Finding the proof of a named theorem is far easier :). Here is one. I'll copy it here and expand on it, and there is a bit I fear I can't understand yet. Note that I will denote $F^{-1}(A)=\{x:F(x)\cap A\neq\varnothing\}$ in what follows, hence weak measurability is the condition that $F^{-1}(V)$ is measurable for each open set $V$, which is formally identical to measurability of maps.
Proof:
For any weakly measurable $G$ and any open cover $\mathcal{U}=(U_n)_{n\in\mathbb{N}}$ of $Y$ we define $G_{\mathcal{U}}:X\to2^Y$ another closed-valued weakly measurable map by:
$$G_{\mathcal{U}}(x)=\overline{G(x)\cap U_n} \qquad \text{where: }x\in G^{-1}(U_n)\smallsetminus\bigcup_{\nu=1}^{n-1}G^{-1}(U_\nu).$$
Let us immediately check that this map is well-defined. Then the reference lists three properties, including measurability.
- For all $x\in X$, there exists $n:x\in G^{-1}(U_n)$. Indeed, there is a $y\in G(x)$, and the $U_n$'s cover $Y$, so $y$ belongs to at least one $U_n$, hence $x\in G^{-1}(U_n)$;
- There is at least one $n$ such that $x\in G^{-1}(U_n)\smallsetminus\bigcup_{\nu=1}^{-1}G^{-1}(U_\nu)$; that is the case since, for all $x$, there exists $\overline n(x)=\min\{n\in\mathbb{N}:x\in G^{-1}(U_n)\}$ such that $x\in G^{-1}(U_{\overline n(x)})$ and is in none of the $G^{-1}(U_\nu)$ with $\nu\leq\overline n(x)$;
- The $n$ in 2. is unique; indeed, if $n\neq m$, $G^{-1}(U_n)\smallsetminus\bigcup_{\nu=1}^{n-1}$ is disjoint from $G^{-1}(U_m)\smallsetminus\bigcup_{\nu=1}^{m-1}G^{-1}(U_m)$, since, assuming $n<m$, the former is included in $G^{-1}(U_n)$ which is included in the union removed from the latter.
Hence, $G_{\mathcal{U}}$ is well-defined, since, using the notation in 2., we can term it alternately as:
$$G_{\mathcal{U}}(x)=\overline{G(x)\cap U_{\overline n(x)}}.$$
Then $G_{\mathcal{U}}$ has the following properties:
$$\sup_{x\in X}\operatorname{diam}G_{\mathcal{U}}(x)\leq\sup_{n\in\mathbb{N}}\operatorname{diam}(U_n). \qquad\qquad (i)$$
For all $x\in X$, $G(x)$ is included in $U_{\overline n(x)}$, hence:
$$\operatorname{diam}G(x)\leq\operatorname{diam}(U_{\overline n(x)})\leq\sup_n\operatorname{diam}(U_n),$$
and (i) follows.
$$\forall x\in X,\,G_{\mathcal{U}}(x)\subseteq G(x). \qquad\qquad (ii)$$
$G_{\mathcal{U}}(x)\subseteq\overline{G(x)\cap U_{\overline n(x)}}$. Let $y\in G_{\mathcal{U}}(x)$. Then we have $y_n\to y$ with $\{y_n\}\subseteq G(x)\cap U_{\overline n(x)}\subseteq G(x)$, but $G(x)$ is closed, hence $y\in G(x)$, proving (ii).
$$\forall\text{ open }V\subseteq Y,\,G^{-1}_{\mathcal{U}}(V)=\bigcup_{n\in\mathbb{N}}\left[G^{-1}(V\cap U_n)\smallsetminus\bigcup_{\nu=1}^{n-1}G^{-1}(U_\nu)\right]\text{, which is measurable.} \qquad\qquad (iii)$$
$\supseteq$ Let $y$ be in the right-hand side. Hence, $y\in G^{-1}(V\cap U_{\overline n})\smallsetminus\bigcup_1^{\overline n-1}G^{-1}(U_\nu)$ for some $n$. This set is included in the set appearing in the definition of $G_{\mathcal{U}}$, hence $G_{\mathcal{U}}(x)=\overline{G(x)\cap U_{\overline n}}$, aka $\overline n=\overline n(x)$ with $\overline n(x)$ as in 2. above. In particular, $G_{\mathcal{U}}(x)$ intersects $U_{\overline n}$. Now we need to show it actually intersects $V$. We know $G(x)$ intersects $V$. Actually, we know $G(x)\cap V\cap U_{\overline n}\neq\varnothing$, so in particular $G_{\mathcal{U}}(x)\cap V=\overline{G(x)\cap U_{\overline n}}\cap V\neq\varnothing$, proving $\supseteq$.
$\subseteq$ If $x\in G_{\mathcal{U}}^{-1}(V)$, then $G_{\mathcal{U}}(x)=\overline{G(x)\cap U_{\overline n(x)}}$ intersects $V$. Take $y$ in that intersection. $y$ is the limit point of a sequence $\{y_m\}\subseteq G(x)\cap U_{\overline n(x)}$. But then $y_m$ eventually lies in $V$, hence $V$ intersects $G(x)\cap U_{\overline n(x)}$, hence $x\in G^{-1}(V\cap U_{\overline n(x)})$. The above assumed $y$ was not an isolated point, but if it is, then it is in $G(x)\cap U_{\overline n(x)}$ and there is no need to call upon a sequence $y_m$. By definition of $\overline n(x)$, $x\in X\smallsetminus\bigcup_1^{\overline n(x)-1}G^{-1}(U_\nu)$, hence $x$ is in the RHS of (iii), proving $\subseteq$ and (iii).
So open sets have measurable preimages, since the preimages are countable uninos of measurable sets because $V\cap U_n$ and $U_\nu$ are open, hence by weak measurability their $G^{-1}$'s are measurable. What becomes of $G_{\mathcal{U}}^{-1}(A)$ for $A$ measurable? If $G_{\mathcal{U}}$ were a function, we know there are properties of preimage that make it enough to have preimages of generators of a $\sigma$-algebra measurable to have the map measurable. If the same properties hold for set functions like $G_{\mathcal{U}}$, then it is weakly measurable, since the preimages of open sets are measurable by the above and the open sets generate the Borel $\sigma$-algebra which is the one we are considering for measurability. So we need to prove that:
\begin{gather*}
G^{-1}\left(\bigcup_{i=1}^\infty A_i\right)=\bigcup_{i=1}^infty G^{-1}(A_i), \\
G^{-1}\left(\bigcap_{i=1}^\infty A_i\right)=\bigcap_{i=1}^\infty G^{-1}(A_i), \\
G^{-1}(Y\smallsetminus A)=X\smallsetminus G^{-1}(A),
\end{gather*}
where $G$ is now any set function, perhaps closed-valued, and $G_{\mathcal{U}}$ from above will be weakly measurable.
- If $G(x)$ intersects a union, then it intersects at least one of the $A_i$, hence $x\in G^{-1}(A_i)$, hence $x\in\bigcup G^{-1}(A_i)$; if $x$ is in that union, then it is in one of the $A_i$, meaning $G(x)$ intersects $A_i$, hence it intersects the union;
- If $G(x)$ intersects an intersection, it must intersect all the $A_i$, hence $x\in\bigcap G^{-1}(A_i)$; I leave the vice versa to the reader, who should simply imitate the vice versa of 1.;
- If $G(x)$ intersects a complement… hm, I'm afraid this doesn't hold, because $G(x)$ may intersect both $A$ and its complement; all I need, though, is that $G^{-1}(Y\smallsetminus A)$ be measurable; $X\smallsetminus G^{-1}(A)\subseteq G^{-1}(Y\smallsetminus A)$, so if I prove $G^{-1}(A^C)\smallsetminus(G^{-1}(A))^C$ ($^C$ for complementation in their supersets), then I am done; but how do I prove this? The problem is $\{x:G(x)\cap A\neq\varnothing,G(x)\cap A^C\neq\varnothing\}$; is it measurable? I seem to be unable to prove it.
Now, for every $n\in\mathbb{N}$, we choose a countable open cover $\mathcal{U}_n$ of $Y$ with $\operatorname{diam}(U)\leq\frac1n$ for all $U\in\mathcal{U}_n$,
this is possible since $Y$ is separable, so every open cover admits a countable subcover (cfr here), hence there are countably many balls of radius $\frac1n$ that cover $Y$, for any $n$,
and define:
$$F_1:=F_{\mathcal{U}_1},\quad F_{n+1}:=(F_n)_{\mathcal{U}_{n+1}}.$$
By definition, $F_i(x)$ is closed for all $i$ and $x$. By the above, the $F_i$ are weakly measurable. By definition again, $F_{n+1}(x)=\overline{F_n(x)\cap U_{n,m}}\subseteq\overline{F_n(x)}=F_n(x)$ for all $x$, where $m$ is what I called $\overline n(x)$ in 2., but I called it $m$ to avoid the confusion with $n$, and $U_{n,m}$ is the $m$th element of $\mathcal{U}_n$. Hence, $F_n(x)$ is a decreasing sequence of closed sets for all $x$. Also, by (i), $\operatorname{diam}F_n(x)\leq\frac1n$, meaning those diameters tend to zero for fixed $x$ and $n\to\infty$.
Since $Y$ is complete and, for each $x\in X$, $(F_n(x))_{n\in\mathbb{N}}$ is a decreasing sequence of closed sets whose diameter tends to zero, we know that $\bigcap_{n\in\mathbb{N}}F_n(x)$ is a singleton.
That intersection, being contained in each $F_n(x)$, must have diameter zero, and hence be either a singleton or empty. For all $n$, $F_n(x)$ is nonempty, so we have $y_n\in F_n(x)$. Naturally, since the sequence $F_n(x)$ decreases, if $n\leq m$, $y_n,y_m\in F_n(x)$. Hence:
$$d(y_n,y_m)\leq\operatorname{diam}(U_{\min\{n,m\}})\leq\frac{1}{\min\{n,m\}}\xrightarrow{n,m\to\infty}0.$$
Hence, $y_n$ is Cauchy in $Y$, which is complete, so $y_n\to y$ for some $y\in Y$. Now $y$ is the limit of a sequence which is eventually in each $F_n(x)$, and each $F_n(x)$ is closed, hence $y$ must belong to each $F_n(x)$, proving the intersection of the $F_n(x)$ is a singleton.
The map $f:X\to Y$ defined by:
$$f(x)\in\bigcap_{n\in\mathbb{N}}F_n(x)$$
is obviously a selection for $F$.
Indeed, that big intersection is contained, e.g., in $F_1(x)$, which is contained in $F(x)$ (yeah, in $\overline{F(x)\cap U_{1,n}}$ for some $n$, which is contained etc as I did above in a similar situation).
To conclude, we must prove $f$ is measurable w.r.t. the $\sigma$-algebra $\mathfrak{A}$ on $X$ and the Borel $\sigma$-algebra on $Y$ $\mathcal{B}(Y)($.
To prove that $f$ is $\mathfrak{A}$-$\mathcal{B}(Y)$-measurable, let an open $V\subseteq Y$ be given. We set $V_n:=\{y\in Y:d(y,V^C)>\frac1n\}$. Then $V_n$ is open,
indeed, $V_n^C=\{y:d(y,V^C)\leq\frac1n$, suppose $y_m\to y$ and $\{y_m\}\subseteq V_n^C$, then $d(y,V^C)=\lim_{m\to\infty}d(y_m,V^c)\leq\frac1n$, so $V_n^C$ is closed, so $V_n$ is open,
and we will show:
$$f^{-1}(V)=\bigcup_{n\in\mathbb{N}}F_n^{-1}(V_n),$$
hence $f^{-1}(V)\in\mathfrak{A}$.
$f$ is now a map, so we know that if the preimages of open sets are measurable, as we now are setting out to prove, then the preimages of all Borel sets are measurable, hence $f$ is measurable as desired.
"$\subset$" Let $x\in F^{-1}(V)$ be arbitrary. Then there exists an $n\in\mathbb{N}$ with $d(f(x),V^C)>\frac1n$,
that is what I cannot seem to manage proving: why should it be? I mean, suppose it weren't true. Then $d(f(x),V^C)=0$, and since $V^C$ is closed $f(x)\in V^C$. Then what? Well, $F(x)$ intersects $V$, but nothing seems to imply any of the $F_n(x)$ do…
i.e. $f(x)\in V_n$. Since $f(x)\in F_n(x)$, we deduce $F_n(x)\cap V_n\neq\varnothing$, i.e. $x\in F_n^{-1}(V_n)$. Thus we have $\subset$.
"$\supset$" For $x\in F_n^{-1}(V_n)$ we have $F_n(x)\cap V_n\neq\varnothing$. Since $\operatorname{diam}(F_n(x))\leq\frac1n$, the definition of $V_n$ implies $F_n(x)\cap V^C=\varnothing$,
if there were $y\in F_n(x)\cap V^C$, $d(y,f(x))\leq\frac1n$ because $y\in F_n(x)$ which has diameter less than $\frac1n$, yet $d(y,f(x))>\frac1n$ by definition of $V_n$, contradiction,
hence $f(x)\in F_n(x)\subset V$, which proves $\bigcup_{n\in\mathbb{N}}F_n(x)\subset f^{-1}(V)$.
"which proves" that $x\in f^{-1}(V)$, but $x$ was arbitrary in $F_n^{-1}(V_n)$, hence $\bigcup F_n^{-1}(V_n)\subseteq f^{-1}(V)$.
So modulo getting the missing bits proved, we have the theorem proved.
Now we only need to prove USC implies weakly measurable. I will try after dinner, or leave a comment saying I was unable to conclude.
Update
- The complement thing is useless, since weak measurability only concerns preimages of open sets, which we dealt with already;
- Just as I feared, I got stuck on proving u.s.c. implies w.m.; could anyone help me?
Update 2
I just realised the problematic passage is problematic only due to a typo. Specifically, the passage reads:
"$\subset$" Let $x\in F^{-1}(V)$ be arbitrary. Then there exists an $n\in\mathbb{N}$ with $d(f(x),V^C)>\frac1n$,
and that $F$ should be lowercase. $x\in f^{-1}(V)$ means $f(x)\in V$, but $V$ is open, hence $f(x)\in V$ implies $d(f(x),V^C)>0$, since $V^C$ is closed and $d(f(x),V^c)=0$ would imply $f(x)\in V^C$ which is not the case, and so there must be an $n$ as said above. Thus, the proof no longer has any trouble.
It remains to prove that a USC map is weakly measurable. Hopefully I'll get an answer to that question.
