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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!

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  • $\begingroup$ I suppose you should have a $\mathcal{F}^+(t)$ inside the first display, is that correct? $\endgroup$
    – Kernel
    Commented Jun 8, 2022 at 7:24
  • $\begingroup$ 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? $\endgroup$
    – Kernel
    Commented Jun 8, 2022 at 7:34
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    $\begingroup$ @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 $T> t$?) $\endgroup$
    – Mathaddict
    Commented Jun 8, 2022 at 12:47
  • $\begingroup$ @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. $\endgroup$
    – user140541
    Commented Jun 8, 2022 at 15:21
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    $\begingroup$ @Mathaddict Yes. Consider $T_n:=2^{-n}\lceil 2^n T\rceil$. $\endgroup$
    – user140541
    Commented Jun 8, 2022 at 17:29

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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 $g$.

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  • $\begingroup$ 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. $\endgroup$ Commented Jun 14, 2022 at 16:39
  • $\begingroup$ Indeed, sorry, I forgot to add the last part. $\endgroup$
    – Kernel
    Commented Jun 15, 2022 at 8:00

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