Consider a sequence of random variables $(p_n)$ such that:
(1) $\Pr\bigg(\limsup_{n \to \infty} p_n\leq U\bigg)=1$
(2) $\Pr\bigg(\liminf_{n \to \infty} p_n\geq L\bigg)=1$
where $L,U$ are real numbers. Does these assumptions imply $$ \Pr\bigg(\lim_{n \to \infty}d\big(p_n, \big[L,U\big]\big)= 0\bigg)=1, $$ where $d(p_n, [L,U] ):= \inf \{|p_n - y| : y \in [L,U ] \}$. In other words, I wonder if the limit of the distance exists.
Intuitively, I think it exists. In fact, by (1) and (2), $\inf \{|p_n - y| : y \in [L,U ] \}$ becomes almost surely zero. Could you help me to show this formally?
There is not much probability here: You show that $\limsup_{n\to\infty} p_n≤U$ and $\liminf_{n\to\infty p_n} ≥ L$ together imply $\lim_{n\to\infty} d(p_n, [L,U])$ for a not-random sequence $(p_n)$. This then implies $$ \mathbb P(\lim_{n\to\infty} d(p_n, [L,U])) ≥ \mathbb P\left(\limsup_{n\to\infty} p_n≤U, \liminf_{n\to\infty p_n} p_n ≥ L \right) = 1. $$
To show the non-probabilistic result, note that $d(p_n,[L,U]) = \max(p_n-U, L-p_n, 0)$. But since $\limsup_{n\to\infty} p_n-U, \limsup_{n\to\infty} L-p_n$ and $\limsup_{n\to\infty} 0$ are all 0, this means that the limsup of the maximum, $d(p_n,[L,U])$ must be smaller equal 0. But because the limit is not negative, this already implies that $\lim_{n\to\infty}d(p_n,[L,U]) = 0$.