Proposition. Let \(\mu\) be an additive finite set function on an algebra of sets \(\mathcal A\). Then the following conditions are equivalent:
the function \(\mu\) is countably additive,
the function \(\mu\) is continuous at zero in the following sense: if \((A_n)_{n\in \mathbb N}\subseteq \mathcal A\) is a decreasing sequence of sets ( \(A_{n+1}\subseteq A_n\) for all \(n\in \mathbb N\)) with \(\bigcap_{n\in\mathbb N}A_n=\varnothing\), then \[\lim_{n\to \infty}\mu(A_n)=0.\]
Proof of (a) \(\Longrightarrow\) (b). Let \(\mu\) be countably additive and let \((A_n)_{n\in \mathbb N}\subseteq \mathcal A\) decrease monotonically to the empty set. Set \(B_n = A_n \setminus A_{n+1}\in \mathcal A\). The sets \(B_n\) are disjoint and their union is \(A_1\). Note that \[\mu(A_1)=\sum_{k\in\mathbb N}\mu(B_k)<\infty.\]
Hence \[\lim_{n\to\infty}\mu(A_n)=\lim_{n\to\infty}\sum_{k=n}^{\infty}\mu(B_k)=0\] and (b) follows.
Proof of (b) \(\Longrightarrow\) (a). Let \((B_n)_{n\in \mathbb N}\subseteq \mathcal A\) be a sequence of disjoint sets such that \(\bigcup_{n\in\mathbb N}B_n=B\in \mathcal A\). Define \[A_n=B\setminus\bigcup_{k=1}^nB_k\in \mathcal A.\] It is clear that \((A_n)_{n\in \mathbb N}\subseteq \mathcal A\) forms a decreasing sequence of sets such that \(\bigcap_{n\in\mathbb N}A_n=\varnothing\). By (b) we have \(\lim_{n\to \infty}\mu(A_n)=0\). By finite additivity this in turn means that \[\lim_{n\to \infty}\sum_{k=1}^n\mu(B_k)=\mu(B).\] Here we also used finiteness of \(\mu\).$$\tag*{$\blacksquare$}$$
Definition. Let \((X_{\alpha})_{\alpha \in A}\) be an indexed collection of nonempty sets, \(X=\prod_{\alpha \in A}X_{\alpha}\), and \(\pi_{\alpha}:X \to X_{\alpha}\) the coordinate maps. If \(\mathcal{A}_{\alpha}\) is a \(\sigma\)-algebra on \(X_{\alpha}\) for each \(\alpha\in A\), the product \(\sigma\)-algebra on \(X\) is the \(\sigma\)-algebra generated by \[\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}.\] We denote this \(\sigma\)-algebra by \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\). In other words, \[\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha} =\sigma\left(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}\right).\] If \(A=\{1,\ldots,n\}\), we write \(\mathcal{A}_1 \otimes \mathcal{A}_2 \otimes \ldots \otimes \mathcal{A}_n\).
Proof. Let \[\mathcal{E}=\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}.\]
By the definition \(\sigma(\mathcal{E})=\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\).
Obviously \(\mathcal{E} \subseteq \mathcal{F} \subseteq\sigma(\mathcal{F})\), which implies \(\sigma(\mathcal{E}) \subseteq \sigma(\mathcal{F})\).
It is easy to see that \(\mathcal F\subseteq \sigma(\mathcal{E})\), hence \(\sigma(\mathcal F)\subseteq \sigma(\mathcal{E})\), which gives \(\sigma(\mathcal F)= \sigma(\mathcal{E})\) as desired. $$\tag*{$\blacksquare$}$$
Let \(\{(X_\alpha, \mathcal{A}_\alpha, \mu_\alpha):\alpha\in A\}\) be a family of probability spaces.
Remark.
By the previous proposition we know that \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}=\sigma(\mathcal F),\) where \[ \mathcal F= \Big\{\bigcap_{j=1}^k\pi_{\alpha_j}^{-1}[E_{\alpha_j}]: E_{\alpha_1} \in \mathcal{A}_{\alpha_1},\ldots, E_{\alpha_k} \in \mathcal{A}_{\alpha_k},\ \alpha_1, \ldots, \alpha_k\in A,\ k\in\mathbb N\Big\}.\]
In other words, \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\) is generated by sets of the form \(\prod_{\alpha\in A} A_\alpha\), where only finitely many \(A_\alpha\) may differ from \(X_\alpha\). Sets of such a form are called cylidrical or cylinders.
Our goal is to define a product measure \(\mu\) on \(X=\prod_{\alpha\in A} X_\alpha\) paired with the \(\sigma\)-algebra \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\) such that \[\mu(E)=\bigotimes_{j=1}^k\mu_{\alpha_j}(E_{\alpha_j}) \quad \text{ if } \quad E=\bigcap_{j=1}^k\pi_{\alpha_j}^{-1}[E_{\alpha_j}]\in \mathcal F\]
We start with a countable set \(A\). Let \[\mathcal{E}_n = \{C \times \prod_{j=1}^{\infty}X_{n+j}: C \in \bigotimes_{i=1}^n \mathcal{A}_i\}\]
Then \[\mathcal A=\bigcup_{n\in \mathbb N} \mathcal{E}_n\] is an algebra, and it is not difficult to show that \[\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}=\sigma(\mathcal A).\]
We define a set function \(\mu:\mathcal{A}\to [0, 1]\) by setting \[\mu(A)=\mu_1 \otimes \ldots \otimes \mu_n (C) \quad \text{ for } \quad A=C \times \prod_{j=1}^{\infty}X_{n+j} \in \mathcal E_n.\]
We show that \(\mu\) is well-defined on \(\mathcal{A}\).
If \(A=C \times \prod_{j=1}^{\infty}X_{n+j} \in \mathcal E_n\) is regarded as an element of \(\mathcal{E}_k\) with \(k > n\), then it can be written as a set \[E=(C \times X_{n+1}\times \ldots\times X_{k}) \times\prod_{j=1}^{\infty}X_{k+j} \in \mathcal E_k.\] Then we have \[\mu(E)=\mu_1 \otimes \ldots \otimes \mu_k (C\times\prod_{j=n+1}^kX_j)=\mu_1 \otimes \ldots \otimes \mu_n (C)=\mu(A),\] since \[\mu_{n+1}(X_{n+1}) =\ldots=\mu_k(X_k)= 1.\]
\(\mu\) is countably additive on \(\bigotimes_{i=1}^n \mathcal{A}_i\) for each \(n\in\mathbb N\), thus it finitely additivity on \(\mathcal A\).
We now show that \(\mu\) is a premeasure on \(\mathcal A\).
Let \(\{(X_\alpha, \mathcal{A}_\alpha, \mu_\alpha):\alpha\in A\}\) be a countable family of probability spaces. The set function \(\mu\) on the algebra \(\mathcal{A}\) is a premeasure and hence it can be uniquely extended to a measure on \(\sigma(\mathcal A)=\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\).
Proof. It suffices to show that \(\mu\) is countably additive on the algebra \(\mathcal{A}\). Then by the Carathéodory extension theorem it can be uniquely extended to a countably additive measure on \(\sigma(\mathcal A)=\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\).
To prove that \(\mu\) is countably additive on the algebra \(\mathcal{A}\) we take a decreasing sequence \((A_n)_{n\in \mathbb N}\subseteq \mathcal A\) of sets such that \(\bigcap_{n\in\mathbb N}A_n=\varnothing\), we have to show that \[\lim_{n\to\infty}\mu(A_n)=0.\]
Suppose for a contradiction that \(\mu(A_n) >\epsilon >0\) for all \(n\in\mathbb N\). We will show that \(\bigcap_{n\in\mathbb N}A_n\neq\varnothing\), which is impossible!
Let \(\mathcal{A}^{(n)}\) denote the algebra of sets in \(\prod_{i=n+1}^\infty X_i\) defined by analogy with \(\mathcal{A}\) and let \(\mu^{(n)}\) be the set function on \(\mathcal{A}^{(n)}\) corresponding to the product of the measures \(\mu_{n+1}, \mu_{n+2}, \ldots\) by analogy with \(\mu\).
By the properties of finite products it follows that, for every set \(A \in \mathcal{A}\) and every fixed \((x_1, \ldots, x_n) \in \prod_{i=1}^n X_i\), the section \[A^{x_1, \ldots, x_n} = \Big\{ (z_{n+1}, z_{n+2}, \ldots ) \in \prod_{i=n+1}^\infty X_i: (x_1, \ldots, x_n, z_{n+1}, \ldots ) \in A \Big\}\] belongs to \(\mathcal{A}^{(n)}\) and the function \[\prod_{i=1}^n X_i\ni (x_1, \ldots, x_n) \mapsto \mu^{(n)} (A^{x_1, \ldots, x_n})\] is measurable with respect to \(\bigotimes_{i=1}^n \mathcal{A}_i\).
Let \[B_1^k=\{x_1\in X_1: \mu^{(1)}(A_k^{x_1}) > \epsilon /2\}.\]
Then \(B_1^k \in \mathcal{A}_1\) and \(\mu_1(B_1^k) > \epsilon/2\), which follows by Fubini’s theorem for finite products and the inequality \(\mu(A_k)>\epsilon\) for all \(k\in \mathbb N\).
Indeed, \(A_k = C_m \times \prod_{j=m+1}^{\infty}X_{j}\) for some \(m\in \mathbb N\), whence one has \(\mu(A_k) = \bigotimes_{i=1}^m \mu_i(C_m)\). By Fubini’s theorem we obtain \[\epsilon < \mu(A_k) \leq \mu_1(B_1^k) + \frac{\epsilon}{2} \mu_1(X_1 \backslash B_1^k) \leq \mu_1(B^k_1) + \epsilon/2\] which yields the necessary estimate.
The sequence of sets \((B_1^k)_{k\in\mathbb N}\) is decreasing and \[B_1=\bigcap_{k\in\mathbb N}B_1^k\neq\varnothing,\] since \(\mu_1\) is a measure on \(X_1\) and \(\mu_1(B_1^k) > \epsilon/2\) for all \(k\in\mathbb N\).
We fix an arbitrary point \(x_1 \in B_1\) and repeat the described procedure for the decreasing sets \(A_k^{x_1}\) in place of \(A_k\). This is possible, since \(\mu^{(1)}(A_k^{x_1}) > \epsilon/2\). We obtain a point \(x_2 \in X_2\) such that \(\mu^{(2)}(A_k^{x_1, x_2}) > \epsilon/4\) for all \(k\in \mathbb N\). We continue this process inductively.
After the \(n\)-th step we obtain \((x_1, \ldots, x_n) \in \prod_{i=1}^n X_i\) such that \(\mu^{(n)}(A_k^{x_1, \ldots x_n}) > \epsilon 2^{-n}\) for all \(k\in \mathbb N\). Our construction can be continued, yielding a point \(x=(x_1, \ldots, x_n, \ldots)\) belonging to all \(A_k\).
Indeed, fix \(k\in \mathbb N\) and write \(A_k = C_m \times \prod_{j=m+1}^{\infty}X_{j}\) for some \(m\in \mathbb N\). The set \(A_k^{x_1, \ldots, x_m}\neq\varnothing\), so there exists \((z_{m+1}, z_{m+2}, \ldots) \in \prod_{i=m+1}^\infty X_i\) such that \[(x_1, \ldots, x_m, z_{m+1}, z_{m+2}, \ldots ) \in A_k,\] implying \((x_1, \ldots, x_m)\in C_m\) and since \((x_{m+1}, x_{m+2}, \ldots) \in \prod_{j=m+1}^{\infty}X_{j}\), thus \((x_1, \ldots, x_m, x_{m+1}, x_{m+2}, \ldots) \in A_k\) and we are done. $$\tag*{$\blacksquare$}$$
Proof. By the definition \(\sigma(\mathcal{E})=\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\), where \[\mathcal{E}=\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}.\]
Obviously \(\mathcal{E} \subseteq \mathcal{F} \subseteq\sigma(\mathcal{F})\), which implies \(\sigma(\mathcal{E}) \subseteq \sigma(\mathcal{F})\).
It is easy to see that \(\mathcal F\subseteq \sigma(\mathcal{E})\), hence \(\sigma(\mathcal F)\subseteq \sigma(\mathcal{E})\), which gives \(\sigma(\mathcal F)= \sigma(\mathcal{E})\) as desired. $$\tag*{$\blacksquare$}$$
We fix an arbitrary uncountable index set \(A\).
By the previous proposition we know that \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}=\sigma(\mathcal F),\) where \[\mathcal F= \Big\{\bigcap_{j=1}^\infty\pi_{\alpha_j}^{-1}[E_{\alpha_j}]: E_{\alpha_1} \in \mathcal{A}_{\alpha_1}, E_{\alpha_2} \in \mathcal{A}_{\alpha_2},\ldots, \text{ and } (\alpha_j)_{j\in\mathbb N}\subseteq A\Big\}.\]
Thus \[\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}= \bigcup_{(\alpha_j)_{j\in\mathbb N}\subseteq A}\sigma(\mathcal F_{(\alpha_j)_{j\in\mathbb N}}),\] where \[\mathcal F_{(\alpha_j)_{j\in\mathbb N}}= \Big\{\bigcap_{j=1}^\infty\pi_{\alpha_j}^{-1}[E_{\alpha_j}]: E_{\alpha_1} \in \mathcal{A}_{\alpha_1}, E_{\alpha_2} \in \mathcal{A}_{\alpha_2},\ldots\Big\}.\]
It is easy to verify that \(\bigcup_{(\alpha_j)_{j\in\mathbb N}\subseteq A}\sigma(\mathcal F_{(\alpha_j)_{j\in\mathbb N}})\) is a \(\sigma\)-algebra.
Let \(\{(X_\alpha, \mathcal{A}_\alpha, \mu_\alpha):\alpha\in A\}\) be a family of probability spaces. Then there exists a product measure on the product space \((\prod_{\alpha \in A}X_{\alpha}, \bigotimes_{\alpha \in A}\mathcal{A}_{\alpha})\).
Proof. By the previous discussion we define \(\mu:\bigotimes_{\alpha \in A} \mathcal{A}_\alpha\to [0, 1]\) by \[\mu(A)=\bigotimes_{n\in \mathbb N} \mu_{\alpha_n}(A), \quad \text{ if } \quad A\in \sigma(\mathcal F_{(\alpha_n)_{n\in\mathbb N}})\] for some \((\alpha_n)_{n\in\mathbb N}\subseteq A\), since \[\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}= \bigcup_{(\alpha_j)_{j\in\mathbb N}\subseteq A}\sigma(\mathcal F_{(\alpha_j)_{j\in\mathbb N}}).\] By the previous theorem \(\mu\) is well-defined measure. It will be denoted by \(\bigotimes_{\alpha \in A}\mu_{\alpha}\) and called the product measure on \((\prod_{\alpha \in A}X_{\alpha}, \bigotimes_{\alpha \in A}\mathcal{A}_{\alpha})\). $$\tag*{$\blacksquare$}$$
A family \(\mathcal K\) of subsets of a set \(X\) is called a compact class for \(X\) if for any \((K_n)_{n\in \mathbb N}\subseteq \mathcal K\) with \(\bigcap_{n\in\mathbb N}K_n=\varnothing\), there is \(m\in\mathbb N\) such that \(\bigcap_{n=1}^mK_n=\varnothing\).
Let \(\mu\) be a nonnegative additive set function on an algebra \(\mathcal A\). Suppose that there exists a compact class \(\mathcal K\) approximating \(\mu\) in the following sense: for every \(A\in \mathcal A\) and every \(\varepsilon>0\), there exist \(K_{\varepsilon}\in \mathcal K\) and \(A_{\varepsilon}\in \mathcal A\) such that \(A_{\varepsilon}\subseteq K_{\varepsilon}\subseteq A\) and \(\mu(A\setminus A_{\varepsilon}) < \varepsilon\). Then \(\mu\) is countably additive.
In particular, this is true if the compact class \(\mathcal K\) is contained in \(\mathcal A\) and for any \(A\in \mathcal A\) one has the equality \[\mu(A)=\sup\{\mu(K): K\subseteq A \text{ and } K\in \mathcal K\}.\]
Proof. Suppose that \((A_n)_{n\in \mathbb N}\subseteq \mathcal A\) is a decreasing sequence of sets such that \(\bigcap_{n\in\mathbb N}A_n=\varnothing\), we have to show that \(\lim_{n\to\infty}\mu(A_n)=0.\)
Fix \(\varepsilon>0\), then for every \(n\in\mathbb N\) there are \(K_n\in\mathcal K\) and \(B_n\in\mathcal A\) so that \(B_n\subseteq K_n\subseteq A_n\) and \(\mu(A_n\setminus B_n)<\varepsilon 2^{-n}\).
It is clear that \(\bigcap_{n\in\mathbb N}K_n\subseteq \bigcap_{n\in\mathbb N}A_n=\varnothing.\) Hence there is \(m\in\mathbb N\) such that \(\bigcap_{n=1}^mK_n=\varnothing\), then \(\bigcap_{n=1}^mB_n=\varnothing\). Note that \[A_m=\bigcap_{n=1}^mA_n\subseteq \bigcup_{n=1}^m(A_n\setminus B_n).\]
Indeed, if \(x\in A_m\) then \(x\in A_n\) for all \(n\le m\). If \(x\not\in \bigcup_{n=1}^m(A_n\setminus B_n)\), then \(x\not\in A_n\setminus B_n\) for all \(n\le m\), which means that \(x\in B_n\) for all \(n\le m\), hence \(x\in \bigcap_{n=1}^mB_n=\varnothing\), which is impossible!
Hence \(\lim_{m\to\infty}\mu(A_m)=0\), since \[\qquad\qquad\mu(A_m)\le \sum_{n=1}^m\mu(A_n\setminus B_n)\le \sum_{n=1}^m\varepsilon 2^{-n}\le\varepsilon. \qquad\qquad \tag*{$\blacksquare$}\]
Proposition. Let \(\mathcal K\) be a compact class of subsets of a set \(X\). Then the class \(\mathcal K_{<\sigma}\) of finite unions of elements from \(\mathcal K\) is a compact class. Moreover, the class \(\mathcal K_{<\sigma\delta}\) of countable intersections of finite unions of elements from \(\mathcal K\) is a compact class as well.
Proof. It suffices to prove that \(\mathcal K_{<\sigma}\) is a compact class. Let \[A_i=\bigcup_{n=1}^{m_i}K_{i, n},\] where \(K_{i, n}\in \mathcal K\), be such that \(\bigcap_{i=1}^kA_i\neq\varnothing\) for every \(k\in\mathbb N\). Our aim is to show that \[\bigcap_{i\in \mathbb N} A_i\neq\varnothing.\]
Let \(M=\prod_{i\in\mathbb N}([1, m_i]\cap\mathbb Z)\). Let \[M_k=\Big\{\nu=(\nu_i)_{i\in\mathbb N}\in M: \bigcap_{n=1}^{k}K_{i, \nu_i}\neq\varnothing\Big\}.\]
Note that \(M_k\neq\varnothing\) for all \(k\in \mathbb N\), which follows from the relation \[\bigcup_{\nu\in M}\bigcap_{i=1}^kK_{i, \nu_i}=\bigcap_{i=1}^k A_i\neq\varnothing.\]
Moreover, \(M_{k+1}\subseteq M_k\) for all \(k\in \mathbb N\). We prove that \(\bigcap_{k\in \mathbb N} M_k\neq\varnothing.\)
Then taking \(\nu\in \bigcap_{k\in \mathbb N} M_k\) we see that \(\nu\in M_k\) for all \(k\in \mathbb N\), hence \(\bigcap_{n=1}^{k}K_{i, \nu_i}\neq\varnothing\), and consequently \(\bigcap_{n=1}^{\infty}K_{i, \nu_i}\neq\varnothing\) by the compactness of \(\mathcal K\). This implies that \(\bigcap_{i\in \mathbb N} A_i\neq\varnothing\), since \[\bigcap_{n=1}^{\infty}K_{i, \nu_i}\subseteq \bigcup_{\nu\in M}\bigcap_{i=1}^kK_{i, \nu_i}=\bigcap_{i=1}^k A_i\neq\varnothing \quad\text{ for all } \quad k\in \mathbb N.\]
To prove that \(\bigcap_{k\in \mathbb N} M_k\neq\varnothing\), we pick \[\nu^{(k)}=(\nu_n^{(k)})_{n\in\mathbb N}\in M_k \quad \text{ for every } \quad k\in \mathbb N.\]
Since \(1\le \nu_n^{(k)}\le m_n\) for all \(n, k\in \mathbb N\), then there exist infinitely many indices \(k\in \mathbb N\) such that the numbers \(\nu_1^{(k)}= \mu_1\) for some \(\mu_1\in [1, m_1]\).
By induction, we construct a sequence \(\mu=(\mu_i)_{i\in\mathbb N}\in M\) such that, for every \(n\in \mathbb N\), there exist infinitely many \(k \in \mathbb N\) with the property that \[\qquad\qquad\nu_i^{(k)}= \mu_i \quad \text{ for all }\quad i\in\{1,\ldots, n\}. \qquad\qquad {\color{purple}(*)}\]
We show that \[\mu\in \bigcap_{k\in \mathbb N} M_k.\] We fix an arbitrary \(n\in \mathbb N\) and pick an integer \(k>n\) satisfying (*), then \(\nu^{(k)}=(\mu_1,\ldots, \mu_n, \nu_{n+1}^{(k)}, \nu_{n+2}^{(k)},\ldots)\in M_k\subseteq M_n\). This means that \(\mu=(\mu_1,\ldots, \mu_n, \mu_{n+1}, \mu_{n+2},\ldots)\in M_n\), since the membership in \(M_n\) is determined by the first \(n\) coordinates of a sequence. $$\tag*{$\blacksquare$}$$
Let \(\{(X_\alpha, \mathcal{A}_\alpha):\alpha\in A\}\) be a family of measurable spaces. Suppose that for every \(\alpha \in A\) we are given a compact class \(\mathcal{K}_\alpha\) for \(X_\alpha\). Then, the class \(\mathcal C_{<\sigma\delta}\) of countable intersections of finite unions of cylindrical sets from \[\mathcal C=\Big\{\Big(\prod_{\alpha\in F}K_{\alpha}\Big) \times \Big(\prod_{\beta\in A\setminus F } X_\beta\Big): K_\alpha \in \mathcal{K}_\alpha, \ \alpha\in F\subseteq A, \ |F|<\infty \Big\}\] is compact as well.
Proof. By the previous proposition it suffices to verify the compactness for \[\mathcal C_0=\Big\{K_\alpha \times \prod_{\beta\in A\setminus\{\alpha\} } X_\beta: K_\alpha \in \mathcal{K}_\alpha, \ \alpha\in A\Big\},\] since \(\mathcal C=\{C_1\cap\ldots\cap C_n: C_1,\ldots, C_n\in \mathcal C_0, n\in \mathbb N\}\).
Let \((C_i)_{i\in \mathbb N}\subseteq \mathcal C_0\) be a family of cylinders such that \[C_i=K_{\alpha_i}^{(i)} \times \prod_{\beta\in A\setminus\{\alpha_i\} } X_\beta\in \mathcal C_0\] with \(K_{\alpha_i}^{(i)} \in \mathcal{K}_{\alpha_i}\).
Setting \(S = \{ \alpha_i:i\in\mathbb N \}\), note that \[\bigcap_{i\in\mathbb N}C_i=(\prod_{\alpha \in S} Q_\alpha) \times (\prod_{\beta \in S^c} X_\alpha), \quad \text{ where } \quad Q_\alpha = \bigcap_{\substack{i \in\mathbb N\\ \alpha_i = \alpha}} K_{\alpha_i}^{(i)}.\]
If \(\bigcap_{i\in\mathbb N}C_i=\varnothing\), then so one of the sets \(Q_\alpha=\varnothing\). By the compactness of the class \(\mathcal{K}_\alpha\), there exists \(n\in\mathbb N\) such that \(K_\alpha^{(1)} \cap \ldots \cap K_\alpha^{(n)} = \varnothing\).
Then we have \[\qquad\qquad C_1 \cap \ldots \cap C_n = \varnothing. \qquad\qquad\tag*{$\blacksquare$}\]
In many problems of measure theory and probability theory and their applications, one has to construct measures on products of measurable spaces that are more complicated than product measures.
Now we prove the principal result in this direction: the Kolmogorov extension theorem on consistent probability distributions.
The original Kolmogorov’s result was concerned with measures on products of real lines. We present an abstract formulation that goes back to E. Marczewski.
Let \(T \not = \varnothing\). Suppose that a family \(\{(\Omega_t, \mathcal{B}_t):t\in T\}\) of measurable spaces is given. For every nonempty set \(\Lambda \subset T\), we denote by \[\Omega_\Lambda=\prod_{t\in \Lambda}\Omega_t, \quad \text{ and } \quad \mathcal{B}_\Lambda=\bigotimes_{t\in \Lambda}\mathcal{B}_t,\] the corresponding product space and product \(\sigma\)-algebra, respectively.
Let \(T \not = \varnothing\) and let \(\{(\Omega_t, \mathcal{B}_t, \mu_t):t\in T\}\) be a family of probability spaces.
Suppose that for every finite sets \(\Lambda_1 \subseteq \Lambda_2\), the image of the measure \(\mu_{\Lambda_2}\) under the projection from \(\Omega_{\Lambda_2}\) to \(\Omega_{\Lambda_1}\) coincides with \(\mu_{\Lambda_1}\), i.e. \[ \mu_{\Lambda_2}\Big(E\times \prod_{t\in\Lambda_2\setminus\Lambda_1}\Omega_t\Big)=\mu_{\Lambda_1}(E) \quad\text{ for every } \quad E\in \mathcal B_{\Lambda_1}.\]
Moreover, suppose that for every \(t \in T\), the measure \(\mu_t\) on \(\mathcal{B}_t\) has an approximating compact class \(\mathcal{K}_t \subseteq \mathcal{B}_t\).
Then, there exists a probability measure \(\mu\) on \((\prod_{t \in T} \Omega_t, \bigotimes_{t \in T} \mathcal{B}_t)\) such that the image of \(\mu\) under the natural projection from \(\Omega\) to \(\Omega_\Lambda\) is \(\mu_\Lambda\) for each finite set \(\Lambda \subset T\), i.e. \[ \mu\Big(E\times \prod_{t\in T\setminus\Lambda}\Omega_t\Big)=\mu_{\Lambda}(E) \quad\text{ for every } \quad E\in \mathcal B_{\Lambda}.\]
Proof. We construct a probability measure \(\mu\) on \((\prod_{t \in T} \Omega_t, \bigotimes_{t \in T} \mathcal{B}_t)\), which is called the projective limit of the measures \(\mu_\Lambda\) for finite \(\Lambda\subseteq T\).
Every set \(B \in \mathcal{B}_\Omega\) can be identified with the cylindrical set \(C_\Lambda = B \times \prod_{t \in T \backslash \Lambda} \Omega_t\). It is clear that the set \[\mathcal A=\Big\{B \times \prod_{t \in T \backslash \Lambda} \Omega_t: B\in \mathcal B_{\Lambda}, \ \Lambda\subseteq T, \ |\Lambda|<\infty\Big\},\] forms an algebra generated by the semi-algebra of finite products \[\mathcal S=\Big\{\Big(\prod_{\alpha\in \Lambda}B_{\alpha}\Big) \times \Big(\prod_{t\in A\setminus \Lambda } \Omega_t\Big): B_\alpha \in \mathcal{B}_\alpha, \ \alpha\in \Lambda\subseteq T, \ |\Lambda|<\infty \Big\}.\]
On the algebra \(\mathcal{A}\), we define the set function by \[\mu(C_\Lambda)= \mu_\Lambda(B), \quad \text{ whenever } \quad C_\Lambda = B \times \prod_{t \in T \backslash \Lambda} \Omega_t.\]
The consistency condition yields that \(\mu\) is well-defined, i.e. \(\mu(C_\Lambda)\) is independent of the representation of \(C_\Lambda\). Indeed, suppose that \[C_\Lambda = B \times \prod_{t \in T \backslash \Lambda} \Omega_t = B' \times \prod_{t \in T \backslash \Lambda'} \Omega_t=C_{\Lambda'}\] for some finite \(\Lambda\subseteq \Lambda'\) and \(B\in\mathcal B_{\Lambda}\) and \(B'\in\mathcal B_{\Lambda'}\). This implies that \[B'=B \times \prod_{t \in \Lambda' \backslash \Lambda}\Omega_t,\] hence by the consistency condition we obtain that \[\mu(C_{\Lambda'})=\mu_{\Lambda'}(B')=\mu_{\Lambda'}\Big(B \times \prod_{t \in \Lambda' \backslash \Lambda}\Omega_t\Big)=\mu_{\Lambda}(B)=\mu(C_{\Lambda}).\]
We verify the countable additivity of the set function \(\mu\) on the algebra \(\mathcal{A}\). Note that every cylindrical set from \(\mathcal A\) can be approximated from inside by countable intersections of finite unions of products from \(\mathcal S\).
By the previous lemma the class \(\mathcal C_{<\sigma}\) of finite unions of cylindrical sets from \[\mathcal C=\Big\{\Big(\prod_{\alpha\in \Lambda}K_{\alpha}\Big) \times \Big(\prod_{t\in T\setminus \Lambda } \Omega_t\Big): K_\alpha \in \mathcal{K}_\alpha, \ \alpha\in \Lambda\subseteq A, \ |\Lambda|<\infty \Big\}\] is compact. We prove that \(\mu\) on the algebra \(\mathcal A\) is approximated by the compact class \(\mathcal C_{<\sigma}\).
We show that for every \(B=(\prod_{\alpha\in \Lambda}B_{\alpha}) \times (\prod_{t\in T\setminus \Lambda } \Omega_\beta)\in \mathcal S\), and for every \(\varepsilon>0\) there exists \(K=(\prod_{\alpha\in \Lambda}K_{\alpha}) \times (\prod_{t\in T\setminus \Lambda } \Omega_\beta)\in \mathcal C\) so that \[\mu(B\setminus K)<\varepsilon.\]
For every \(\alpha\in \Lambda\) it suffices to find \(K_\alpha \in \mathcal{K}_{\alpha}\) such that \(K_\alpha \subseteq B_\alpha\) and \(\mu_{\alpha}(B_\alpha \backslash K_\alpha) < \varepsilon n^{-1}\). Then we see that \[B \backslash K \subset \bigcup_{\alpha\in \Lambda} \big( (B_{\alpha} \backslash K_{\alpha}) \times \prod_{t \in \Lambda\setminus\{\alpha\}} \Omega_t \big).\]
This implies that \[\mu(B \backslash K) \leq \sum_{\alpha\in \Lambda} \mu_{\Lambda} ((B_\alpha \backslash K_\alpha) \times \prod_{t\in \Lambda\setminus\{\alpha\}} \Omega_t) = \sum_{\alpha\in \Lambda} \mu_\alpha(B_\alpha \backslash K_\alpha) < \varepsilon.\] which completes the proof. $$\tag*{$\blacksquare$}$$
Remark.
It is clear that the Kolmogorov extension theorem is applicable if \(\mu_\Lambda\) are consistent Radon measures.
There is a very interesting construction (due to Andersen–Jessen) of a consistent family of measures that has no Kolmogorov extension.
In particular, the Andersen–Jessen example shows that in (KET), in contrast to (IPMT), the assumption that the measures \(\mu_t\) on \(\mathcal{B}_t\) have approximating compact classes \(\mathcal{K}_t \subseteq \mathcal{B}_t\) cannot be omitted.
Remark.
Let \(\{(X_n, {\rm Bor}(X_n), \mu_n):{n\in\mathbb N}\}\) be a family of probability Borel spaces, where \((X_n)_{n\in\mathbb N}\) is a sequence of second countable metric spaces. If \(X=\prod_{n\in \mathbb N}X_n\), then \[{\rm Bor}(X)=\bigotimes_{n\in\mathbb N}{\rm Bor}(X_n),\] and the product measure \(\bigotimes_{n\in\mathbb N}\mu_n\) is in fact defined on \({\rm Bor}(X)\).
In general case, if \(\{(X_\alpha, {\rm Bor}(X_\alpha), \mu_\alpha):{\alpha\in A}\}\) is a family of probability Borel spaces and \(X=\prod_{\alpha\in A}X_\alpha\), the best what can be said is that \[\bigotimes_{\alpha\in A}{\rm Bor}(X_\alpha)\subseteq {\rm Bor}(X),\] and in fact \({\rm Bor}(X)\) may be much larger than the left hand side.
Let \(X=\mathbb R^{[0, \infty)}=\prod_{t\in [0, \infty)}X_t\), where \(X_t=\mathbb R\) for all \(t\in [0, \infty)\). In this case \({\rm Bor}(X)\) is strictly larger than the product \(\sigma\)-algebra, and \[\bigotimes_{t\in [0, \infty)}{\rm Bor}(X_t)\subset {\rm Bor}(X).\]
One can easily show that \(C([0, \infty), \mathbb R)\), the space of all real-valued continuous functions on \([0, \infty)\), is an \(\mathcal F_{\sigma\delta}\) set, thus \[C([0, \infty), \mathbb R)\in {\rm Bor}(X). \qquad \textcolor{red}{\text{Justify this!}}\] But on the other hand \[C([0, \infty), \mathbb R)\not\in \bigotimes_{t\in [0, \infty)}{\rm Bor}(X_t). \qquad \textcolor{red}{\text{Justify this!}}\]