# Jumping times on Borel sets away from zero are stopping times

The following comes from some remarks of Philip Protter at page 26 of the book Stochastic integration and Differential equations that I have not been able to prove yet.

Let $X$ a Levy process, under a filtration satisfying the usual conditions. If $\Lambda$ is a Borel set in $\mathbb{R}$ bounded away from zero (that is $0 \notin \bar{\Lambda}$), then the jumping times

\begin{align} &T_{\Lambda}^{1} = \lbrace t \geq : \Delta X_{t} \in \Lambda \rbrace \\ &\vdots \\ &T_{\Lambda}^{n} = \lbrace t > T_{\Lambda}^{n-1} : \Delta X_{t} \in \Lambda \rbrace \end{align} are stopping times.

My attempt Since the filtration satisfies the usual conditions, we only need to prove that $\lbrace T_{\Lambda} < t \rbrace \in \mathcal{F}_{t}$.

Let $\epsilon := d(0, \Lambda) >0$ and $M:= ( - \infty, - \epsilon] \cup [\epsilon, \infty)$, I am trying to prove

\begin{align} \lbrace T_{\Lambda} < t \rbrace = \left( \bigcup_{r \in [0, t) \cap \mathbb{Q}} \lbrace \Delta X_{r} \in \Lambda \rbrace \right) \cap \lbrace T_{M} < t\rbrace \end{align} If we can prove this equation, we are done. This is due to the fact that $\lbrace T_{M} \leq t\rbrace \in \mathcal{F}_{t}$ since \begin{align} \lbrace T_{M} < t\rbrace = \bigcap_{n} \bigcup_{r,s \in [0, t+1/n)\\ \vert r-s \vert < 1/n} \lbrace \vert X_{s} - X_{r} \vert > \epsilon \rbrace \end{align}

We know that the "$\supset$" is the easy part, but the "$\subset$" part is the only part that I need to prove. I was trying to prove this by contradiction, and seems that it is the best way.

If $w \in \lbrace T_{M} < t\rbrace$ and $w \notin \lbrace T_{M} < t\rbrace$ is a contradiction. This can be done using lemmas of discontinuities and the fact that $d(0, \Lambda) >0$. However the part $w \in \lbrace T_{M} < t\rbrace$ and $w \notin \left( \bigcup_{r \in [0, t) \cap \mathbb{Q}} \lbrace \Delta X_{r} \in \Lambda \rbrace \right)$ is the difficult one.

Any hint will be welcome.