Measurability of a set

Question

This question is from Karatzas's Brownian Motion and Stochastic Calculus page 108.

Let $W_t$ be standard one-dimensional Brownian motion then it concludes that the set $$\Lambda(m)=\left\{(t,\omega)\in[0,\infty)\times\Omega;\max_{(t-(1/m))^+\leq s\leq t}W_s(\omega)=W_t(\omega)=\min_{t\leq s\leq t+(1/m)}W_s(\omega)\right\}$$ is in $\cal B([0,\infty))\otimes\cal F$.

I think it need to prove $f(t,\omega)=\max_{(t-(1/m))^+\leq s\leq t}W_s(\omega)$ is measurable, but can't figure it out. Or maybe there's other way. Any help, thanks!

Answer 1

In order to prove measurability of $\Lambda(m)$, it suffices to show that the processes

$$(t,\omega) \mapsto \max_{(t-1/m)^+ \leq s \leq t} W_s(\omega) \qquad \quad (t,\omega) \mapsto \min_{t \leq s \leq t+1/m} W_s(\omega)$$

are progressively measurable. Since any continuous (adapted) process is progressively measurable, we are done if we can show that these processes are continuous. We only prove it for the first one, the second one is similar.

Lemma The mapping $$t \mapsto g(t,\omega):= \max_{(t-1/m)^+ \leq s \leq t} W_s(\omega)$$ is continuous for almost all $\omega \in \Omega$.

Remark: What we actually prove is that if $f: [0,\infty) \to \mathbb{R}$ is continuous, then $t \mapsto \sup_{(t -1/m)+ \leq s \leq t} f(s)$ is continuous. If you know this statement, then you are already done.

Proof: Fix $\omega \in \Omega$ such that $t \mapsto W_t(\omega)$ is continuous and let $T>0$. We are going to prove continuity of $g(\cdot,\omega)$ at $t=t_0 \in [0,T]$. For simplicity of notation, let us assume $t \geq 1/m$. Because of the uniform continuty of $[0,T] \ni t \mapsto W_t(\omega)$, there exists for any $\epsilon>0$ some $\delta>0$ such that

$$|t-s| \leq \delta, t,s \in [0,T] \Rightarrow |W_t(\omega)-W_s(\omega)| \leq \epsilon \tag{1}$$

Now choose any $t \geq 0$ such that $t_0 \leq t \leq t_0+\delta$. By the continuity of $t \mapsto W_t(\omega)$ there exists some $t_0^* \in [t_0-1/m,t_0]$ such that $g(t_0,\omega)= W(t_0^*,\omega)$ and $t^* \in [t-1/m,t]$ such that $g(t,\omega)= W(t^*,\omega)$. We consider some cases separately:

• $t^* \in [t-1/m,t_0]$ and $t_0^* \in [t-1/m,t_0]$: In this case, we have $$ g(t,\omega) = \sup_{t-1/m \leq s \leq t_0} W_s(\omega) = g(t_0,\omega)$$ and therefore $$|g(t,\omega)-g(t_0,\omega)|=0.$$
• $t^* \in [t-1/m,t_0]$ and $t_0^* \in [t_0-1/m,t-1/m]$: We have $$ g(t,\omega) = \sup_{t-1/m \leq s \leq t} W_s(\omega) \geq W(t-1/m,\omega) \stackrel{(1)}{\geq} W(t_0^*,\omega)-\epsilon \tag{2}$$ and therefore $$\begin{align*} |g(t,\omega)-g(t_0)| &= \left| \sup_{t-1/m \leq s \leq t_0} W_s (\omega)- \sup_{t_0-1/m \leq s \leq t-1/m} W_s(\omega) \right| \\ &= \sup_{t_0-1/m \leq s \leq t-1/m} W_s(\omega)- \sup_{t-1/m \leq s \leq t_0} W_s(\omega) \\ &= W(t_0^*,\omega) - \sup_{t-1/m \leq s \leq t_0} W_s (\omega) \\ &\stackrel{(2)}{\leq} W(t_0^*,\omega)- (W(t_0^*,\omega)-\epsilon) = \epsilon \end{align*}$$
• $t^* \in [t_0,t]$ and $t_0^* \in [t-1/m,t_0]$: As $$g(t_0,\omega) \geq W(t_0,\omega) \stackrel{(1)}{\geq} W(t^*,\omega) - \epsilon \tag{3}$$ and $$g(t,\omega) = \sup_{t-1/m \leq s \leq t} W_s(\omega) \geq \sup_{t_0-1/m \leq s \leq t} W_s(\omega) = W(t_0^*,\omega) = g(t_0)$$ we get $$|g(t,\omega)-g(t_0,\omega)| = g(t,\omega)-g(t_0,\omega) = W(t^*)-g(t_0,\omega) \stackrel{(3)}{\leq} W(t^*,\omega)-(W(t^*,\omega)-\epsilon) =\epsilon.$$
• $t^* \in [t_0,t]$ and $t_0^* \in [t_0-1/m,t-1/m]$: We have $g(t,\omega) \geq W(t-1/m,\omega)$ and therefore $$g(t_0,\omega)-g(t,\omega) \leq W(t^*,\omega) - W(t-1/m,\omega) \stackrel{(1)}{\leq \epsilon}.$$ Similarly, $g(t_0,\omega) \geq W(t_0,\omega)$ implies $$g(t,\omega)-g(t_0,\omega) \leq W(t^*,\omega)-W(t_0,\omega) \stackrel{(1)}{\leq} \epsilon.$$ Consequently, $$|g(t,\omega)-g(t_0,\omega)| \leq \epsilon.$$

Exactly the same reasoning applies for $t$ such that $t_0 - \delta \leq t \leq t_0$. This finishes the proof.