Weak Markov property implies strong Markov property for discrete time

From Klenke, p. 356:

Theorem 17.14: If $I \subset [0,\infty)$ is countable and closed under addition, then every Markov process $(X_n)_{n\in I}$ with distributions $(\mathbf{P}_x)_{x\in E}$ has the strong Markov property.

Proof: Let $f\colon E^I \rightarrow \mathbb{R}$ be measurable and bounded. Then, for every $s \in I$, the random variable $1_{\{\tau=s\}} \mathbf{E}_x[f((X_{s+t})_{t\in I}) | \mathcal{F}_\tau ]$ is measurable with respect to $\mathcal{F}_s$. Using the tower property of the conditional expectation and Theorem 17.9 in the third equality, we thus get

\begin{align*} \mathbf{E}_x\bigl[f\bigl((X_{\tau+t})_{t\in I} \bigr) \big| \mathcal{F}_\tau \bigr] &= \sum_{s\in I} \mathbf{1}_{\{\tau=s\}} \mathbf{E}_x \bigl[ f \bigl( (X_{s+t})_{t\in I} \bigr) \big| \mathcal{F}_\tau \bigr] \\ & = \sum_{s\in I} \mathbf{E}_x\bigl[\mathbf{1}_{\{\tau=s\}} \mathbf{E}_x \bigl[ f \bigl((X_{s+t})_{t\in I} \bigr) \big| \mathcal{F}_s\bigr] \big| \mathcal{F}_\tau \bigr]\\ & = \sum_{s\in I} \mathbf{E}_x \bigl[ \mathbf{1}_{\{\tau=s\}} \mathbf{E}_{X_s} \bigl[ f \bigl( (X_t)_{t\in I} \bigr) \big| \mathcal{F}_\tau \bigr] \\ & = \mathbf{E}_{X_\tau}\bigl[f \bigl( (X_t)_{t\in I}\bigl)\bigr]\, . \end{align*}

In the last step, why can the conditional probability be eliminated?

Since $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, we have $$\mathbb{E}_x(g(X_{\tau}) \mid \mathcal{F}_{\tau}) = g(X_{\tau})$$ for any measurable function $g: E \to \mathbb{R}$. Use this for $g(x) := \mathbb{E}_x(f((X_t)_{t \in I})$.
• Thank you! One question: isn't it $g\colon E\rightarrow \mathbb{R}$? – Ystar May 12 '15 at 14:33