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.