Proving Subsets of $\mathbb R$ are Borel Sets Let $(E_n)$ be a sequence of Borel Sets. Assume $A$ and $B$ are subsets of $\mathbb R$ such that $A$ consists all of $x \in \mathbb R$ which belong to infinitely many of the sets $E_n$, and $B$ consists of all $x \in \mathbb R$ which belong to all but a finite number of the sets $E_n$. I want to show that both $A$ and $B$ are also Borel sets. First, I want to somehow prove
$$
A = \limsup_{n\to\infty }E_n = \bigcap_{m=1}^\infty\left(\bigcup_{n=m}^\infty E_n\right)\\B = \liminf_{n\to\infty }E_n = \bigcup_{m=1}^\infty\left(\bigcap_{n=m}^\infty E_n\right),
$$
that is, $A$ is the limit superior of $(E_n)$, and $B$ is the limit inferior of $(E_n)$ but I am not sure how. 
 A: If $x\in A$, then $x$ belongs to each set $\bigcup_{n\geqslant m}E_n$. This means that $x$ belongs infinitely many $E_n$'s and, reciprocally, if $x$ belongs infinitely many $E_n$'s, then $x$ belongs to each $\bigcup_{n\geqslant m}E_n$ and therefore to $A$. So, $A$ is indeed the set of all real numbers $x$ which belong to infinitely many $E_n$'s.
On the other hand, since the Borel sets form a $\sigma$-algebra and each $E_n$ is a Borel set, each set $\bigcup_{n\geqslant m}E_n$ is a borel set too, and therefore their intersection (which is $A$) is also a Borel set.
THe case of $B$ is similar.
A: Suppose that $x$ it's only in a finite number of the $E_n$, that means that there is a $n\in \mathbb{R}$ such that if $m>n \Rightarrow x \not\in E_m$ so $x \not\in \bigcup\limits_{m=n+1}^{\infty} E_m$ so $x \not\in \bigcap\limits_{i=1}^{\infty}\bigcup\limits_{m=i}^{\infty} E_m$. That should give you the intuition for the argument. To prove the rest of the argument, remember that in a $\sigma$-algebra, the union and intersection is well behaved. 
