Let $Y$ be absolute continuous random variable. Let $\left\{ a_{n}\right\} _{n=1}^{\infty}$ be a sequence with $\lim_{n\rightarrow\infty}a_{n}=L$. Let $X_{i}$ be the indicator of the event $Y\leq a_{i} \left(X_{i}=1\left(Y\leq a_{i}\right)\right)$ and $X=1\left(Y\leq L\right). $
Is it true that $X_{i}\xrightarrow{\,\,a.s\,\,}X?$ or $X_{i}\xrightarrow{\,\,p\,\,}X?$
A full explanation would be appreciated
No. Take the easy sequence $$a_n = -\frac{1}{n}$$ hence $L=0$ and $$Y \equiv 0$$ which is obviously absolutely continuous.
Then $$X = 1_{\{Y \le L\}} = 1_{\{0 \le 0\}} \equiv 1$$
but $$X_n = 1_{\{Y \le -\frac{1}{n}\}} \equiv 0$$
Hence in general $(X_n)_{n\in\Bbb N}$ neither converges almost surely nor in probability to $X$.
But if your sequence $(a_n)_{n\in\Bbb N}$ is monotonous decreasing we have for each $\omega$ that $$\lim_{n\to\infty} 1_{\{Y(\omega) \le a_n\}} = 1_{\{Y(\omega) \le L\}}$$ and so $$\lim_{n\to\infty} X_n = X$$ for all $\omega$ hence $$X_n \to X \text{ almost surely and in probability }$$