The following is a problem from a text I (in portuguese) I am reading about Critical Point Theory:
Let $\Omega \subset \Bbb{R}^N$ be an open set, $f: \overline \Omega \times \Bbb{R} \longrightarrow \Bbb{R}$ be a Carathéodory function and $F(x, t) = \int_0^t f(x, s) \ ds$. Show that for every $u : \Omega \longrightarrow \Bbb{R}$ measurable the map $F(\cdot, u(\cdot))$ is measurable.
Recall that $f$ is a Carathéodory function if (a) $x \mapsto f(x, s)$ is measurable for every $s$ and (b)$s \mapsto f(x, s)$ is continuous for almost every $x \in \Omega$.
I proceeded as follows:
If $u$ is measurable there is a sequence of simple functions $(u_n)$ such that $u_n(x) \to u(x)$ for every $x \in \Omega$. Write each $u_n$ as follows: $$ u_n = \sum_1^{k_n} z_{ni} \chi_{E_{ni}}. $$ Then $$ F(x, u_n(x)) = \int_0^{\sum_1^{k_n} z_{ni} \chi_{E_{ni}}(x)} f(x, s) \ ds = \int_0^{z_{nj}} f(x, s) \ ds \\ = \sum_1^{k_n} \left( \int_0^{z_{ni}} f(x, s) \ ds \right) \chi_{E_{ni}}(x) = \sum_i^{k_n} F(x, z_{ni}) \chi_{E_{ni}}(x) $$
Now, if each $F(x, z_{ni})$ is measurable then we have a sequence of measurable functions converging pointwise to $F(\cdot, u(\cdot))$, and we are done.
My question is:
Is the above argument correct? If yes, how to show that $F(x, z_{ni})$ is measurable? If not, how does one solve this exercise?
Thanks in advance and kind regards.
EDIT I have made some progress with a hint by Gláucio Terra:
By Tonelli's Theorem, we only have to show that $f$ is measurable with respect to the product $\sigma$-algebra $\mathcal{L} \otimes \mathcal{B}$, where $\mathcal{L}$ is the Lebesgue $\sigma$-algebra in $\Bbb{R}^N$ and $\mathcal{B}$ is the Borel $\sigma$-algebra in $\Bbb{R}$. We also assume the Lebesgue $\sigma$-algebra on the image $\Bbb{R}$. Then $f^+$ and $f^-$ are measurable. By Tonelli's Theorem, the maps $\int_0^{z_{ni}}f^+(x, s) \ ds$ and $\int_0^{z_{ni}}f^-(x, s) \ ds$ are measurable and so also is $$ F(x, z_{ni}) =\int_0^{z_{ni}}f^+(x, s)- f^-(x, s) \ ds, $$
To show that $f$ is measurable, Gláucio suggested addapting Exercise 2.11 in Folland's Real Analysis. I tried as follows:
For $n \in \Bbb{N}$, define $f_n$ as follows. Given $i \in \Bbb{Z}$, let $a_i^n = i/n$ and define $$ f_n(x, s) = \sum_{i = - \infty}^{\infty} \frac{f(x, a_{i+1}^n) (s - a_i^n) - f(x, a_i^n)(s - a_{i + 1}^n)}{a_{i+1}^n - a_i^n} \chi_{[a_i^n, a_{i + 1}^n]}(s). $$ Note that, for all $i, n$, $$ (x, s) \mapsto x \mapsto f(x, a_{i}^n) $$ is a composition of $\mathcal{L} \times \mathcal{B}$-measurable maps and hence is measurable, and the map $(x, s) \mapsto (s - a_i^n)$ is continuous and hence $\mathcal{L} \otimes \mathcal{B}$-measurable. Hence $f_n$ is measurable, and $f_n \to f$ almost everywhere.
Now, to say that $f$ is measurable we need conevrgence everywhere. How to overcome this difficulty?
To show that $F(x, z_{n_{i}})$ is measurable: as $x\mapsto f(x,s)$ is measurable, we can write $$f(x,s)=\sum\limits^{\infty}_{k=1}r_{k}\cdot \chi_{A_{k}}(x)$$ for some $r_{k}\geq 0$ and $A_{k}$ measurable sets. So we have
$$F(x,z_{n_{i}})=\int\limits^{z_{n_{i}}}_{0}f(x,s)ds=\sum\limits^{\infty}_{k=1}r_{k}\cdot \chi_{A_{k}}(x)\int\limits^{z_{n_{i}}}_{0}ds=\sum\limits^{\infty}_{k=1}(r_{k}z_{n_{i}})\cdot \chi_{A_{k}}(x). $$ Then $F(x,z_{n_{i}})$ is measurable, as we wanted to demonstrate.