Theorem 18.9.3 (Kolmogorov’s Extension Theorem).label Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite,
- (a)
Every finite measure on $\prod_{j \in J}X_{j}$ is regular.
- (b)
$X_{j}$ is Hausdorff.
- (c)
$X_{j}$ is separable.
Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i}\to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_{J} \circ \pi_{J}^{-1}$.
Proof [Theorem 1.14, Bau14]. Let
then $\alg$ is an algebra. For any $B \in \bigotimes_{j \in J}\cb_{X_j}$, define $\mu_{0}(\pi_{J}^{-1}(B)) = \mu_{J}(B)$, then $\mu_{0}: \alg \to [0, 1]$ is well-defined and finitely additive by the consistency of $\bracs{\mu_J}$.
To show that $\mu_{0}$ is a premeasure, it is sufficient to show that for any $\{\pi_{J_n}^{-1}(B_{n})\}_{1}^{\infty}$ with $\pi_{J_n}^{-1}(B_{n}) \downto \emptyset$, $\mu_{0}(\pi_{J_n}^{-1}(B_{n})) \downto 0$. To this end, suppose for contradiction that $\limv{n}\mu_{0}(\pi_{J_n}^{-1}(B_{n})) = \eps > 0$.
By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_{n} = [n]$ for all $n \in \natp$. By Lemma 18.9.2, there exists $\seq{K_n}$ such that:
- (1)
$K_{n} \subset \prod_{j = 1}^{n} X_{j}$ is compact.
- (2)
$K_{n} \subset B_{n}$.
- (3)
$K_{n+1}\subset K_{n} \times X_{n+1}$.
- (4)
$\mu(K_{n}) \ge \eps/2$.
Let $N \in \natp$ and $x \in \prod_{j = 1}^{N} X_{j}$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_{n})$. By compactness and (4), there exists $x_{N+1}\in X_{N+1}$ such that $(x_{1}, \cdots, x_{N}, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_{n})$. Thus there exists $x \in \prod_{i \in I}X_{i}$ such that $x \in \pi_{[n]}^{-1}(K_{n})$ for all $n \in \natp$, and
$\square$