\lecture{3}{2023-04-13}{}
\begin{notation}
    Let $\cB_n$ denote $\cB(\R^n)$.
\end{notation}
\begin{goal}
    Suppose we have a probability measure $\mu_n$ on $(\R^n, \cB(\R^n))$
    for each $n \in \N$.
    We want to show that there exists a unique probability measure
    $\bP^{\otimes}$ on $(\R^\infty, \cB_\infty)$
    (where the $\sigma$-algebra $\cB_{\infty}$ still needs to be defined),
    such that
    \[
        \bP^{\otimes}\left( \prod_{n \in \N}  B_n \right)
        = \prod_{n \in \N} \mu_n(B_n)
    \]
    for all $\{B_n\}_{n \in  \N}$, $B_n \in \cB_1$.
\end{goal}
\begin{remark}
    $\prod_{n \in \N} \mu_n(B_n)$ converges, since $0 \le \mu_n(B_n) \le 1$
    for all $n$.
\end{remark}

First we need to define $\cB_{\infty}$.
This $\sigma$-algebra must contain all ``boxes'' $\prod_{n \in \N} B_n$
for $B_i \in  \cB_1$.
We simply take the smallest $\sigma$-algebra with this property:
\begin{definition}
    \[
        \cB_\infty \coloneqq  \sigma \left( \left\{\prod_{n \in \N} B_n :
        \forall  n .~ B_n \in  \cB(\R)\right\}  \right).
    \]
\end{definition}
\begin{question}
    What is there in $\cB_\infty$?
    Can we identify sets in $\cB_\infty$
    for which we can define the desired product measure easily?
\end{question}
Let $\cF_n \coloneqq  \{ C \times \R^{\infty} | C \in \cB_n\}$.
It is easy to see that $\cF_n \subseteq  \cF_{n+1}$
and using that $\cB_n$ is a $\sigma$-algebra, we can show that $\cF_n$
is also a $\sigma$-algebra.
Now, for any $C \subseteq  \R^n$ let $C^\ast \coloneqq  C \times \R^{\infty}$.
Note that $C \in \cB_n \implies C^\ast \in \cF_n$.
Thus $\cF_n = \{C^\ast : C \in  \cB_n\}$.
Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq  (\mu_1 \otimes \ldots \otimes \mu_n)(C)$.
It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}).


Recall the following theorem from measure theory:
\begin{theorem}[Caratheodory's extension theorem] % 2.3.3 in the notes
    \label{caratheodory}
    Suppose $\cA$ is an algebra (i.e.~closed under finite union)
    und $\Omega \neq  \emptyset$.
    Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$
    are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq  \cA $
    then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N}  \bP(A_n)$).
    Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$,
    where $\cF = \sigma(\cA)$.
\end{theorem}
\begin{proof}
    See theorem 2.3.3 in Stochastik.
\end{proof}

Define $\cF = \bigcup_{n \in \N} \cF_n$. Then $\cF$ is an algebra.
We'll show that if we define $\lambda: \cF \to  [0,1]$ with
$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
then $\lambda$ is countably additive on $\cF$.
Using \autoref{caratheodory},  $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.

We want to prove:
\begin{claim}
        \label{claim:sF=Binfty}
        $\sigma(\cF) = \cB_\infty$.
    \end{claim}
\begin{claim}
        \label{claim:lambdacountadd}
        $\lambda$ is countably additive on $\cF$.
\end{claim}

\begin{refproof}{claim:sF=Binfty}
    Consider an infinite dimensional box  $\prod_{n \in \N} B_n$.
    We have
     \[
         \left( \prod_{n=1}^N B_n \right)^\ast \in \cF_n \subseteq \cF
    \]
    thus
    \[
    \prod_{n \in \N} B_n = \bigcap_{N \in \N} \left( \prod_{n=1}^N B_n \right)^\ast \in \sigma(\cF).
    \]
    Since $\sigma(\cF)$ is a $\sigma$-algebra, $\cB_\infty \subseteq  \sigma(\cF)$. This proves ``$\supseteq$''.
    For the other direction we'll show $\cF_n \subseteq \cB_\infty$ for all $n$.
    Let  $\cC \coloneqq  \{ Q  \in  \cB_n | Q^\ast \in \cB_\infty\}$.
    For $B_1,\ldots,B_n \in \cB$, $B_1 \times \ldots \times  B_n \in \cB_n$
    and $(B_1 \times \ldots \times B_n)^\ast \in \cB_\infty$.
    We have $B_1 \times  \ldots \times  B_n \in  \cC$.
    And $\cC$ is a $\sigma$-algebra, because:
    \begin{itemize}
        \item $\cB_n$ is a $\sigma$-algebra
        \item  $\cB_\infty$ is a $\sigma$-algebra,
        \item $\emptyset^\ast = \emptyset$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$.
    \end{itemize}
    Since $\cC \subseteq  \cB_n$ is a $\sigma$-algebra
    and contains all rectangles,
    it holds that $\cC = \cB_n$.
    Hence  $\cF_n \subseteq  \cB_\infty$ for all $n$,
    thus $\cF \subseteq  \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
    $\sigma(\cF) \subseteq  \cB_\infty$.
\end{refproof}
For the proof of \autoref{claim:lambdacountadd},
we are going to use the following:
\begin{fact}
    \label{fact:finaddtocountadd}
    Suppose $\cA$ is an algebra on $\Omega \neq \emptyset$,
    and suppose $\bP: \cA \to [0,1]$ is a finitely additive
    probability measure.
    Suppose whenever $\{B_n\}_n$ is a sequence of sets from $\cA $
    decreasing to $\emptyset$ it is the case that
    $\bP(B_n) \to 0$. Then $\bP$ must be countably additive.
\end{fact}
\begin{proof}
    Let $(A_n)_{n\in \N}$ be a sequence of disjoint,
    measurable sets with
    $A \coloneqq \bigcup_{n} A_n \in \cA$.
    Let $A'_n \coloneqq  A \setminus \bigcup_{i=1}^n A_i$.
    Then we have $ \bP[A] = \bP[A'_n] + \sum_{i=1}^{n} \bP[A_i]$ for all $n$.
    Thus
    \[
        \bP[A] - \lim_{n \to \infty} \bP[A'_n]
        = \lim_{n \to \infty} \sum_{i=1}^{n} \bP[A_i].
    \]
    Since $\bigcap_{n \in \N} A'_n = \emptyset$, we have
    $\lim_{n \to \infty} \bP[A'_n] = 0$,
    hence
    \[
        \bP\left[\bigcup_{i \in \N} A_i\right]
        = \bP[A]
        = \sum_{i \in \N} \bP[A_i].
    \]
\end{proof}
\begin{refproof}{claim:lambdacountadd}
    Let us prove that $\lambda$ is finitely additive.
    We have $\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$ and
    $\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
    Suppose that $A_1, A_2 \in \cF$ are disjoint.
    Then pick some $n$ such that $A_1, A_2 \in \cF_n$.
    Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$
    and $C_2^\ast = A_2$.
    Then $C_1$ and $C_2$ are disjoint
    and $A_1 \cup  A_2 = (C_1 \cup  C_2)^\ast$.
    Hence
    \[
        \lambda(A_1 \cup  A_2) = \lambda_n(A_1 \cup  A_2)
        = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup  C_2)
        = \lambda_n(C_1) + \lambda_n(C_2)
    \]
    by the definition of the finite product measure.
    \phantom\qedhere
\end{refproof}
