# If $Z$ is $\sigma(X,Y)$-measurable, is there a measurable $f$ with $Z=f(X,Y)$?

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 $$Z,f$$?)

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 $$Z$$.
• My problem with the Banach space case is that there is no $\operatorname{lim sup}$. Nov 17, 2018 at 10:49
• Something is wrong with your definition of your $\sigma$-algebra. Why $C\in\mathcal B(\mathbb R^2)$? Nov 17, 2018 at 10:54