Conditions under which a linear functional on a space of measures must be integration of a function 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.
 A: 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.
