Càdlàg process and a predictable stopping time

Question

Let $A_t$ be a predictable càdlàg process, and let $a > 0$. Define $$ T = \inf \{t \geq 0 : \Delta A_t > a \} \ .$$ I wish to show that $T$ is a predictable stopping time, so we must show that there exists a sequence of stopping times $T_n \nearrow T$ and $T_n < T$ when $T > 0$. My guess is that the sequence is given by $$T_n = \inf \left\{ t \geq 0 : \Delta A_t > a - \frac{1}{n} \right\} \ .$$ To show that this sequence is increasing, let $\varepsilon > 0$. By the definition of the infimum, we know that there exists $t < T_{n+1} + \varepsilon$, such that $t$ satisfies $$ \Delta A_t > a - \frac{1}{n+1} \ .$$ We have $$ \Delta A_t > a - \frac{1}{n+1} > a - \frac{1}{n} \ ,$$ so $t \geq T_n$, and thus we have the inequality $T_n \leq T_{n+1} + \varepsilon$. Since this holds for all $\varepsilon > 0$, we have $T_n \leq T_{n+1}$. One can also show, in the same way, that $T_n \leq T$ for all $n$. I would like to show that $$ \lim_{n \to \infty} T_n = T,$$ however I am having problems since $A_t$ is only càdlàg. In particular, we have an increasing sequence $T_n$, yet $A_t$ is right continuous, so I am unable to use the right continuity. Any hints?

Answer 1

$T=\inf\{t\ge 0:\Delta A_t>a\}$ is a predictable stopping time: Since $A$(and $\Delta A$) is a predictable process, $B\stackrel{\text{def}}=\{(t,\omega)\in \mathbb{R}_+\times\Omega:\Delta A_t(\omega)>a\}$ is a predictable set, denote it by $B\in\mathscr{P}$. Since $A$ is càdlàg and $a>0$, there are at most finite numbers of $t$, for which $\Delta A_t>a$, in every finite interval $[0,N]$, this means that $\Delta A_T>a$ on $\{T<\infty\}$ and $$ [\![T]\!](=\{(T(\omega),\omega)\in \mathbb{R}_+\times\Omega\})\subset B.$$ Hence $[\![T]\!]=[\![0,T]\!]\cap B\in \mathscr{P}$, since both $[\![0,T]\!]$ and $B$ are predictable sets. Therefore $T$ is a predictable stoping time.