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KET is often used to construct stochastic processes in continuous time when the state space is $\Bbb R^d$. As far as I am familiar with its proof, it uses standard monotonic class-like arguments together with Caratheodory Extension Theorem. Neither of the two latter theorems requires any topological conditions. Moreover, I have not been able to find where KET uses that the space is $\Bbb R^d$ rather than just some measurable space.

Q: is it true that KET holds for general measurable spaces, and if not - where is the topology of $\Bbb R^d$ crucial in the proof? Also, maybe any counterexample (for exsitence or uniqueness) is known?


I also asked a related question here but didn't get any answer.

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    $\begingroup$ Check the Theorems 14.35,14,36 in A.Klenke's book "Probability theory. A Comprehensive Course." The state space can be quite general (for example a Polish space), but there are some restrictions. $\endgroup$ – Liviu Nicolaescu Jan 11 '13 at 16:16
  • $\begingroup$ As to where the topology is crucial in "the proof", perhaps you'd like to point to a particular proof you have in mind? The topology (or something similar) must certainly be used, but it might show up in different ways depending on how the proof is written. $\endgroup$ – Nate Eldredge Apr 3 at 20:49
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The KET fails for general measurable spaces, the classical example can be found in a paper by Andersen and Jessen. Topological assumptions are necessary so that the resulting measure is not only finitely additive but countably additive. There exists a quasi-topological condition of measure spaces, perfectness, that is sufficient. A probability space $(\Omega,\sigma,\mu)$ is perfect if for every random variable $f:\Omega\to\mathbb{R}$, there exists a Borel set $B\subseteq f(\Omega)$ with measure one under the distribution $\mu\circ f^{-1}$. A proof of KET under the assumption that the marginal measures are perfect due to Lamb is given here. The strategy of the proof is to employ an existence result for regular conditional probability spaces and the construct the proces for them using the Ionescu-Tulcea theorem.

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  • $\begingroup$ Thanks a lot, Michael! Will you consider also putting this answer on a linked MSE question? $\endgroup$ – Ilya Jan 11 '13 at 16:32
  • $\begingroup$ Sadly, both links are broken. $\endgroup$ – arsmath Apr 3 at 18:41
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    $\begingroup$ @arsmath Thanks, it should work now. $\endgroup$ – Michael Greinecker Apr 3 at 18:58

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