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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 $0$.]

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    $\begingroup$ Might want to clarify: I'm guessing you want a nontrivial positive countably additive measure. By "positive measure" one usually means "nonnegative measure", and since you're assuming $F \ge 0$, the trivial measure $m \equiv 0$ always works. As you mentioned, you can use Hahn-Banach to a nontrivial positive finitely additive measure such that $\int_\Gamma fdm \le F(f)$: consider $f \equiv 1$ and the subspace $S = \{\lambda F(f) : \lambda \in \mathbb{R}\}$. Then $\lambda f \mapsto \lambda$ then extends to a nontrivial (since $F(f) \ge 1$) continuous linear functional dominated by $F$. $\endgroup$
    – Dan
    Commented Sep 27, 2012 at 17:42
  • $\begingroup$ Thanks, now I corrected the statement... Yes, I mean a sigma additive measure; I wrote positive instead of nonnegative meaning that I wanted $\mathfrak{m} (\Gamma ) >0$ $\endgroup$
    – Simo_the_Wolf
    Commented Sep 28, 2012 at 14:37
  • $\begingroup$ To be more clear, an abstract version could be: giving a functional F like this on a space X, and a subspace $A \subseteq X^*$, where $X^*$ means the algebraic dual, what condition on $F$ should I have to get the existence of a nonzero element of $A$ that is less than or equal to $F$ ? $\endgroup$
    – Simo_the_Wolf
    Commented Sep 28, 2012 at 14:43

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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.

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  • $\begingroup$ Thanks!! I put my question in general terms because I tought it was helpful, but in fact, in my "real" problem I have also the condition that $F( 1) = 1$ where $1(x) = 1$ for all $x \in \Gamma$. $\endgroup$
    – Simo_the_Wolf
    Commented Sep 30, 2012 at 16:32
  • $\begingroup$ And yes, by the way I can look at $\Gamma$ as a topological space (actually also a metric space, if I want) and the $\sigma$-algebra is that of Borelian sets... $\endgroup$
    – Simo_the_Wolf
    Commented Sep 30, 2012 at 16:34
  • $\begingroup$ In that case, if $A \subset C(\Gamma)$, where $C(\Gamma)$ is the space of continuous functions on $\Gamma$, and say $\Gamma$ is a compact Hausdorff space, then your conclusion follows from Hahn-Banach, since the dual $C(\Gamma)^*$ consists only of countably additive measures. But without assuming $A \subset C(\Gamma)$, we may be stuck. You could still find a nontrivial countably additive measure $m$ such that $\int f dm \le F(f)$ for all $f \in A \cap C(\Gamma)$, but from that we can't deduce an inequality for $f \in C(\Gamma) \backslash A$ without a stronger continuity assumption on $F$. $\endgroup$
    – Dan
    Commented Sep 30, 2012 at 17:24

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