Any stopping time $S_n$ in $\{S_n\leq T\}$ must be constant. Let the filtration by defined by $\mathcal{F}_t= \sigma (\{T\leq s\}_{s\leq t})$.
Let $S_n$ be a stopping time, with respect to the above filtration, and $T$ be a random time with values in $[0,\infty[$.
The book I'm reading states that we must have $S_n$ constant in the event $\{S_n < T\}$. Why is that?
 A: (Implicit: There is a measurable space $(\Omega,\mathcal F)$, and an $\mathcal F$-measurable function $T:\Omega\to[0,\infty[$.)
Another way to describe $\mathcal F_t$ would be as the $\sigma$-algebra generated by the random variable $\omega\mapsto T(\omega)\wedge t$. This means that if $S$ is an $(\mathcal F_t)_{t\ge 0}$-stopping time, then for each $t\ge 0$, the indicator $1_{\{S\le t\}}$ is a Borel measurable function (call it $I_t$) of $T\wedge t$ taking, values in $\{0,1\}$. Thus, for each $\omega\in\Omega$, and $t\ge 0$,
$$
1_{\{S\le t\}}(\omega) = I_t(T(\omega)\wedge t).\qquad(\ddagger)
$$
If $S$ is not constant  on $\{S<T\}$ then there exist $\omega_1$ and $\omega_2$ in $\{S<T\}$ such that $S(\omega_1)<S(\omega_2)$. Choose $t\in]S(\omega_1),S(\omega_2)[$. Using this $t$ in ($\ddagger$) with each of $\omega_1$ and $\omega_2$ you get
$$
1=1_{\{S\le t\}}(\omega_1)=I_t(T(\omega_1)\wedge t) = I_t(t)
$$
but
$$
0=1_{\{S\le t\}}(\omega_2)=I_t(T(\omega_2)\wedge t) = I_t(t)
$$
yielding a contradiction. It follows that $S$ is constant on $\{S<T\}$.
