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Let $(\Omega, \mathcal{H}), (E, \mathcal{E})$ and $(F,\mathcal{F})$ be measurable spaces. Let $(E \times F, \mathcal{E} \otimes \mathcal{F})$ be a product space. Define the following three functions:

  1. $X:(\Omega,\mathcal{H}) \rightarrow (E,\mathcal{E})$
  2. $Y: (\Omega, \mathcal{H} \rightarrow (F,\mathcal{F})$
  3. $Z = (X,Y): (\Omega, \mathcal{H}) \rightarrow (E \times F, \mathcal{E} \otimes \mathcal{F})$

Now, I am trying to show that the following: $X:(\Omega,\mathcal{H}) \rightarrow (E,\mathcal{E})$ and $Y:(\Omega, \mathcal{H}) \rightarrow (F,\mathcal{F})$ are measurable $\Leftrightarrow (X,Y): (\Omega,\mathcal{H}) \rightarrow (E \times F, \mathcal{E} \otimes \mathcal{F})$ is measurable.

Here is my current work:

$(\Rightarrow):$

Since X and Y are both measurable, we have \begin{align*} Z^{-1}(E \times F) &= \{w \in \Omega: Z(w) \in E \times F\}\\ &= \{w \in \Omega:(X(w),Y(w)) \in E \times F\}\\ &= \{w \in \Omega: X(w) \in E\text{ and }Y(w) \in F\}\\ &= \{w \in \Omega: X(w) \in E \} \cap \{w \in \Omega: Y(w) \in F\}\\ &= X^{-1}(E) \cap Y^{-1}(F), \end{align*} so the function $Z(w)$ is measurable for any $w \in \Omega$.

$(\Leftarrow):$ On this part I'm stuck so I just wrote out my assumptions and what I want to show:

We assume $Z(w)$ is measurable, so we have $Z^{-1}(A) \in \mathcal{H}$ for all $A \in \mathcal{E} \times \mathcal{F}.$ We need to show that X and Y are measurable, i.e. $X^{-1}(A) \in \Omega$ for all $B \in \mathcal{E}$ and $Y^{-1}(C) \in \Omega$ for all $C \in \mathcal{F}.$

How can I finish the rest of this proof? Unfortunately, I couldn't find anything like of this in my book (Probability and Stochastics by Cinlar)

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  • $\begingroup$ You should really not use $E$ and $F$ to denote arbitrary sets in $\mathcal{E}$ and $\mathcal{F}$. This looks like it's causing you some confusion. Also what is $w$ and what does it mean when you say $Z(w)$ is measurable or that $X^{-1}(w)\in\Omega$? $\endgroup$ Mar 18, 2015 at 7:18
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    $\begingroup$ Yeah well, you can't use $E$ to denote the state space and at the same time be arbitrary sets of your sigma-algebra. $\endgroup$ Mar 18, 2015 at 7:23
  • $\begingroup$ $w$ is some element in the set $\Omega$. E and F are also sets, which together with their $\sigma$-algebras make up the two measurable spaces. Where am I using them as arbitrary sets of my sigma-algebra? $\endgroup$
    – Olorun
    Mar 18, 2015 at 7:34
  • $\begingroup$ If $w\in\Omega$ then quantities such as $X^{-1}(w)$ makes no sense. You're using $E$ and $F$ as arbitrary elements in the statement "$Z^{-1}(E\times F)\in\Omega$ for all $E\in\mathcal{E}$, $F\in\mathcal{F}$", no? Actually, the statement "$Z^{-1}(E\times F)\in \Omega$" makes no sense since $Z^{-1}(E\times F)$ is a subset of $\Omega$ (this has nothing to do with measurability). That $Z$ is measurable means $Z^{-1}(A)\in\mathcal{H}$ for all $A\in\mathcal{E}\otimes\mathcal{F}$. $\endgroup$ Mar 18, 2015 at 7:40
  • $\begingroup$ Ahh, I agree..I have updated my question accordingly. $\endgroup$
    – Olorun
    Mar 18, 2015 at 7:54

1 Answer 1

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Hint. Given $e \in \mathcal E$ (that is $e$ is a subset $e \subseteq E$, note, as Stefan commented, that you named the whole set $E$ already) we have $$ Z^{-1}(e \times F) = \{ \omega \in \Omega \mid Z(\omega) \in e \times F \} = \{\omega \in \Omega \mid X(\omega) \in e \} $$ (as $Y(\omega) \in F$ holds because of $Y \colon \Omega \to F$). Along the same lines $$ Z^{-1}(E \times f) = Y^{-1}(f), \quad f \subseteq F $$

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  • $\begingroup$ I'm a little bit confused - is your hint for the $\Leftarrow$ part of the proof? $\endgroup$
    – Olorun
    Mar 18, 2015 at 7:59
  • $\begingroup$ Yes it is ... if $Z$ is measurable, $Z^{-1}(e \times F) \in \mathcal H$ for each $e \in \mathcal E$. As $X^{-1}(e) = Z^{-1}(e \times F)$ this implies ... $\endgroup$
    – martini
    Mar 18, 2015 at 8:05
  • $\begingroup$ this implies that $X^{-1}(e) \in \mathcal{H}$. Then I could do the same for $Y^{-1}(f)$. Hope it understood it correctly.. Also, should I write $e \subseteq E$ or $e \in E$? Since technically $e$ is a subset, using $\in$ feels a little wrong. $\endgroup$
    – Olorun
    Mar 18, 2015 at 8:25
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    $\begingroup$ $e$ is a subset of $E$, but an element of $\mathcal E$. $\endgroup$
    – martini
    Mar 18, 2015 at 9:02

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