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 $f(\emptyset) = \{0\}$.