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I dropped the $n\rightarrow\infty$ in the title as it was exceeding the character limit.

In the book I'm currently reading, the author claims that $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ and $\limsup\limits_{n\rightarrow\infty} E_{n_0+n} = \limsup\limits_{n\rightarrow\infty} E_{n}$ where $\mu(E_{n_{0}}) < \infty$ and $E_n$ is a decreasing sequence? ($\mu$ is an arbitrary measure!)


My attempt at a proof is as follows:

Let $x \in \liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n_0+n}$

Then $x \in \bigcap_{k\geq n}E_{n_0+n_{0}'}$ for some $n_{0}' \in \mathbb{N}$

But $\bigcap_{k\geq n}E_{n_0+n_{0}'} \subset \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ and so $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} \subset \liminf\limits_{n\rightarrow\infty} E_{n}$.

I however couldn't prove the reverse inclusion in my attempt below:

Let $x \in \liminf\limits_{n\rightarrow\infty} E_{n} = \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n}$

Then $x \in \bigcap_{k\geq n}E_{n_{0}''}$ for some $n_{0}'' \in \mathbb{N}$

I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $\bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n_0+n}$ from $\bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n}$.

I suspect the proof for $\limsup\limits_{n\rightarrow\infty} E_{n_0+n} = \limsup\limits_{n\rightarrow\infty} E_{n}$ is similar and would also fall apart at the same point as my proof for $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ ($n_{0} \neq n_{0}''$) and so I didn't attempt it.

Perhaps I'm missing something about how $n_0$ was chosen such that $\mu(E_{n_{0}}) <\infty$...


Original Text:

The main part of my question comes from here:

enter image description here

Theorem 1.26:

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And finally, Lemma 1.7:

enter image description here


Hopefully someone can shed some light!

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    $\begingroup$ If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i \ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part? $\endgroup$ – Mike Earnest Jan 4 at 1:01
  • $\begingroup$ Hey @MikeEarnest, I agree that $x\in\liminf\limits_{n\rightarrow\infty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $i\geq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $x\in\liminf\limits_{n\rightarrow\infty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $i\leq n_{0}' \in \mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $\liminf\limits_{n\rightarrow\infty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' \leq j \leq n_{0}$ . $\endgroup$ – Darius Jan 4 at 1:56
  • $\begingroup$ $I_{\lim \sup A_n}=\lim \sup I_{A_n}$ and $I_{\lim \inf A_n}=\lim \inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x \in A$, $0$ otherwise]. $\endgroup$ – Kabo Murphy Jan 4 at 6:20
  • $\begingroup$ @MikeEarnest, I just realized I was being a dummy! Thanks for your help! $\endgroup$ – Darius Jan 5 at 2:40
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Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.

Then $\liminf_{n \to \infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, \ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.

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