Lemma 18.9.2 ([Lemma 1.17, Bau14]).label Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$,
- (a)
Every finite measure on $\prod_{j = 1}^{n} X_{j}$ is regular.
- (b)
$X_{n}$ is Hausdorff.
- (c)
$X_{n}$ is separable.
Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
- (d)
For each $n \in \nat$, $B_{n} \in \cb_{\prod_{j = 1}^n X_j}$.
- (e)
For each $n \in \nat$, $B_{n+1}\subset B_{n} \times X_{n+1}$.
- (f)
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 $n \in \natp$,
- (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)
$\mu(K_{n}) \ge \eps/2$.
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 $\mu_{[n]}(K_{n}) \ge \eps/2$.$\square$