Does there exist a non-deterministic $\mathbb{R}$- valued Lévy process $(X_t)_{t \in [0, \infty)}$ such that there exist a $t_0 \in (0, \infty) $ and $R>0$ such that
$$ P(0 < X_{t_0} < R)=1 $$
i.e. $X_{t_0}$ is almost surely positive and bounded by $R$.
Since Lévy processes are more or less generated by infinitely divisible random variables, this rases the question if there exist a non-deterministic $\mathbb{R}$- valued infinitely divisible random variable (i.e. it's density measure $\mu_X$ is infinitely divisible with respect convolution) such that there exist $R>0$ with $ P(0 < X_{t_0} < R)=1 $; equivalently, $\mu_X((0,R))=1$.
The question is motivated by the problem here: translation invariance of expectation value of hit counting variable for Lévy process
I'm seeking to construct a counterexample with a $\mathbb{R}$- valued Lévy process $(X_t)_{t \in [0, \infty)}$ such that there exist $s,a,u>0$ with
$$\mathbb{E}[M_u(a,s)]= \mathbb{E}[M_0(a,s)] $$
where $M_u(a,s)$ the random variable counting the number of tiles $[k \cdot a, (k+1) \cdot a]$ in $I(a):= \{[k \cdot a, (k+1) \cdot a] \ : \ k \in \mathbb{Z} \} $ hit by $(X_t)$ at some time $t \in [u,u+s]$.
Obviously, finding such Lévy process $(X_t)_{t \in [0, \infty)}$
containing an a.s. positive and bounded member $X_{t_0}$ as above would do the job.
An answer to your actual question: No.
Thus, suppose that $(X_t)$ is a Lévy process with $X_0=0$ and $0<X_{t_0}<R$ a.s., for some $t_0>0$ and some finite $R>0$.
I assume without loss of generality that $t_0=1$.
Claim: $X_t>0$ for all $t>0$ a.s.
Indeed, for positive integer $n$, $P(X_n>0)\ge\left(P(X_1>0)\right)^n=1$. Consider now $X_{1/2}$. If we had $P(X_{1/2}\le 0)>0$, then we would have $P(X_1\le 0)\ge \left( P(X_{1/2}\le 0)\right)^2>0$, a contradiction. Therefore $P(X_{1/2}>0)=1$ as well. Proceeding recursively, $P(X_{2^{-k}}>0)=1$ for all $k=1,2,\ldots$. In combination with the first sentence of this paragraph this shows that $P(X_{k2^{-n}}>0)=1$ for all positive integer $k$ and $n$; by countable additivity, $$ P(X_s>0,\hbox{ for all dyadic rational }s>0)=1.\qquad (1) $$ Because the paths of a Lévy process are right continuous, we deduce from (1) that $$ P(X_t\ge 0,\forall t>0)=1.\qquad (2) $$ It is known that for a non-negative Lévy process, the Laplace transform $E(\exp(-\lambda X_t))$, $\lambda>0$, takes the form $\exp(-t\psi(\lambda)$, where $$ \psi(\lambda) =b\lambda +\int_{(0,\infty)}(1-e^{-\lambda x})\nu(dx),\qquad (3) $$ for a constant $b\ge 0$ and a measure $\nu$ such that $\int_{(0,\infty)}(x\wedge 1)\nu(dx)<\infty$. Using (3), $$ P(X_t=0)=\lim_{\lambda\to+\infty}e^{-t\psi(\lambda)}.\qquad (4) $$ From (1), the limit in (4) is $0$ if $t$ is a dyadic rational, so in fact it must be $0$ for every $t>0$. Thus $$ P(X_t>0,\forall t>0)=1.\qquad (5) $$
By virtue of the Claim, The process $(X_t)$ has non-decreasing sample paths; as such it admits a Lévy-Ito decomposition. In particular, $$ X_1=b+\sum_{0<s\le 1} \Delta X_s, $$ where $\Delta X_s:=X_s-X_{s-}\ge 0$, and the sum in (6) is a countable sum over those times $s\in (0,1]$ for which $\Delta X_s>0$. The crucial point here is that for $\epsilon>0$ the number $J(\epsilon)$ of $s\in(0,1]$ such that $\Delta X_s >\epsilon$ has the Poisson distribution with parameter $\nu(\epsilon,\infty)$. Assuming that $(X_t)$ is non-deterministic, we can choose $\epsilon$ such that $\nu(\epsilon,\infty)>0$. Now choose a positive integer $K$ such that $K>R/\epsilon$. Then $P(J(\epsilon)=K)>0$ and on the event $\{J(\epsilon)=K\}$ we have $$ X_1\ge\epsilon\cdot J(\epsilon)=\epsilon\cdot K > R. $$ That is, $P(X_1>R)>0$. There is no non-deterministic Lévy process with $P(0<X_{t_0}<R)=1$ for even one time $t_0$.
Consequently, there is no no-trivial positive infinitely divisible random variable that is (a.s.) bounded.