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


*

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

*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?

*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?

*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?
 A: *

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


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


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

*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}

*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}
