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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.

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  • $\begingroup$ There have been many questions about limsup & liminf of sets here before. Some appear on the Related list. Have you looked to see whether any answer your questions? $\endgroup$ – Gerry Myerson Apr 19 '20 at 3:08
  • $\begingroup$ @GerryMyerson Yes, I did. Also when typing in the question I checked the suggestions given. Most of the related concepts are about limit superior of sequences or if they are about measure theory, then it is about more advanced concepts such as Lebesgue measure, which I at this point am not familiar with. $\endgroup$ – user672245 Apr 19 '20 at 3:13
  • $\begingroup$ You can use \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
  • $\begingroup$ The word is limit, not limes. Limes are green citrus fruits. Please edit your question. $\endgroup$ – gen-ℤ ready to perish Apr 20 '20 at 18:59
  • $\begingroup$ Read it in Latin and all is good. $\endgroup$ – user672245 Apr 21 '20 at 4:32
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Say for illustration that $A_1,A_2,A_3,\ldots$ are all subsets of $\mathbb{R}^2$ and $x$ is some point in $\mathbb{R}^2$. Mark the position of $x$ and start drawing the sets $A_1,A_2,A_3,\ldots$ on a piece of paper. Then:

  • If $x$ is eventually "trapped" by this sequence of sets, i.e. in the drawing process, you get to some set $A_k$ so that $x$ will be inside $A_k$ and all the sets that follow ($A_{k+1}$, $A_{k+2}$ and so on), then $x$ is in $\liminf_{n \to \infty}A_n$.
  • If $x$ is not necessarily trapped by the sequence, but it never escapes it completely, it is in $\limsup_{n \to \infty}A_n$. In other words, any time you get to a set $A_k$ that does not include $x$, then as you continue drawing $A_{k+1}$, $A_{k+2}$, you eventually get to a set that will include $x$. In other words, the sequence never permanently separates from x.

So $\liminf_{n \to \infty} A_n$ and $\limsup_{n \to \infty} A_n$ are in some sense lower and upper bounds: $\liminf_{n \to \infty} A_n$ contains all points that are eventually trapped by the sequence, and $\limsup_{n \to \infty} A_n$ contains all points that never escape it. Obviously, by this logic, $\liminf_{n \to \infty} A_n \subseteq \limsup_{n \to \infty} A_n$, since points that are eventually trapped by the sequence never escape the sequence.

As for proving $\limsup_{n \to \infty} A_n = \bigcap_{n \geq 1} \bigcup_{m \geq n} A_m$, it's all about interpreting $\bigcap_{n \geq 1} \bigcup_{m \geq n} A_m$ correctly. It may help to think of what it means to not be in this set.

To say that $x$ is in $\bigcap_{n \geq 1} \bigcup_{m \geq n} A_m$ means that for any $n \geq 1$ you choose, $x$ will belong to $\bigcup_{m \geq n} A_m$, which is the same as saying that $x$ is in at least one of the sets $A_n, A_{n+1},A_{n+2},\ldots$. What would it mean for this not to be true? It would mean that you can find some $n \geq 1$ so that $x$ is in none of the sets $A_{n}, A_{n+1}, A_{n+2}, \ldots$. But if $x$ is not in any of the sets $A_{n}, A_{n+1}, A_{n+2}, \ldots$, it cannot be in infinitely many sets in the sequence $A_1,A_2,\ldots$, which means that it cannot be in $\limsup_{n \to \infty} A_n$. You can reverse this argument to see that if $x$ is not in $\limsup_{n \to \infty} A_n$, then it is not in $\bigcap_{n \geq 1} \bigcup_{m \geq n} A_m$. And this will prove that $\limsup_{n \to \infty} = \bigcap_{n \geq 1} \bigcup_{m \geq n} A_m$, as desired.

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  • $\begingroup$ So you proved the contrapositive? $\endgroup$ – user672245 Apr 19 '20 at 5:35
  • $\begingroup$ Yes exactly. This is how I would do this at least. $\endgroup$ – spalein Apr 19 '20 at 5:41
  • $\begingroup$ So as I currently understand it for the limit superior it can happen that the point $x$ is not in finitely many $A_{k}$. So i.e. if we take the sequence of sets $A_{1}, A_{2}, \ldots $ and then the point $x$ is not in $A_{1}$ and $A_{2}$, but is in the remaining sets in the sequence, then it is in the limit superior? I believe my confusion came from mixing the phrase $\textit{for infinitely many}$ $n$ with the phrase $\textit{for all}$ $n$ i.e. intersection. $\endgroup$ – user672245 Apr 19 '20 at 8:33
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    $\begingroup$ I see. Note that there are three phrases here: 1) For all $n$, 2) For all $n$ starting at some $n_0$ (i.e. for all $n \geq n_0$) and 3) For infinitely many $n$. The first means that x would be in $A_1,A_2,\ldots$. The second means that you can find some $n_0$ so that $x$ is in $A_{n_0},A_{n_0+1},\ldots$. For example, this would be true if $x$ is in $A_{10},A_{11},A_{12},\ldots$. The third means that you never run out of sets $x$ belongs to, which is e.g. true if $x$ is in $A_{10},A_{20},A_{30},\ldots$. $\endgroup$ – spalein Apr 19 '20 at 16:24
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    $\begingroup$ The first statement is the strongest, and is the intersection as you say. The second characterizes the limit inferior and the third characterizes the limit superior. So in the example you gave, $x$ would be in the limit inferior (which means it is also in the limit superior, since the limit superior is larger): If $x$ is eventually in all remaining sets, it is definitely going to be in infinitely many sets, but the reverse is not necessarily true (see the two examples I gave above). That's why statement 2) is stronger than statement 3), and liminf is smaller than limsup. $\endgroup$ – spalein Apr 19 '20 at 16:27

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