Let an be stopping times and let the filtration on which they are defined be right continuous. Then

- ,
- ,
- and

are stopping times where .

For the first we have and both the latter are in by the definition of a stopping time.

Similarly for the second .

For the fourth we have since .

The third is slightly trickier. For , if and only if for some rational , we have . We can thus we can find such that . Writing we also have . Thus we have if and only if there exist and such that and and . In other words

By right continuity (Protter 2004 Theorem 1) of the filtration, we know the terms on the right hand side are in and so that the whole right hand side is in . We thus know that the left hand side is in and using right continuity again that therefore must be a stopping time.

# Bibliography

Protter, P.E. 2004. *Stochastic Integration and Differential Equations: Version 2.1*. Applications of Mathematics. Springer. http://books.google.co.uk/books?id=mJkFuqwr5xgC.

3 and 4 require that your filtration is only defined for positive times. This suggests that they are not particularly natural.

On the other hand, 4 does not need a continuity condition for, unless I am very much mistaken, for all is equivalent to for all . To prove this note that and take the intersection over all rationals .

I edited your comment to make them render more nicely. Let me know if I inadvertently changed the meaning.

I don’t really understand your first point. Aren’t the (implicit) filtrations defined on ? Or perhaps this is not what you meant.

For the second point, I am inclined to agree but in which case why does Protter feel the need for right continuity in order to prove the equivalence of the two definitions of stopping times: and ? Perhaps I shall post a question on mathoverflow.

Thank you for making the LaTeX render correctly. How can I correctly access LaTeX functionality myself?

I think I was very much mistaken and ${ T \le t } \in \mathcal{F}_t$ for all $t$ implies ${ T < t } \in \mathcal{F}_t$, but the converse does not hold unless the filtration is right continuous. I'm still somewhat surprised that the addition result requires it though.

Regarding the first point, in the general case filtrations can be defined for any totally ordered index set, though we normally take subsets of $\mathbb{N}$ or intervals of $\mathbb{R}$. A filtration defined on all of $\mathcal{R}$ is a perfectly natural object to work with but neither 3 nor 4 will hold for it.