Infinite Product Measures and Kolmogorov Extension

Put $H=F\cup G$ and let $p_F,p_G$ be the coordinate projections from ${\prod }_H{E}_i$. Then $C:=p_F^{-1}(A)\cap p_G^{-1}(B)$ is measurable in ${⨂}_H{\mathcal{E}}_i$.

Directly from restriction of coordinates, ${\pi }_F^{-1}(A)\cap {\pi }_G^{-1}(B)={\pi }_H^{-1}(C)$, a finite-coordinate cylinder. The family is nonempty because it contains $E$, so it is a pi-system.

The empty-support cylinders give $\mathrm{\varnothing },E\in {\mathcal{A}}_I$. Also $E\setminus {\pi }_F^{-1}(A)={\pi }_F^{-1}(({\prod }_F{E}_i)\setminus A)\in {\mathcal{A}}_I$.

For $H=F\cup G$, pull both bases to ${\prod }_H{E}_i$; their union is measurable and its inverse image is ${\pi }_F^{-1}(A)\cup {\pi }_G^{-1}(B)$. Thus ${\mathcal{A}}_I$ has the three algebra operations.

Fix $x^{\ast }\in E$. If ${\pi }_F^{-1}(A)={\pi }_G^{-1}(B)$, pull both sets to $H=F\cup G$. Their lifted bases are equal: otherwise a point of their symmetric difference, combined with the coordinates of $x^{\ast }$ outside $H$, would distinguish the cylinders. Hence consistency gives ${\mu }_F(A)={\mu }_H(p_{H,F}^{-1}A)={\mu }_H(p_{H,G}^{-1}B)={\mu }_G(B)$.

For disjoint cylinders $C_r={\pi }_{F_r}^{-1}(A_r)$ with $r<m$, pull all bases to $H={\bigcup }_{r<m}{F}_r$. They are disjoint measurable sets, so finite additivity of ${\mu }_H$ yields ${\mu }_0({\bigcup }_r{C}_r)={\sum }_r{\mu }_0(C_r)$. The union is in the cylinder algebra by [F2].

Index the given coordinates by $0,1,\dots$. Each coordinate space is nonempty since its measure has total mass one. Countable choice in [F4] supplies a point of their product, and [F5] gives the finitely additive cylinder law. Finite product laws are consistent: projecting away a factor preserves the rectangle values because that factor has mass one, and uniqueness in [F1] gives equality on the finite product sigma-algebra. This also justifies regrouping any finite block into two factors.

For a cylinder $C$ and a prefix $p=(x_0,\dots ,x_{m-1})$, define $h_C(p)$ using only finite products: represent $C$ on coordinates $0,\dots ,k-1$ with $k\ge m$, fix the prefix in its measurable base, and measure the remaining section with ${⨂}_{m\le j<k}{\mu }_j$. For $k=m$ this is the indicator of membership in the base. Enlarging $k$ leaves the value unchanged by the mass-one rectangle formula; two representations agree after enlargement, using the product point of step 1.1. Thus $h_C(p)$ is well-defined and lies in $[0,1]$. By choosing $k\ge m+1$, [F2] gives a measurable function $z\mapsto h_C(p,z)$ and $h_C(p)={\int }_{E_m}{h}_C(p,z)d{\mu }_m(z)\mathrm{.}$ Inclusion of cylinders gives pointwise inequality of these functions by finite-product monotonicity. No measure on an infinite tail is used.

Let $C_n\downarrow \mathrm{\varnothing }$ and suppose their masses do not tend to zero. Finite additivity makes the masses decreasing in $[0,1]$, so there is $\eta >0$ with ${\mu }_0(C_n)\ge \eta$ for all $n$. Call a prefix $p$ of length $m$ admissible if $h_{C_n}(p)\ge \eta \mathrm{/}{2}^m$ for every $n$; the empty prefix is admissible. Given one, put $t=\eta \mathrm{/}{2}^{m+1}$ and $D_n=\{z\in E_m:h_{C_n}(p,z)\ge t\}\mathrm{.}$ These sets are measurable and decrease. Step 2.1 and the bound $0\le h_{C_n}\le 1$ yield $2t\le h_{C_n}(p)\le {\mu }_m(D_n)+t{\mu }_m(E_m\setminus D_n)\le {\mu }_m(D_n)+t\mathrm{.}$ Hence ${\mu }_m(D_n)\ge t$, and [F3] implies ${\mu }_m({\bigcap }_n{D}_n)\ge t>0$. Any point in this intersection extends $p$ to an admissible prefix.

The admissible finite prefixes form a nonempty set. Relate a prefix to its admissible one-coordinate extensions; step 3.1 proves this relation is entire. Dependent choice in [F6], starting at the empty prefix, gives a compatible chain, whose union is a point $x$ of the original coordinate product. For each $n$, choose a finite initial block determining $C_n$. Once that block is fixed, $h_{C_n}$ is its membership indicator; admissibility makes it positive, so $x\in C_n$. This contradicts the empty intersection. Thus ${\mu }_0(C_n)\to 0$. Using all original coordinates also handles cylinders with empty or finitely many active coordinates; there is no terminal-prefix exception.

If disjoint cylinders $A_j$ have union a cylinder $A$, then $R_N=A\setminus {\bigcup }_{j<N}{A}_j$ decreases to the empty set and remains in the cylinder algebra. Finite additivity gives ${\mu }_0(A)={\sum }_{j<N}{\mu }_0(A_j)+{\mu }_0(R_N)\mathrm{.}$ Step 4.1 sends the remainder to zero, proving countable additivity for every such union. Hence ${\mu }_0$ is a premeasure.

Define ${\mu }_0({\pi }_F^{-1}(A))=({⨂}_{n\in F}{\mu }_n)(A)$. The finite product marginals are consistent, so [F1] applies.

By [F2], ${\mu }_0$ extends to a measure $\mu$ on the generated cylinder sigma-algebra. Since the empty cylinder is $E$ and has value $1$, $\mu$ is a probability measure.

Let $\nu$ be another probability measure with the stated marginals. The family $\mathcal{D}=\{A\in {\mathcal{C}}_{\mathbb{N}}:\mu (A)=\nu (A)\}$ is a lambda-system: it contains the whole space because both measures have mass one, is closed under relative complements of nested members, and is closed under increasing countable unions by continuity from below. The two measures agree on every cylinder, and cylinders are a pi-system by Finite-coordinate cylinders form a $\pi$-system, so [F3] gives ${\mathcal{C}}_{\mathbb{N}}\subseteq \mathcal{D}$. Hence $\mu =\nu$.

For $A_j\in {\mathcal{E}}_{n_j}$ and the distinct indices fixed above, [F1] gives $\mathbb{P}({\bigcap }_j\{X_{n_j}\in A_j\})={\prod }_j{\mu }_{n_j}(A_j)$; taking one coordinate gives the asserted law.

The right side is ${\prod }_j\mathbb{P}(X_{n_j}\in A_j)$, which is exactly [F2]. Since the finite family was arbitrary, all coordinates are independent.

Apply [F1] with $E_n=S$ and ${\mu }_n=\nu$ for every $n$, and write $X_n$ for its coordinate maps.

By [F2], every $X_n$ has law $\nu$ and every finite subfamily is independent; this is the required independent-copy sequence.

If $n=0$, the product is the singleton with its full two-element sigma-algebra by [F5,F6]; the unique metric on a singleton is complete and its singleton set is dense, so it is Polish and standard Borel. If a factor is empty, the product is empty and its sigma-algebra is $\{\mathrm{\varnothing }\}$; the empty metric is complete and the empty set is dense in the empty topology, so the empty space is Polish and standard Borel. These settle both boundary cases.

For any finite family of sigma-algebras, their recursive product is the sigma-algebra generated by all full measurable rectangles ${\prod }_{j<n}{A}_j$. This follows by induction from [F6]. The zero and one cases follow directly from the definitions. At the successor stage, let $\mathcal{T}$ be generated by the full $(n+1)$-rectangles. For any measurable last-coordinate set $B$, the class of $A$ in the preceding product space for which $A\times B\in \mathcal{T}$ is a sigma-algebra: it contains the whole space since the whole-space rectangle times $B$ is a full rectangle, complements subtract from that rectangle, and countable unions commute with the product by $B$. It contains every preceding full rectangle, so the inductive hypothesis gives every preceding product-measurable $A$. Thus all recursive generating rectangles belong to $\mathcal{T}$. Conversely every full rectangle is a recursive generating rectangle because its preceding rectangle is measurable. This proves equality at the successor and hence at every finite stage.

Assume $n\ge 1$ and all factors nonempty. For each $j<n$, [F1] provides a Polish presentation $h_j:E_j\to P_j$; use only finite choice [F2] to select these finitely many presentations, compatible complete metrics $d_j$, nonempty countable dense subsets $D_j$, and surjective enumerations of the $D_j$ from [F4]. The product map $h(x{)}_j=h_j(x_j)$ is a bijection from ${\prod }_{j<n}{E}_j$ onto $P={\prod }_{j<n}{P}_j$, with inverse obtained by the finitely many $h_j^{-1}$. The product is nonempty by finite choice.

Put ${\delta }_j(x,y)=\min \{1,d_j(x,y)\}$. This is a metric: symmetry and separation are inherited, and for nonnegative $a,b$, $\min (1,a+b)\le \min (1,a)+\min (1,b)$ proves the triangle inequality from that of $d_j$. Balls of radius less than $1$ are unchanged, so its topology agrees with that of $d_j$. A ${\delta }_j$-Cauchy sequence is $d_j$-Cauchy by testing every tolerance less than $1$; its $d_j$-limit is also its ${\delta }_j$-limit. Hence ${\delta }_j$ is complete. Define $D(x,y)={\sum }_{j<n}{\delta }_j(x_j,y_j)$ on $P$. Finite sums of the metric inequalities prove that $D$ is a metric.

The metric $D$ induces the product topology. For every $x$ and $\epsilon >0$, the product of coordinate ${\delta }_j$-balls of radius $\epsilon \mathrm{/}n$ lies in $B_D(x,\epsilon )$ by the finite sum inequality. Conversely, given a basic product-open neighbourhood ${\prod }_{j<n}{U}_j$ of $x$, choose positive coordinate radii $r_j$ with $B_{{\delta }_j}(x_j,r_j)\subseteq U_j$; a single positive radius smaller than every $r_j$ makes its $D$-ball lie in that rectangle. These are only finite selections. Both neighbourhood comparisons establish equality of the two topologies.

If $(x^{(k)}{)}_k$ is $D$-Cauchy, each coordinate sequence is ${\delta }_j$-Cauchy since ${\delta }_j(x_j^{(k)},x_j^{(l)})\le D(x^{(k)},x^{(l)})$. Its unique limit $x_j$ exists by step 3.1. The finite tuple $x=(x_j{)}_{j<n}$ lies in $P$. Given $\epsilon >0$, each coordinate error is eventually less than $\epsilon \mathrm{/}n$, and the maximum of the finitely many thresholds makes $D(x^{(k)},x)<\epsilon$. Thus $D$ is complete.

In each factor, the balls for ${\delta }_j$ centred in $D_j$ with positive rational radii form a countable basis. Indeed, for $x\in U$ choose $\epsilon >0$ with $B_{{\delta }_j}(x,\epsilon )\subseteq U$; choose $a\in D_j$ with ${\delta }_j(x,a)<\epsilon \mathrm{/}4$ and a rational $r$ between ${\delta }_j(x,a)$ and $\epsilon -{\delta }_j(x,a)$. Then $x\in B_{{\delta }_j}(a,r)\subseteq U$ by the triangle inequality. Centres and radii have countable enumerations, so [F4] enumerates this basis. Finite products of the basis members form a countable product basis: a tuple of their natural indices is in ${\mathbb{N}}^n$, and every basic neighbourhood contains one of these boxes. Every product-open set is therefore the union of the members of this fixed countable rectangle basis contained in it; no point-indexed selection is made.

The product of the chosen dense sets is countable: the coordinate enumerations give a surjection from ${\mathbb{N}}^n$ onto ${\prod }_{j<n}{D}_j$, and the domain is at most countable by [F4]. It is dense in $P$, because a nonempty basic rectangle meets each $D_j$, and finite choice selects a point in the product of those intersections. Hence $P$ is separable and, by the topology and completeness proved above, Polish.

Write $\mathcal{S}={⨂}_{j<n}\mathcal{B}(P_j)$. Every basic rectangle from step 5.2 belongs to $\mathcal{S}$, so every open subset of $P$ belongs to $\mathcal{S}$ as a countable union. Minimality of the Borel sigma-algebra gives $\mathcal{B}(P)\subseteq \mathcal{S}$. For the reverse inclusion, fix $j<n$ and let ${\mathcal{C}}_j=\{A\subseteq P_j:{\pi }_j^{-1}(A)\in \mathcal{B}(P)\}$. Preimages preserve complements and countable unions, so this is a sigma-algebra, and it contains every open $A$ by the projection property in [F5]. Hence it contains $\mathcal{B}(P_j)$ by [F6]. Every Borel rectangle is the finite intersection of these Borel cylinders, and is therefore Borel in $P$. Step 1.2 now gives $\mathcal{S}\subseteq \mathcal{B}(P)$, proving both inclusions.

The bijection $h$ of step 2.1 carries each full measurable rectangle in $\prod E_j$ to a full Borel rectangle in $P$, and conversely, by the exact measurable-isomorphism property of every $h_j$ in [F1]. Since images and inverse images under a bijection preserve complements and countable unions, step 1.2 identifies the transported finite product sigma-algebra with ${⨂}_{j<n}\mathcal{B}(P_j)$; step 6.2 identifies this with $\mathcal{B}(P)$. The Polish space $P$ and this bijection are thus a Polish presentation of the given product. Together with step 1.1 this proves the unchanged Statement, without AC or Countable Choice.

The empty $P$ is immediate, so suppose $P\mathrm{\ne }\mathrm{\varnothing }$ and enumerate a countable dense set as $(x_m)$. For each $r\ge 1$, continuity from below chooses a finite union $U_r$ of $2^{-r-2}$-balls centered at the $x_m$ with $\mu (P\setminus U_r)<\epsilon 2^{-r-2}$. Countable choice makes these choices simultaneously.

The set $K_0={\bigcap }_r\overline{U_r}$ is closed. Countable subadditivity gives $\mu (P\setminus K_0)<\epsilon \mathrm{/}2$. For every scale, one of the finite closed-ball covers from step 1.1 covers $K_0$; choosing one point of $K_0$ from each nonempty member of that finite cover and doubling the radius gives a finite net with centres in $K_0$. Hence $K_0$ is totally bounded and [F3] makes it compact.

Every open $G$ contains a compact $K\subseteq G$ losing less than $\epsilon$: for $G=P$ use $K_0$; otherwise intersect $K_0$ with the increasing closed sets $\{x:d(x,P\setminus G)\ge 1\mathrm{/}n\}$. Their union is $K_0\cap G$, so [F2] gives one with the required loss.

Let $\mathcal{R}$ be the Borel sets which, for every $\delta >0$, have compact $K\subseteq A\subseteq G$ with $G$ open and $\mu (G\setminus K)<\delta$. Step 3.1 puts every open set in $\mathcal{R}$. The tight compact set from step 2.1 shows that complements remain in $\mathcal{R}$: from $K\subseteq A\subseteq G$, use $K_0\setminus G\subseteq P\setminus A\subseteq P\setminus K$ and bound the loss by $\mu (P\setminus K_0)+\mu (G\setminus K)$.

For pairwise disjoint $A_j\in \mathcal{R}$, continuity from below makes the measure of the tail of ${\bigcup }_j{A}_j$ arbitrarily small. A finite union of compact inner approximants handles that tail, while the union of the open outer approximants has loss bounded by the summable errors. Thus $\mathcal{R}$ is a lambda-system. Open sets are a pi-system generating the Borel sigma-algebra, so [F4] gives $\mathcal{B}(P)\subseteq \mathcal{R}$. The compact inner approximant for $A$ proves the statement.

By AC, choose once and for all, for every $i\in I$, a Polish presentation $h_i:E_i\to P_i$ witnessing that $(E_i,{\mathcal{E}}_i)$ is standard Borel. Pull the topology and a complete compatible metric of $P_i$ back to $E_i$. The finite product topologies now used below come from these same coordinate presentations, so every restriction map between them is continuous.

Let $C_n\downarrow \mathrm{\varnothing }$ be cylinders and suppose their cylinder masses stayed above $\eta >0$. Enumerate the countable union of their finite supports. For each $n$, choose a nested finite support $H_n$ that contains both the original support of $C_n$ and the first $n$ active coordinates, and view $C_n$ as a cylinder over $H_n$. In the product presentation fixed in step 1.1, inner-regularly replace its $H_n$-base by a compact base, losing less than $\eta 2^{-n-2}$. Finite intersections of the lifted compact cylinders then have positive cylinder mass.

For each $N$, choose a point in the intersection of the first $N$ lifted compact cylinders from step 2.1. For fixed $n$, all points with $N\ge n$ have their $H_n$-coordinates in the compact $n$th base. Sequential compactness successively supplies a subsequence converging on $H_1$, a further subsequence converging on $H_2$, and so on; take the diagonal subsequence. Because every restriction ${\prod }_{i\in H_{n+1}}{E}_i\to {\prod }_{i\in H_n}{E}_i$ is continuous by step 1.1, the successive limits restrict to the earlier limits. They therefore define one point on the union of the active coordinates. Each compact base is closed, so this point lies in every lifted compact cylinder.

Use AC to fill the inactive coordinates. The resulting point lies in every $C_n$, a contradiction. Hence the cylinder law is continuous at the empty set and is a premeasure.

By [F4] it extends to ${\mathcal{C}}_I$ and has total mass one. Two such extensions agree on all cylinders; these are a pi-system, so Dynkin's pi-lambda theorem gives uniqueness on ${\mathcal{C}}_I$, and nowhere larger.

For finite $F$, define ${\nu }_F={⨂}_{i\in F}{\mu }_i$. The rectangle formula shows that projecting ${\nu }_G$ to $F\subseteq G$ gives ${\nu }_F$, so the family is consistent.

Apply [F2] to $({\nu }_F)$. Its conclusion is exactly the stated measure and its cylinder-sigma uniqueness; it does not assert a measure on $\mathcal{P}({\prod }_i{E}_i)$.

Each coordinate map is measurable because inverse images of its measurable sets are one-coordinate cylinders. Thus $(X_i)$ is a process.

For finite $F$, $(X_i{)}_{i\in F}={\pi }_F$, so its pushforward law is ${\mu }_F$ by [F1]. This is precisely the finite-dimensional-law convention.

On $\mathcal{D}=\{C\in {\mathcal{C}}_I:\mu (C)=\nu (C)\}$, complements and disjoint countable unions preserve equality because both measures are probabilities. Thus $\mathcal{D}$ is a lambda-system.

Equal finite-dimensional marginals put every cylinder in $\mathcal{D}$. By [F1] and [F2], ${\mathcal{C}}_I\subseteq \mathcal{D}$, hence $\mu =\nu$.

Let $\mathcal{D}$ be the events determined by some countable coordinate set. Complements preserve the same support. For a countable family in $\mathcal{D}$, countable choice selects one countable support for each event, and [F2] makes their union countable; that union determines the union of the events. Thus $\mathcal{D}$ is a sigma-algebra.

Every finite-coordinate cylinder belongs to $\mathcal{D}$, using its finite support. By generated-sigma minimality and [F1], ${\mathcal{C}}_I\subseteq \mathcal{D}$.