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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    $\begingroup$ Presumably, part of this definition is the requirement that $X_\tau$ is in fact measurable for all stopping times $\tau$. $\endgroup$ – Nate Eldredge Sep 2 '12 at 21:44
  • $\begingroup$ @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? $\endgroup$ – Evan Aad Sep 2 '12 at 21:59
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    $\begingroup$ 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". $\endgroup$ – user940 Sep 2 '12 at 23:27
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ I have to admit i have not verified your answer, since i feel i'm not in a position to do so at my present stage of learning. I hope to remember to revisit this thread in the future after i finish reading chapter 21 in Klenke's book that is promised to deal with the issue of the continuity of paths. For the time being i'm satisfied with your and Nate's comments to my original thread. $\endgroup$ – Evan Aad Sep 4 '12 at 7:15

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.


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