Let the indicator function be defined as $$I(x) \triangleq \begin{cases} 1, & \quad x \geq 0 \\ 0, & \quad x < 0 \end{cases}$$ and $I_{\nu}(x) \in [0,1]$ be a continuous approximation of the $I(x)$ such that $$\lim_{\nu \rightarrow \infty} I_{\nu}(x) = I(x), \quad \forall x$$ $$\lim_{x \rightarrow – \infty} I_{\nu}(x) = 0, \quad \forall \nu$$ $$\lim_{x \rightarrow \infty} I_{\nu}(x) = 1, \quad \forall \nu.$$
Let us define the sequence of connected sets $\{ C_{\nu} \}$, where each $C_{\nu}$ is the set of values assumed by $I_{\nu}(x)$, $\forall x \in \mathbb{R}$ (hence, each element of $C_{\nu}$ belongs to $[0,1]$).
How do I calculate $\liminf_{\nu \rightarrow +\infty} C_{\nu}$ and $\limsup_{\nu \rightarrow +\infty} C_{\nu}$?
My intuition is that $\liminf_{\nu \rightarrow +\infty} C_{\nu} = \limsup_{\nu \rightarrow +\infty} C_{\nu}=\{0,1\}$, but I can't prove it.
Best Answer
Let $I_n(x)$ be $0$ for $x \le -1/n$ and $1$ for $x \ge 1/n$. On $[-/n,0]$ define $I_n(x)$ as $$f_n(x)=n^n(1-1/n)(x+1/n)^n,$$ and on $[0,1/n]$ define it as $$g_n(x)=1-n^{n-1}(1/n-x)^n.$$ One can check that $f_n(0)=g_n(0)=1-1/n$ and that $f(-1/n)=g(1/n)=0,$ so that the rules are not in conflict at interval endpoints and this piecewise function $I_n(x)$ is continuous, so its range is all of $[0,1].$ Furthermore for fixed negative $x$ eventually $I_n(x)=0$ (as soon as $-1/n\ge x$) and similarly for fixed positive $x$ eventually $I_n(x)=1.$ Since at $0$ itself we have $I_n(x)=1-1/n$ we have $\lim_{n \to \infty}I_n(0)=1$ as required.
So this is an example wherein the sequence $C_n$ of sets is the constant sequence each term being the interval $[0,1].$ I wasn't familiar with lim inf /sup of sets but looked it up on Wiki, and for a constant sequence $A,A,\cdots$ each of lim sup and lim inf is just $A$, so that this example shows these limits may be all ofr $[0,1]$, rather than the two point set $\{0,1\}$ mentioned in the post.
Edit: A much simpler $I_n(x)$ sequence: $0$ for $x<-1/n,$ linear from $0$ to $1-1/n$ on $[-1/n,0],$ then linear from $1-1/n$ to $1$ on $[0,1/n],$ and $1$ for $x>1/n$. The above definition was made when I thought one would need the high powers in order to ensure the right properties, but these follow from the shrinking domain $[-1/n,1/n]$ on which $I_n$ is not $0$ or $1$.