Does the neighborhood of sequence of bounded uniform density have positive measure? 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 here:
Does a neighborhood of a sequence of positive density have positive measure?
 A: While the factorial grows too quickly to be useful here, the idea behind DanielWainfleet's answer to your previous question still works. Define $t_{2^n+k}=2^n$ for $n\ge0$ and $1\le k\le2^n$. Observe that $\frac12n\le t_n\le n$ for all $n$, so certainly your condition holds. However,
$$m\left(\bigcup_{j=1}^{2^n+k}[t_j-1,t_j+1]\right)=m\left(\bigcup_{i=0}^n[2^i-1,2^i+1]\right)\le2(n+1)\le C\log(2^n+k)$$
and so $\frac1Nm\left(\bigcup_{n=1}^N[t_n-1,t_n+1]\right)\to0$.
Using the same principle, you can show that it's not even enough for $\frac{t_n}n\to c$. In that case, let $t_n=k^2$ for $k^2\le n<(k+1)^2$. Then
$$\frac1{N^2}m\left(\bigcup_{n=1}^{N^2}[t_n-1,t_n+1]\right)=\frac1{N^2}m\left(\bigcup_{k=1}^N[k^2-1,k^2+1]\right)\le\frac 2N.$$
The problem is we need something like the the following condition: there exists $k$ and $\delta>0$ such that $\max\{t_{n+1},\ldots,t_{n+k}\}\ge t_n+\delta$ for all $n$ sufficiently large. In all examples provided so far, no such $k$ exists. Assumptions on $\{t_n\}$ that imply this condition likely require finer asymptotics, such as $t_n=cn+o(1)$.
