# 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!

• I suppose you should have a $\mathcal{F}^+(t)$ inside the first display, is that correct? Commented Jun 8, 2022 at 7:24
• I think here it is really just a matter of writing all the definitions as explicitly as possible. To get that $B_T$ is $\mathcal{F}^+_T$ measurable, you need to show that $[B_T\in I] \in \mathcal{F}^+_T$ for all interval. Now, use the definition of the latter $\sigma$-field. You will have to go one definition further now. Can you see where I am going? Commented Jun 8, 2022 at 7:34
• @Kernel So is the following correct: I have to show that $\{B(T) \in I\} \in \mathcal{F}^+(t)$. So if $\{B(T) \in I\} = A$ then we have, if $T \leq t$ then, $A \in \mathcal{F}^+(t)$, further $\{T \leq t\} \in \mathcal{F}^+(t)$ by definition, so $A \cap \{T \leq t\} \in \mathcal{F}^+(t)$. Hence, $A \in \mathcal{F}^+(T)$. Somehow I feel that I am missing something here. (I guess, I am not considering the case when $T> t$?) Commented Jun 8, 2022 at 12:47
• @Mathaddict You may ask one of the authors here directly :) BTW, showing that $B(T)$ is $\mathcal{F}(T)$-measurable is easy if you approximate $T$ by simple functions from above.
– user140541
Commented Jun 8, 2022 at 15:21
• @Mathaddict Yes. Consider $T_n:=2^{-n}\lceil 2^n T\rceil$.
– user140541
Commented Jun 8, 2022 at 17:29

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 $$g$$.
• This proves that $B_T$ is $\mathcal{F}^+(T)-$measurable, the same method can show that the map from $\Omega$ to $C[0,\infty)$ that takes $\omega$ to $t \mapsto B_{t \wedge T}(\omega)$ is $\mathcal{F}^+(T)-$measurable when $C[0,\infty)$ is equipped with the topology of uniform convergence on compact sets and the corresponding Borel $\sigma$-algebra. Commented Jun 14, 2022 at 16:39