In Protter's Stochastic Integration, Chapter I, Theorem 7's proof says that if $X$ is adapted and cadlag, then you can use Theorem 4 to say $T(\omega):=\inf\{t>0:|\Delta X_t|>1/n\}$ is a stopping time. Here $\Delta X_t:=X_t-X_{t-}$ is the jump of the cadlag process.
Although the fact $T(\omega)$ is a stopping time has been proven in many places, none of such proofs use Protter's Theorem 4 e.g. The jumping times of a càdlàg process are stopping times. and The jump of cadlag process is indistinguishable from the zero process.
Protter's Theorem 4 says that if $\Lambda$ is a closed set, then $T(\omega):=\inf\{t>0:X_t(\omega)\in\Lambda\text{ or } X_{t-}\in\Lambda\}$ is a stopping time.
Let me paraphrase Protter's proof: "Restrict $X$ to $t\in[0,t_0]$. The set $\{t>0:|\Delta X_t|>1/n\}$ is finite a.s. Therefore, by Theorem 4, $T(\omega)$ [as defined above] is a stopping time." Now, Protter doesn't give more details. My question is, how does he use Theorem 4? I need help but first this is an attempt at the proof:
The fact that the set of discontinuities of a cadlag process is finite (when considering only $t\in[0,t_0]$) is very well known. While this is a statement about the time variable $t$, $\Lambda$ in Protter's Theorem 4 concerns the space of $X_t(\omega)$. However, notice that the a.s.-finiteness of $\{t>0:|\Delta X_t|>1/n\}$ implies $\{X_t:|\Delta X_t|>1/n\}$ and $\{X_{t-}:|\Delta X_t|>1/n\}$ are also finite, and hence closed, a.s. However, Theorem 4 requires closedness, not a.s.-closedness. How can I keep going?