Does almost sure convergence imply such probability converging to 0? We have observations $Y_i, X_{i,1},\cdots, X_{i,p}\ (i=1, \cdots, n)$. From the observations, we manage to get an estimator $\hat{a}_n$, which converges to a constant $a$ almost surely, i.e., $\hat{a}_n{\to}_{a.s.} a$. We also know that $X_{i,1}$ is continuously distirbuted. Then can I show that $P(\hat{a}_n<X_{i,1}<a)\to 0$ ?
If not, can I add some other conditions so that $P(\hat{a}_n<X_{i,1}<a)\to 0$ becomes true?
Any help would be appreciated, thank you very much!
 A: Presumably the $X_{i,j}$ are independent samples from the same distribution.
In fact it's true with $X_{i,1}$ replaced by any $X$.
It suffices to show that given $\epsilon > 0$, there is $N$ such that for all $n > N$, $\mathbb P(\hat{a}_n < X < a) < \epsilon$.
There is $\delta > 0$ such that
$\mathbb P(a - \delta < X < a) < \epsilon/2$: this follows from the fact that $$0 = \mathbb P(\emptyset) = \mathbb P\left(X \in \bigcap_{m=1}^\infty 
 (a-\frac1m, a)\right) = \lim_{m \to \infty} \mathbb P(X \in (a-\frac1m, a))$$
As $\hat{a}_n \to a$ almost surely, it also does so in distribution, so there is $N$ such that for all $n > N$,
$\mathbb P(|\hat{a}_n - a| \ge  \delta) < \epsilon/2$.
Now if $\hat{a}_n < X < a$ we must have either $\hat{a}_n \le a - \delta$ (which implies $|\hat{a}_n - a| \ge \delta$) or $\hat{a}_n > a - \delta$ (which implies $a - \delta < X < a$), so for all $n > N$,
$$\mathbb P(\hat{a}_n < X < a) \le \mathbb P(|\hat{a}_n - a| \ge \delta) + \mathbb P(a - \delta < X < a) < \epsilon$$
