example of totally inaccessible stopping time.: Why $S_n$ must be constant on {$S_n$
Can you explicitly explain why any s.t. $S_n$ must be constant on $\{S_n<T\}$?
I understand that $S_n$ to be a s.t. $\{S_n\leq t\}\in F_t$ and $F_t$ is generated by sets of omegas for which ${T\leq s}$ for any $s\leq t$. Can't understand how they are related.
 A: This is example 6.2.5 from Stochastic Calculus and Applications by
Cohen & Elliott (http://dx.doi.org/10.1007/978-1-4939-2867-5).
I can't give a precise proof, and I don't know what Cohen and
Elliott had in mind, but I think I can illustrate why this is the case.
Since $\mathcal{F}_t$ is generated by the nested sets $\{T\le u\}$
for $u\le t$, and sigma algebras are closed under complements, it


*

*contains $\{T>t\}$, and

*the only subsets of $\{T>t\}$ in $\mathcal{F}_t$ differ from it by
sets of measure zero.


Since $S_n$ is a stopping time, $\{S_n \le t\}\in \mathcal{F}$.
Suppose $\{S_n \le t\} \nsubseteq \{T \le t\}$.  Then $\{S_n \le t\}
\cap \{T > t\}$ is nonempty.  If it has positive measure, then it must
be the case (by point 2 above) that $\{S_n \le t\} \cap \{T > t\} = \{T > t\}$ up to a
set of measure zero.  In particular, then $S_n \le t$ (because $\{T >
t\}$ and $\{T \le t\}$ partition $\Omega$).
Let $u$ be the infimum of the set of $t$ for which $\{S_n \le t\}
\cap \{T > t\}$ is of positive measure.  Then $S_n \le u$, and for $u'<u$, it
must be the case that $\{S_n \le u'\} \subset \{T \le u'\}$, so in
fact, $S_n = u$ on $\{T>u\}$ and the latter set in fact equals $\{S_n
< T\}$.
