In Serfozo's book, the Strong Markov Property is stated as ($*$): \begin{align*} \newcommand{\cprob}[2]{\mathrm{P}\!\left[#1 \mid #2 \right]} \newcommand{\prob}[1]{\mathrm{P}\!\left[#1\right]} & \cprob{X_{\tau+1}=j_1,\ldots,X_{\tau+m}=j_m}{X_0=i_0,\ldots,X_{\tau-1}=i_{\tau-1},X_\tau=i}\\ &= \cprob{X_1=j_1,\ldots,X_m=j_m}{X_0=i}\tag{$*$}. \end{align*} where $\tau$ is a stopping time of the DTMC $(X_n)_{n\geq0}$. I am confused about one point in the proof.
Proof. Write $(*)$ as \begin{align*} \cprob{A_\tau}{B_\tau} = \cprob{A_0}{X_0=i} \end{align*} for \begin{align*} A_n&=\{X_{n+1}=j_1,\ldots,X_{n+m}=j_m\}\\ B_n&=\{X_{0}=i_0,\ldots,X_{n-1}=i_{n-1},X_{n}=i\}. \end{align*} Now condition on $\tau$ so that by law of total probability we have \begin{align*} \cprob{A_\tau}{B_\tau} = \sum_{n=0}^\infty \cprob{A_\tau}{B_\tau,\tau=n}\cdot\cprob{\tau=n}{B_\tau}. \end{align*} Then \begin{align*} \cprob{A_\tau}{B_\tau,\tau=n} = \frac{\prob{A_n,B_n,\tau=n}}{\prob{B_n,\tau=n}}. \end{align*} But $\tau$ is a stopping time, so the event $\{\tau=n\}$ is determined by $X_0,\ldots,X_n$. Thus $B_n = B_n\cap\{\tau=n\}$ when $\prob{B_n,\tau=n}>0$. This fact simplifies the preceding display so that \begin{align*} \cprob{A_\tau}{B_\tau,\tau=n} = \frac{\prob{A_n,B_n}}{\prob{B_n}}. \end{align*} Now we can express the above display as multiplication of transition probabilities so that, cancelling terms, we obtain \begin{align*} \frac{\prob{A_n,B_n}}{\prob{B_n}} = P_{i,j_1}\cdots P_{j_{m-1},j_m} = \cprob{A_0}{X_0=i}. \end{align*} At this point, Serfozo ends the proof. I am confused about what happens to the term $\cprob{\tau=n}{B_\tau}$. Is it 0 except for when the events $B_\tau$ and $B_n$ coincide (in which case it is 1)?
I find your book here Basics of Applied Stochastic Processes. The reason for your question is given below.
According to Definition 25 on p. 16, for any finite $n$, the event $\{\tau= n\}$ is a function of the history $X_0,X_1,...,X_n$ up to time $n$. That is to say, the indicator function $1_{\{\tau=n\}}$ of the event that $T = n$ is a (measurable) function of the variables $X_0,X_1 , ..., X_n$, i.e., $1_{\{\tau=n\}}=g(X_0,X_1,...,X_n)$.
Let $\Omega$ be the sample space. Note that the sets $\{ \tau =k\}, k=0,1,2,...$, form a partition of $\Omega$ (mutually disjoint, and their union is $\Omega$).
Note that \begin{align*} X_0(\omega)=i_0,...,X_{k-1}(\omega)=i_{k-1}, X_k(\omega)=i \ \ (\Leftrightarrow \omega\in B_k);\tag{$1$}\\ 1=1_{\{\tau=k\}}(\omega)= g(X_0(\omega), X_1(\omega),...,X_k(\omega)) \ \ (\Leftrightarrow \omega\in \{\tau=k\}). \tag{$2$} \end{align*} We can see from eq. (1) and (2) that for a fixed $k$, if there is some $\omega \in B_k\cap \{\tau=k\}$, then all those $\omega'\in B_k$ belong to $\{\tau=k\}$ (Notice that $(i_0,i_1,...,i_{k-1},i)$ is a singleton in the product space $\mathbb{N}\times \cdots\times \mathbb{N}$!). That is to say, if $B_k\cap \{\tau=k\}\neq \varnothing$, then $B_k\subset \{\tau=k\}$, so that $B_k\cap \{\tau=k\}=B_k$.
Let \begin{align*} &S_1= \{n: B_\tau\cap \{\tau=n\}=B_n\cap \{\tau=n\}\neq \varnothing\},\\ &S_2=S_1^c . \end{align*} Hence, for all those $n\in S_1$, we have \begin{align*} \mathbb{P}(A_\tau \mid B_\tau, \tau=n) = \frac{\mathbb{P}(A_n,B_n,\tau=n)}{\mathbb{P}(B_n,\tau=n)} =\frac{\mathbb{P}(A_n,B_n)}{\mathbb{P}(B_n)}=\mathbb{P}(A_0\mid X_0=i), \end{align*} and for those $n\in S_2$, $\mathbb{P}( B_\tau, \tau=n)=0$. That is to say, \begin{equation} \mathbb{P}(A_\tau \mid B_\tau, \tau=n) =\begin{cases} \mathbb{P}(A_0\mid X_0=i), & n\in S_1;\\ 0, &n\in S_2. \end{cases} \end{equation}
Therefore, \begin{align*} \mathbb{P}(A_\tau\mid B_\tau)&= \sum_{n=0}^\infty \mathbb{P}(A_\tau\mid B_\tau,\tau=n) \ \mathbb{P}(\tau=n\mid B_\tau)\\ &=\sum_{n\in S_1} \mathbb{P}(A_\tau\mid B_\tau,\tau=n) \ \mathbb{P}(\tau=n\mid B_\tau) +\sum_{n\in S_2} \mathbb{P}(A_\tau\mid B_\tau,\tau=n) \ \mathbb{P}(\tau=n\mid B_\tau)\\ &= \mathbb{P}(A_0\mid X_0=i)\sum_{n\in S_1} \mathbb{P}(\tau=n\mid B_\tau)+0, \end{align*} We need to calculate the term $\sum_{n\in S_1}\mathbb{P}(\tau=n\mid B_\tau)$. Note that for those $n\in S_2$, according to our definition, $B_\tau \cap \{\tau=n\}=B_n\cap \{\tau=n\}=\varnothing$, so that \begin{equation} \mathbb{P}(\tau=n\mid B_\tau) = \frac{\mathbb{P}(\tau=n, B_\tau)}{\mathbb{P}(B_\tau)}= \frac{\mathbb{P}(\tau=n, B_n)}{\mathbb{P}(B_\tau)}= 0. \end{equation} Thus, \begin{align*} \sum_{n\in S_1}\mathbb{P}(\tau=n\mid B_\tau)= \sum_{n=0}^\infty \mathbb{P}(\tau=n\mid B_\tau) = \mathbb{P}\left( \left. \sum_{n=0}^\infty\{\tau=n\} \ \right|\ B_\tau\right) = \mathbb{P}\left( \Omega\mid\ B_\tau\right)=1. \end{align*} Hence we have done.