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 $\mu_{[n]}(K_{n}) \ge \eps/2$.$\square$