The jump of cadlag process is indistinguishable from the zero process

Question

Let $X$ be adapted and cadlag. If $\Delta X_T1_{T<\infty}=0$ a.s. for each stopping time $T$, then $\Delta X$ is indistinguishable from the zero process.

Protter proves the above statement by the following arguments:

Define stopping times for each $n$ inductively:

$T^{n,1} = \inf\{t >0 : |\Delta X_t|>\frac{1}{n}\},$

$T^{n,k}=\inf\{t >T^{n,k-1} : |\Delta X_t|>\frac{1}{n}\}.$

Then $T^{n,k}>T^{n,k-1}$ a.s. on $\{T^{n,k-1}<\infty\}$. Moreover $$\{|\Delta X_t|>0\}=\bigcup_{n,k}\{|\Delta X_{T^{n,k}}1_{T^{n,k}<\infty}|>0\}.\tag{1}$$

I am trying to understand this argument. We are trying to establish the LHS of (1) has measure $0$, and using the assumption, it suffices to establish $(1)$.

We know $$\{|\Delta X_t|>0\}=\bigcup_{n}\{\omega: \exists t>0, s.t. |\Delta X_t|>\frac{1}{n}\}.\tag{2}$$ Comparing (2) with (1), we know it suffices to show $$\{\omega: \exists t>0, s.t. |\Delta X_t|>\frac{1}{n}\}=\bigcup_{k}\{|\Delta X_{T^{n,k}}1_{T^{n,k}<\infty}|>0\}.$$

Then I don't know how to proceed. In particular, I don't know what the state of $\Delta X_{T^{n,k}}$ since $\Delta X_t$ does not have any continuity. Moreover, I don't understand why "$T^{n,k}>T^{n,k-1}$ a.s. on $\{T^{n,k-1}<\infty\}$" is true and how to use it.

Answer 1

Let $t_0\in (0,\infty)$. We prove the result on $[0,t_o]$. X is cadlag $\implies\{t:|\Delta X_t|>0\}$ is countable a.s. $$\cup_{n=1}^\infty \{t:|\Delta X_t|>1/n\}=\{t:|\Delta X_t|>0\}.$$ Now, define inductively $T^{n,k}$ as you have done above. The definition of $T^{n,k}$ implies $T^{n,k}>T^{n,k-1}$ a.s. on $\{T^{n,k-1}< \infty \}.$ Protter says these are stopping times as a result of Theorem 4, but $\Delta X$ is not cadlag. So, thereom 4 specifically does not apply. He mentions the Debut theorem after that. I consulted Medvegyev's Stochastic Integration Theory for a discussion of the Debut Theorem. $\Delta X$ is progressively measurable since $X$ is cadlag. The Debut theorem implies they are stopping times. Finally, the last line of the proof follows by noting that the countable union of sets of measure zero has measure zero.