Mathematics Asked by user531706 on November 2, 2021
Let $Omega$ be a compact Hausdorff space in $mathbb{C}^2$. Let $sigma_Omega$ be the Borel sigma algebra on $Omega$. Let $f: Omegalongrightarrowpartial mathbb{D}$ be a non constant continuous function. Let $sigma_{partial mathbb{D}}$ be the Borel sigma algebra on $partial mathbb{D}$(Unit circle on the complex plane). Now consider the sigma algebra $sigma_0={f^{-1}(A): ;Ain sigma_{partial mathbb{D}}}subset sigma_Omega$.
Now consider the function $g: Omegalongrightarrowmathbb{C}$ as $g(z_1,z_2)=z_1^mz_2^n(bar{z_1})^p(bar{z_2})^q$, where $m,n,p,qinmathbb{N}.$
Now as $g$ is continuous, it is clearly measurable w.r.t $sigma_Omega$ but. will it be measurable w.r.t $sigma_0$?
Is there any additional condition that can be put on $f$ , so that the above holds?
As a subset of a Hausdorff space, $Omega$ will automatically be Hausdorff, so I guess you just mean that $Omega$ is an arbitrary compact subset of $mathbb{C}^2$, right? Anyway, the answer to the question is obviously no, because $g$ is measurable with respect to $sigma_0 = sigma(f)$ (the $sigma$-algebra generated by $f$) if and only if $g=phi circ f$ with some Borel function $phi$ (not sure if this theorem has a name, but it is in every textbook on measure and probability, at least for real-valued functions), and this fails for almost every example where $f$ is not one-to-one. On the other hand, if $Omega$ is a circle and $f$ is a homeomorphism, then the claim is easily seen to be true (but not very interesting or surprising.)
Answered by Lukas Geyer on November 2, 2021
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