Let an be stopping times and let the filtration on which they are defined be right continuous. Then
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.
Protter, P.E. 2004. Stochastic Integration and Differential Equations: Version 2.1. Applications of Mathematics. Springer. http://books.google.co.uk/books?id=mJkFuqwr5xgC.