1}\geq \lim \sup_{n\to \infty}t_{n}/n\geq \lim \inf_{n\to \infty}t_{n}/n\geq c_{2}>0$ where $c_{1}\geq c_{2}>0$ are positive constants.

Does it follow that $\lim \inf_{N\to \infty}\dfrac{m(\cup^{N}_{n=1}[t_{n}-1,t_{n}+1])}{N}>0$ where $m$ is Lebesgue measure.

Remark: Yesterday I asked whether this was true assuming only $\lim \inf_{n\to \infty}t_{n}/n>0$. The answer to this question was no, as answered by DanielWainfleet.

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    $\begingroup$ It's not good to change the original question. You should leave the original one, for which you received an answer, and then add the new one, with something like "added later" in front of it. $\endgroup$
    – zhw.
    Jul 24 '17 at 20:45
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NO. Let $t_1=1.$ For $j\in \mathbb N,$ let $t_n=(j+1)!$ for $j!< n\leq (j+1)!.$ Then $t_n/n\geq 1=t_{n!}/n!$ for all $n.$ We have $$\frac {m(\cup_{n=1}^{N!}[t_n-1,t_n+1])}{N!}=$$ $$=\frac {m(\cup_{j=1}^N[j!-1,j!+1])}{N!}\leq \frac {2N}{N!}.$$

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  • $\begingroup$ Thanks! I edited the question to reflect what I was actually interested in...I actually want $t_{n}/n$ to be bounded between two positive constants. $\endgroup$
    – Yellow Pig
    Jul 24 '17 at 15:57
  • $\begingroup$ Thank you for your comment. Some members re-write a Q after an A without saying so. The new version is interesting, but I don't have an A yet. $\endgroup$ Jul 24 '17 at 18:32
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The answer to the new question is also no.

For $m=1,2,\dots$ define $t_n = 2^m$ for $2^{m-1}\le n < 2^m.$ So the sequence $(t_n)$ looks like

$$2,4,4,8,8,8,8, 16,16,16,16,16,16,16,16, \dots $$

Then $1\le t_n/n\le 2$ for all $n.$ But note that

$$\bigcup_{n=1}^{2^m} (t_n-1,t_n+1) = \bigcup_{k=1}^{m} (t_{2^k}-1,t_{2^k}+1).$$

The measure of the last set is $2m.$ If we divide that by $2^m$ we get a sequence rapidly approaching $0.$

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