Proof of Kolmogorov Extension Theorem requires inner regularity of measures?

Question

If you already know the proof of the Kolmogorov extension theorem, skip to the last paragraph.

A theorem of Kolmogorov says the following:

Let $\mathcal{I}$ be an index set, and suppose for each finite tuple $I= (i_1,...,i_k)$ we have a probability measure on $(\mathbb{R}^k, \mathcal{B})$, where $\mathcal{B}$ are the Borel sets. Suppose also that these probability measures satisfy some consistency conditions:

1. $P_{\sigma I}=P_{I}\circ \sigma'$ for $\sigma$ a permutation of $I$ and $\sigma':\mathbb{R}^k\rightarrow \mathbb{R}^k$ the corresponding linear isomorphism.

2. $P_I= P_J\circ \pi^{-1}$ whenever $J$ extends $I$ and $\pi: \mathbb{R}^{|J|}\rightarrow \mathbb{R}^{|I|}$ is the projection.

Then there exists a probability space $\Omega$ and random variables $\{X_i:\Omega \rightarrow \mathbb{R}\}_{i\in \mathcal{I}}$ such that for each tuple $I$, the "joint" random variable $$X_I: \Omega \rightarrow \mathbb{R}^k, \omega \mapsto (X_{i_1}(\omega),...,X_{i_k}(\omega))$$ pushes the measure of $\Omega$ forward to $P_I$.

The proof proceeds by taking $\Omega = \Pi_{i=\mathcal{I}}\mathbb{R}$ and defining the measure of the "cylinder sets" $$C= A \times \Pi_{i\notin I}\mathbb{R}$$ (where $A\subset \mathbb{R}^{|I|}$) in the obvious way: $$P(C):=P_I(A).$$ The consistency conditions guarantee this doesn't depend on which multi-index you use. So you get a finitely additive "measure" on the algebra of cylinder sets and you want to extend this to the $\Sigma$ algebra generated by the cylinder sets. You can do this so long as the following is true:

Suppose $C_1\supset C_2\supset...$ is a decreasing sequence of cylinder sets with $P(C_j)>\delta>0$ for all $j$. Then $\cap_j C_j\neq \emptyset$.

To prove this latter statement, Kolmogorov replaces each $$C_j = A_j \times \Pi \mathbb{R}$$ ($A_j \subset \mathbb{R}^{|I|}$) with $$K_j\times \Pi \mathbb{R}$$ where $K_j\subset A_j$ is compact and has measure very close to $A_j$, say $P_I(A_j-K_j)\leq \delta/2^j$. The argument continues from there, but

My Question is how do we know we can find these compact sets approximating $A_j$ from within? A priori these measures $P_I$ needn't be inner regular; they're just some Borel measures with total measure 1.

Answer 1

There is the following general statement:

Any finite Borel measure on a Polish space is inner regular.

Since $\mathbb{R}^I$ is a Polish space, this implies that $P_I$ is inner regular for any finite index set $I$. For a proof see for instance Bauer, Lemma 26.2, or this pdf file, Corollary 2.5+Theorem 2.6