• $T>0$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\in[0,\:T]}$ be a complete filtration on $(\Omega,\mathcal A)$
  • $M$ be an almost surely continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$ almost surely and $$N:=\left\{\omega\in\Omega:X(\omega)\text{ is not continuous}\right\}\cap\left\{X_0\ne0\right\}$$
  • $n\in\mathbb N$, $\tau_0^n:=0$ and $$\tau_k^n:=\inf\left\{t\in(t_{k-1}^n,T]:\left|M_t-M_{\tau_{k-1}^n}\right|=\frac1{2^n}\right\}\wedge T\;\;\;\text{for }k\in\mathbb N$$ with $\inf\emptyset:=\infty$
  • $\mathcal M^2_c$ denote the set of square-integrable almost surely continuous $\mathcal F$-martingales $Y$ with $Y_0=0$ almost surely equipped with $$\left\|Y\right\|_{\mathcal M^2}^2:=\operatorname E\left[\sup_{t\in[0,\:T]}|Y_t|^2\right]$$

We can show that $\tau_k^n$ is an $\mathcal F$-stopping for all $k\in\mathbb N$ with $$\tau_k^n\uparrow T\;\;\;\text{for }k\to\infty\tag1$$ on $\Omega\setminus N$. Now, let $$V_t^n:=\left\{\begin{array}{{{\displaystyle}}l}\displaystyle\sum_{k\in\mathbb N}1_{\left(\tau_{k-1}^n,\:\tau_k^n\right]}(t)M_{\tau_{k-1}^n}&&\text{on }\Omega\setminus N\\0&&\text{on }N\end{array}\right\}\;\;\;\text{for }t\in[0,T]$$ and $$Q^n_t:=\left\{\begin{array}{{{\displaystyle}}l}\displaystyle\sum_{k\in\mathbb N}\left|M_{\tau_{k-1}^n\:\wedge\:t}-M_{\tau_k^n\:\wedge\:t}\right|^2&&\text{on }\Omega\setminus N\\0&&\text{on }N\end{array}\right\}\;\;\;\text{for }t\in[0,T]$$ as well as $$(V^n\cdot M)_t:=\left\{\begin{array}{{{\displaystyle}}l}\displaystyle\sum_{k\in\mathbb N}M_{\tau_{k-1}^n}\left(M_{\tau_k^n\:\wedge\:t}-M_{\tau_{k-1}^n\:\wedge\:t}\right)&&\text{on }\Omega\setminus N\\0&&\text{on }N\end{array}\right\}\;\;\;\text{for }t\in[0,T]\;.$$

Assume that $M$ is bounded (and hence an $\mathcal F$-martingale). We can show that $V^n\cdot M$ is a continuous $\mathcal F$-martingale with $$\left\|V^m\cdot M-V^n\cdot M\right\|_{\mathcal M^2}\xrightarrow{m,\:n\:\to\:\infty}0$$ and hence (since $\mathcal M_c^2$ is a complete semi-normed space) there is a $X\in\mathcal M_c^2$ with $$\left\|V^n\cdot M-X\right\|_{\mathcal M^2}\xrightarrow{n\to\infty}0\;.$$ This implies that $$\sup_{t\in[0,\:T]}\left|(V^n\cdot M)_t-X_t\right|\xrightarrow{\text{in probability}}0\;\;\;\text{for }n\to\infty\;.$$ Now, let $$[M]:=M^2-2X\;.$$ Since $$M^2=2V^n\cdot M+Q^n\;,$$ we obtain $$\sup_{t\in[0,\:T]}\left|Q^n_t-[M]_t\right|=2\sup_{t\in[0,\:T]}\left|(V^n\cdot M)_t-X_t\right|\xrightarrow{\text{in probability}}0\;\;\;\text{for }n\to\infty\;.\tag 1$$

Why can we conclude from $(1)$ that $[M]$ is almost surely increasing?

This claim is made in the proof of the existence of the covariation $[M]$ of $M$ presented in the book of Kallenberg


1 Answer 1


Consider $0<s<t\le T$. There exists $n_0$ (depending on $s,t$, and $\omega$) such that for all $n\ge n_0$ there exists $k$ such that $s<\tau_k^n(\omega)<t$, in which case $Q^n_s(\omega)\le Q_t^n(\omega)$. By (1) there is a subsequence $(n_j)$ such that $Q_t^{n_j}$ converges to $[M]_t$, uniformly in $t\in[0,T]$, a.s. This pointwise convergence implies that $t\mapsto [M]_t$ is non-decreasing, a.s.


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