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I am having trouble understanding part of Proposition 5.13. For context, Proposition 5.13 seeks to prove that in order to show a stochastic process has the Markov property, we can replace conditions (c) and (d) below

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with the following condition:

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where $U_t$ is defined as below:

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However, I am struggling to show the following lines in their proof:

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Specifically, the last line, where they state that "if the function $\alpha(y)$ were measurable we would be able to conclude..."

My work: By the definition of conditional probabilities, it suffices to show $$ \int \mathbb{1}_{X_{t + s} \in \Gamma} \cdot \mathbb{1}_{\hbox{arbitrary event in $\mathcal{F}_s$}} d\omega = \int \mathbb{P}^{X_s(\omega)}[Z_{t + s} \in \Gamma] \cdot \mathbb{1}_{\hbox{arbitrary event in $\mathcal{F}_s$}} $$ where $\{Z_{t}\}$ is an independent stochastic process with the same distribution as $\{X_{t}\}$.

Intuitively, (c) implies that the statement holds true for any single value $X_s(\omega) = c$, but how does the author use the measurability of $\alpha(y)$ to extend to arbitrary events in $\mathcal{F}_s$?

Been stuck for a while, so any insight would be deeply appreciated :)

Stephen Jiang
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  • Do you know what $$ \mathbb P^x{X_{t+s}\in\Gamma|X_s=y} $$ means and how we recover the conditional probability $\mathbb P^x{X_{t+s}\in\Gamma|X_s}$ from this function of $y,?$ Some hints. – Kurt G. Aug 08 '24 at 08:12
  • @KurtG. Oh, is this correct? For any outcome $\omega$, $P^{x}{X_{t + s} \vert X_s}(\omega) = P^{x}{X_{t + s} \vert X_s = X_s(\omega) } = \alpha(X_s)(\omega)$. Generally, $\alpha(X_s)$ may not be measurable with respect to $\mathcal{F}_s$, but if $\alpha$ is measurable, $\alpha(X_s)$ must also be. – Stephen Jiang Aug 08 '24 at 12:24
  • This sounds right. K&S circumvent the non measurability of $\alpha$ by using a version $g$ of it but this was not your question. – Kurt G. Aug 08 '24 at 15:50

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