I'm trying to understand the proof of the existence of the covariation presented in the book of Kallenberg and got some questions about the involved objects:
Let
- $T>0$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration on $(\Omega,\mathcal A)$
- $(M_t)_{t\in[0,\:T]}$ be an almost surely continuous $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$ almost surely
- $n\in\mathbb N$, $$\tau_0^n:=0$$ and $$\tau_k^n:=\inf\left\{t>\tau_{k-1}^n:\left|M_t-M_{\tau_{k-1}^n}\right|=\frac1{2^n}\right\}\wedge T\;\;\;\text{for }k\in\mathbb N$$
Question 1:$\;\;\;$How can we show that $\tau_k^n$ is an $\mathcal F$-stopping time for all $k\in\mathbb N$?
Now, let $$V_t^n:=\sum_{k\in\mathbb N}1_{\left(\tau_{k-1}^n,\:\tau_k^n\right]}(t)M_{\tau_{k-1}^n}\;\;\;\text{for }t\in[0,T]\tag 1$$ and $$Q_t^n:=\sum_{k\in\mathbb N}\left|M_{\tau_{k-1}^n\:\wedge\:t}-M_{\tau_k^n\:\wedge\:t}\right|^2\;\;\;\text{for }t\in[0,T]\;.\tag 2$$
Question 2:$\;\;\;$Why do the series in $(1)$ and $(2)$ converge?
Q1: (Assume the filtration satisfies the usual conditions.) Consider, more generally, a real-valued adapted process $X$ with $[0,T]\mapsto X_t$ continuous a.s., and a stopping time $\sigma\le T$ such that $X_\sigma=0$. Then for a fixed real $b>0$, $\tau:=\inf\{t>\sigma:X_t=b\}\wedge T$ is a stopping time. With this fact in hand, a simple induction show that $\tau_1^n, \tau_2^n,\ldots$ are stopping times.
To prove the assertion, we can (because of the "usual conditions") assume that $t\mapsto X_t(\omega)$ is continuous for every $\omega\in\Omega$. (Else, let $B:=\{\omega:t\mapsto X_t(\omega)$ fails to be continuous at some $t\in[0,T]\}$. Then $B\in\mathcal F_0$ and $\operatorname P(B)=0$. The process $X'_t(\omega):=1_{B^c}(\omega)X_t(\omega)$ is indistinguishable from $X$ and has everywhere continuous sample paths. The function ${\tau_k^n}'$ defined as was $\tau_k^n$ but in terms of $X'$ will be shown to be a stopping time in what follows, and because $\operatorname P(\tau_k^n={\tau_k^n}')=1$, you see that $\tau_k^n$ is also a stopping time.) Define $Y_t:=\sup_{0\le s\le t}1_{\{\sigma\le s\}}X_s$. By path continuity, this is the same as $\sup_{0\le s\le t,s\in\Bbb Q}1_{\{\sigma\le s\}}X_s$, which shows that $Y$ is an adapted process. Finally, $\{\tau\le t\}=\{Y_t\ge b\}\in\mathcal F_t$, for each $t\in[0,T)$, so $\tau$ is indeed a stopping time.
Q2: Fix $n\in\Bbb N$. Each of the sums (1) and (2) has only finitely many non-zero terms, a.s., because the assumed path-continuity of $M$ implies that $\operatorname P(\cup_k\{\tau_k^n=T\})=1$. (For a sample point $\omega\in\Omega$ with $\tau^n_1(\omega)<\tau^n_2(\omega)<\cdots<\tau_k^n(\omega)<\tau^n_{k+1}(\omega)<\cdots<T$ for all $k$, the limit $\lim_{t\uparrow T}M_t(\omega)$ fails to exist.)