How can we show that this is a stopping time?

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
• $T>0$
• $I:=(0,T]$
• $(\mathcal F_t)_{t\in\overline I}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$
• $M:\Omega\times\overline I\times\Lambda\to\mathbb R$ such that $M(x)$ is a continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ for all $x\in\Lambda$ and $$A(x,y):=[M(x),M(y)]\;\;\;\text{for }x,y\in\Lambda,$$ where $[M(x),M(y)]$ denotes the quadratic covariation of $M(x)$ and $M(y)$

Now, let $\beta\in(0,1]$, $$\left\|f\right\|_{C^{0+\beta}(K)}:=\sup_{\stackrel{x,\:y,\:x',\:y'\:\in\:K}{x\:\ne\:x',\:y\:\ne\:y'}}\frac{\left|f(x,y)-f(x',y)-f(x,y')+f(x',y')\right|}{\left|x-x'\right|^\beta\left|y-y'\right|^\beta}$$ for $K\subseteq\Lambda$ and $f:\Lambda\times\Lambda\to\mathbb R$. Let $K\subseteq\Lambda$ be compact and assume $$\left\|A_t\right\|_{C^{0+\beta}(K)}<\infty\;\;\;\text{almost surely for all }t\in\overline I\tag1,$$ where $A_t=A(\;\cdot\;,t,\;\cdot\;,\;\cdot\;)$. Moreover, assume that $\overline I\ni t\mapsto A(\omega,t\;\cdot\;,\;\cdot\;)$ is continuous with respect to $\left\|\;\cdot\;\right\|_{C^{0+\beta}(K)}$ for all $\omega\in\Omega$.

Let $C>0$. How can we show that $$\tau:=\inf\left\{t\in\overline I:\left\|A_t\right\|_{0+\beta}\ge C\right\}$$ is a $\mathcal F$-stopping time?

Let $t\in\overline I$. The imposed continuity assumptions implies that $$\left\{\tau\le t\right\}=\left\{\sup_{s\:\in\:[0,\:t]}\left\|A_s\right\|_{C^{0+\beta}(K)}\ge C\right\}.\tag1$$

My problem is that I don't see why the set on the right-hand side belongs to $\mathcal F_t$. By the continuity assumption, this reduces to the question why $\left\|A_s\right\|_{C^{0+\beta}(K)}$ is $\mathcal F_s$-measurable for all $s\in\overline I$.

Maybe we're able to prove that by showing that the supremum in the definition of $\left\|\;\cdot\;\right\|_{C^{0+\beta}(K)}$ is equal to the supremum over a dense subset of $K$ (which exists, since $K$ is separable).

Answer 1

By the Cauchy-Schwartz inequality, $$ \big|[M(x),M(y)]_t - [M(x'),M(y)]_t - [M(x),M(y')]_t + [M(x'),M(y')]_t\big|\\ = \big|[M(x)-M(x'),M(y)-M(y')]_t\big|\le \left([M(x)-M(x')]_t[M(y)-M(y')]_t\right)^{1/2}, $$ so the supremum in the definition of $A$ is attained for $\{x,x'\}=\{y,y'\}$. Even more important, by the triangle inequality $$ |[M(x)]^{1/2}_t - [M(x')]^{1/2}_t|\le [M(x)-M(x')]_t^{1/2}, $$ so $[M(\cdot)]^{1/2}_t$, hence$[M(\cdot)]_t$, is almost surely continuous (even Hölder continuous). Then $A$ is (jointly) continuous as well: $$ |A_t(x,y)-A_t(x',y')|\le\big|[M(x),M(y)]_t - [M(x'),M(y)]_t\big|\\ + \big|[M(x'),M(y)]_t - [M(x'),M(y')]_t\big|\\ \le \sup_{K}[M(\cdot)]_t^{1/2} \big([M(x)-M(x')]_t^{1/2} + [M(y)-M(y')]_t^{1/2}\big)\\ \le \sup_{K}[M(\cdot)]_t^{1/2} ||A_t||_{C^{0+\beta}(K)}^{1/2}\big(|x-x'|^{\beta} + |y-y'|^{\beta}\big). $$ So indeed, the supremum is attained over some fixed countable set, which implies the desired meausurability of $A_t$.