Compact Approximation for Product-Space Extensions

Why Compact Approximation Enters Product Extensions

To extend consistent finite-dimensional probability laws to a countable product, one first needs a countably additive set function on the algebra of cylinder sets. Finite-dimensional consistency gives finite additivity, but countable additivity requires a further check. A useful criterion is continuity from above: if cylinder sets decrease to the empty set, their assigned probabilities must tend to zero.

Compactness helps establish this condition. If a decreasing sequence of cylinders had probabilities bounded away from zero, we will approximate each cylinder from inside by a compact cylinder and restrict coordinates to suitably chosen compact sets. The resulting finite intersections are nonempty compact subsets of one compact product. Their intersection must be nonempty, contradicting the assumption that the original cylinders have empty intersection.

Let \(S_i\) be a Polish space for each \(i\in\mathbb{N}\), meaning a separable space with a complete metric that generates its topology, and let \(S=\prod_{i=1}^{\infty}S_i\). For each finite \(F\subset\mathbb{N}\), suppose \(\mu_F\) is a Borel probability measure on \(S_F=\prod_{i\in F}S_i\). Assume consistency: whenever \(F\subset G\) are finite and \(B\subset S_F\) is Borel,

A cylinder is a set \(\pi_F^{-1}(B)=\{x\in S:x_F\in B\}\), where \(F\) is finite, \(B\) is Borel in \(S_F\), and \(\pi_F\) is the coordinate projection. The cylinder sets form an algebra \(\mathcal{A}\): complements and finite unions can be represented using a common finite set of coordinates. Consistency defines a function \(p:\mathcal{A}\to[0,1]\) by \(p(\pi_F^{-1}(B))=\mu_F(B)\). This is well-defined, and it is finitely additive; both facts follow by lifting any finite collection of representations to the union of their coordinate sets and using consistency.

Compact Sets in the Coordinates

Each coordinate marginal \(\mu_{\{i\}}\) is a Borel probability on a Polish space. Such probabilities are tight: for every \(\varepsilon>0\), there is a compact \(K_i\subset S_i\) with \(\mu_{\{i\}}(K_i)>1-\varepsilon\). Here is the standard tightness argument. For each positive integer \(r\), separability gives a countable cover by open balls of radius \(2^{-r}\). Since the increasing finite unions of these balls cover \(S_i\), continuity from below for the measure \(\mu_{\{i\}}\) gives a finite union \(U_r\) with \(\mu_{\{i\}}(U_r)>1-\varepsilon 2^{-r-1}\). The closed set \(K_i=\bigcap_r\overline{U_r}\) has measure at least \(1-\sum_r\varepsilon 2^{-r-1}>1-\varepsilon\). It is totally bounded because, at every scale \(2^{-r}\), it is contained in the closure of a finite union of balls of that radius; it is complete as a closed subset of \(S_i\). A complete, totally bounded metric space is compact.

We also use the standard inner regularity theorem for Borel probabilities on Polish spaces: if \(\lambda\) is such a probability and \(B\) is Borel, then for every \(\eta>0\) there is a compact \(L\subset B\) with \(\lambda(B\setminus L)<\eta\). This standard regularity result applies to each finite product \(S_F\), which is Polish. The proof below uses inner regularity only for these finite-dimensional probabilities.

Continuity from Above for the Cylinder Premeasure

Proof. We first prove continuity from above. Since \(A_n\) decrease, \(p(A_n)\) is a nonincreasing sequence in \([0,1]\), so it has a limit \(\ell\ge0\). Suppose, for contradiction, that \(\ell>0\), and put \(\delta=\ell/2\). Then \(p(A_n)\ge\delta\) for every \(n\).

Write \(A_j=\pi_{F_j}^{-1}(B_j)\), with \(F_j\subset\mathbb{N}\) finite and \(B_j\subset S_{F_j}\) Borel. Choose compact coordinate sets \(K_i\subset S_i\) so that \(\mu_{\{i\}}(K_i^c)<\delta_i\), where \(\delta_i=\delta 2^{-i-2}\). Thus \(\sum_i\delta_i=\delta/4\). For each \(j\), inner regularity gives a compact \(L_j\subset B_j\) with \(\mu_{F_j}(B_j\setminus L_j)<\eta_j\), where \(\eta_j=\delta 2^{-j-2}\); hence \(\sum_j\eta_j=\delta/4\).

Let \(C_j=\pi_{F_j}^{-1}(L_j)\), a compact-coordinate cylinder contained in \(A_j\). For each \(k\), consider the constraints \(C_1,\ldots,C_k\) together with \(x_i\in K_i\) for each \(i\in G_k=\bigcup_{j=1}^kF_j\). These are conditions on finitely many coordinates, so their probability is computed using the appropriate finite-dimensional law. Starting from \(A_k\), the loss from imposing all \(K_i\), \(i\in G_k\), is at most \(\sum_{i\in G_k}\delta_i\): for each \(i\), consistency bounds the probability of \(x_i\notin K_i\) by \(\mu_{\{i\}}(K_i^c)<\delta_i\). The additional loss from imposing \(C_j\), for \(1\le j\le k\), is at most \(\sum_{j=1}^k\eta_j\), because \(A_k\subseteq A_j\) and \(p(A_j\setminus C_j)<\eta_j\). Finite subadditivity therefore bounds the probability of all these constraints below by

Consequently the compact set \(Q_k=(\prod_iK_i)\cap\bigcap_{j=1}^kC_j\) is nonempty: the positive finite-dimensional probability just obtained supplies compatible values on \(G_k\), and the other coordinates can be chosen in their nonempty \(K_i\). Each \(Q_k\) is closed in the compact product \(\prod_iK_i\), and \(Q_{k+1}\subseteq Q_k\). The compact intersection property gives \(\bigcap_kQ_k\ne\varnothing\). Every point in this intersection belongs to every \(C_j\), hence to every \(A_j\), contradicting \(\bigcap_jA_j=\varnothing\). Thus \(\ell=0\), proving continuity from above.

It remains to verify the premeasure criterion. Let \(E_n\in\mathcal{A}\) be disjoint and suppose \(E=\bigcup_{n=1}^{\infty}E_n\in\mathcal{A}\). Define \(R_N=E\setminus\bigcup_{n=1}^NE_n\). These sets belong to \(\mathcal{A}\), decrease, and have empty intersection. Finite additivity gives \(p(E)=\sum_{n=1}^Np(E_n)+p(R_N)\). By continuity from above, \(p(R_N)\to0\); taking \(N\to\infty\) proves \(p(E)=\sum_{n=1}^{\infty}p(E_n)\). Conversely, if \(p\) is a premeasure and \(A_n\downarrow\varnothing\), then \(A_1\) is the disjoint union of \(A_n\setminus A_{n+1}\) for \(n\ge1\); countable additivity implies \(p(A_n)\to0\). This proves the stated equivalence. \(\square\)

The theorem supplies the continuity condition needed before applying a measure-extension result to the cylinder algebra. It does not itself construct the probability measure on the full product sigma-algebra.

A Common Pitfall: Positive Measures Do Not Force an Intersection

A tempting but invalid argument is: “Every \(A_n\) has positive measure, so the decreasing intersection must be nonempty.” On \([0,1]\) with Lebesgue measure \(m\), take \(A_n=(0,1/n)\). Each set is nonempty and \(m(A_n)=1/n>0\), but \(\bigcap_nA_n=\varnothing\). The measures are not bounded below by a fixed positive number: \(1/n\to0\). The proof for product cylinders needs precisely a uniform lower bound \(p(A_n)\ge\delta>0\), as well as compact approximation with total error smaller than that bound. Compactness then applies to nested compact subsets, not merely to the original measurable sets.