Conditional independence given elementary events implies conditional independence given $\sigma $-algebra

Question

Proposition: Let $X$ be a continuous markov chain with discrete state space $S$. Let $Z$ be the corresponding jump chain and $\left\{ {{W_i},i \in \mathbb{N}} \right\}$ its holding times.

Let ${\mathcal{H}_n} = \sigma \left( {{Z_0}, \ldots ,{Z_n};{W_1}, \ldots ,{W_n}} \right)$. For $j \in S$ and $u > 0$ we have $$\mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|{\mathcal{H}_n}} \right) = \mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|{Z_n}} \right)$$. In other words, $\left( {{Z_{n + 1}},{W_{n + 1}}} \right)$ and ${{\mathcal{H}_n}}$ are conditionally independent given ${{Z_n}}$.

Proof: It's sufficient to show that $$\mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|{Z_0} = {i_0}, \ldots ,{Z_n} = {i_n},\left( {{W_1}, \ldots ,{W_n}} \right) \in B} \right) = \mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|{Z_n} = {i_n}} \right)$$ for any ${i_0}, \ldots ,{i_n} \in S$ and $B \in \mathcal{B}\left( {{\mathbb{R}^n}} \right)$.

My question: Why is it sufficient to show that instead of $$\mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|\left( {{Z_0}, \ldots ,{Z_n}} \right) \in \Sigma ,\left( {{W_1}, \ldots ,{W_n}} \right) \in B} \right) = \mathbb{P}\left( {{Z_{n + 1}} = j,{W_{n + 1}} > u|{Z_n} \in {\Sigma _n}} \right)$$

for any $\Sigma \subseteq S^{n+1}$, ${\Sigma _n} \subseteq S$ and $B \in \mathcal{B}\left( {{\mathbb{R}^n}} \right)$?

EDIT: Assuming I go this way $$\mathbb{P}\left( {A|{Z_1}} \right) = \sum\limits_{{z_1}} {\mathbb{P}\left( {A|{Z_1} = {z_1}} \right){1_{{Z_1} = {z_1}}}} = \sum\limits_{{z_1}} {\mathbb{P}\left( {A|\left( {{Z_0},{Z_1}} \right) = \left( {{z_0},{z_1}} \right),\left( {{W_0},{W_1}} \right) \in B} \right){1_{{Z_1} = {z_1}}}} $$ by assuming the first condition holds, how would I proceed? Is it necessary to use the assumption about operating with discrete random vectors?

Answer 1

I see no difference between the two expressions. One says something about $Z_0=i_0,Z_1=i_1,\cdots Z_n=i_n$ for all $i_k\in S$ and the other says $(Z_0,...Z_n)=\Sigma$ for all $\Sigma\subset S^{n+1}$ (assuming you have a typo). You're on a discrete set, so obviously the two imply each other by writing out $\Sigma$ as a union over all tuples in it and using countable additivity.