Let $\Omega$ be a compact Hausdorff space in $\mathbb{C}^2$. Let $\sigma_\Omega$ be the Borel sigma algebra on $\Omega$. Let $f: \Omega\longrightarrow\partial \mathbb{D}$ 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 $\sigma_0=\{f^{-1}(A): \;A\in \sigma_{\partial \mathbb{D}}\}\subset \sigma_\Omega$.
Now consider the function $g: \Omega\longrightarrow\mathbb{C}$ as $g(z_1,z_2)=z_1^mz_2^n(\bar{z_1})^p(\bar{z_2})^q$, where $m,n,p,q\in\mathbb{N}.$
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.)