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  1. Suppose there are three measurable spaces $(\Omega, \mathbb{F})$, $(S_i, \mathbb{S}_i), i=1,2$, and two measurable mappings $f_i: \Omega \rightarrow S_i, i=1,2$. Is the mapping $f$ defined as $f(\omega):=(f_1(\omega), f_2(\omega))$ a measurable mapping from $(\Omega, \mathbb{F})$ to $(\prod_{i=1}^2 S_i, \prod_{i=1}^2 \mathbb{S}_i)$, where $\prod_{i=1}^2 \mathbb{S}_i$ is the product sigma algebra of $\mathbb{S}_i, i=1,2$?
  2. Suppose there are four measurable spaces $(\Omega_i, \mathbb{F}_i), i=1,2$, $(S_i, \mathbb{S}_i), i=1,2$, and two measurable mappings $f_i: \Omega_i \rightarrow S_i, i=1,2$. Is the mapping $f$ defined as $f(\omega_1, \omega_2):=(f_1(\omega_1), f_2(\omega_2))$ a measurable mapping from $(\prod_{i=1}^2 \Omega_i, \prod_{i=1}^2 \mathbb{F}_i)$ to $(\prod_{i=1}^2 S_i, \prod_{i=1}^2 \mathbb{S}_i)$?
  3. In Part 1 and Part 2, conversely, if $f$ is a measurable mapping, will $f_i, i=1,2$ be measurable mappings?
  4. Can the statements in Part 1,2 and 3 be generalized to any collection of $(S_i, \mathbb{S}_i) i \in I$ and $(\Omega_i, \mathbb{F}_i) i \in I$?

Thanks and regards! Are there some websites or books that address these questions?

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1 Answer 1

up vote 3 down vote accepted
  1. The product $\sigma$-algebra on $S_1\times S_2$ is generated by sets of the form $A\times B$, with $A\in \mathbb{S}_1$ and $B\in\mathbb{S}_2$. The function $f$ is measurable if and only if the inverse image of a measurable set is measurable, and it suffices to check that the inverse image of a measurable set in a specific generating set for the $\sigma$-algebra is measurable. So it suffices to see if $f^{-1}(A\times B)$, with $A$ and $B$ as above, is measurable.

    If $x\in f^{-1}(A\times B)$, then $f(x)\in A\times B$, so $f_1(x)\in A$ and $f_2(x)\in B$. Thus, $x\in f_1^{-1}(A)\cap f_2^{-1}(B)$. Conversely, if $x\in f_1^{-1}(A)\cap f_2^{-1}(B)$, then $f(x)\in A\times B$. So $f^{-1}(A\times B) = f_1^{-1}(A)\cap f_2^{-1}(B)$. Is this set in $\mathbb{F}$?

  2. Same idea: what is $f^{-1}(A\times B)$? $(x,y)\in f^{-1}(A\times B)$ if and only if $x\in f_1^{-1}(A)$ and $y\in f_2^{-1}(B)$, so $f^{-1}(A\times B) = f_1^{-1}(A)\times f_2^{-1}(B)$. Is this set in the product $\sigma$-algebra $\mathbb{F}_1\times\mathbb{F}_2$?

  3. Are the projection maps $\pi_i\colon (S_1\times S_2,\,\mathbb{S}_1\times\mathbb{S}_2)$ measurable? Is the composition of measurable functions measurable? What are $\pi_i\circ f$?

I'll leave 4 to you.

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Thanks! In Part 4: (1) for questions similar to Part 1, if your proof in Part 1 can still apply here, since measurability of subsets is preserved only under countable intersection of measurable sets, if the collection of component sets are uncountable, then $f$ may not be measurable? (2) for questions similar to Part 2, if your proof in Part 2 can still apply here, I would like to ask when $I$ is infinite, $\forall A_i \in \mathbb{S}_i$, is $\prod_{i \in I} A_i$ still measurable wrt the product $\sigma$-algebra? If yes, then is the answer here also yes? –  Tim Feb 21 '11 at 5:03
    
(3) for questions in Part 4 similar to Part 3, I guess your proof in Part 3 can still apply, and therefore the answer is the statements in Part 3 can be generalized to any collection of measurable spaces? –  Tim Feb 21 '11 at 5:06
    
@Tim: In (1), notice that the sigma algebra, even in the infinite case, is generated by products in which only finitely many components are not the entire space. So you only need to worry about finitely many sets, not even countably many. This is not the "box" $\sigma$-algebra, but the product $\sigma$-algebra. Do it carefully. For (2), not necesarily. Again: the product sigma algebra is not the "box" sigma algebra. It is generated by sets of the form $\prod A_i$ where $A_i=\Omega_i$ for all but finitely many $i$. So, you have to be careful there. That is: be careful, and try it. –  Arturo Magidin Feb 21 '11 at 5:17

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