Let me start with a simple example:
Let $f_n:[0,1]\to[-1,1],x\mapsto \sin 2\pi nx$. For each $x\in[0,1]$, consider the sequence $\lbrace f_n(x):n\ge1\rbrace$ and denote by $F(x)$ the set of points of accumulation of $\lbrace f_n(x):n\ge1\rbrace$. This induces a map $F:[0,1]\to\mathcal{K}([-1,1])$, where $\mathcal{K}(X)$ is the set of nonempty compact subsets of $X$.
For example $F(x)=[-1,1]$ if $x$ is irrational, $F(q/p)=\lbrace \sin\frac{2\pi k}{p}:0\le k\le p-1 \rbrace$ if $(q,p)$ is coprime. In particular $F$ is measurable (in fact lower semicontinuous).
As suggested by Nate, $\mathcal{K}(X)$ is equiped by the Hausdorff distance and the induced topology. More precisely, $D(K,L)=\inf\lbrace\delta>0: K\subset B(L,\delta)\text{ and }L\subset B(K,\delta)\rbrace$.
About limsup: for a sequence of compact sets $K_n\subset X$, $\limsup\limits_{n\to\infty} K_n=\bigcap_{N\ge1}\overline{\bigcup_{n\ge N}K_n}$.
Now let $X,Y$ be two compact metric space and $f_n:X\to Y$ be continuous functions. This induces a map $F:X\to \mathcal{K}(Y)$, where $F(x)$ is the set of points of accumulation of $\lbrace f_n(x):n\ge1\rbrace$. We also denote $F=\limsup f_n$.
- What can we say about such $F$? Is it still measurable?
Assume that there exists a Borel subset $X_0\subset X$ on which the limit $f_n(x)$ exists, say $f(x)$. Then we can land on earth and define $f:X_0\to Y$. In this case we can ask
- Is $f$, defined on $X_0$, measurable?
Thank you!
Let's put this in a more general frame:
Let $X,Y$ be compact metric space and $F_n:X\to \mathcal{K}(Y)$ be a sequence of continuous maps (w.r.t. the topology induced by Hausdorff distance). Let $F(x)=\limsup\limits_{n\to\infty} F_n(x)$ be defined as above (2). Then $F(x)$ is well defined for every $x\in X$ and hence we get a map $F:X\to \mathcal{K}(Y)$.
- Is $F$ a Borel measurable map with respect to the induced topology?
I am trying to mimic Leonid's approach and characterize $F^{-1}\mathcal{U}$ for open sets $\mathcal{U}\subset\mathcal{K}(Y)$.
$F(x)\in \mathcal{U}$ iff $D(F(x),\mathcal{U}^c)>1/k$ for some $k\ge1$.
And $D(F(x),\mathcal{U}^c)>1/k$ iff $\exists N\ge1$ such that $D(\overline{\bigcup_{n\ge N}F_n(x)},\mathcal{U}^c)>1/k$ for all $n\ge N$.
Then I am stuck..