I'm working at optimal control and I came across this situation.
We have a set of functions that consists of all the measurable functions with values in $U \subset \mathbb{R}^m$ a.e., where $U$ is compact. Consider a function $$F:\mathbb{R} \times \mathbb{R}^m \to \mathbb{R}^n$$ $$(t,u) \mapsto F(t,u)$$ such that $F$ is continuous in $u$ for each $t$ and such that $F$ is measurable in $t$ for each $u$. Clearly the set of functions defined on a finite interval $I$ $$K = \{u \mid u(t) \in U \; a.e. \; \mathrm{in} \; I \}$$ is closed in $\Vert .\Vert_1$. Here, as usual, by $\Vert .\Vert_p$ of a function $f : \mathbb{R} \to \mathbb{R}^n$ we mean $$\Vert f\Vert_p^p = \int_I \sum_{i=1}^n \mid f^i(t) \mid^p dt$$ where $f^i$ are the coordinate functions of $f$.
$\textbf{The problem is}$ Prove that the set of functions $$H = \{v \mid v(t) = F(t,u(t)) \; u \in K \}$$ is closed too.
We have also the hypotesis that $$\Vert F(t,u) \Vert \leq m(t)$$ with $m$ a function that is integrable on the finite intervals of $\mathbb{R}$.
I tried, but I can't prove it. Does someone have any ideas? Thanks to everyone who wants to help me!
P.S. Note that if $F$ is injective in $u$ a.e. then the statement is clearly true, and note also that if $m=1$ the statement is clearly true either.
$\textbf{Edit}$
Maybe I found something.
Let $u_n$ be the sequence of functions as above. Then all these functions are in $L^2(I)$, thanks to the fact that $U$ is compact. The sequence is also bounded, clearly. $L^2$ is reflexive, so we can use the following well known theorem of functional analysis
$\textbf{Theorem}$ Let $E$ be a reflexive normed space and $x_n$ a bounded sequence. Then there exists a sub-sequence converging weakly.
In our case, we can then extract a subsequence $u_{n_k}$ converging weakly to something, and so it must be pointwise convergent to some function in K, thanks to the closure of $U$.
(Clearly this doesn't work as weakly convergence doesn't implie pointwise convergence.)
$\textbf{Edit}$
If $u$ was a function with real values (not in $\mathbb{R}^m$) we could just take $\limsup u_n(t)$. Is there something similar we can do in general here?
$\textbf{Edit}$
Maybe we could define $f(s) = \lim_{n \to \infty} F(s,u_n(s))$ and then consider the multivalued function $$G(t) = \left\{u \in U \mid f(t) = F(t,u) \right\}$$
Now if we can prove that there exists at least one measurable selection of $G(t)$ we could take that selection to prove the closure of $H$! Now $G(t)$ is closed valued and non empty, so by a known theorem there exists a measurable selection if $G$ is measurable... so can someone prove that $G$ is measurable?
$\textbf{Definition}$ A set valued map $G: \Omega \to \mathbb{R}^n$ is measurable if for every closed set $C \subset \mathbb{R}^n$ the set $$\left\{t \in \Omega \mid G(t) \cap C \neq \emptyset \right\}$$ is measurable.