Exit time of $\alpha$-stable process to an open set and to its closure

Question

Let $W$ be a symmetric $\alpha$-stable process with its generator $-(-\Delta)^{\alpha/2}$ for some $\alpha \in (0, 2]$ under $\mathbb P$. Let $\mathbb P^x$ be the probability measure induced by a process $$X(t) = x + t + W(t)$$ starting from $x$, and we set $$\hat \zeta = \inf\{t>0: X(t) \notin (-1, 1)\}, \quad \zeta = \inf\{t>0: X(t) \notin [-1, 1]\}.$$ My question is that

[Q.] Is $\mathbb P^x (\hat \zeta = \zeta) = 1$ valid for $\alpha \in (0, 1)$?

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The answer should be positive if $\alpha \ge 1$. In fact, one can use strong Markov property and time-shift operator to obtain $$\mathbb P^x(\hat \zeta = \zeta) = \mathbb P^x (\zeta \circ \theta_{\hat \zeta} = 0) = \mathbb E^x [ \mathbb P^{X(\hat \zeta)} (\zeta = 0)] = 1.$$ The last equality is true, since $\zeta = 0$ $\mathbb P^{x}$-a.s for all $x\notin (-1, 1)$ when $\alpha\ge 1$.

However, when $\alpha <1$, $x = -1$ is not regular to $[-1, 1]^c$, i.e. $\zeta>0$ holds almost surely in $\mathbb P^{-1}$. Therefore, the above argument does not work any more, unless $\mathbb P^x(X(\hat \zeta) = -1) = 0$, which is not clear to me.

Thanks.

Answer 1

If $\alpha < 1$, then $X$ has bounded variation and positive drift, and therefore it does not creep downwards; see Theorem 7.11 in Kyprianou's book Introductory Lectures on Fluctuations of Lévy Processes with Applications. This means that $$\mathbb{P}^x(X(\tau_{-1}) \ne -1) = 1,$$ for $x > -1$, where $$\tau_{-1} = \inf \{t > 0 : X(t) \leqslant -1\}.$$ In particular $$\mathbb{P}^x(X(\hat\zeta) \ne -1) = 1,$$ and consequently $$\mathbb{P}^x(\hat\zeta = \zeta) = 1.$$

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Edited: As pointed out by kenneth, the above argument uses $\tau_{-1}$ defined with the weak inequality, while Theorem 7.11 in Kyprianou's book has a strict inequality, that is, $$\tau_{-1^-} = \inf \{t > 0 : X(t) < -1\}.$$ Getting from there to here is not immediate, though!

The proof of Theorem 7.11(i) involves the following reasoning: outside of an event of probability zero, $X(\tau_{-1^-}) = -1$ if and only if $\inf_{s \in [0, t]} X(s) = -1$ for some $t > 0$, and the latter event has zero probability due to the fact that the descending ladder-height process has no drift (see Theorem 5.9 in the book).

The same argument actually works for $X(\tau_{-1}) = -1$, and the proof is in fact slightly simpler. Unfortunately, I do not have time now to write up the details.