Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.