How to show that $\{ \lim\limits_{k} ~X_k \leq x\} = \bigcap_{\varepsilon \in \mathbb{Q}^+}\bigcup_{K}\bigcap_{k \geq K} \{X_k < x+\varepsilon \}$? In one of my exercises, the solution uses another way to describe the limit.
We have a sequence of random variables $X_1, X_2,...$
$$\{ \lim\limits_{k} ~X_k \leq x\} = \bigcap_{\varepsilon  \in \mathbb{Q}^+}\bigcup_{K}\bigcap_{k \geq K} \{X_k < x+\varepsilon  \}$$
Can anyone explain it to me?
I know that for example: $\liminf\limits_{n \rightarrow \infty} ~X_n = \bigcup_{n \geq 1} \bigcap_{k \geq n} X_k$
So why is $\lim$ defined like this?
 A: Let $A=\{ \lim_nX_n\leqslant x\}$ then $\forall \epsilon>0, \exists n_0 \in \mathbb{N},$   such that $X_n - \lim_n X_n < \epsilon , \forall n \geqslant n_0 \Rightarrow $ $X_n < \lim_nX_n + \epsilon \leqslant x+ \epsilon$.
Thus $\forall \epsilon>0, \exists n_0 \in \mathbb{N},$ such that $X_n<x+ \epsilon , \forall n \geqslant n_0 \Rightarrow x \in \bigcap_{\epsilon>0} \bigcup_{n_0 \in \mathbb{N}} \bigcap_{n \geqslant n_0} \{X_n<x+ \epsilon\}=B$.
Now let $x   \in B= \bigcap_{\epsilon>0} \bigcup_{n_0 \in \mathbb{N}} \bigcap_{n \geqslant n_0} \{X_n<x+ \epsilon\}$, thus  $\forall \epsilon>0, \exists n_0 \in \mathbb{N},$ such that $X_n<x+ \epsilon , \forall n \geqslant n_0 \Rightarrow \limsup X_n \leqslant x \Rightarrow \lim_nX_n \leqslant x$ thus $x \in A$.
The trick here is to interpret the quantifiers $\exists, \forall$ as union and intersection respectively.
Also form your post i assumed that the limit of $X_n$ exists that's why i passed from limsup to the limit of $X_n$
A: Wlog $\epsilon\in \mathbb Q^+.$
Now if we parse $S:=\bigcap_{\varepsilon  \in \mathbb{Q}^+}\bigcup_{K}\bigcap_{k \geq K} \{X_k < x+\epsilon  \}$ we have, step by step:
$1). x\in S$ if and only if $x\in S_K=\bigcup_{K}\bigcap_{k \geq K} \{X_k < x+\epsilon  \} \ \textit {for each}\  \epsilon\in \mathbb Q^+  $   
$2). x\in S_K$ if and only if there is an integer $K$ such that $x\in \bigcap_{k \geq K} \{X_k < x+\epsilon  \} $ and this happens if and only if
$3).\ x\in \textit {each of }\ \{X_K < x+\epsilon  \},\{X_{K+1} < x+\epsilon  \},\cdots .$ 
Combining $1), 2), 3)$ we have $x\in S$ if and only if for each $\epsilon\in \mathbb Q^+  ,$ there is an integer $K$ such that $X_k\le x+\epsilon  $ whenever $k\ge K.$ But this implies that $\lim X_k\le x$ and so $x\in \{ \lim\limits_{k} ~X_k \leq x\}.$
A similar argument shows that $x\in \{ \lim\limits_{k} ~X_k \leq x\}\Rightarrow x\in S$ and therefore that $\{ \lim\limits_{k} ~X_k \leq x\}=S.$
