Let $X(.)$ be a (strictly) $\alpha$-stable process (with $\alpha \in (1,2)$). Assume also that $X(.)$ is spectrally positive (its Lévy measure is concentrated in $[0,+\infty)$).
I am looking for a result that qualitatively says that the set of jumps heights of $X(.)$ is unbounded. More formally, define $J_t(x(.)) := \sup_{0\leq s \leq t} \{\vert x(s) - x(s^-)\vert\}$. Is it true that
\begin{equation} J_t(X(.)) \stackrel{t\rightarrow\infty}{\longrightarrow} + \infty\qquad \mathrm {a.s.} \end{equation} or any other suitable convergence?
Denote by $N$ the jump measure of the Lévy process, i.e. $$N_t(B) := N([0,t] \times B) := \sharp \{s \in [0,t]; \Delta X_s := X_s-X_{s-} \in B\},$$ and by $\nu$ its Lévy measure. It is widely known that $(N_t(B))_{t \geq 0}$ is a Poisson process with intensity $\nu(B)$. In particular, we have
$$\mathbb{P}(N_t(B) >0) = 1- \mathbb{P}(N_t(B)=0)= 1-e^{-\nu(B) t}.$$
For any set $B$ such that $0<\nu(B)<\infty$ this implies
$$\mathbb{P}(\exists s \in [0,t]: \Delta X_s \in B) = \mathbb{P}(N_t(B) >0) \stackrel{t \to \infty}{\to} 1.$$
Applying this for $B = [n,n+1)$, we get
$$\mathbb{P}(\exists t \geq 0: \Delta X_t \in [n,n+1)) = 1.$$
Hence,
$$\mathbb{P}(\forall N \geq 1 \exists t \geq 0: \Delta X_t \geq N) = 1.$$
This shows that the jump heights are (almost surely) unbounded.
Remark: The proof applies to any Lévy process with unbounded Lévy measure.