Given a function $f : \mathbb{R} \to \mathbb{R}$, we can compute its infimum over $A$ for all the Borel measurable $A \subset \mathbb{R}$.
I am wondering when we can deduce in the other direction, i.e. given a functional $g : \mathcal{B} \to \mathbb{R}\cup\{-\infty\}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra of $\mathbb{R}$, when can we say there exsits a function $f : \mathbb{R} \to \mathbb{R}$ such that $g(A) = \inf_{x\in A}f(x)$?
Obvious conditions are that $$\text{if }A \subset B, \text{then }g(A) \geq g(B)$$ and that $$g(A\cup B) = \min\{g(A), g(B)\}$$.
But are these conditions sufficient for existence of $f$? How to complete in order to guarantee existence?
If considering all measurable sets are too complicated, how about the case where only $\{g([a,b]), a< b\}$ is given, how to find the function $f$ such that $g([a,b]) = \inf_{x\in[a,b]}f(x)$?
It's a bit vague, if any modification can lead to interesting conclusion, please feel free to modify