58 lines
2.7 KiB
TeX
58 lines
2.7 KiB
TeX
\section{Product $\sigma$-Algebras}
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\label{section:product-sigma-algebra}
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\begin{definition}[Product $\sigma$-Algebra]
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\label{definition:product-sigma-algebra}
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Let $\bracs{(X_i, \cm_i)}_{i \in I}$ be measurable spaces, then the \textbf{product} $\sigma$-algebra $\bigotimes_{i \in I}\cm_i$ on $\prod_{i \in I}X_i$ is the $\sigma$-algebra generated by $\seqi{\pi_i}$.
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\end{definition}
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\begin{lemma}
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\label{lemma:rectangles}
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Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let
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\[
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\ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}
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\]
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then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$.
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\end{lemma}
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\begin{proof}
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\begin{enumerate}
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\item[(P1)] For any $j \in I$, $\emptyset = \pi_j^{-1}(\emptyset) \in \ce$.
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\item[(P2)] Let
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\[
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\bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce
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\]
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Assume without loss of generality that $J = J'$, then
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\[
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\bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce
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\]
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\item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then
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\[
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\braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J}
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\underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}
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\]
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\end{enumerate}
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\end{proof}
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\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}]
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\label{proposition:product-sigma-algebra-metric}
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Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then:
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\begin{enumerate}
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\item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$.
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\item If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
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Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
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\[
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U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
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\]
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The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
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\end{proof}
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