# Showing almost sure equality of two stopping times

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 $$\{T<t\}=\{T'<t\} \quad\text{a.s.}$$ for every $t\geq 0$. My question is, how do we from this conclude that $T=T'$ a.s.?

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What does {T<t}={T′<t} a.s. means for you ? you mean that their symmetric difference has probability 0 ? Or you mean that there exist an event of probability one such that T<t iff T'<t for every t>=O ? – saposcat Dec 14 '12 at 16:29
Or maybe, for every t>=0, there exist an event of probability 1 such that T<t iff T'<t ? – saposcat Dec 14 '12 at 16:31
$A=B$ a.s. means $A\Delta B$ is a $P$-nullset. – Stefan Hansen Dec 14 '12 at 16:33

If for every $t\geqslant 0$, $$\{T<t\}=\{T'<t\} \quad\text{a.s.}$$ Then for every $t\geqslant 0$, the symetric difference between $\{T<t\}$ and $\{T'<t\}$ has a probability $0$. If $A_t$ is $\Omega\setminus(\{T<t\}\Delta\{T'<t\})$, then $\forall \omega\in A_t$ you have $T(\omega)<t$ if and only if $T'(\omega)<t$.
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 $\forall q\in \mathbb{Q}_+$, $T(\omega)<q$ if and only if $T'(\omega)<q$, which means that $$T(\omega)=T'(\omega)$$.
As $P(A)=1$, this means that $T=T'$ a.s.