Show that $M_{t}=\max_{0\leq s\leq t}B_{s}$ is adapted to the natural Brownian filtration.

Question

Let $\left(B_{t}\right)_{t\geq0}$ a Brownian motion. Show that $M_{t}=\max_{0\leq s\leq t}B_{s}$ is adapted to the natural Brownian filtration.

Remark: I try the following:

It suffices to show that $M_{t}^{-1}\left(\left(-\infty,c\right]\right)\in\mathcal{F}_{t}$ where $\left(\mathcal{F}_{t}\right)_{t\geq0}$ is the natural Brownian filtration. Note that

\begin{eqnarray*} M_{t}^{-1}\left(\left(-\infty,c\right]\right) & = & \left\{ \omega\left|M_{t}\left(\omega\right)\leq c\right.\right\} \\ & = & \left\{ \omega\left|\max_{0\leq s\leq t}B_{s}\left(\omega\right)\leq c\right.\right\} \\ & = & \bigcap_{0\leq s\leq t}\left\{ \omega\left|B_{s}\left(\omega\right)\leq c\right.\right\} \end{eqnarray*}

We know that $\left\{ \omega\left|B_{s}\left(\omega\right)\leq c\right.\right\}\in \mathcal{F}_{t}$ because $\left(\mathcal{F}_{t}\right)_{t\geq0}$ is the natural Brownian filtration, but $\bigcap_{0\leq s\leq t}\left\{ \omega\left|B_{s}\left(\omega\right)\leq c\right.\right\}$ is uncountable intersection, What can I say? $\bigcap_{0\leq s\leq t}\left\{ \omega\left|B_{s}\left(\omega\right)\leq c\right.\right\}\in \mathcal{F}_{t}$?

Answer 1

If $I$ is uncountable and $A_i \in \mathcal{F}_t$, then it does in general not follow that $\bigcap_{i \in I} A_i \in \mathcal{F}_t$. We only know that $\mathcal{F}_t$ is stable under countable intersections.

Recall the following lemma:

Let $g: [0,t] \to \mathbb{R}$ be a continuous function. Then $$\max_{s \in [0,t]} g(s) = \max_{s \in [0,t] \cap \mathbb{Q}} g(s).$$

For a proof see e.g. this question (set $f:=-g$ and use that $\max (-f) = - \min f$.)

Applying the above lemma to $g(s) := B_s(\omega)$ for each fixed $\omega \in \Omega$, we get

$$\max_{s \in [0,t]} B_s = \max_{s \in [0,t] \cap \mathbb{Q}} B_s$$

and therefore

$$M_t^{-1}((-\infty,c]) = \bigcap_{s \in [0,t] \cap \mathbb{Q}} \underbrace{\{B_s \leq c\}}_{\in \mathcal{F}_t}.$$

This shows that $M_t^{-1}((-\infty,c]) \in \mathcal{F}_t$ for any $c \in \mathbb{R}$ (since it is a countable intersection of $\mathcal{F}_t$-measurable sets).