The following is of Philip Protter at page 26 of the book [*Stochastic integration and Differential equations*](https://doi.org/10.1007/978-3-662-10061-5) that I have not been able to prove yet.


Let $X$ be a Lévy process, and $\Lambda$ a borel set in $\mathbb{R}$ away from $0$ (that is, $0 \notin \bar{\Lambda}$), then $\nu(\Lambda) = E[N\_{1}^{\Lambda}] < \infty$. Where $N\_{t}^{\Lambda} = \sum\_{0 < s \leq t} 1\_{\Lambda}(\Delta X\_{s})$.
Protter says that it follows from Theorem 34, but for me it means I have to prove first that the jumping times have independent and stationary increments something that I am not sure.


**My attempt**
I assume that the jumping times $T\_{\Lambda}^{1} := \lbrace t >0 : \Delta X\_t \in \Lambda \rbrace, \cdots, T\_{\Lambda}^{n}:= \lbrace t > T\_{\Lambda}^{n-1}: \Delta X\_{t} \in \Lambda \rbrace,\dotsc$ have independent and stationary increments. Therefore, taking into account that $ T\_{\Lambda}^{n} = T\_{\Lambda}^{1} + \sum\_{k=1}^{n} T\_{\Lambda}^{k} - T\_{\Lambda}^{k-1} $ we get


\begin{align}
\nu(\Lambda) &= E[N\_{1}^{\Lambda}] \\
&=\sum\_{n=1}^{\infty}P[T\_{\Lambda}^{n} \leq 1] \\
&\leq \sum\_{n=1}^{\infty}P[T\_{\Lambda}^{1} \leq 1, \dotsc, T\_{\Lambda}^{n} - T\_{\Lambda}^{n-1} \leq 1 ] \\
&\leq \sum\_{n=1}^{\infty}(P[T\_{\Lambda}^{1} \leq 1])^{n} < \infty,
\end{align}
and the last part depends on the fact $P[T\_{\Lambda}^{1} \leq 1] < 1$ something that I am not quite sure. Apparently, this is not at all an intuitive fact.


Any hint will be welcome.



Lévy measure of borel sets away from $0$
It seems that the times $(T\_{\Lambda}^{n})$ does not have independent and stationary increments for any Borel set $\Lambda$ away from zero. Fortunately, by the same arguments made by @saz the proof can be done.


**Proof of the problem:** Because $\Lambda$ is away from zero, there exists a $C > 0$ such that $\Lambda \subset \Gamma\_{C}$, where $\Gamma\_{C}:= (-\infty, C] \cup [C, \infty) $. If we denote 


\begin{align}
T\_{\Gamma\_{C}}^{1} := \inf\lbrace t: X\_{t} \in \Gamma\_{C}\rbrace \cdots T\_{\Gamma\_{C}}^{n} := \inf\lbrace t > T\_{\Gamma\_{C}}^{n-1}: X\_{t} \in \Gamma\_{C} \rbrace \cdots ,
\end{align}


these are stopping times that are stationary and have independent increments as @saz proved in one of these related questions. Therefore, we get 


\begin{align}
P(T\_{\Gamma\_{C}}^{n} < t) \leq \dfrac{E[e^{-T\_{\Gamma\_{C}}^{n}}]}{e^{-t}} = \dfrac{(E[e^{-T\_{\Gamma\_{C}}^{1}}])^{n}}{e^{-t}} \leq e^{t} \epsilon^{n}
\end{align}
for some $\epsilon $, $0 \leq \epsilon < 1$. 


Now, using the equality
\begin{align}
N\_{t}^{\Lambda} = \sum\_{0 < s \leq t}1\_{\Lambda}(\Delta X\_{s}) = \sum\_{n=1}^{\infty} 1\_{ \lbrace T\_{\Lambda}^{n} \leq t \rbrace}
\end{align} 
we have 
\begin{align}
N\_{t}^{\Lambda} = \sum\_{n=1}^{\infty} 1\_{ \lbrace T\_{\Lambda}^{n} \leq t \rbrace} = \sum\_{0 < s \leq t}1\_{\Lambda}(\Delta X\_{s}) \leq \sum\_{0 < s \leq t}1\_{\Gamma\_{C}}(\Delta X\_{s}) = \sum\_{n=1}^{\infty} 1\_{ \lbrace T\_{\Gamma\_{C}}^{n} \leq t \rbrace} 
\end{align}
As a result,
\begin{align}
\nu(\Lambda) &= E[N\_{1}^{\Lambda}] \\
&=\sum\_{n=1}^{\infty}P[T\_{\Lambda}^{n} \leq 1] \\
&\leq \sum\_{n=1}^{\infty}P[T\_{\Gamma\_{C}}^{n} \leq 1 ] \\
&\leq \sum\_{n=1}^{\infty}(P[T\_{\Gamma\_{C}}^{1} \leq 1 ])^{n} \\
&\leq e\sum\_{n=1}^{\infty} \epsilon^{n} < \infty
\end{align}


