Finite-alphabet product and probability compactness

Enumerate the sites and order the finite alphabet. From any sequence in $X$, retain an infinite subsequence constant at $i_1$, then a further infinite subsequence constant at $i_2$, and so on. At each stage use the first alphabet value in the fixed order occurring infinitely often in the current sequence, and enumerate its retained indices increasingly. This is a specified recursive rule, not a sequence of arbitrary dependent choices. Taking the diagonal subsequence makes each fixed coordinate eventually constant. For $\epsilon >0$, choose $K$ with $2^{-K}<\epsilon$. Two sufficiently late diagonal terms agree at the first $K$ sites, so their metric distance is at most $2^{-K}<\epsilon$ and they converge to the coordinatewise limit.

If $[F,B]=[G,D]$, pass to $H=F\cup G$ and use consistency to express both ${\nu }_F(B)$ and ${\nu }_G(D)$ as the ${\nu }_H$-probability of the same subset of $A^H$; hence ${\mu }_0([F,B])={\mu }_0([G,D])$. To check finite additivity, represent any finite collection of cylinders on the common finite coordinate set $H$; disjointness and unions then reduce to finite additivity of ${\nu }_H$. The cylinder $X$ has mass one, so ${\mu }_0$ is a normalized finitely additive probability.

Write the given probability sequence as $({\mu }_k{)}_{k\ge 1}$. The cylinder algebra is countable: there are countably many finite subsets of ${\mathbb{Z}}^d$, and each finite coordinate set has only finitely many subsets of the finite set $A^F$. Enumerate it as $C_1,C_2,\dots$ (allow repetitions if it is finite). Specify a convergent-subsequence selector for any sequence $(u_k)$ in $[0,1]$ as follows. Start with the closed interval $I_0=[0,1]$ and all indices. Bisect the current interval and retain its left closed half if infinitely many current indices have values there; otherwise retain its right closed half. The retained index set is infinite because the two halves cover the previous interval. After stage $r$, let $k_r$ be the least retained index larger than $k_{r-1}$, with $k_0=0$. The intervals are nested with diameter $2^{-r}$, so any two selected values from stages at least $r$ differ by at most $2^{-r}$. By [F3] the bounded selected sequence has a subsequential limit; the diameter bound then makes the whole selected sequence converge to that limit. No witness subsequence from [F3] is selected by the rule: each interval, retained set and index is uniquely specified. Starting with $a_0(k)=k$, apply this selector to $({\mu }_{a_{j-1}(k)}(C_j){)}_{k\ge 1}$ and let $a_j$ be the resulting increasing subsequence of $a_{j-1}$. Ordinary deterministic recursion defines all $a_j$ without dependent choice. The diagonal indices $n_j=a_j(j)$ are strictly increasing, since $a_{j+1}(j+1)\ge a_j(j+1)>a_j(j)$; for each fixed cylinder their probabilities converge along this diagonal. Limits of finite sums preserve normalization and finite additivity, hence all finite-coordinate consistency identities.

Every cylinder is clopen: if $F\subseteq \{i_1,\dots ,i_K\}$, then the ball of radius $2^{-K}$ around any $\sigma \in [F,B]$ lies in $[F,B]$, and its complement is a finite union of cylinders. Conversely, for any $\epsilon >0$, choose $K$ with $2^{-K}<\epsilon$; the cylinder fixing the first $K$ coordinates of $\sigma$ lies in the $\epsilon$-ball about $\sigma$. Thus the countable cylinder family is a basis for the metric topology and generates its Borel sigma-algebra. To prove compactness, suppose an open cover has no finite subcover. Starting with $X$, successively split the current cylinder by the next coordinate into finitely many subcylinders; at least one remains without a finite subcover, and choose the first such cylinder in a fixed order on $A$. The resulting nested cylinders determine one configuration $\sigma$. Some cover member contains $\sigma$ and, being open, contains a basic cylinder fixing finitely many coordinates of $\sigma$. A sufficiently late nested cylinder is contained in that member, contradicting its selection. Thus every open cover has a finite subcover.

If a cylinder $C$ is a disjoint union of countably many cylinders $C_n$, compactness of $C$ gives a finite subcover from the open cover $(C_n)$. Disjointness forces every other $C_n$ to be empty, so finite additivity gives countable additivity whenever the union is in $\mathcal{C}$; the cylinder weight is therefore a premeasure. Apply [F1] to extend it under the stated Countable Choice assumption. The cylinder family is a pi-system generating $\sigma (\mathcal{C})$, so [F2] gives uniqueness.

The extension in step 3.1 gives a probability measure with those cylinder values. Local functions are finite linear combinations of cylinder indicators. Every continuous $f$ is uniformly approximated by local functions. If uniform continuity failed, [A1] would choose, for each $K$, one pair agreeing on the first $K$ coordinates whose $f$-values differ by at least a fixed $\epsilon >0$. These are independent nonempty sets of pairs indexed by $K$, so countable choice suffices. Step 1.1 gives a convergent subsequence of the first components; along the same indices the second components converge to the same configuration, since their distances are at most $2^{-K}$. Continuity then contradicts the fixed separation of their $f$-values. Thus for large $K$, fixing the coordinates after the first $K$ to any fixed reference tail gives a local approximation with uniform error at most $\epsilon$. Cylinder convergence therefore gives convergence of integrals first for local functions and then for every continuous $f$.