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Let $(X_n,\mathcal{X}_n)$, $n=1,2,\ldots$ be measurable spaces. Define $Y_n = \prod_{k=1}^n X_k$ and let $\mathcal{Y}_n$ be the corresponding product $\sigma$-algebra. Similarly let $Y=\prod_{k=1}^\infty X_k$ and $\mathcal{Y}$ the corresponding $\sigma$-algebra.

Let $\mu_n$ be probability measures on $(Y_n,\mathcal{Y}_n)$ which are consistent in the sense that the projection of $\mu_{n+1}$ on $Y_n$ is $\mu_n$.

The Kolmogorov Extension Theorem (KET) states that under additional conditions on the spaces $X_n$ there exists a probability measure $\mu$ on $Y$ such that its projection on each $Y_n$ is $\mu_n$.

I have come across different versions of the KET which impose different conditions on the $X_n$. My question is whether there exist necessary and sufficient conditions which characterize spaces where consistent probability measures on finite product spaces can always be extended to the infinite product?

The extension problem can be generalized by looking at a fixed set $Y$ and considering an increasing sequence of $\sigma$-algebras $\mathcal{Y}_n$ and measures $\mu_n$ on $(Y,\mathcal{Y}_n)$ such that $\mu_{n+1}$ and $\mu_n$ agree on $\mathcal{Y}_n$. The extension problem then is to find a probability measure on $\sigma(\cup_{k=1}^\infty \mathcal{Y}_k)$ which agrees with $\mu_n$ on each of the $\mathcal{Y}_n$? Once again, are there necessary and sufficient conditions for this problem?

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  • $\begingroup$ You need to specify in what terms you want the condition to be stated; any condition is, trivially, necessary and sufficient for itself. $\endgroup$ – Iosif Pinelis Jul 20 '16 at 11:58
  • $\begingroup$ @IosifPinelis I'm looking for restrictions on the spaces $(X_n,\mathcal{X}_n)$ that give the ability to carry out the required extension from their finite products to their infinite products. For example, a commonly cited version of the KET requires the $X_n$ to be compact metrizable and $\mathcal{X}_n$ to be the corresponding Borel $\sigma$-algebra. That some restrictions is required follows from a counterexample of Sparre Andersen and Jessen gymarkiv.sdu.dk/MFM/kdvs/mfm%2020-29/mfm-25-4.pdf $\endgroup$ – Jyotirmoy Bhattacharya Jul 20 '16 at 12:49
  • $\begingroup$ I now surmise that you are looking for a necessary and sufficient condition of the form "for each $n$, the pair $(X_n,\mathcal{X}_n)$ has a certain property, say $P_n$". I see no reason to believe that any condition of this form could be necessary and sufficient -- but it appears very difficult to rule out all such conditions, for all possible sequences $(P_n)$ of properties; apparently, such a task is much more difficult than to construct the counterexample by Sparre Andersen and Jessen. $\endgroup$ – Iosif Pinelis Jul 20 '16 at 14:13
  • $\begingroup$ @IosifPinelis Even a necessary and sufficient condition for the simple case where all spaces are the same and satisfy some property $P$ would be more than what I know now. Also, I should clarify, that my purpose in asking this question was just to find out if there is something already known along these lines that I had missed out while searching the literature. $\endgroup$ – Jyotirmoy Bhattacharya Jul 20 '16 at 14:22
  • $\begingroup$ In the case when all the spaces are the same space $(X,\mathcal{X})$, you don't seem to anyhow specify in what terms you want the necessary and sufficient condition to be stated. Then, as indicated in my first comment, one can just use the trivial, tautological necessary and sufficient condition: a measure with given projections exists if and only if it exists. $\endgroup$ – Iosif Pinelis Jul 20 '16 at 15:11
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Gnedenko and Kolgomorov defined perfect measures for this purpose. Perfectness is a sufficient condition: If all the marginal measures (projections to $X_n$) are perfect then the resulting finitely additive measure on the product is countably additive. In this form perfectness is not necessary. But it is necessary in slightly modified formulations. For example, a probability measure $\mu$ is perfect if and only if for every probability measure $\nu$ every product of one copy of $\nu$ and any number of copies of $\mu$ is countably additive. This can be deduced from Ryll-Nardzewski's paper "On quasi-compact measures" in Fund. Math. 40 (1953). See also section 454 in Fremlin's Measure Theory.

If the condition you are looking for is on the measurable spaces rather than the measures on them, the corresponding condition is "every countably additive measure on the space is perfect".

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