# 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 $$\mathrm{E}_x\left[\left.f\left(\left(X_{\tau+t}\right)_{t\in I}\right)\space\right|\mathcal{F}_\tau\right]=\mathrm{E}_{X_\tau}\left[f\left(X\right)\right]:=\intop_{E^I}\kappa\left(X_\tau,\mathrm{d}y\right)f\left(y\right)$$

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Presumably, part of this definition is the requirement that $X_\tau$ is in fact measurable for all stopping times $\tau$. –  Nate Eldredge Sep 2 '12 at 21:44
@NateEldredge: Thanks. Is this trait, namely that $X_\tau$ be measurable for all a.s. finite stopping times $\tau$, exhibited by common continuous Markov processes $X$, such as Poisson or Brownian Motion? –  Evan Aad Sep 2 '12 at 21:59
On page 194, Klenke tells you: "Here one needs assumptions on the regularity of the paths $t\mapsto X_t(\omega)$; for example, right continuity". –  Byron Schmuland Sep 2 '12 at 23:27
@ByronSchmuland: I meant "continuous" as in "having a continuous index set", but that was an awkward sentence, granted. Thanks for pointing out the note on p. 194. –  Evan Aad Sep 3 '12 at 5:07

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.