How does a sequence of sets in a $\sigma$-algebra satisfying $\bigcup_n\bigcap_{k\geq n}E_k=\bigcap_n\bigcup_{k\geq n}E_k$ look like?

Question

Suppose that for some finite measure space $(X,\mathfrak{M},\mu)$, there is a sequence of sets $\{E_k\}$ in $\mathfrak{M}$ such that $$\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty E_k=\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty E_k$$

How does such a sequence of sets look like? Can anyone give an example?

EDIT:

Bonus: does this mean that $\lim_{n\to\infty}\mu(E_n)$ exists?

Answer 1

Note that $x\in\bigcup_{n\ge 1}\bigcap_{k\ge n}E_k$ if and only if there is an $n$ such $x\in E_k$ for every $k\ge n$: $x$ is eventually in the sets $E_k$. On the other hand, $x\in\bigcap_{n\ge 1}\bigcup_{k\ge n}E_k$ if and only if $x$ is in infinitely many of the sets $E_k$: $x$ is frequently in the sets $E_k$. Clearly

$$\bigcup_{n\ge 1}\bigcap_{k\ge n}E_k\subseteq\bigcap_{n\ge 1}\bigcup_{k\ge n}E_k\;:$$

a point that is in every $E_k$ with $k$ sufficiently large is certainly in infinitely many of them.

Thus, you’re looking for sequences such that if $x$ is in infinitely many of the sets $E_k$, then $x$ is in all of the $E_k$ from some point on. The simplest way to ensure that is to make the sequence $\langle E_k:k\ge 1\rangle$ eventually constant. That is, choose it so that there is some $n$ such that $E_k=E_n$ for all $k\ge n$.

Answer 2

The limit does exist. Note

$\mu(\liminf_{n \to \infty} E_n)\le \liminf_{n\to\infty} \mu(E_n) \le \limsup_{n\to\infty} \mu(E_n) \le \mu(\limsup_{n\to\infty} E_n)$.

The problem tells us $\liminf_{n \to \infty} E_n = \limsup_{n\to\infty} E_n$ so $\lim_{n \to \infty} \mu(E_n)$ must exist.