Meaning of "with probability approaching $1$ as $n\rightarrow \infty$" I have a question on the meaning of this claim:
Consider a sequence of real valued random variables $\{X_n\}_{n\in \mathbb{N}}$ and some constants $a,b$. Then we have
$$
\sqrt{n}a\leq X_{n}\leq \sqrt{n}b \text{ }\text{ with probability approaching $1$ as $n\rightarrow \infty$}
$$

Specifically, I am confused on the meaning of "with probability approaching $1$ as $n\rightarrow \infty$" and an help to clarify it would be extremely helpful.

I know that writing $X_n$ converges to $X$ with probability approaching $1$ as $n\rightarrow \infty$ is equivalent to $X_n\rightarrow_{a.s.}X$, i.e., 
$$
\mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} X_n(\omega)=X(\omega)\})=1
$$
but the claim above contains inequalities. Do they mean
$$
\mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}b]\leq 0\})=1
$$
and
$$
\mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}a]\geq 0\})=1
$$
?
 A: It means that
$$
\lim _{n\to\infty} P(\{\omega\mid \sqrt{n}a\leq X_n(w)\leq \sqrt{n} b\})=1
$$
A: No. It's this one: $$
\lim _{n\to\infty} P(C_n)=1
$$
where
$$C_n := \{ \sqrt{n}a\leq X_n(w)\leq \sqrt{n} b \}$$
$$= \{ \sqrt{n}a\leq X_n(w)\} \cap \{X_n(w)\leq \sqrt{n} b \}$$
$$:= A_n \cap B_n$$
Thus, $$C_n \subseteq A_n, C_n \subseteq B_n$$
$$\to P(C_n) \le P(A_n), P(C_n) \le P(B_n)$$
$$\to \lim_n P(C_n) \le \lim_n P(A_n), \lim_n P(C_n) \le \lim_n P(B_n)$$
Thus when you say

$$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow
 \infty} [X_n(\omega)-\sqrt{n}b]\leq 0\})=1 $$ and $$
 \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow
 \infty} [X_n(\omega)-\sqrt{n}a]\geq 0\})=1 $$

, you are saying something weaker namely merely that $\lim_n P(A_n) = 1 = \lim_n P(B_n)$. Here, it could be that $\lim_n P(C_n) < 1$.
Intuitively, I guess that there isn't a common way that $n$ approaches $\infty$. Do you know/Have you heard of Cauchy principal value ?
Anyhoo, a similar problem arises in comparing
$$P(\bigcap_{n \in T} A_n) = 1$$
$$P(A_n) = 1 \ \forall n \in T$$
Do you know modification vs indistinguishable?
