Defining the conditional distribution of $Z$ as $E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$ I've been reading the first section Furstenberg's Noncommuting Random Products and I am confused with how he is defining conditional distribution.
Here he is considering a group $G$ acting on a space $M$. For a $M$-valued random variable $Z$, he defies the distribution $E^{*}[Z]$ of $Z$ as the functional on $C_{b}(M)$ (bounded continuous real valued functions on $M$) 
$$E^{*}[Z](f)=E[f(Z)]$$
I interpret this as integrating $f$ with respect to the distribution of $Z$. 
He then considers $Z$ to be a random variable on some $\Omega$ with $\sigma$-algebra $\mathcal{F}$, and defines the conditional distribution $$E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$$
and states that $E^{*}{[Z|\mathcal{F}]}$ 
is itself a random variable with values in the space of probability measures on $M$. 
I'm confused because $E[f(Z)|\mathcal{F}]$ looks to be the conditional expectation of $f(Z)$ with respect to $\mathcal{F}$ which is  itself a random variable on $\Omega$. I'm also not seeing how  $E^{*}[Z|\mathcal{F}]$ is itself a random variable.
Pardon my ignorance.
 A: $\newcommand{\M}{\mathcal M}$
$\newcommand{\G}{\mathcal G}$
$\newcommand{\F}{\mathcal F}$
$\newcommand{\P}{\mathsf P}$
$\newcommand{\E}{\mathsf E}$
Suppose that $M$ is a Polish (i.e., complete separable metrizable) space with the Borel sigma-algebra $\M$ over it. Let $Z$ be an $M$-valued random variable (r.v.) defined on a probability space $(\Omega,\G,\P)$. Let $\F$ be a sub-sigma-algebra of $\G$. The key here is that then there exists a so-called regular conditional probability distribution (Theorems 1.13 and 1.17, and Remark 1.7 on pp. 8--9) $\mu_Z\colon \Omega\times\M\to[0,1]$ such that 


*

*for each $\omega\in\Omega$, the map $\M\ni B\mapsto \mu_{Z;\omega}(B):=\mu_Z(\omega,B)$ is a probability measure and 

*for each $B\in\M$, $\P(Z\in B|\F)=\mu_Z(\cdot,B)$ almost surely (a.s.). 
So, for each $\omega\in\Omega$ and each $f\in C_b(M)$, we can introduce $\mu_Z(f)(\omega):=\int_M f\,d\mu_{Z;\omega}$; then it is easy to see that 
$$\mu_Z(f)=\E(f(Z)|\F)=E^*[Z|\F](f). 
$$
a.s. So, we can identify $E^*[Z|\F]$ with $\mu_Z$, which can in turn be identified with the random probability measure $\Omega\ni\omega\mapsto\mu_Z(\omega,\cdot)$. 
