14.10 Kolmogorov’s Extension Theorem
Definition 14.10.1 (Consistent). Let $\bracs{(\Omega_i, \cf_i)}_{i \in I}$ be a family of measurable spaces. For each $J \subset J' \subset I$, denote $\pi_{J}: \prod_{i \in I}X_{i} \to \prod_{j \in J}X_{j}$ and $\pi_{J', J}: \prod_{j \in J'}X_{j} \to \prod_{j \in J}X_{j}$ as the projection maps.
Let $\bracs{\mu_J|J \subset I \text{ finite}}$ such that each $\mu_{J}$ is a probability measure on $\prod_{j \in J}\Omega_{j}$, then $\bracs{\mu_J}$ is consistent if for all $J \subset J' \subset I$, $\mu_{J'}= \mu_{J} \circ \pi_{J', J}^{-1}$.
Lemma 14.10.2. Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$,
Every finite measure on $\prod_{j = 1}^{n} X_{j}$ is regularA potential sufficient condition for this is that each $X_{n}$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products..
$X_{n}$ is Hausdorff.
$X_{n}$ is separable.
Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
For each $n \in \nat$, $B_{n} \in \cb_{\prod_{j = 1}^n X_j}$.
For each $n \in \nat$, $B_{n+1}\subset B_{n} \times X_{n+1}$.
There exists $\eps > 0$ such that $\mu_{[n]}(B_{n}) > \eps$ for all $n \in \natp$.
Then there exists $\seq{K_n}$ such that for every $n \in \natp$,
$K_{n} \subset \prod_{j = 1}^{n} X_{j}$ is compact.
$K_{n} \subset B_{n}$.
$K_{n+1}\subset K_{n} \times X_{n+1}$.
$\mu(K_{n}) \ge \eps/2$.
Proof. Let $n \in \natp$. By (c), $\mu_{[n]}$ is a Borel probability measure. Thus $\mu_{[n]}$ is regular by (a). Thus there exists $C_{n} \subset \prod_{j = 1}^{n} X_{j}$ compact such that $C_{n} \subset B_{n}$ and $\mu_{[n]}(B_{n} \setminus C_{n}) < \eps/2^{n+1}$. Let
Since each $X_{j}$ is Hausdorff, $K_{n} \subset \prod_{j = 1}^{n} X_{j}$ is compact with $K_{n} \subset B_{n}$ and $K_{n+1}\subset K_{n} \times X_{n+1}$. Moreover,
Thus $\mu_{[n]}(K_{n}) \ge \eps/2$.$\square$
Theorem 14.10.3 (Kolmogorov’s Extension Theorem, [Theorem 1.14, Bau14]). Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite,
Every finite measure on $\prod_{j \in J}X_{j}$ is regular.
$X_{j}$ is Hausdorff.
$X_{j}$ 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. 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 14.10.2, there exists $\seq{K_n}$ such that:
$K_{n} \subset \prod_{j = 1}^{n} X_{j}$ is compact.
$K_{n} \subset B_{n}$.
$K_{n+1}\subset K_{n} \times X_{n+1}$.
$\mu(K_{n}) \ge \eps/2$.
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