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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.

saz
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foubw
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  • Judging by the former title, it seems that you were able to find the proof in your book. Please include the part of the proof you don't understand to your question. – zhoraster Nov 09 '15 at 17:13
  • I published the proof as requested – foubw Nov 10 '15 at 00:34
  • No, the last equality does in general not hold true. Just consider $$X_t(\omega) := \begin{cases} 1-t, & t \in [0,1), \ 1, & t \geq 1 \end{cases}$$ and $C:={0}$. Then $\inf_{r \in \mathbb{Q} \cap [0,1)} d(X_r,C)=0$, but $\tau_C = \infty$. – saz Mar 30 '16 at 08:05

1 Answers1

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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".

saz
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  • But it is not true that the set ${(r_n)_{n\in\mathbb{N}}:r_n\in\mathbb{Q}\forall n}$, which is $\mathbb{Q}^{\mathbb{N}}$, is countable. Even its subset ${0,1}^{\mathbb{N}}$ is uncountable, for the binary expansion gives a surjection from ${0,1}^{\mathbb{N}}$ onto the interval $[0,1]$. – Chuwei Zhang Apr 02 '16 at 01:56
  • @ChuweiZhang Ah, damn, you are right... somehow I was thinking about the union of countable sets and not the cartesian product. – saz Apr 02 '16 at 05:34
  • @saz could you better explain this final step: as ${\tau_{K_n} \leq t} \in \mathcal{F}_t$ by the above theorem? which is the "above theorem"? – Buddy_ Nov 09 '19 at 17:46
  • @Alias__ The result by Roger & Williams which I stated some lines earlier. – saz Nov 09 '19 at 18:42