\lecture{4}{}{End of proof of Kolmogorov's consistency theorem}

To finish the proof of \autoref{claim:lambdacountadd},
we need the following:
\begin{fact}
    \label{lec4fact1}
    Suppose $\{x_k^{(n)}\}_{n \in \N}$
    is a bounded sequence of real numbers for each $k \in \N$.
    Then there exists a strictly increasing sequence
    of natural number $\{n_i\}_{i\in \N}$
    such that for all $k \in \N$
    the series $\{x_k^{(n_i)}\}_{i \in \N}$
    converges.
\end{fact}
\begin{proof}
    We'll use a diagonalization argument.
    For $S \subseteq  \N$ infinite,
    we say that  a sequence of real number, $(x_n)_{n \in \N}$,
    \vocab[Convergence along a subset]{converges along $S$},
    if
    \[
    \lim_{\substack{n \to \infty\\n \in S}} x_n
    \]
    exists.

    Let $S_1$ be such that $\{x_1^{(n)}\}_{n \in \N}$
    converges along $S_1$. Such an $S_1$ exists by
    Bolzano-Weierstraß.
    We proceed recursively.
    Suppose we have already chosen $S_1,\ldots, S_{k-1}$.
    Consider $\{x_k^{(n)}\}_{n \in  S_{k-1}}$.
    By Bolzano-Weierstraß, there exists $S_k \subseteq  S_{k-1}$
    such that $\{x_{k}^{(n)}\}_{n \in S_{k-1}}$
    converges along $S_k$.
    For an infinite subset $T \subseteq  \N$ and $\nu \in \N$
    let $\# \nu(T)$ denote the  $\nu$-th smallest
    element of $T$.
    Let
    \[
    S \coloneqq  \{\#\nu(S_k) : k \in \N\}.
    \]
    Since $S_{k+1} \subseteq  S_k$,
    we have $\#(k+1)(S_{k+1}) > \#k(S_{k+1}) \ge  \# k (S_k)$.
    Hence $S$ is infinite.
    Each $\{x_k^{(n)}\}_{n \in \N}$
    converges along $S$,
    since all but finitely many elements of $S$
    belong to $S_k$.
\end{proof}

\begin{lemma}
    \label{lem:intersectioncompactsets}
    Let $\{K_n\}_{n \in \N}$ be a sequence of compact sets
    $K_n \subseteq  \R^{l_n}$ for some $l_n$.
    Suppose for all  $n$
     \[
    \bigcap_{i=1}^n K_i^\ast \neq \emptyset.
    \]
    Then
    \[
    \bigcap_{i\in \N}  K_i^\ast \neq  \emptyset.
    \]

\end{lemma}
\begin{refproof}{lem:intersectioncompactsets}
    We know from analysis
    that if $\{K_n\}_{n \in \N}$ is a sequence of compact sets
    such that the intersection of finitely many of them is non-empty,
    then
    \[
    \bigcap_{n \in \N}  K_n \neq \emptyset.
    \]
    Here, different $K_n$ may have different dimensions $l_n$,
    but we can view them as subsets of $\R^\infty$
    by applying ${}^\ast$.
    For each $n$, choose $x^{(n)} \in  \bigcap_{i =1}^n K_i^\ast$.
    We can assume $x ^{(n)}_k = 0$ for $k > \max \{l_1,\ldots, l_n\}$.
    For all $k \in \N$ we will show that $\{x_k^{(n)}\}$
    is bounded.
    \begin{itemize}
        \item Case 1: Suppose every $l_n \le k$.
            Then $\{x_k^{(n)}\}_n$ only contains zeros.
        \item Case 2:
            Suppose some $l_{n_0} \ge k$.
            Let $Z$ be the projection of $K_{n_0} \subseteq  \R^{l_{n_0}}$
            onto its $k$-th component.
            $Z$ is a compact subset of $\R$.
            Hence it is bounded.
            For all $n \ge n_0$, we have $x^{(n)} \in K_{n_0}^\ast$
            and $x_k^{(n)} \in Z$,
            so $\{x_k^{(n)}\}_n$ is bounded.
    \end{itemize}

    By \autoref{lec4fact1},
    there is an infinite set $S \subseteq  \N$,
    such that $\{x_k^{(n)}\}_{n \in S}$
    converges for every $k$.
    Let $x_k \coloneqq  \lim_{\substack{n \to \infty\\n \in S}} x_k^{(n)}$.
    Now let $x = (x_1,x_2,\ldots)\in \R^{\infty}$.
    \begin{claim}
        $x \in  \bigcap_{i \in \N} K_i^\ast$.
    \end{claim}
    \begin{subproof}
        Consider $x^{(n)}$ for $n > i$ and $n \in S$.
        Then  $(x^{(n)}_1, \ldots, x^{(n)}_{l_i}) \in K_i$
        and
        \[
            \lim_{\substack{n \to \infty\\n \in S}}
                (x^{(n)}_1, \ldots, x^{(n)}_{l_i})
            = (x_1,\ldots, x_{l_i}).
        \]
        Since $K_i$ is compact,
        it follows that $x \in K_i^\ast$.
    \end{subproof}
\end{refproof}

\begin{refproof}{claim:lambdacountadd}
    In order to apply \autoref{fact:finaddtocountadd},
    we need the following:
    \begin{claim}
        For any sequence $B_n \in \cF$
        with $B_n \xrightarrow{n\to \infty} \emptyset$
        we have $\lambda(B_n) \xrightarrow{n \to \infty} 0$.
    \end{claim}
    \begin{subproof}
        Suppose that $B_1^\ast \supseteq B_2^\ast \supseteq \ldots$
        is a decreasing sequence such that
        $\lim_{n \to \infty} \lambda(B_n^\ast) = \epsilon > 0$.
        For each $n$,
        let $l_n$ be such that $B_n \in \cB_{l_n}$.
        By regularity of Borel probability measures, %
        % TODO see the proof of Caratheodory extension theorem
        given $\epsilon > 0$,
        there exists a compact set $L_n \subseteq  B_n$,
        such that
        \[
            (\mu_1 \otimes \ldots \otimes\mu_n)(B_n \setminus L_n)
            < \frac{\epsilon}{2^{n+1}}
        \]
        We have
        \[
          B_n^\ast \setminus \bigcap_{k=1}^n L_k^\ast %
          \subseteq \bigcup_{k=1}^n \left( B_k^\ast \setminus L_k^\ast \right).
        \]
        Hence
        \begin{IEEEeqnarray*}{rCl}
            \lambda\left( B_n^\ast \setminus \bigcap_{k=1}^n L_k^\ast \right)
            &\le & \lambda\left(\bigcup_{k=1}^n %
                B_k^\ast \setminus L_k^\ast \right) \\
            &\le & \sum_{k=1}^{n} \lambda(B_k^\ast \setminus L_k^\ast)\\
            &\le& \sum_{k=1}^{n} \frac{\epsilon}{2^{k+1}}\\
            &\le & \frac{\epsilon}{2}.
        \end{IEEEeqnarray*}
        By our assumption,
        $\lambda(B_n^\ast) \downarrow \epsilon > 0$.
        Hence $\lambda(B_n^\ast) \ge \epsilon$ for all $n$.
        Thus
        \[
            \lambda\left( \bigcap_{k=1}^n L_k^\ast \right) %
            \ge \epsilon - \frac{\epsilon}{2} = \frac{\epsilon}{2}.
        \]
        In particular, for all $n$
        \[
        \bigcap_{k=1}^n L_k^\ast \neq \emptyset.
        \]
        By \autoref{lem:intersectioncompactsets},
        it follows that
        \[
        \bigcap_{k \in \N} L_k^\ast \neq \emptyset.
        \]
        Since
        \[
        \bigcap_{k \in \N} B_k^\ast \supseteq \bigcap_{k \in \N} L_k^\ast,
        \]
        we have $\bigcap_{k \in \N} B_k^\ast \neq  \emptyset$.
    \end{subproof}
\end{refproof}

The measure $\lambda$ is as desired:
For all $n \in \N$ take some $B_n \in \cB_1$
and let $C_n \coloneqq \prod_{i=1}^n B_i$.
Then $C_n^\ast \downarrow \prod_{i=1}^{\infty} B_i$,
hence
\begin{IEEEeqnarray*}{rCl}
    \lambda\left( \prod_{i=1}^{\infty} B_i  \right)
    &\overset{\text{continuity}}{=}& \lim_{N \to \infty} \lambda(C_N^\ast)\\
    &=&  \lim_{N \to \infty} \lambda_N(C_N^\ast)\\
    &=& \lim_{N \to \infty} \prod_{n=1}^{N} \mu_n(B_n)\\
    &=& \prod_{n \in \N} \mu_n(B_n).
\end{IEEEeqnarray*}

For the definition of $\lambda$
as well as the proof of \autoref{claim:lambdacountadd}
we have only used that $(\lambda_n)_{n \in \N}$
is a consistent family.
Hence we have in fact shown \autoref{thm:kolmogorovconsistency}.
