120 lines
5.3 KiB
TeX
120 lines
5.3 KiB
TeX
\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 $n$.
|
|
We want to show that there exists a unique probability measure $\bP^{\otimes}$
|
|
on $(\R^\infty, \cB_\infty)$ (where $\cB_{\infty}$ 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}
|
|
% $\bP_n = \mu_1 \otimes \ldots \otimes \mu_n$.
|
|
\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 sets $\prod_{n \in \N} B_n$
|
|
for all $B_n \in \cB_1$. We simply define $\cB_{\infty}$ to be the
|
|
$\sigma$-algebra
|
|
Let $\cB_\infty \coloneqq \sigma \left( \{\prod_{n \in \N} B_n | B_n \in \cB(\R)\} \right)$.
|
|
\begin{question}
|
|
What is there in $\cB_\infty$?
|
|
Can we identify sets in $\cB_\infty$ for which we can define the 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 $C^\ast \coloneqq C \times \R^{\infty}$.
|
|
|
|
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}
|
|
|
|
Define $\cF = \bigcup_{n \in \N} \cF_n$. Check that $\cF$ 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{enumerate}[(1)]
|
|
\item $\sigma(\cF) = \cB_\infty$,
|
|
\item $\lambda$ as defined above is countably additive on $\F$.
|
|
\end{enumerate}
|
|
\begin{proof}[Proof of (1)]
|
|
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 $\phi^\ast \phi$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$.
|
|
\end{itemize}
|
|
Thus $\cC \subseteq \cB_n$ is a $\sigma$-algebra and contains all rectangles, hence $\cC = \cB_n$.
|
|
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{proof}
|
|
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}
|
|
Exercise
|
|
|
|
\end{proof}
|
|
\begin{proof}[Proof of (2)]
|
|
Let's prove that $\lambda$ is finitely additive.
|
|
$\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$.
|
|
$\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
|
|
Suppose $A_1, A_2 \in \cF$ 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 $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$.
|
|
$\lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2) = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2) = \lambda_n(C_1) + \lambda_n(C_2)$
|
|
by the definition of the finite product measure.
|
|
|
|
In order to use \autoref{fact:finaddtocountadd},
|
|
we need to show that if $B_n \in \cF$ with $B_n \to \emptyset \implies \lambda(B_n) \to 0$.
|
|
%TODO
|
|
\end{proof}
|
|
|