The strong Markov property with an uncountable index set The following definition of the strong Markov property, from Klenke's book, supposes an index set $I$ that is not necessarily countable. However, it is explicitly mentioned previously (following Lemma 9.23) that for uncountable $I$, $X_\tau$ is not always measurable. So how can i make sense of this definition in case $I$ isn't countable?
Definition 17.12 (p. 350) Let $I\subseteq\left[0,\infty\right]$ be closed under addition. A Markov process $\left(X_t\right)_{t\in I}$ with distributions $\left(\mathrm{P}_x,\space x\in E\right)$ has the strong Markov property iff for every a.s. finite stopping time $\tau$, every bounded $\mathcal{B}\left(E\right)^{\otimes I}-\mathcal{B}\left(\mathbb{R}\right)$ measurable function $f:E^I\rightarrow\mathbb{R}$ and every $x\in E$ we have
$$\mathrm{E}_x\left[\left.f\left(\left(X_{\tau+t}\right)_{t\in I}\right)\space\right|\mathcal{F}_\tau\right]=\mathrm{E}_{X_\tau}\left[f\left(X\right)\right]:=\intop_{E^I}\kappa\left(X_\tau,\mathrm{d}y\right)f\left(y\right)$$
 A: A sufficient condition is that the process $X$ is measurable, that is, the map 
$(\omega,t)\to X(\omega,t)$ should be measurable on the product space 
$(\Omega\times I,{\cal F}\times {\cal I})$ to $(E,{\cal E})$. 
Then if $\tau:(\Omega,{\cal F})\to(I,{\cal I})$ is 
a (measurable) random time,  the composition 
$$\begin{array}{ccccc}
\omega &\to& (\omega,\tau(\omega))&\to& X(\omega,\tau(\omega))\\[3pt]
{\cal F}&& {\cal F}\times{\cal I} &&{\cal E}\end{array}
$$
is measurable.
The joint measurablility of $(\omega,t)\to X(\omega,t)$ is often proved 
by combining  measurability of the slices $\omega\to X(\omega,t)$
 with  some sort of sample path regularity. For instance, left or right continuity is enough. 
A: You don't need $$\tilde X:=\left(X_{\tau+t}\right)_{t\in I}:\Omega\to E^I$$ to be $\mathcal F_\tau$-measurable (which is, more or less, the statement of Lemma 9.23). Please note, that in that case $$\operatorname E_x\left[f\circ\tilde X\mid\mathcal F_\tau\right]=f\circ\tilde X\;.$$ What you need is $\mathcal A$-measurability of $\tilde X$. A sufficient condition would be $\mathbb F$-progressive measurability of $X$, i.e. $$\Omega\times[0,t]\to E\;,\;\;\;(\omega,s)\mapsto X_s(\omega)$$ should be measurable with respect to $\mathcal F_t\otimes\mathcal B\left([0,t]\cap I\right)$, for all $t\in I$. 

Try to show, that in this case, the stopped process $\left(X_t^\tau\right)_{t\in I}$ is $\mathbb F$-progressive measurable, too.

Then, take $B\in\mathcal E$ and $t\in I$ and observe, that since $$\left\{X_\tau\in B\right\}\cap\left\{\tau\le t\right\}=\left\{X_{\tau\wedge t}\in B\right\}\cap\left\{\tau\le t\right\}\;,$$ it's sufficient to show, that $\left\{X_{\tau\wedge t}\in B\right\}\in\mathcal F_t$. But by the previous claim $\left(X_{\tau\wedge t}\right)_{t\in I}$ is $\mathbb F$-adapted. Thus, $X_\tau$ is $\mathcal F_\tau$-measurable. Since $\tau+t$ is a $\mathbb F$-stopping time, too (for all $t\in I$), $\tilde X$ is $\mathcal A$-measurable.
At the end, you might want to notice, that

A sufficient condition for $\mathbb F$-progressive measurability is $\mathbb F$-adaptedness and left- or right-continuity.

