Questions about Understanding The First Part of Billingsley's Proof of Strong Markov Property

Question

I'm reading about the strong markov property in Billingsley. I'm confused about some aspects of the proof:

Theorem (Strong Markov Property): Let $\tau$ be a stopping time and $$B^*_t(\omega) = B_{\tau(\omega)+t}(\omega) - B_{\tau(\omega)}(\omega)$$ Then $(B^*_t)_{t \geq 0}$ is a Brownian motion and independent of $\mathcal{F}_{\tau}$ which is to say: $$\textbf{P}(\{B^*_{t_1},...,B^{*}_{t_k}\} \in A \cap M) = \textbf{P}(\{B^{*}_{t_1},...,B^{*}_{t_k}\} \in A) \textbf{P}(M) = \textbf{P}(\{B_{t_1},...,B_{t_k}\} \in A) \textbf{P}(M)$$ for any $A\subset \mathbb{R}^n$ and for any $M \in \mathcal{F}_{\tau}$

Proof: Supposed first that $\tau$ has countable range $V$ and let $t_0$ be some point in $V$. Since $$\{\omega:B^*_t(\omega)\in A\} = \bigcup_{t_o\in V}\{\omega:B_{t_0+t}(\omega)-B_{t_0}(\omega),\tau(\omega)=t_o\}$$ $B^*_t$ is a random variable. Also if $M \in \mathcal{F}_{\tau}$ and $i$ is a finite index: $$ \bigg( \bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\} \bigg) \cap M = \bigcup_{t_o\in V}\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\}$$ so that $$\textbf{P}(\bigg( \bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\} \bigg) \cap M) = \sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\})$$ If $M \in \mathcal{F}_{\tau}$, then $M \cap \{\tau = t_0\} \in \mathcal{F}_{t_0}$. Further, if $\tau = t_0$, then $B^*_t$ is just a translated Brownian motion so that we can use the usual markov property: $$\sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\}) \\ = \sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\})\textbf{P}(\tau(\omega)=t_o,M\}) \quad (*) \\ = \textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\}) \textbf{P}(M) \quad (**)$$ This proves the first and third equality, to get the second, consider the case $M = \Omega$.

Then there's some work on moving to the continuous case which I haven't gotten to yet. My questions:

1. Why is the first bolded statement showing that $B^*_t$ is a random variable? Is it showing that $B^*_t$ is measurable for any $\tau$?

2. In equation $(*)$, I thought the $t_i$ were fixed, do they vary so they're always "ahead" of $t_0$? If the $t_i$ are fixed, how does one invoke markov property for $t_0$ sufficiently large?

3. This is related to the last question -- for equation $(**)$, for the sum to push through to $\textbf{P}(\tau(\omega)=t_o,M\})$, the $t_i$'s would have to be constant right?

4. I don't understand the final bolded statement. The three equalities that they wish to show must be equivalent for any $M\ \in \mathcal{F}_{\tau}$. How is it sufficient to choose one case?

Answer 1

1. $B_t^*$ is a random variable because for any Borel set $A$, $\{B^*_t\in A\}$ is a countable union of measurable sets:

$$ \bigcup_{t_0\in V}\{(B_{t_0+t}-B_{t_0})\in A,\tau=t_0\}. $$

2. The indices $\{t_i\}$ are fixed but their "independent" of the index set of the original process and you care about $B_{t_i}'=B_{t_i+t_0}-B_{t_0}$.

3. The second line (*) is wrong. The whole argument is as follows: \begin{align} &\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap M) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap M\cap\{\tau=t_0\}) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}(\{(B_{t_1}',\ldots,B_{t_k}')\in A\}\cap M\cap\{\tau=t_0\}) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}((B_{t_1}',\ldots,B_{t_k}')\in A)\mathsf{P}(M\cap\{\tau=t_0\}) \\ &\qquad=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A)\sum_{t_0\in V}\mathsf{P}(M\cap\{\tau=t_0\}) \\ &\qquad=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A)\mathsf{P}(M). \end{align}

4. Taking $M=\Omega$ and using the equality between the first and the third terms, \begin{align} \mathsf{P}((B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A)&=\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap \Omega) \\ &=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A). \end{align}