Filtration related to stopping time of Brownian Motion: Paths are measurable?

Question

Given a stopping time $T:\Omega\rightarrow \mathbb{R}$ there is a related $\sigma$-algebra used in Mörters and Peres book, Brownian Motion defined on p. 42 as $$\mathcal{F}^+(T) = \{A\in \mathcal{A}: A\cap \{T\leq t\}\in \mathcal{F}^+(t)\text{ for all $t\geq 0$}\}$$ where $\mathcal{A}$ is a $\sigma$-algebra and $\mathcal{F}^+(t)$ is a filtration related to Brownian motion via $$\mathcal{F}^+(t) = \bigcap_{s>t}\sigma(B(\tau):0\leq \tau\leq s).$$ Now on p. 43 they claim that $\{B(t): t\leq T\}$ is $\mathcal{F}^+(T)$ measurable. However I'm confused as to what this means. $\mathcal{F}^+(T)$ is clearly a $\sigma$-algebra on $\Omega$ while $\{B(t):t\leq T\}$ is for each $\omega\in \Omega$ a subset of $\mathbb{R}$? Could someone clarify this for me?

Answer 1

Measurability is defined in terms of preimages. In probability we often use more compact notation for the events that we consider, both because it aligns with our intuition and because it alleviates the notation.

We have for example \begin{align*} \sigma ( B(\tau) : 0 \leq \tau \leq s) &= \sigma (\lbrace B(\tau) \in B \rbrace: 0\leq \tau \leq s, B\in\mathcal{B}(\mathbb{R})) \\ &= \sigma (B(\tau)^{-1}(B):0\leq \tau \leq s, B \in \mathcal{B}(\mathbb{R})) \end{align*} and now it is more clear that the $\sigma$-algebra is actually generated by sets in $\Omega$.

Regarding the set $\lbrace B(t):t\leq T\rbrace$ it seems to me that this is defined on the path space in the same way that the process $\lbrace B(t):t\geq 0\rbrace$ would be. So $B(\cdot,\omega)$ is considered a mapping from $\Omega$ to the space of continuous functions $C([0,\infty),\mathbb{R}) $.