Stopping times of Poisson process Given a Poisson process $N$, and let $S_n$ be the $n-$th jump time, i.e.
$$S_n = \inf\{t\mid N_t = n\}$$
Question: is there a way to characterize all stopping times? especially,  can all (or at least, all the bounded) stopping times be described in terms of $S_n$ in some way? 
I'm not sure how to define "being described in terms of $S_n$" exactly, but here is an example, considering the following stopping time: at the 3-rd jump, we will check whether $S_{3}> 10$, if so, we stop, otherwise, stop after 2 more jumps.  Such a stopping time can be written as:  $$T = 1_{S_3>10}S_3 + 1_{S_3\le 10}S_5$$.
[EDIT] My guess,  $T$ is a stopping time if and only if $T$ can be written as
$$ T = 1_{A_0}f_0+1_{A_1} f_1(S_1) + 1_{A_2}f_2(S_1,S_2) + 1_{A_3} f_3(S_1,S_2,S_3) + ...
$$ 
where  $A_i\in \mathcal F_{S_i}$ are mutually exclusive, and $\cup A_i = \Omega$. $f_k(t_1,t_2,...,t_k)$ are deterministic functions, with $f_k(t_1,t_2,...,t_k)\ge t_k$. $A_0 = \Phi$ or $A_0 =\Omega$, $f_0$ is a constant.
[EDIT2] If we define $B_k = \cup\{ C\in \mathcal F_{S_k} \mid 1_CT \in \mathcal F_{S_k}\} $, and let $A_k = B_k -B_{k-1}$. We need to show $1_{A_{k}}T\ge 1_{A_k}S_k$.
I believe the following is true: if $T$ is a stopping time, and $T<S_k$, then $T \in  \mathcal F_{S_{k-1}}$. 
If this is true, it seems that my guess above is correct.
 A: @Jay.H: Even  more is true: If $T$ is a stopping time and $k$ is a positive integer such that  $T<S_k$, then $T\le S_{k-1}$ and consequently $T$ is $\mathcal F_{S_{k-1}}$-measurable.
Consider first the case $k=1$, so we assume that $T<S_1$. Let $\omega$ and $\omega'$ be two sample points, and suppose (without loss of generality) that $0<S_1(\omega)\le S_1(\omega')$. Choose $\epsilon>0$ so small that $T(\omega)<t:=S_1(\omega)-\epsilon$. Clearly $N_s(\omega)=N_s(\omega')=0$ for all $s\in[0,t]$. Because $T$ is a stopping time of the filtration of $N$, it follows from Galmarino's Test (Proving Galmarino's Test) that $T(\omega')=T(\omega)$. Thus $T$ is a non-negative constant. As $\Bbb P[S_1< u]>0$ for all $u>0$, it must be that $T=0$.
For general (fixed) $k$ apply the above observation to the time-shifted filtration $\mathcal G_t:=\mathcal F_{S_{k-1}+t}$, $t\ge 0$. The time $R:=(T-S_{k-1})^+$ is a stopping time of $(\mathcal G_t)$, and $R<S:=S_k-S_{k-1}$. It follows that $R=0$, so $T\le S_{k-1}$. 
