set of points in $E_n$ infinitely often and almost always are in $\sigma$-algebra

Question

Let $\Omega$ be a $\sigma$-algebra on a set $X$, let $E_n \in \Omega, n \in \mathbb{N}$. Prove that

1. $A=$ {$x:x \in E_n$ infinitely often} $\in\Omega$
2. $B=$ {$x:x \in E_n$ almost always} $\in\Omega$

My attempt is

$B= ( \,\bigcup\limits_{i=1}^{\infty} $ $\bigcup\limits_{k \in \mathbb{N}-\{i\}} E_{k}) \,\cup (\,\bigcup\limits_{i, j=1}^{\infty} $ $\bigcup\limits_{k \in \mathbb{N}-\{i, j\}} E_{k}) \,\cup ...\cup ( \,\bigcup\limits_{i_1, i_2, ...im}^{\infty} $ $\bigcup\limits_{k \in \mathbb{N}-\{i\}} E_{k}) \, \cup ...$

where points in the first union appear everywhere except one of $E_n$, points in the second union appear everywhere except two of $E_n$, and so on. Hence $ B$ is a countable union of $E_n$ and it is in $\Omega$.

To get $A$, we first consider $C=\{x:x \in E_n \text{ finitely often}\}$. Let's show $C \in \Omega$.
$C=(\bigcup\limits_{i=1}^{\infty}(E_{i}-\bigcup\limits_{k \in \mathbb{N}-\{i\}}E _k))$ $\cup$ $(\bigcup\limits_{i,j=1}^{\infty}(E_{i}\cap E_j-\bigcup\limits_{k \in \mathbb{N}-\{i,j\}}E _k)) \, \cup ...$ $ \cup (\bigcup\limits_{i_1, i_2, ..., i_m}^{\infty}(\bigcap\limits_{1 \leq l \leq m}E_{i_{l}} -\bigcup\limits_{k \in \mathbb{N}-\{i_1, i_2, ..., i_m\}}E _{k}))\, \cup ...$
where points in the first union appear in exactly one of $E_n$ and points in the second union appear in exactly two of $E_n$, and so on. Hene $C$ is a countable union and intersection of $E_n$ and it is in $\Omega$.

Now $\bigcup\limits_{n}E_n=A \cup C$ is disjoin union. Therefore $A=(\bigcup\limits_{n}E_n) -C \in \Omega$

Any thought on this?

Answer 1

The set $A$ is often denoted as $\limsup E_n$. Note that $$ A=\limsup E_n=\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty E_n. $$ Further $x\in A$ iff for every $k$, there exists $n\ge k$ such that $x\in E_n$. This is precisely what it means for $x\in E_n$ infinitely often.

Next the set $B$ is often denoted as $\liminf E_n$. Note that $$ B=\liminf E_n=\bigcup_{k=1}^\infty\bigcap_{n=k}^\infty E_n. $$ Further $x\in B$ iff there exists $k$ such that for all $n\geq k$, $x\in E_n$. This is precisely what it means for $x\in E_n$ almost always (sometimes people call this all but finitely often).