Infinite product probability v.s. Kolmogorov's extension theorem Let $ (\Omega_i, \mathcal{F}_i, \mu_i), i \in I $ be a family of probability spaces, where the index set may be infinite or even uncountable. For subsets $ K \subseteq J \subseteq I $, define the projection map $ \pi_{J \to K}: \Omega_J \to \Omega_K $ by $ \pi_{J \to K}( (x_i)_{i \in J} ) = (x_i)_{i \in K} $. Abbreviate $ \pi_{I \to \{i\}} $ as $ \pi_i $.
It is a fact that we can always construct a unique product probability space $ ( \prod_i \Omega_i, \bigotimes_i \mathcal{F}_i, \bigotimes_i \mu_i) $, where $ \bigotimes_i \mathcal{F}_i $ is the $\sigma$-algebra generated by the projection maps $ \pi_i $, and the probability measure $ \bigotimes_i \mu_i $ satisfies
$$ \left( \bigotimes \mu_i \right) \left( \prod_{i \in J} A_i \times \Omega_{I \setminus J} \right)
= \prod_{i \in J} \mu_i(A_i) $$
where $J$ is a finite subset of $I$ and $ A_i \in \mathcal{F}_i $ for $i \in J$. Basically it means $ \bigotimes_i \mu_i $ behaves as a finite product measure on product cylinders.
Next consider a more general problem. Suppose we have not only a probability measure $\mu_i$ for each $i \in I$ but also a probability measure $\mu_J$ on $(\Omega_J, \mathcal{F}_J)$ for each finite subset $J \subseteq I$, and the $\mu_J$'s satisfy the consistency condition:
$$ (\pi_{J \to K})_*\mu_J = \mu_K \quad \text{for any finite subsets } K \subseteq J \subseteq I $$
The problem is, can we construct a probability measure $\mu$ on $ (\prod_i \Omega_i, \bigotimes_i \mathcal{F}_i) $ that also satisfies the consistency condition:
$$ (\pi_{I \to J})_*\mu = \mu_J \quad \text{for any finite subset } J \subseteq I \text{ ?} $$
The famous Kolmogorov's extension theorem guarantees the existence of such a probability measure under some regularity conditions. At minimum, we require the followings:

*

*for each $i \in I$, the space $\Omega_i$ is a Hausdorff topological space

*for each $i \in I$, the $\sigma$-algebra $\mathcal{F}_i$ is the Borel $\sigma$-algebra $\mathcal{B}(\Omega_i)$

*for each finite subset $J \subseteq I$, the probability measure $\mu_J$ is an tight Borel measure, i.e. for any Borel set $ E \in \mathcal{B}(\Omega_J) $,
$$ \mu_J(E) = \sup\{ \mu_J(K) : K \subseteq E, K \text{ compact} \} $$
Questions:
Why do we need tightness in Kolmogorov's extension theorem while the product probability does not require any condition? Could you pinpoint the exact step where tightness is used in the proof of Kolmogorov's extension theorem? Could you give an example that the Kolmogorov's extension fails to exist?
 A: The crucial step in the extension process is to go from a finitely to a countably additive measure. This boils down to showing that for a decreasing measurable sequence $A_n \searrow \emptyset$, i.e. with   $\bigcap_{n=1}^\infty A_n = \emptyset$ you have $\lim_n P(A_n)=0$.
For a countable product space with associated product measure this may be shown because of the simple geometry of the product. In
J A Wellner the proof is to assume that $\liminf_n P(A_n)\geq \epsilon>0$ and then inductively construct a point in the intersection of all the $A_n$'s, thus contradicting the assumption of an empty intersection. I think it does not carry over to an uncountable product.
This procedure using an inductive construction  is not available for a consistent family of measures. The approach here is to assume the existence of a compact class ${\cal C}$, i.e. a class of measurable subsets having the finite intersection property: Given a sequence $(C_n)$ in the class with $\bigcap_{n\geq 1} C_n=\emptyset$ there is $N$ for which $C_1\cap ...\cap C_N=\emptyset$.   If your finitely additive measure is tight w.r.t. ${\cal C}$ then it is countably additive. You may find a very nice exposition in K C Border
This last reference also discusses a counter-example by Andersen and Jessen from 1948.
