Hi everyone,


my problem seems quite simple: I have a set $\Gamma$ along with a nice $\sigma$-algebra $\mathscr{B}$. Then I have a vector space of bounded measurable functions $A \subset \mathscr{B}\_{\infty} ( \Gamma )$, and a convex function $F : A \to [0, + \infty]$ which is lower semicontinuous with respect to pointwise convergence, and it satisfies:


* $F(0)=0$;
* $F( \lambda f ) = \lambda F( f)$ for every $f \geq 0$, $\lambda \geq 0$.
* $F( f) \geq \inf\_{\Gamma} f$


Can I find a nonnegative (sigma-additive) measure $\mathfrak{m}$ on $\Gamma$ such that $\int\_{\Gamma} f d \mathfrak{m} \leq F(f)$ and $ \mathfrak{m} (\Gamma) >0 $? 


[This reminds me a bit Hahn-Banach theorem: recalling Daniell's integral I'm asking a linear functional $L$ that stays between $F$ and $\inf\_{\Gamma} f$ (that is concave) and with additional condition that $L(f\_n) \to 0$ everytime that I have a sequence $f\_n$ decreasing to $0$.] 



Existence of a measure under certain condition
No. Some counterexamples follow: First, fix $V \in \mathcal{B}$, and let $A$ be the set of $\mathcal{B}$-measurable functions which vanish on $V$. Let $F \equiv 0$. Then $F$ satisfies all of the assumptions, since $\inf\_\Gamma f \le 0$ for all $f \in A$.


Stranger examples show that even assuming $F(f) > 0$ for some $f$ won't save you. Suppose $\Gamma = \mathbb{R}$, with Lebesgue $\sigma$-algebra $\mathcal{B}$. Let $A$ be the set of bounded Lebesgue-measurable functions with compact support. Then again $\inf\_\Gamma f \le 0$ for all $f \in A$. Let $\Lambda$ denote the set of finitely additive measures on $\mathcal{B}$ such that $\int f d\lambda = 0$ for all bounded continuous functions $f$ and $\lambda \in \Lambda$. By Theorem 3.4 of Yosida-Hewitt, $\Lambda \backslash \{0\} \neq \emptyset$. Define $F(f) := \sup\_{\lambda \in \Lambda}\int f d\lambda$ (alternatively, $F(f) := |\int f d\lambda|$ for certain fixed $\lambda \in \Lambda \backslash \{0\}$ would work). Then $F(f) \ge 0$ for all $f \in A$, and $F(f) > 0$ for certain discontinuous $f \in A$. But if $m \ge 0$ is a countably additive measure with $\int f dm \le F(f)$ for all $f \in A$, then $\int f dm = 0$ for all nonnegative bounded continuous $f \in A$, which implies $m \equiv 0$.


Yosida & Hewitt, "Finitely additive measures": <http://www.ams.org/journals/tran/1952-072-01/S0002-9947-1952-0045194-X/S0002-9947-1952-0045194-X.pdf>


I suppose the moral of the story is that the set of countably additive measures is not a very well-behaved subset of $\mathcal{B}\_\infty^\*(\Gamma) = ba(\Gamma)$. With this in mind, I suspect that if $\Gamma$ is a nice enough topological space with $\mathcal{B}$ its Borel $\sigma$-algebra, and if you assume there exists a continuous function $f \in A$ for which $F(f) > 0$, then your conclusion should hold.


