Measurable-set-valued random variables and their relation to conditional probabilities Let $(\Omega, \mathcal F, P)$ be a probability space.
I am seeking references on measurable-set-valued random variables. That is, I am interested in functions $X$ from $\Omega$ into $\mathcal F$. In order for such $X$ to be measurable, $\mathcal F$ must be equipped with a sigma-field. My first question, then, is:

Is there a natural sigma-field for $\mathcal F$? Will $2^{\mathcal F}$ do?

Here is some partial motivation for the first question, which leads to a second question. Suppose that $X$ takes countably many values $F_1, F_2,...$ in $\mathcal F$. Suppose, moreover, that $F_n = \{\omega: X(\omega) = F_n\}$, so that $\{F_n\}$ is a countable partition of $\Omega$. Then, viewed as a function of $\omega$, $P(\cdot \mid X(\omega))$ is just a version of the regular conditional probability $P(\cdot \mid \mathcal A)$, where $\mathcal A$ is the sigma-field generated by the $F_n$.
In general, it won't be the case that $X$ takes only countably many values, however. And it won't be the case that the values $F$ of $X$ satisfy $F = \{X=F\}$. But $X$ always generates a sub-sigma-field $\sigma(X)$ of $\mathcal F$ in the usual way. Thus, a second question:

What is the general relationship between the conditional probability $P(A \mid \sigma(X))$ and $P(A \mid X(\omega))$, viewed as a function of $\omega$?

Note that the quantity $P(A \mid X(\omega))$ is an elementary conditional probability, i.e. it is equal to $P(A \cap X(\omega))/P(X(\omega))$.
(For the purposes of addressing the last question, let us stipulate that $P(A \mid B)=P(A)$ whenever $P(B)=0$.)
 A: I would like to answer the first questions in the more general context where $X\colon (\Omega,\mathcal F,P)\to\tilde{\Omega}$ is any map. Your case $\tilde{\Omega}=\mathcal F$ is just a special case.
If $X$ is fixed and you want to make it measurable, one natural choice for the $\sigma$-algebra $\mathcal A_X$ on $\tilde{\Omega}$ is "the largest $\sigma$-algebra on $\tilde{\Omega}$ such that $X$ is measurable", i.e.
$$
\mathcal A_X := \{A\subseteq \tilde{\Omega}\mid X^{-1}(A)\in\mathcal F \}
$$
(it is a standard exercise to show that this actually defines a $\sigma$-algebra).
Concerning your second question:
Let us first fix $\omega\in\Omega$ and let $x_\omega := X(\omega)\in\mathcal F$. Then
\begin{align*}
P(A|\sigma(X))(\omega)
&=
P(A|X=x_\omega)
=
\frac{P(A\cap \{X=x_\omega)\} }{P(X=x_\omega)},
\\
P(A|X(\omega))
&=
\frac{P(A\cap x_\omega)}{P(x_\omega)}.
\end{align*}
Therefore, one condition that makes the two concepts agree, is 
$$
\{X=x_\omega \} = x_\omega\qquad \forall \omega,
$$
which is exactly your condition $F = \{X=F\}$ (note that the latter does not have to hold for all $F\in\mathcal F$, but only for $F\in{\rm ran}(X)$ in the range of $X$).
All of this also works without your stipulation at the end of your question as long as all objects exist (conditioning on events of measure zero is often well-defined).
Now, if this condition does not hold (as you write "it won't be the case that the values $F$ of $X$ satisfy $F = \{X=F\}$"), I don't see any natural relation between these two concepts, since the values in $F$ have nothing to do with those $\omega$ that map to $F$.
One interesting criterion that I thought through is the weaker condition
$$
\omega\in X(\omega)\qquad \forall \omega,
$$
which implies $\{X=x_\omega \} \subseteq x_\omega\ \forall \omega$. But even then there is nothing one can say about the relation of the two conditional probabilities, because conditioning on some event $A$ and conditioning on some event $B$ can give any two values in $[0,1]$ even for $A\subseteq B$.
