Finiteness of the number of big jumps of a Lévy process on a finite interval Let $Y(t)$, $0 \leq t \leq 1$, be a Lévy process. Denote by $\{ \Delta Y_i \}$ the jumps of the process. I would like to show that the set $\{i: |\Delta Y_i| > r\}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.
Thank you.
Regards,
Ivan
 A: Let $(Y_t)_{t \in [0,1]}$ be a Lévy process and denote by $\Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t \in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$$\{t \in [0,1]; |\Delta Y_t(\omega)|>r\}$$
is a finite set for (almost) all $\omega \in \Omega$. This follows from the following lemma (applied to the sample paths $[0,1] \ni t \mapsto Y_t(\omega)$ for fixed $\omega \in \Omega$):

Lemma Let $f:[0,1] \to \mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $$\{t \in [0,1]; |\Delta f(t)|>r\}$$ is finite for each $r>0$.

Proof:


*

*Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n \in \mathbb{N}} \subseteq [0,1]$ such that $|\Delta f(t_n)|>r$ for each $n \in \mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k \in \mathbb{N}}$ such that $t_{n(k)} \to s$ for some $s \in [0,1]$.

*Since $f$ is right-continuous at $s$, there exists $\delta_1>0$ such that $$|f(s)-f(t)| < \frac{r}{4}$$ for any $t \in [s,s+\delta_1]$. This means in particular that there cannot exist $t \in [s,s+\delta_1]$ such that $|\Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $\delta_2>0$ such that $|\Delta f(t)| \leq r$ for all $t \in [s-\delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
