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

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Ahh, Ilya: I see my mistake. I should learn to read better. Sorry! – Ravi Jan 30 '12 at 10:41
What do you look for, a specific $X$ and a $\sigma$-algebra or for every $X$ and $\sigma$-algebra... – Asaf Karagila Jan 30 '12 at 10:46
You need to add the assumption that $f$ is always nonempty. Otherwise, the answer is trivially no. – Michael Greinecker Jan 30 '12 at 11:11
If $f(\emptyset) \neq \{ 0 \}$, we can choose $\mu_f(B) = \infty$ for every $B \in \mathscr F \setminus {\emptyset}$. – Alexander Thumm Jan 30 '12 at 11:47
If $0 < a \in f(\emptyset)$, then $\infty = \sum_{n\in\mathbb N} a \in f( \bigcup_{n\in\mathbb N} \emptyset) = f(\emptyset)$, so $\infty \in f(B) = f(B \cup \emptyset)$ for every $B \in \mathscr F$. – Alexander Thumm Jan 30 '12 at 13:00

1. 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 $\infty\in f(A)$.
2. 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 $f(\{x\})\neq\{0\}$.
Why shouldn't you have countably many atoms? Let $\mathcal{C}=\{C_n\}$ be a countable set of atoms and let $\{c_n\}$ be a selection from $\mathcal{C}$. Then letting $\mu(B)=\sum_{n:c_n\in B}1/2^n$ gives you a well defined probability measure. – Michael Greinecker Jan 30 '12 at 22:28