I am currently following Jürgen Elstrodt "Maß- und Integrationstheorie", 7th edition. On p. 9 he defines the limes superior $\overline{\lim}_{n \rightarrow \infty}$ and limes inferior $\underline{\lim}_{n \rightarrow \infty}$ by
Def. Let $(A_{n})_{n \geq 1}$ be a sequence of subsets of a set $X$. We define the limes superior by
$$\overline{\lim_{n \rightarrow \infty}} A_{n} := \{ x \in X : x \in A_{n} \text{ for infinitely many } n \in \mathbb{N} \setminus \{ 0 \} \},$$
and the limes inferior by
$$\underline{\lim_{n \rightarrow \infty}} A_{n} := \{ x \in X : \text{ there exists } n_{0}(x) \in \mathbb{N} \setminus \{ 0 \} \text{, such that } x \in A_{n} \text{ for all } n \geq n_{0}(x) \}.$$
How should I think of these concepts? Is there some intuition behind it? According to just the words (and what I know about $\sup$, $\inf$, and $\lim$) the limes superior would seem to be something like a limit from above and in particular, the least upper bound limit. And similarly, the limes inferior would then be the greatest lowest bound limit.
Then, the following three relations are presented without proof
$$\overline{\lim_{n \rightarrow \infty}} A_{n} = \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}, $$
$$\underline{\lim_{n \rightarrow \infty}} A_{n} = \bigcup_{n = 1}^{\infty} \bigcap_{k = n}^{\infty} A_{k}, $$
$$\underline{\lim_{n \rightarrow \infty}} A_{n} \subseteq \overline{\lim_{n \rightarrow \infty}} A_{n}.$$
So I tried proving these and I am uncertain whether I am on the right track or even perhaps if I completed the proof:
We prove $\overline{\lim_{n \rightarrow \infty}} A_{n} = \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$ by showing $\overline{\lim_{n \rightarrow \infty}} A_{n} \subseteq \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$ and $\overline{\lim_{n \rightarrow \infty}} A_{n} \supseteq \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$.
We prove first $\overline{\lim_{n \rightarrow \infty}} A_{n} \subseteq \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$. Let $x \in \overline{\lim_{n \rightarrow \infty}} A_{n}$. This means that $x \in X$ such that $x \in A_{n}$ for infinitely many $n \in \mathbb{N} \setminus \{ 0 \}$.
We want to show that $x \in \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$, which is true if for all $n \in \mathbb{N} \setminus \{ 0 \}$ we have $x \in \bigcup_{k = n}^{\infty} A_{k}$, which in turn is true if for all $n \in \mathbb{N} \setminus \{ 0 \}$ there exists $k \geq n$ (or $k = n$? or is it $k \in \{ n, n + 1, n + 2, \ldots \}$?) such that $x \in A_{k}$.
And this is where I am stuck on this part. We have that $x \in X$ such that $x \in A_{n}$ for infinitely many $n \in \mathbb{N} \setminus \{ 0 \}$ and we want to show that for all $n \in \mathbb{N} \setminus \{ 0 \}$ there exists $k \geq n$ (or $k = n$? or is it $k \in \{ n, n + 1, n + 2, \ldots \}$?) such that $x \in A_{k}$. Perhaps these statements are the same and I am just not understanding it (linguistically or logically).
Now, for the other inclusion, i.e. we show $\overline{\lim_{n \rightarrow \infty}} A_{n} \supseteq \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$. Let $x \in \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} A_{k}$. This means that for all $n \in \mathbb{N} \setminus \{ 0 \}$ there exists $k$ (again here I am confused as to $k \geq n$ or $k = n$) such that $x \in A_{k}$.
We want to show that $x \in \overline{\lim_{n \rightarrow \infty}} A_{n}$, which is true if $x \in X$ such that $x \in A_{n}$ for infinitely many $n \in \mathbb{N} \setminus \{ 0 \}$.
So now we have that for all $n \in \mathbb{N} \setminus \{ 0 \}$ there exists $k$ (again here I am confused as to $k \geq n$ or $k = n$) such that $x \in A_{k}$ and we want to show that $x \in \overline{\lim_{n \rightarrow \infty}} A_{n}$, which is true if $x \in X$ such that $x \in A_{n}$ for infinitely many $n \in \mathbb{N} \setminus \{ 0 \}$.
I believe that if I understand how to prove the first relation, then I will be able to do the second one.
\limsup
and\liminf
to product $\limsup$ and $\liminf$. They are easier to parse than the overline and underline, especially when the “$n\to\infty$” appears as in-line instead of under the $\lim$ sign. $\endgroup$ – Arturo Magidin Apr 19 '20 at 3:40