Csdrhrt

Let

• 
  $(Omega_i,mathcal A_i)$ be a measurable space


• $X:Omega_1toOmega_2$


• $Z:Omega_1tomathbb R$

It's easy to show that $Z$ is $sigma(X)$-measurable if and only if there is a $mathcal A_2$-measurable $f:Omega_2tomathbb R$ with $$Z=f(X).tag1$$ Now, suppose $Y:Omega_1toOmega_3$. Are we able to show that if $Z$ is $sigma(X,Y)$-measurable, then there is a $mathcal A_2otimesmathcal A_3$-measurable $g:Omega_2timesOmega_3tomathbb R$ with $$Z=g(X,Y)tag2?$$

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

If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ 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 $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.