Suppose I have a space of cumulative distribution functions on $\mathbb{R}$ denoted $\mathcal{F}$. There is a linear functional $A$ that maps to some subset of $\mathbb{R}$ whose domain is $\mathcal{F}$ and $A$ is sup-norm continuous in $\mathcal{F}$ (i.e. if a sequence $\{F_n\}_{n=1} ^\infty$ in $\mathcal{F}$ converges in the sup-norm then $\{A[F_n]\}_{n=1} ^\infty$ converges).
What I want to know is the following. Can I represent $A$ as follows: $A[F]=\lim_{k\to\infty} \int a_k(x) dF(a)$ for some sequence of functions $\{a_k\}_{k=1} ^\infty$?
I suppose what I'm really asking is a question about the dual of $L_\infty$ and whether there is a particular dense subset of this space, but I'm not sure. Any help greatly appreciated.
I think the following proof works, but I'm not sure.
Let $B_b$ be a functional that returns $B_b[F](z)=\frac{1}{b}\int \phi (\frac{z-x}{b}) dF(x)$ for some constant $b$ where $\phi$ is the standard normal density function. Clearly for any $b$ $B_b[F]$ is $L_1$ with the Lebesgue measure and hence by Radon-Nikodym there must be a function $a$ such that $A[B_b[F]]=\int a(x) B_b[F](x) dx$, and so we have:
$A[B_b[F]]=\int a(x) \frac{1}{b}\int \phi (\frac{x-z}{b}) dF(z) dx= \frac{1}{b}\int \int a(x) \phi (\frac{x-z}{b}) dxdF(z)$
And so we see that:
$A[F]=\lim_{b\to 0} \frac{1}{b}\int \int a(x) \phi (\frac{x-z}{b}) dxdF(z)$
I've not been very careful coming up with that so some of the steps might be wrong.