Lévy measure of borel sets away from $0$ The following is  of Philip Protter at page 26 of the book Stochastic integration and Differential equations 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.
 A: 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}
