sequence of stopping times

Question

I have a problem understanding symbol $\tau_{n} \nearrow T$ which is for example used in definition of local martingale - for case where $T$ is finite. This symbol usually stands for sequence of stopping times which satisfy: $\mathbb{P}(\{ \omega : \tau_{n}(\omega)\leq \tau_{m}\ \text{for}\ m>n,\ m,n\in \mathbb{N} \})=1,$ so $(\tau_{n})_{n=0}^{\infty}$ is almost sure increasing and $\mathbb{P}(\{ \omega : \lim_{n \rightarrow \infty }\tau_{n}=\infty \})=1,$ so $(\tau_{n})_{n=0}^{\infty}$ almost sure diverges. For cases where we work on unbounded intervals this is fine. Now the case in question is when we are intested in bounded interval - say $[0, T).$ Does it make sense to require $\mathbb{P}(\{ \omega : \lim_{n \rightarrow \infty }\tau_{n}=T \})=1$ since $\tau_{n}$ can take value of $\infty$? For instance in here there is extra condition which requires that $\tau_{n}<T$ almost sure apart from simultaneous requirement that the sequence of stopping times is divergent and attains limit of $T$ which only causes more confusion.

Answer 1

You made statements which are not true.

This symbol usually stands for sequence of stopping times which satisfy: $\mathbb{P}(\{ \omega : \tau_{n}(\omega)\leq \tau_{m}\ \text{for}\ m>n,\ m,n\in \mathbb{N} \})=1$

This is ok but why do you think that $\mathbb{P}(\{ \omega : \lim_{n > \rightarrow \infty }\tau_{n}=\infty \})=1$ also has to hold?

Actually just $\mathbb{P}(\{ \omega : \lim_{n \rightarrow \infty }\tau_{n}=T \})=1$ is given (what's only the same if $T=\infty$).

Also the statement "since $\tau_n$ can take value of $\infty$" is wrong a.s. if $\tau_{n} \nearrow T$ is given for $T<\infty$.

E.g. take $\tau_n \nearrow \infty$ and consider $$\sigma_n := \tau_n \wedge T$$

Then obviously $\sigma_n \nearrow T, \sigma_n \le T$ and so $\sigma_n$ is bounded and especially never $\infty$