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Let

  • $(\Omega_i,\mathcal A_i)$ be a measurable space
  • $X:\Omega_1\to\Omega_2$
  • $Z:\Omega_1\to\mathbb R$

It's easy to show that $Z$ is $\sigma(X)$-measurable if and only if there is a $\mathcal A_2$-measurable $f:\Omega_2\to\mathbb R$ with $$Z=f(X).\tag1$$ Now, suppose $Y:\Omega_1\to\Omega_3$. Are we able to show that if $Z$ is $\sigma(X,Y)$-measurable, then there is a $\mathcal A_2\otimes\mathcal A_3$-measurable $g:\Omega_2\times\Omega_3\to\mathbb R$ with $$Z=g(X,Y)\tag2?$$

(By the way: Is it possible to generalize $(1)$ to Banach space valued strongly measurable $Z,f$?)

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1 Answer 1

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If $Z=I_{X^{-1}(A)\cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{A\times B}$. Now $\{C\in \mathcal B(\mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) \text {for some measurable} f:\mathbb R^{2} \to \mathbb R\}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $\mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $\sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n \to Z$ then $Z=\lim\sup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.

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  • $\begingroup$ My problem with the Banach space case is that there is no $\operatorname{lim sup}$. $\endgroup$
    – 0xbadf00d
    Nov 17, 2018 at 10:49
  • $\begingroup$ Something is wrong with your definition of your $\sigma$-algebra. Why $C\in\mathcal B(\mathbb R^2)$? $\endgroup$
    – 0xbadf00d
    Nov 17, 2018 at 10:54
  • $\begingroup$ You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d $\endgroup$ Nov 17, 2018 at 11:36

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