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.

conditional lawof $Z$, which is a law-valued random variable. Morally, for a given $\omega \in \Omega$, assume that you already have all the information on $\mathcal{F}$ (at point $\omega$): then it remains some uncertainty on $\omega$, and the value of $E^*[Z|\mathcal{F}]$ at $\omega$ is the law of $Z$ when you consider that remaining uncertainty. So, $E^*[Z|\mathcal{F}]$ maps $\Omega$ to the dual space of $C_b(M)$. $\endgroup$