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?

allsuch 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