Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Ahh, Ilya: I see my mistake. I should learn to read better. Sorry! –  Ravi Donepudi 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
1  
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
2  
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 Answer 1

To extend Alexander's observation:

  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 $f(\{x\})\neq\{0\}$.

  3. 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.

share|improve this answer
1  
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
    
@Michael: This is true, I will come up with a better reasoning tomorrow. –  Asaf Karagila Jan 31 '12 at 0:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.