Let $g:\mathbb R\to\mathbb R$, $g\in V$
Define functional $F:V \times\mathbb R\to\mathbb R$ $$(g,x)\mapsto F(g,x)$$
Since $\forall \text{operator } T\ \exists\mu$ s.t. $ T(f)=\int_{\mathbb R} f \, d\mu$, where $\mu$ is a complex Borel measure,
Is it possible to prove that $\forall F\ \exists\mu$ s.t. $ F(g,x)=\int_{-\infty}^x g \, d\mu$? or any similar forms.
On what kind of text I could learn more no this topic?
After some reading I find a possible way of doing this:
For any operator $F$ there exists another operator $T_x:V\to V$ s.t. $F(g,x)=T_x(g)$ then, if $T_x$ is continuous and linear, $$T_x(g)=\int_\mathbb R gd\mu$$ where $\mu:\mathbb R\to\mathbb R$ is a Bounded variation (BV) function.
I am not sure that, in this case, whether $\mu$ has to be a complex Borel measure or just BV function is enough?
"Functional Analysis (Methods of Modern Mathematical Physics) by Reed & Simon" is your book. I've actually readed it a couple of days ago, it's in the chapter IV, sections 4 and 5. The theorem that you are talking sounds like the Riesz-Markov theorem and it works perfectly when your $V$ is in fact $C(X)$ with $X$ a compact set.