Let $X$ be a metric space, $\nu,\mu$ be Borel measures on $X$, $f:X\times \mathbb{R}\rightarrow [0,\infty)$ be a measurable function. Under what conditions is the integral functional $$ \begin{aligned} F_f: L^1(X) & \rightarrow [0,\infty] \\ g&\mapsto \int_{x \in X} f(x,g(x))d\mu(x) \end{aligned} $$ continuous? I'm particularly interested in the case where $X$ is $\mathbb{R}^n$ or is a topological manifold modelled thereon.

(Does the cituatoin simplify when $F_f$ is instead considered on $C(X)$ with the uniform topology and $X$ were a compact space?)

I expect that $f$ should at-least be Carath'{e}odory; i.e.: measurable in its first argument and continuous in its second, and probably some sort of growth condition such as $\int_{x\in X} \sup_{y \in \mathbb{R}}f(x,y)d\mu(x)<\infty$ to ensure that $F_f$ is finite-valued...

Let $X$ be a metric space, $\nu,\mu$ be Borel measures on $X$, $f:X\times \mathbb{R}\rightarrow [0,\infty)$ be a measurable function. Under what conditions is the integral functional $F_f$, defined by: $$ \begin{aligned} F_f: L^1(X) & \rightarrow [0,\infty] \\ g&\mapsto \int_{x \in X} f(x,g(x))d\mu(x) \end{aligned} $$ continuous?

I expect that $f$ should at-least be Carath'{e}odory; i.e.: measurable in its first argument and continuous in its second, and probably some sort of growth condition such as $\int_{x\in X} \sup_{y \in \mathbb{R}}f(x,y)d\mu(x)<\infty$ to ensure that $F_f$ is finite-valued...

I'm particularly interested in the case where $X$ is $\mathbb{R}^n$ or is a topological manifold modelled thereon.

(Does the situation simplify when $F_f$ is instead considered on $C(X)$ with the uniform topology and $X$ were a compact space?)