The Kolmogorov Extension Theorem says, essentially, that one can get a process on $\mathbb{R}^T$ for $T$ being an arbitrary, non-empty index set, by specifying all finite dimensional distributions in a "consistent" way. My favorite formulation of the consistency condition can be found here. Now for the case in which $T$ is countable, this has already be shown by P. J. Daniell (see for example here or here). So I would like to know what the extension to uncountable index sets brings. Events like "sample paths are continuous" are not in the $\sigma$-algebra. In a rather critical paper on Kolmogrov's work on the foundation of probability, Shafer and Vovk write about the extension to uncountable index sets: "This greater generality is merely formal, in two senses: it involves no additional mathematical complications and it has no practical use." My impression is that this sentiment is not universally shared, so I would like to know:

How is the Kolmogorov Extension Theorem applied in the construction of stochastic processes in continuous time? Especially, how are the constructed probabilities transferred to richer measurable spaces?

  • $\begingroup$ I don't know if this is in any way what you're looking for, but Fremlin's measure theory, Chapter 45, Volume 4 Part I contains quite a few related constructions and also some remarks towards applications to processes. $\endgroup$ – t.b. Dec 27 '11 at 14:43
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    $\begingroup$ If I understand your question correctly, a classical example will be construction of Brownian motion from finite Gaussian distributions. It is described in a very brief and rigorous way in Oksendal's book, Section 2.2 $\endgroup$ – Ilya Dec 27 '11 at 17:35
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    $\begingroup$ Thank you, but Oksendal constructs a process on $R^{[0,\infty)}$ with the finite dimensional distributions of BM and then mentions that one can assume the process to be supported on the space of continuous functions. He does not prove this. I have to check what Fremlin writes on the issue. $\endgroup$ – Michael Greinecker Dec 27 '11 at 18:33
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    $\begingroup$ Sorry, I looked in the wrong place in the book by Oksendal. I've seen now that he uses the continuity theorem. So this gives an answer to both questions. I'm still courious about other applications, since it is not that hard to construct BM "by hand" using weak convergence theory. $\endgroup$ – Michael Greinecker Dec 29 '11 at 19:14
  • $\begingroup$ Nice that it answers your question - but your last formulation of the question in the last comment confuses me $\endgroup$ – Ilya Jan 2 '12 at 19:35

Assume that you have a set of finite-dimensional distributions $(\mu_S)_{S\in A}$, where $A$ is the set of finite subsets of, say, $\mathbb{R}$, and assume that you would like to argue for the existence of a stochastic process $X$ with càdlàg (right-continuous with left limits) paths such that the family of finite-dimensional distributions of $X$ is $(\mu_S)_{S\in A}$. Kolmogorov's extension theorem allows you to split this problem into two parts:

  1. Establishing the existence of a measure on $(\mathbb{R}^{\mathbb{R}_+},\mathbb{B}^{\mathbb{R}_+})$ with the appropriate finite-dimensional distributions (here, the extension theorem is invoked).

  2. Using the measure constructed above, argue for the existence of a measure on $D[0,\infty)$ - the space of functions from $\mathbb{R}_+$ to $\mathbb{R}$ which are right-continuous with left limits - with the same finite-dimensional distributions.

One example of where this is a viable proof technique is in the theory of continuous-time Markov processes on general state spaces. For convenience, consider a complete, separable metric space (E,d) endowed with its Borel $\sigma$-algebra $\mathbb{E}$. Assume given a family of probability measures $(P(x,\cdot))_{x\in E}$, where $P(x,\cdot)$ is a probability measure on $(E,\mathbb{E})$. We wish to argue for the existence of a càdlàg continuous-time Markov process with values in $E$ and $(P(x,\cdot))_{x\in E}$ as its transition probabilities.

The following argument is given in Rogers & Williams: "Diffusions, Martingales and Markov Processes", Volume 1, Section III.7, and uses the two steps outlined above. First, the Kolmogorov extension theorem is invoked to obtain a measure $P$ on $(E^{\mathbb{R}_+},\mathbb{E}^{\mathbb{R}_+})$ with the desired finite-dimensional distributions. Letting $X$ be the identity on $E^{\mathbb{R}_+}$, $X$ is then a "non-regularized" process with the desired finite-dimensional distributions. Afterwards, in the case where the family of transition probabilities satisfy certain regularity criteria, a supermartingale argument is applied to obtain a càdlàg version of $X$. This supermartingale argument could not have been immediately applied without the existence of the measure $P$ on $(E^{\mathbb{R}_+},\mathbb{E}^{\mathbb{R}_+})$: Without this measure, there would be no candidate for a common probability space on which to define the supermartingales applied in the regularity proof. Thus, it is not obvious how to obtain the same existence result without the Kolmogorov extension theorem.

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    $\begingroup$ But is it clear that the uncountable version of Kolmogorov is essential? Another typical approach would be to construct a measure $P$ on the countable product $(E^\mathbb{Q}, \mathbb{E}^\mathbb{Q})$, and then show that $P$-almost every element of $E^\mathbb{Q}$ extends to a cadlag path $\mathbb{R} \to E$, which one calls $X$. This has the benefit that one is working with standard Borel spaces throughout. I haven't looked at Rogers and Williams, but is there some reason why this wouldn't work here? $\endgroup$ – Nate Eldredge Apr 12 '12 at 13:04
  • $\begingroup$ @NateEldredge have you found and answer for this question of yours? $\endgroup$ – Ilya Jan 23 '13 at 20:37
  • $\begingroup$ @Ilya: No, I haven't, but I haven't looked very hard either. $\endgroup$ – Nate Eldredge Jan 23 '13 at 21:47
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    $\begingroup$ @NateEldredge: I agree with you that seemingly in this case, the uncountable version is not essential, merely convenient. I personally don't know of any applications where the uncountable version is essential. $\endgroup$ – Alexander Sokol Feb 5 '13 at 16:04

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