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Let $X$ be a measurable space, and let $M(X)$ be the vector space of finite signed measures on $X$. Are there natural conditions on a linear functional $f:M(X)\rightarrow\mathbb{R}$ that are equivalent to the existence of a measurable function $g:X\rightarrow\mathbb{R}$ such that $f(\mu)=\int gd\mu$ for all $\mu\in M(X)$?

In particular, where $P(X)\subseteq M(X)$ consists of the probability measures on $X$, and $\text{proj}:P(P(X))\rightarrow P(X)$ is given by $\text{proj}(\mathbb{P})(A)=\int_{P(X)}\mu(A)d\mathbb{P}(\mu)$ for measurable $A\subseteq X$ (so that $\text{proj}$ flattens probability measures over probability measures over $X$ into probability measures over $X$ in the natural way. $P(X)$ should be given the smallest $\sigma$-algebra such that $\mu\mapsto\mu(A)$ is measurable for all measurable $A\subseteq X$; I think this is the $\sigma$-algebra generated by the topology of strong convergence, as in https://en.wikipedia.org/wiki/Convergence_of_measures#Strong_convergence_of_measures), suppose that $\forall\mathbb{P}_1,\mathbb{P}_2\in P(P(X))$ if $\forall x\in\mathbb{R}\,$ $\mathbb{P}_1(\{\mu|f(\mu)\geq x\})\geq\mathbb{P}_2(\{\mu|f(\mu)\geq x\})$ then $f(\text{proj}(\mathbb{P}_1))\geq f(\text{proj}(\mathbb{P}_2))$. Does this (possibly together with some pathology-excluding condition like that $X$ is Polish, or that $f$ is measurable or continuous with respect to the topology of strong convergence) imply that $f$ must be integration of a measurable function $g$?

Edit: I've been given a good equivalent condition to $f$ being integration of a function, but I'm still curious whether the other condition I proposed is also equivalent to it (obviously not in conjunction with the assumption that $f$ is continuous in the topology of strong convergence), or whether there is a counterexample.

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  • $\begingroup$ In general, if $M$ is a linear space (here $M=M(X)$) and $F$ a linear subspace of the algebraic dual $M'$ of $M$, then any $f\in M'$ is in $F$ if and only if it is continuous w.r.to the weak topology $\sigma(M,F)$. (that is almost a tautology yet maybe a starting point) $\endgroup$ Jan 22 '17 at 8:53
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    $\begingroup$ @Alex Mennen: The natural measure-theoretic condition coming from the theory of the Giry monad ( link.springer.com/chapter/10.1007/BFb0092872 ) is that a bounded measurable function $P(X) \to \mathbb{R}$ arises from integration against a bounded measurable function $X \to \mathbb{R}$ iff $$ f(\mathrm{proj}(\mathbb{P})) = \int_{P(X)}f d\mathbb{P} $$ for all $\mathbb{P} \in P(X)$. Is the condition you give in the question intended to rephrase this condition, or did you come up with it independently? $\endgroup$ Jan 25 '17 at 8:13
  • $\begingroup$ Ah, thanks. That condition clearly implies the condition I gave (which I came up with independently), but it doesn't seem obvious whether the reverse implication holds. $\endgroup$ Jan 26 '17 at 23:41
  • $\begingroup$ I should have given more context for this question. The VNM theorem provides conditions for when a binary relation $\preceq$ on P(X) is given by $x\preceq y\iff f(x)\leq f(y)$ for some affine function $f$. This is normally interpreted as conditions under which $x\preceq y$ iff $U$ has higher expected value on y than on x, for some function $U$ on $X$, because those are equivalent if $X$ is finite (or if we only pay attention to finitely-supported probability measures on $X$). I want to strengthen the theorem so the conclusion is what the theorem is usually interpreted as saying, for any $X$. $\endgroup$ Jan 26 '17 at 23:47
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If you work in terms of topologies on M, then the way to go is to use the theory of dualities from functional analysis. The basic theory of these things can be found in books called "Topological Vector Spaces", such as Schaefer's or Bourbaki's.

In this case, the relevant duality is $$ \langle \mu, f \rangle = \int_X f d\mu $$ as a mapping $M(X,\Sigma) \times \mathcal{L}^\infty(X,\Sigma) \rightarrow \mathbb{R}$, where $\mathcal{L}^\infty(X,\Sigma)$ is the bounded measurable functions (not modulo anything). The topology $\sigma(M,\mathcal{L}^\infty)$ on $M(X,\Sigma)$ is defined to be the coarsest such that $\langle -,f\rangle$ is continuous for all $f \in \mathcal{L}^\infty(X,\Sigma)$.

A basic theorem (IV.1.2 in Schaefer) shows that for any $\sigma(M,\mathcal{L}^\infty)$-continuous linear functional $\phi : M(X,\Sigma) \rightarrow \mathbb{R}$ there exists $f \in \mathcal{L}^\infty(X,\Sigma)$ such that $\phi = \langle -, f \rangle$. So there definitely exists such a topology (a comment by Pietro Majer makes the same suggestion).

Now, we can also consider $\sigma(M,Y)$ for any subset $Y \subseteq \mathcal{L}^\infty$ separating the points of $M(X,\Sigma)$. This topology coincides with the topology of pointwise convergence, considering elements of $M$ as functions $Y \rightarrow \mathbb{R}$. Additionally, if $E$ is the linear span of $Y$, then $\sigma(M,Y) = \sigma(M,E)$. The topology of "strong convergence" is the topology of pointwise convergence on elements of $\Sigma$ (measurable subsets) or equivalently on their linear span, the simple functions. Therefore a functional $\phi : M(X,\Sigma) \rightarrow \mathbb{R}$ is continuous with respect to strong convergence iff there exists a simple function $f : X \rightarrow \mathbb{R}$ such that $\phi = \langle -, f \rangle$. Therefore continuity in with respect to strong convergence is, fittingly, stronger than needed.

I should add one more important fact to qualify what I said in the last paragraph. The norm-boundedness of $P(X,\Sigma) \subseteq M(X,\Sigma)$ and the norm-density of simple functions in $\mathcal{L}^\infty(X,\Sigma)$ imply that $\sigma(M,\mathcal{L}^\infty)$ and the topology of strong convergence agree on $P(X,\Sigma)$. I can explain more about why this is if pushed.

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  • $\begingroup$ I realize I forgot to address the part of the question about the $\sigma$-algebra on $P(X)$. You can get some of the distance with that by using the theory of the Giry monad. $\endgroup$ Jan 22 '17 at 10:09
  • $\begingroup$ Ah, thanks. I had somehow talked myself into thinking that the countably supported measures were closed in the topology of strong convergence, and hence that requiring $f$ to be continuous in that topology would be a weak condition, but I see now that that is far from true. $\endgroup$ Jan 22 '17 at 23:02
  • $\begingroup$ To clarify your last remark, does that mean that if $f$ is bounded and measurable but not simple, then ⟨−,f⟩ is discontinuous in the topology of strong convergence, but its restriction to nonnegative measures is continuous? $\endgroup$ Jan 22 '17 at 23:03
  • $\begingroup$ It is the boundedness of the set of probability measures that I am using, not the positivity (so $\langle -, f\rangle$ is also continuous on the set of signed measures of variation $\leq 1$). I do not know if $\langle -, f \rangle$ will be continuous when restricted to positive measures. $\endgroup$ Jan 22 '17 at 23:13
  • $\begingroup$ Ah, I was thinking that since, ignoring the origin, the space of positive measures is the product of the space of probability measures with the interval $(0,\infty)$, and $\langle-,f\rangle$ scales linearly with respect to the latter, that continuity on probability measures would imply continuity on positive measures. But one could make the same argument for continuity on signed measures of variation $1$ implying continuity on signed measures, so I guess something must be wrong with it. $\endgroup$ Jan 22 '17 at 23:29

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