The following is a remark of Philip Protter at page 26 of the book Stochastic integration and Differential equation that I have not been able to proved yet.

Let $\Lambda$ be a borel set in $\mathbb{R}$ bounded away from $0$ (that is, $0 \notin \bar{\Lambda}$}. For a Lévy process $X$ define $T_{\Lambda}^{1} := \lbrace t >0 : \Delta X_t \in \Lambda \rbrace, \cdots, T_{\Lambda}^{n}:= \lbrace t > T_{\Lambda}^{n-1}: \Delta X_{t} \in \Lambda \rbrace$. How can you prove $\lim_{n \to \infty} T_{\Lambda}^{n} \stackrel{a.s}{=} \infty$?

My attempt I'm trying to prove that $(T_{\Lambda}^{n})_{n \geq 1} $ has independent and stationary increments.

We know that $T_{\Lambda}^{n}$ is a stopping time for $n=1,2,...$, and taking into account that $T_{\Lambda}^{1} > 0$ a.s (because of the fact that $0 \notin \Lambda $ and the process is a cadlag process), we have $P( T_{\Lambda}^{1} > 1/m ) > 0$, for each $m \in \mathbb{N} \setminus \lbrace 0 \rbrace $.

Now for each $m \in \mathbb{N} \setminus \lbrace 0 \rbrace $ let $A_{n}^{(1/m)} := \lbrace T_{\Lambda}^{n}- T_{\Lambda}^{n-1} > 1/m \rbrace$ (with $A_{1}^{(1/m)} := \lbrace T_{\Lambda}^{1} > 1/m \rbrace$ ) , by the assumption we have that $\sum_{n=1}^{\infty} P(A_{n}^{(1/m)})= \lim_{n \to \infty} nP(A_{1}^{(1/m)}) = \infty$. Therefore, by Borel Cantelli's theorem we get that $ P(\varlimsup_{n \to \infty} A_{n}^{(1/m)} ) = 1 $ for each $m$. Thefore, using monotone convergence theorem for $(\varlimsup_{n \to \infty} A_{n}^{(1/m)})_{m \geq 1}$ we should get the result.

I would really appreciate any hint.


Acutally the assertion is a direct consequence of the fact that Lévy processes have càdlág sample paths. Recall the following elementary statement

Lemma If $f: [0,\infty) \to \mathbb{R}$ is a càdlàg function, then $$\sharp\{t \in [0,T]; |\Delta f(t)| \geq \epsilon\} < \infty$$ for any $\epsilon>0$ and $T>0$.

Using this statement one can easily deduce that the sequence of stopping times converges almost surely to infinity.

Since $0 \notin \bar{\Lambda}$ there exists $\epsilon>0$ such that

$$\Lambda \subseteq B :=\{x \in \mathbb{R}; |x| \geq \epsilon\}.$$

Now suppose that $\lim_{n \to \infty} T_{\Lambda}^n(\omega)<\infty$ for some $\omega \in \Omega$. As $\Lambda \subseteq B$, this implies $T:=\lim_{n \to \infty}T_B^n(\omega)<\infty$; in particular, $$\sharp\{t \in [0,T]; |\Delta X_t(\omega)| \geq \epsilon\} = \infty.$$ It follows from the above lemma that $\omega \mapsto X_t(\omega)$ is not càdlàg. Since $(X_t)_{t \geq 0}$ has almost surely càdlàg sample paths we conclude

$$\mathbb{P}(\lim_n T_{\Lambda}^n < \infty) \leq \mathbb{P}(t \mapsto X_t \, \, \text{is not càdlàg}) = 0.$$

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  • $\begingroup$ Thanks. Nice approach saz. I was too focused on my approach that I did not pay attention to easier ways to prove that result. Only comment is $\Gamma = \Lambda$ $\endgroup$ – Ivan Mar 27 '18 at 23:54
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    $\begingroup$ @Ivan Oh, right, thanks. If you find the answer helpful, you can accept/upvote it by clicking on the tick/arrow next to it. $\endgroup$ – saz Mar 28 '18 at 7:26

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