Indicator function and liminf and limsup

Can anyone please explain why the following is true? And what is the intuition behind it?

$$\chi_A(x) = \begin{cases}1 &, x \in A\\ 0 &, x \notin A.\end{cases}$$

Then we have

$$\chi_{\liminf A_n}(x) = \liminf \chi_{A_n}(x); \quad \chi_{\limsup A_n}(x) = \limsup \chi_{A_n}(x)$$

for all $x\in X$.

Thank you.

Solution 1:

Fix a sequence $(A_n \mid n \in \mathbb N)$ of sets.

By definition $\liminf A_n = \bigcup_{m \in \mathbb N} \bigcap_{l \ge m} A_l$. In other words: $x \in \liminf A_n$ if and only if there is an $m \in \mathbb N$ such that $x \in A_l$ for all $l \ge m$. Thus $\chi_{\liminf_{n \in \mathbb N} A_n}(x) = 1$ if and only if there is some $m \in \mathbb N$ such that for all $l \ge m$ we have that $x \in A_l$.

Now consider a sequence $(x_n \mid n \in \mathbb N)$ of real numbers $x_n$. By definition $\liminf_{n \in \mathbb N} x_n = \sup_{m \in \mathbb N} \inf_{l \ge m} x_l$. We are interested in the case where $x_n = \chi_{A_n}(x)$ for some fixed $x$. Then $\liminf_{n \in \mathbb N} \chi_{A_n}(x) = 1$ if and only if there is some $m \in \mathbb N$ such that for all $l \ge m$ we have that $\inf_{l \ge m} \chi_{A_n}(x) = 1$. This in turn holds if and only if there is some $m \in \mathbb N$ such that for all $l \ge m$ we have that $x \in A_l$. Comparing these two yields $\chi_{\liminf_{n \in \mathbb N} A_n} (x) = 1$ if and only if $\liminf_{n \in \mathbb N} \chi_{A_n}(x) = 1$.

This also yields $\chi_{\liminf_{n \in \mathbb N} A_n} (x) = 0$ if and only if $\liminf_{n \in \mathbb N} \chi_{A_n}(x) = 0$ and consequently $\chi_{\liminf_{n \in \mathbb N} A_n} (x) = \liminf_{n \in \mathbb N} \chi_{A_n}(x)$. The argument for $\limsup$ is analogous.

Regarding the intuition: $\liminf_{n \in \mathbb N} A_n$ contains precisely those element of $\bigcup_{n \in \mathbb N} A_n$ which are members of all but finitely many $A_n$'s. For example $x$ may be in some but not all of the sets $A_1, \ldots, A_{100}$. But for $l > 100$ we always have that $x \in A_l$. Such an $x$ appears in all the $A_n$'s except for finitely many (namely some of the sets $A_1, \ldots, A_{100}$) and thus $x \in \liminf_{n \in \mathbb N} A_n$.

In contrast, $\limsup_{n \in \mathbb N} A_n$ contains precisely those elements of $\bigcup_{n \in \mathbb N} A_n$ which are members of infinitely many of the $A_n$'s. It may be the case that such a member also misses from infinitely many of the $A_n$'s and therefore $\liminf_{n \in \mathbb N} A_n$ may be different from $\limsup_{n \in \mathbb N} A_n$. But we always have that $\liminf_{n \in \mathbb N} A_n \subseteq \limsup_{n \in \mathbb N} A_n$.