Quasi-left-continuity of Feller-processes as in Proposition 7, p. 21, of Bertoin, Lévy processes

Question

The statement of Prop.7, p.21, of Lévy processes, Bertoin, asserts the quasi-left-continuity of a Lévy process. The proof relies only on its property to be a Feller-process and having cádlág paths. So one can read the statement as:

Let $(X_t)_{t\ge 0}$ be a Feller-process with cádlág paths. Then, if $T_n$ is a sequence of stopping times with $T_n\leq T_{n+1}$ and $T:= \lim_{n\to\infty}T_n$, we have $$\lim_{n\to\infty} X_{T_n} = X_T$$ almost surely on $\{T<\infty\}$.

The first sentence of the proof reads:

With no loss of generality we may assume that $T<\infty$ and $T_n < T$ fo all $n$ a.s..

Question: How can this step be made more precise? I don't see how we can directly pass from the situation in the statement to the situation of $T<\infty$ and $T_n <T$.

After this assumption one ends up with the situation of the definition of quasi-left-continuity as in this post and the proof of the quasi-left-continuity in the post is in accord with the proof in Bertoins book. Therefore, this means that the a-priori stronger property from the statement above is equivalent to quasi-left-continuity as defined in the post.

Answer 1

Partial answer

Assume that the result is already established for stopping times taking only finite values. Let $T$ be any stopping time. Then given an integer $t \ge 1$, the result applies for non-decreasing sequence of stopping times $(T_n \wedge t)_{n \ge 1}$ which converges to $T \wedge t = \min(T,t)$. On the event $[T<t]$, we have $T_n<t$ so $X_{T_n \wedge t} = X_{T_n}$ for all large enough $n$, so we deduce that $X_{T_n} \to X_T$ almost surely. Hence $X_{T_n} \to X_T$ almost surely on the event $[T<+\infty] = \bigcup_{t \in \mathbb{N}} [T<t]$, which is the general result to be proved.

I do not see how to reduce the proof to the case where all the equalities $T_n<T$ are almost sure.

Answer 2

Define new stopping times $T_n^*$ and $T^*$ by $$ T_n^*=\cases{T_n,& on $\{T_n<T\}$;\cr \infty,& on $\{T_n\ge T\}$,\cr} $$ and $$ T^*=\cases{T,& on $\Gamma$;\cr \infty,& on $\Gamma^c$,\cr} $$ where $\Gamma:=\cap_n\{T_n<T\}$. Then $\lim_n T^*_n=T^*$ and $T^*_n<T^*$ for all $n$ on $\{T<\infty\}$.

Answer 3

The other anwers gave me an idea: We start with the following statement:

Statement: For stopping times with $T_n\leq T_{n+1}$ a.s. and $T:= \lim_{n\to\infty} T_n < \infty$ and $T_n < T$ almost surely, we have that $$\lim_{n\to\infty}X_{T_n}=X_T.$$

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Now let $T_n$ be stopping times with $T_n\leq T_{n+1}$ and $T:=\lim_{n\to\infty} T_n \leq t$, where $t$ is a constant number $t<\infty$.

Then we can define similar to John Dawkins answer: $$T_n^* := T_n 1_{\{T_n < T\}} + \left(t+1 - \frac 1 n \right) 1_{\{T_n = T\}}$$ and $$T^* := T 1_\Gamma + \left( t+1\right) 1_{\Gamma^c},$$ where $\Gamma = \bigcap_{n\in\Bbb N} \{T_n<T\}$.

We have $T^* <\infty$, $T_n^* < T^*$, $\lim_{n\to\infty} T_n^* = T^*$ a.s.. $T_n^*,T^*$ are stopping times (proof at the bottom of the post) Now we can apply the statement of the assumption and get:

$$\lim_{n\to\infty} X_{T_n^*} = X_{T^*}$$

This means that $$\lim_{n\to\infty}X_{T_n} = \lim_{n\to\infty}X_{T_n^*} 1_\Gamma + \lim_{n\to\infty}X_{T} 1_{\Gamma^c}\\ = X_{T^*} 1_\Gamma + X_{T}1_{\Gamma^c} = X_T$$

Therefore we now have the statement:

Statement 2: For stopping times with $T_n\leq T_{n+1}$ a.s. and $T:= \lim_{n\to\infty} T_n \leq t$ for a $t <\infty$ almost surely, we have that $$\lim_{n\to\infty}X_{T_n}=X_T.$$

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Now as in Christophe Leuridan's answer: Let $T_n\leq T_{n+1}$ a.s. and $T:= \lim_{n\to\infty} T_n$. Then we can define $T_n^t = T_n \wedge t$ with $t\in\Bbb N$. We have $T_n^t \leq T_{n+1}^t$ and $\lim_{n\to\infty} T_n^t = T^t := T\wedge t$. We can apply the result of Statement 2 and get:

$$\lim_{n\to\infty} X_{T_n^t} = X_{T^t}$$ which proves that $\lim_{n\to\infty} X_{T_n} = X_T$ on $\{T< t\}$, therefore this is true on $\{T<\infty\} = \bigcup_{t\in\Bbb N}\{T < t\}$.

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Let $s\geq 0$. Then $$\{T_n^*\leq s\} = (\{T_n \leq s\}\cap \{T_n < T\} )\cup \left(\left\{t+1-\frac1 n \leq s\right\} \cap \{T_n = T\}\right)\\ = \begin{cases}\{T_n \leq s\}\cap \{T_n < T\} &: t +1 -\frac 1 n > s\\ (\underbrace{\{T_n \leq s\}\cap \underbrace{\{T_n < T\}}_{\in \mathcal{F}_{T_n \wedge T} \subset \mathcal{F}_{T_n} }}_{\in \mathcal F_s} ) \cup \underbrace{\{T_n = T\}}_{\in \mathcal{F}_{T_n \wedge T} \subset \mathcal{F}_{t} \subset \mathcal{F}_{s} } ) &: t + 1 -\frac 1 n \leq s \end{cases}$$ Let $s\leq t$. Since $\Gamma \in \mathcal F_T$ we have $$\{T^* \leq s\} = \{T\leq s\} \cap \Gamma \in \mathcal F_s$$