s23-probability-theory/inputs/lecture_04.tex

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\lecture{4}{}{End of proof of Kolmogorov's consistency theorem}
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To finish the proof of \yaref{claim:lambdacountadd},
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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}
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By \yaref{lec4fact1},
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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}
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In order to apply \yaref{fact:finaddtocountadd},
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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 $n$
\[
\bigcap_{k=1}^n L_k^\ast \neq \emptyset.
\]
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By \yaref{lem:intersectioncompactsets},
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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 $\lambda$
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as well as the proof of \yaref{claim:lambdacountadd}
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we have only used that $(\lambda_n)_{n \in \N}$
is a consistent family.
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Hence we have in fact shown \yaref{thm:kolmogorovconsistency}.