2
$\begingroup$

Lemma 1.19 in Jacod and Shiryaev's Limit Theorems for Stochastic Processes states the following:

Any stopping time $T$ on the completed stochastic basis $(\Omega,\mathcal{F}^P,\mathbf{F}^P,P)$ is a.s. equal to a stopping time on $(\Omega,\mathcal{F},\mathbf{F},P)$.

Here $\mathbf{F}=(\mathcal{F}_t)_{t\geq 0}$ is a filtration and $(\Omega,\mathcal{F}^P,\mathbf{F}^P,P)$ denotes the $P$-completion of $(\Omega,\mathcal{F},\mathbf{F},P)$. They construct a stopping $T'$ with respect to $(\Omega,\mathcal{F},\mathbf{F},P)$ which satisfies $ \{T for every $t\geq 0$. My question is, how do we from this conclude that $T=T'$ a.s.?

  • 0
    $A=B$ a.s. means $A\Delta B$ is a $P$-nullset.2012-12-14

1 Answers 1

2

If for every $t\geqslant 0$, $ \{T Then for every $t\geqslant 0$, the symetric difference between $\{T and $\{T' has a probability $0$. If $A_t$ is $\Omega\setminus(\{T, then $\forall \omega\in A_t$ you have $T(\omega) if and only if $T'(\omega).

Then, you can set $A=\cap_{t\in \mathbb{Q}} A_t$ $A$ is an element of $\mathcal{F}^P$. You have $P(A)=1$ (denombrable intersection of almost sure events). And $\forall \omega\in A$, you have that $\forall q\in \mathbb{Q}_+$, $T(\omega) if and only if $T'(\omega), which means that $ T(\omega)=T'(\omega) $.

As $P(A)=1$, this means that $T=T'$ a.s.