# Limsup of continuous functions between metric spaces

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).

1. As suggested by Nate, $\mathcal{K}(X)$ is equiped by the Hausdorff distance and the induced topology. More precisely, $D(K,L)=\inf\lbrace\delta>0: K\subset B(L,\delta)\text{ and }L\subset B(K,\delta)\rbrace$.

2. 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..

• I suppose that by measurable you mean "Borel measurable", since we must have some relation between measurability and topology. – user31373 Jun 29 '12 at 15:40
• I assume $\mathcal{K}(Y)$ is equipped with the Hausdorff distance? This makes it a compact metric space and so we could ask whether $F$ is Borel. By the way, the notation $F = \limsup f_n$ seems confusing to me. – Nate Eldredge Jun 29 '12 at 17:04
• Nate makes a good point: the notation $\lim\sup$ already has a meaning in the context of set-valued maps, so using it to mean the set of accumulation points is apt to lead to confusion. – user31373 Jun 29 '12 at 17:32
• @NateEldredge and LeonidKovalev: What about the following explanation? We lift each $x\mapsto f_n(x)$ to a map $x\mapsto F_n(x)=\lbrace f_n(x) \rbrace$. Then the $F$ defined above is exactly the limsup of $F_n$. – Pengfei Jun 30 '12 at 2:07
• @Pengfei That's exactly the confusing point, because the set of cluster points is not $\limsup \{f_n(x)\}$. For example, if $f_n(x)=x+1/n$, then $f_n(0)\to 0$, but $\limsup \{f_n(x)\}=\limsup \{1/n\}=\varnothing$, since no point belongs to infinitely many of the sets (singletons) $\{f_n(x)\}$. – user31373 Jun 30 '12 at 15:53

The second part is easy: we are looking at the pointwise limit of $f_n$ restricted to $X_0$. For every open subset $U\subset Y$ we have $$f^{-1}(U)=\bigcup_{k=1}^\infty \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty \{x\in X_0 : \forall n\ge N \ d(f_n(x),U^c)>1/k\}$$ and since the sets under the intersection are Borel, $f^{-1}(U)$ is a Borel set.
• Thank you! I just added the definition of Hausdorff distance. So $F$ is a point-valued map from $X$ to $\mathcal{K}(Y)$ (also a compact metric space). The Borel measurability of $F$ is defined in the usual way. – Pengfei Jun 30 '12 at 2:21