The following is a standard theorem on the theory of Nemytskii operators, but whose proof I am finding it hard to grasp. Could someone explain it to me?
Theorem: If $\Omega \subset \Bbb{R}^N$ is an open set and $f: \Omega \times \Bbb{R} \longrightarrow \Bbb{R}$ is Carathéodory, then the map $x \mapsto f(x, u(x))$ is measurable for every measurable $u : \Omega \longrightarrow \Bbb{R}$.
A Carathéodory function is such that (a) $x \mapsto f(x, s)$ is measurable for every fixed $s$ and (b) $s \mapsto f(x, s)$ is continuous for almost every fixed $x$.
The argument of the proof is as follows:
Let $u_n(x)$ be a sequence of simple functions converging a.e. to $u(x)$. Then each $x \mapsto f(x, u_n(x))$ is measurable by (a). Now (b) implies that $f(x, u_n(x)) \to f(x, u(x))$ a.e., which gives measurability.
What I do not understand:
- How (a) implies that $f(x, u_n(x))$ is measurable?
Here, measurability is in the Lebesgue sense.
The text I am following is de Figueiredo's Lectures on the Ekeland Variational Principle with Applications and Detours.
Thanks in advance and kindest regards.
EDIT
I think I got this, and I would very much appreciate any comments with regard to the correctness of the following argument:
For each $n$, we have $$ u_n = \sum_1^{k_n} \alpha_{nj} \chi_{E_{nj}} $$ Hence $$ f(x, u_n(x)) = \sum_1^{k_n} f(x, \alpha_{nj}) \chi_{E_{nj}} $$ which is a sum of measurable functions, and hence is measurable.