Existence of the measure without Axiom of Choice Let $X$ be any set and $\mathscr F$ be a $\sigma$-algebra of its subsets, so $(X,\mathscr F)$ is a measure space. The function 
$$
\mu:\mathscr F\to[0,\infty]
$$
is called a measure if 
$\quad 1.$ $\mu(\emptyset) = 0$,
$\quad 2.$ for any sequence $(B_n)_{n\in\mathbb N}$ such that $B_i\cap B_j = \emptyset$ it holds that
$$
\mu\left(\bigcup\limits_{n\in\mathbb N}B_n\right) = \sum\limits_{n\in\mathbb N}\mu(B_n).
$$
Let us consider a set-valued function $f:\mathscr F\to\mathscr P([0,\infty])$ where $\mathscr P$ denotes the powerset. Suppose that
$\quad 1^*.$ $0\in f(\emptyset)$
$\quad2^*.$ for any sequence $(B_n)_{n\in\mathbb N}$ such that $B_i\cap B_j = \emptyset$ and any sequence $x_n\in f(B_n)$ it holds that
$$
x:=\sum\limits_{n\in\mathbb N}x_n\in f\left(\bigcup\limits_{n\in\mathbb N}B_n\right).
$$
$\quad3^*.$ for any $B\in\mathscr F$ the set $f(B)$ is not empty.
The question is: does there exist a measure $\mu_f$ such that
$$
\mu_f(B)\in f(B)
$$
for any set $B\in\mathscr F$. I wonder if the question can be answered assuming Axiom of Choice and without this assumption.
Remark 1: clearly if $f(B)$ is a singleton for any $B\in\mathscr F$, which satisfies both of assumptions above, the measure $\mu_f$ exists, $\mu_f = f$.
Remark 2: thanks to Alexander, in the case when $f(\emptyset)$ contains a positive element, we can take $\mu_f(B) = \infty$ for any all $B\in\mathscr F\setminus\{\emptyset\}$. So the only unconsidered case is $f(\emptyset) = \{0\}$.
 A: To extend Alexander's observation:


*

*If $A\in\mathscr F$ can be split into infinitely many disjoint sets $A_n$ then every sum from the sets $f(A_n)$ must converge or else $\infty\in f(A)$. In particular this means that if for more than finitely many $k$'s we have $\dfrac{1}{k}\le x\in f(A_k)$ then we can construct the harmonic series. However since the index of the $A_k$'s was more or less arbitrary this means that only finitely many of them can have any nonzero elements, or else $\infty\in f(A)$.

*Reiterating the above argument gives that either every nonempty $A$ has $\infty\in f(A)$ or $A$ can be decomposed into countably many atoms (e.g. singletons) out of which only finitely many $\{x\}\subseteq A$ have a nontrivial measure, that is $f(\{x\})\neq\{0\}$.

*From this follows that either all subsets can be assigned an infinite measure, or that there are finitely many atoms which have a nonzero image, choose any representatives from these sets and define the measure as a finite sum of atomic measures.
