I am looking at the proof of the lemma below. But I don't get the final part. Here the filtrations are assumed right continuous. So they show that $\{T<t\}=\{T'<t\}$ a.s. for all $t \in \mathbb{R}^+$. But how do we get that the two random times must then be equal a.s.?
After giving some thought I think this may be a valid proof. I would appreciate verification. Denote $N= \bigcup_{r \in \mathbb{Q}} \{\{T<r\}=\{T'<r\}\} $. Then $P(N)=1$. Now suppose we have $T'(\omega) \neq T(\omega)$. WLOG suppose $T'(\omega)=s<T(\omega)=t$. Then take some rational $r \in (s,t)$. Then we have $\omega \in \{T'<r\}$ but $\omega \notin \{T<r\}$. Hence $\omega \in N^c$. Thus we have $T=T'$ on $N$ and so a.s.
If $P[T\not= T']>0$, then either there is a rational $r$ such that $P[T\ge r>T']>0$ or there is a rational $r$ such that $P[T'\ge r>T]>0$. But $$ \{T\ge r>T'\}\subset \{T<r\}\Delta\{T'<r\}, $$ where $\Delta$ indicates symmetric difference. But $P[\{T<r\}\Delta\{T'<r\}]=0$ because $\{T<r\}=\{T'<r\}$ a.s. Therefore $P[T\ge r>T']=0$. Likewise $P[T'\ge r>T]=0$.