Let $\Omega$ be a compact Hausdorff space in $\mathbb{C}^n$. Let $\sigma_\Omega$ be the Borel sigma algebra on $\Omega$. Let $\zeta: \Omega\longrightarrow\partial \mathbb{D}$ 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 $\sigma_\zeta=\{{\zeta}^{-1}(A): \;A\in \sigma_{\partial \mathbb{D}}\}\subset \sigma_\Omega$.
Now let $f\in L^1(\Omega, \sigma_\Omega, \mu)$ and lets define a new measure $f_\mu$ on $(\Omega,\sigma_\zeta)$ as $f_{\mu}(A)=\int_A f d\mu$. It is easy to see that for $A\in \sigma_\zeta $, ${\mu}(A)=0$ implies $f_{\mu}(A)=0$, i.e $f_{\mu}(A)$ is absolutely continuous with the restriction of $\mu$ to $\sigma_\zeta$, so by the Radon Nikodym theorem there exists a $g\in L^1 (\Omega, \sigma_\zeta, \mu)$ such that $\int_A f d\mu =\int_A g d\mu$ for every $A\in \sigma_\zeta$. Lets call this $g$ as the conditional expectation of $f$ and denote it as $E(f|\sigma_\zeta)$.
Now suppose $h,k\in L^1(\Omega, \sigma_\Omega, \mu)$ are bounded. Then will it be true that $E(hk|\sigma_\zeta)=E(h|\sigma_\zeta)E(k|\sigma_\zeta).$ If not, then is there a condition (preferably if and only if) under which the same would hold?
An answer in Measure Theory terms would be really appreciated.