Consider a function $f: X \rightarrow (0, \infty)$ whose domain $X$ is a standard Borel space. Suppose $f$ is upper semi-analytic, i.e. for every $\lambda \geq 0$ the set $\{x \in X : f(x) > \lambda\}$ is analytic.
Question: Is it possible to construct a non-negative real-valued function $g$ on $X$ such that $f+g$ is Borel-measurable?
Thoughts: Let $\mathbb{Q}$ denote the set of all rational numbers. Since $x \in X$ satisfies $f(x) + g(x) > \lambda$ if and only if there exists a $q \in \mathbb{Q}$ satisfying $f(x) > q > \lambda - g(x)$, we can write
$$ \{x \in X : f(x) + g(x) > \lambda\} = \bigcup_{q \in \mathbb{Q}}\{x \in X : f(x) > q\}\cap\{x \in X : g(x) > \lambda - q\}. $$
So, $f+g$ would be Borel-measurable if the sets $\{x \in X : f(x) > q\} \cap \{x \in X : g(x) > \lambda - q\}$ are all Borel. Note that, since $f$ is upper semi-analytic, the sets $\{x \in X : f(x) > q\}$ are analytic.