How $\{B(t): t \leq T\}$ is $\mathcal{F}^+-$measurable in Brownian Motion, with $T$ a stopping time I am reading the book Brownian Motion by Yuval Peres and Peter Morters. In the book the following $\sigma$-algebra is defined for a given stopping time $T$,
$$\mathcal{F}^+(T)=\{A \in \mathcal{A}: A \cap \{T \leq t\} \in \mathcal{F}^+(t) \text{ for all } t \geq 0\}$$
Where, $\mathcal{F}^+(s) = \bigcap_{t > s} \sigma (B(q): 0 \leq q \leq t)$.
It is mentioned later, without proof, that the random path $\{B(t): t \leq T\}$ is $\mathcal{F}^+(T)-$measurable for any stopping time $T$. I don't understad how to prove this using the definition, I have to show that for any set in the borel sigma algebra of the function space of continuous functions the inverse image of any $B(s), s \leq T$ lies in $\mathcal{F}^+(T)$. Any help is appreciated. Thanks!
 A: Following up on my comment, we will show that given $A=[B_T \in I]$ for any interval $I$, we have $A \cap [T\le t] \in \mathcal{F}^+(t)$ for all $t$. To take this to a Borel set of $\mathbb{R}$ rather than an interval, one can use standard measurable theoretic arguments.
First, as $T$ is a stopping time, we have $[T\le t] \in \mathcal{F}(t) \subset \mathcal{F}^+(t) $, this is the definition of stopping time.
Now, as $T\le t$, we have that $(B_T)(\omega) = B^t_{\cdot}(\omega) \circ T(\omega)$ where $B^t_s(\omega):= B_{t \wedge s}(\omega)$.
As $t \mapsto B_t$ is continuous (and therefore measurable) and $\{B^t_s: s \ge 0\}$ is measurable in $\mathcal{F}^+(t)$.
Using that given two functions $f: \mathbb{R}\times \Omega \mapsto \mathbb{R}$ and $g: \Omega \mapsto \mathbb{R}$, the composition $F(\omega)=f(g(\omega),\omega)$ is also measurable, we complete the answer. One can prove this last statement by taking $g$ to be a simple function first, then taking an approximation via simple function for general $g$.
