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:
Theorem 1.26:
And finally, Lemma 1.7:
Hopefully someone can shed some light!
Best Answer
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.