# Levy 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 proved yet.

Let $X$ a Levy 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 $T_{\Lambda}^{1} := \lbrace t >0 : \Delta X_t \in \Lambda \rbrace, \cdots, T_{\Lambda}^{n}:= \lbrace t > T_{\Lambda}^{n-1}: \Delta X_{t} \in \Lambda \rbrace,...$ 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, ..., 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.

Since the Lévy process $(X_t)_{t \geq 0}$ has right-continuous paths with probability $1$, we can choose $t>0$ such that $\mathbb{P}(T_{\Lambda}^1 \leq t) < 1$. Following the reasoning in your question we find

$$t \nu(\Lambda) = \mathbb{E}(N_t^{\Lambda}) \leq \sum_{n \geq 1} (\mathbb{P}(T_{\Lambda}^1 \leq t))^n < \infty$$

and so $\nu(\Lambda)<\infty$.

• Thanks @saz for the insights. – Ivan Apr 2 '18 at 23:48

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}