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Let $(X,\mathscr F)$ be some measurable space and $Y$ be a finite set with a $\sigma$-algebra $2^Y$. Let the map $$ f:X\to Y $$ be $\mathscr F|2^Y$-measurable. Consider sets $X^\mathbb N$ and $Y^\mathbb N$ endowed with product $\sigma$-algebras and extend $$ f':X^\mathbb N\to Y^\mathbb N,\quad f'(x_1,x_2,\dots) = (f(x_1),f(x_2),\dots). $$ Is it true that $f'$ is measurable? If yes, how can I show that? It seems to be an easy problem, but I guess I am missing some point.

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up vote 5 down vote accepted

Yes, it is. One way to see it is to recall that the product $\sigma$-algebra on $Y^\mathbb{N}$ is generated by "cylinder sets" of the form $$A_1 \times A_2 \times \dots \times A_n \times Y \times Y \times \cdots.$$ It is clear that $f'^{-1}$ of such a set is measurable. But the collection $$\{A \subset Y^\mathbb{N} : f'^{-1}(A) \in \mathcal{F}^\mathbb{N}\}$$ is a $\sigma$-algebra. Therefore, it contains all the sets in the product $\sigma$-algebra, which means $f'$ is measurable.

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Do you mean that this is just an application of Theorem 1.3.1 (p. 13) here? That if inverse of the generating class is a subclass of a $\sigma$-algebra of the domain, then the map is measurable? – Ilya Jun 26 '12 at 16:46
@Ilya: Yes, exactly. – Nate Eldredge Jun 26 '12 at 17:15
Thanks a lot for a quick reply - I imagined this problem slightly more difficult than it is. – Ilya Jun 26 '12 at 17:16

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