# [Math] Stopping times and hitting times for càdlàg processes

probability theorystochastic-calculusstochastic-processesstopping-times

I can't find the proof of the following lemma in any book:

LEMMA: If $X=\{X_t\}_{t\in T}$ is adapted and right continuous, then for every closed set $C \subset E$, the variable $\tau_{C}:=\inf\{t\in T:X_t\in C\}$ is a stopping time.

I know that this lemma is a particular case of Debut theorem. But in my opinion exists an easier proof of the lemma, anyone know some reference?

I found the following proof but in my opinion it isn't correct:

Proof: Set $d(\centerdot,\centerdot)$ the distance in $E$, it's easy to prove that for every $D\subset E$ the function $x\mapsto d(x,D)$ is continuous from $E$ to $\mathbb{R}$.
So, if $X$ is a right continuous process, for every $\omega \in \Omega$ the real function $u\mapsto d(X_{u}(\omega),C)$ is right continuous and is zero in every and only points $u\geq 0$ s.t. $X_u(\omega)\in C$. we can write:
$$\{\tau_{C}\leq t\}=\{\exists u \in T\cap[0,t]:d(X_u,C)=0\}=\{\inf_{u\in(T\cap[0,t]\cap\mathbb{Q})\cup\{t\}}d(X_u,C)=0\}$$

I can't understand the last equality. In my opinion you need that the infimum is a minimum in order to claim this equality, but I can't understand why this is true.

I agree that the "proof" is not correct. Using the argument sketched in the proof, it is possible to show that

$$T_C := \inf\{t; X_{t} \in C \, \text{or} \, X_{t-} \in C\}$$

is a stopping time, but not that

$$\tau_C := \inf\{t; X_t \in C\}$$

is a stopping time.

I'm aware of two proofs which do not rely on Début's theorem. There is a proof in a book by Rogers & Williams ("Diffusions, Markov Processes and Martingales, Vol.1") which applies to right-continuous processes, but it requires a sound knowledge of ordinals. In Itô's book ("Stochastic Processes") there is a more elementary (or, at least, more probabilistic) proof, but he assumes additionally that the process is a Markov process.

Rogers & Williams (Lemma II.75.1) prove the following statement

Let $E$ be a separable space and $K \subseteq E$ compact. Moreover, let $(X_t)_{t \geq 0}$ be a right-continuous process and suppose that $(X_t)_{t \geq 0}$ is adapted to a filtration $(\mathcal{F}_t)_{t \geq 0}$ which satisfies the usual conditions. Then $\tau_K$ is an $\mathcal{F}_t$-stopping time.

The proof does not rely on Début's theorem and is not too lengthy (roughly 1 page), but I wouldn't call it elementary (knowledge of ordinals is required).

If $E$ is a "nice" space, say $E= \mathbb{R}^d$, then this statement implies that

$$\tau_C := \inf\{t; X_t \in C\}$$

is a stopping time for any closed set $C \subseteq \mathbb{R}^d$. This follows from the fact that $K_n := C \cap \{x \in \mathbb{R}^d; |x| \leq n\}$, $n \in \mathbb{N}$, is a sequence of compact sets satisfying $K_n \uparrow C$; hence,

$$\{\tau_C \leq t\} = \bigcup_{n \in \mathbb{N}} \{\tau_{K_n} \leq t\} \in \mathcal{F}_t$$

as $\{\tau_{K_n} \leq t\} \in \mathcal{F}_t$ by the above theorem.

Itô (Section 2.10) shows the following result:

Let $(X_t)_{t \geq 0}$ be a càdlàg Markov process and $C$ a closed set. Then $\tau_C$ is a stopping time with respect to the completed (canonical) filtration.

The idea of the proof is as follows: Choose a sequence of open sets $(G_n)_{n \in \mathbb{N}}$ such that $G_1 \supseteq G_2 \supseteq \dots$ and $C = \bigcap_{n \in \mathbb{N}} G_n$. Since $(X_t)_{t \geq 0}$ has càdlàg sample paths and $G_n$ is open, it is well-known (and simple to prove) that

$$\tau_{G_n} := \inf\{t \geq 0; X_t \in G_n\}$$

is for each $n \in \mathbb{N}$ is a stopping time. Since $\tau_{G_1} \leq \tau_{G_2} \leq \ldots$, we know that $\tau := \lim_{n \to \infty} \tau_{G_n}$ exists and that $\tau$ is a stopping time. Then Itô proceeds to show that $(X_t)_{t \geq 0}$ is quasi-left continuous at $\tau$ on a (suitable chosen) set $A$ (using a result which is not difficult to prove; here comes the Markovian nature into play) and deduces that $\tau = \tau_C$ almost surely.

Remark: Recall that the assertion is simple to prove if $(X_t)_{t \geq 0}$ has left-continuous sample paths. Roughly speaking, Itô's proof shows that we may replace "left-continuity" but "quasi-left continuity".