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Let $(\Omega, \mathcal A)$ be a measure space. Consider the product measure space $(\Omega^{\mathbb N}, \mathcal A^{\mathbb N})$ and denote by $\pi_n : \Omega^{\mathbb N} \to \Omega$ the $n$-th projection map. Let $\mathcal A_n := \sigma(\pi_i : i \le n)$, then the $\{ A_n \}$ are a filtration of $\mathcal A^{\mathbb N}$.

Does $\bigcup_n \mathcal A_n$ generates $\mathcal A^{\mathbb N}$?

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  • $\begingroup$ How do you define $\mathcal{A}^{\mathbb{N}}$? $\endgroup$ – saz Jan 24 '15 at 20:15
  • $\begingroup$ The product measure space, i.e. the $\sigma$-algebra generated by all sets $B_1 \times B_2 \times B_3 \times \ldots$, with $B_i \in \mathcal A$, but such that only for finitely many indices we have $B_i \ne \Omega$, see for example Defintion 7.9 here: math.ucla.edu/~biskup/275b.1.13w/PDFs/Standard-Borel-Spaces.pdf $\endgroup$ – StefanH Jan 24 '15 at 20:20
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Now I see this is quite obvious... but did not saw it 3 hours ago. So here is a proof. As each $\mathcal A_n \subseteq \mathcal A$ we have $\sigma\left( \bigcup_n \mathcal A_n \right) \subseteq \mathcal A$. This works for each filtration. For the other inclusion, note that $\mathcal A^{\mathbb N}$ is generated by the countable product of sets from $\mathcal A$ such that only finitely many are $\ne \Omega$, said differently by sets of the form $$ \pi_{i_1}^{-1}(B_1) \cap \pi_{i_2}^{-1}(B_2) \cap \ldots \cap \pi_{i_k}^{-1}(B_k) $$ for $i_1 < i_2 < \ldots < i_k$ and $B_j \in \mathcal A$ for $j = 1,\ldots, k$. This generating family is contained in $\bigcup \mathcal A_n$, for if $B$ is a generating set, then for some $k$ we have $$ B = B_1 \times B_2 \times \ldots \times B_k \times \Omega \times \Omega \ldots = \pi_1^{-1}(B_1) \cap \pi_2^{-1}(B_2) \cap \times \cap \pi_2^{-1}(B_k) \in \mathcal A_k $$ for $B_i \in \mathcal A$, that it is contained in $\mathcal A_k$ follows by the definition that it is precisely the $\sigma$-algebra generated by the projections up to and including $k$. Therefore the generating sets of $\mathcal A$ are contained in $\bigcup_n \mathcal A_n$ and the monotonicity of $\sigma(\ldots)$ implies $\mathcal A \subseteq \sigma(\bigcup \mathcal A_n)$.

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