Continuous-time stopping times and stopped sigma-algebras

Let $(\mathrm{\Omega },\mathcal{F},P)$ be a probability space and let $({\mathcal{F}}_t{)}_{t\ge 0}$ be a continuous-time filtration Continuous-time filtrations and all-pairs martingales.

1. A map $\tau :\mathrm{\Omega }\to [0,\mathrm{\infty }]$ (infinite values allowed) is a stopping time for $({\mathcal{F}}_t)$ when $\{\tau \le t\}\in {\mathcal{F}}_t\text{for every }t\ge 0.$
2. For a stopping time $\tau$ its stopped sigma-algebra is ${\mathcal{F}}_{\tau }:=\{A\in \mathcal{F}:A\cap \{\tau \le t\}\in {\mathcal{F}}_t\text{ for every }t\ge 0\}\mathrm{.}$

The following facts are used below, and each is a direct set computation.

(a) ${\mathcal{F}}_{\tau }$ is a sigma-algebra: $\mathrm{\varnothing }\cap \{\tau \le t\}=\mathrm{\varnothing }\in {\mathcal{F}}_t$; for $A\in {\mathcal{F}}_{\tau }$ one has $A^c\cap \{\tau \le t\}=\{\tau \le t\}\setminus (A\cap \{\tau \le t\})\in {\mathcal{F}}_t$; and for a sequence $A_n\in {\mathcal{F}}_{\tau }$ the union satisfies $\left({\bigcup }_n{A}_n\right)\cap \{\tau \le t\}={\bigcup }_n(A_n\cap \{\tau \le t\})\in {\mathcal{F}}_t$. All operations are literal, not modulo null sets, and no completeness of the filtration is assumed. (b) If $\tau \equiv t_0<\mathrm{\infty }$ is deterministic, then $A\cap \{\tau \le t\}$ is $A$ for $t\ge t_0$ and $\mathrm{\varnothing }$ for $t<t_0$, so ${\mathcal{F}}_{\tau }={\bigcap }_{t\ge t_0}{\mathcal{F}}_t={\mathcal{F}}_{t_0}$, where the last equality follows because the intersection includes its least member ${\mathcal{F}}_{t_0}$. If $\tau \equiv \mathrm{\infty }$, then every test event $\{\tau \le t\}$ is empty and ${\mathcal{F}}_{\tau }=\mathcal{F}$. (c) Suppose the filtration is right-continuous. Then $\tau$ is a stopping time if and only if $\{\tau <t\}\in {\mathcal{F}}_t$ for every $t>0$. Indeed $\{\tau <t\}={\bigcup }_{n\ge 1}\{\tau \le t-\frac{1}{n}\}$ (with the terms for $t-\frac{1}{n}<0$ read as the empty set, since $\tau \ge 0$), which gives the forward implication; conversely $\{\tau \le t\}={\bigcap }_{n\ge 1}\{\tau <t+\frac{1}{n}\},$ and ${\bigcap }_n{\mathcal{F}}_{t+1\mathrm{/}n}={\bigcap }_{s>t}{\mathcal{F}}_s={\mathcal{F}}_t$ by right-continuity together with monotonicity of the filtration. (d) Under the same right-continuity assumption, for $A\in \mathcal{F}$ one has $A\in {\mathcal{F}}_{\tau }$ if and only if $A\cap \{\tau <t\}\in {\mathcal{F}}_t$ for every $t>0$: the forward implication follows from $A\cap \{\tau <t\}={\bigcup }_n(A\cap \{\tau \le t-\frac{1}{n}\})$ and the converse from $A\cap \{\tau \le t\}={\bigcap }_n(A\cap \{\tau <t+\frac{1}{n}\})$ together with ${\bigcap }_n{\mathcal{F}}_{t+1\mathrm{/}n}={\mathcal{F}}_t$.

Because the two versions of each test are interchangeable exactly when the filtration is right-continuous, the convention is recorded here once: the Brownian strong Markov theorem is stated for the usual augmentation, which is right-continuous, and the raw-filtration statements use the non-strict test directly. No choice principle is used by this definition.

Durrett, Section 7.3, and Sousi, Section 6.4, define stopping times by $\{\tau \le t\}\in {\mathcal{F}}_t$ and record the strict-test form for right-continuous filtrations. The stopped sigma-algebra convention is fixed here because the strong Markov theorem and the dyadic ceiling argument both consume the containment ${\mathcal{F}}_{\tau }\subseteq {\mathcal{F}}_{{\tau }_n}$ for ${\tau }_n\downarrow \tau$.