Prove that probability of convergence of sequence of independent r.v. is either $0$ or $1$.
Proving the convergence is $0$ is straightforward, using B-C lemma. But for proving it to be $1$, we need events to be independent. So can someone prove that part.
This is given as exercise in K.L. Chung book after explaining and proving different versions of Borel-Cantelli lemma, so please don't use Kolmogorov 0-1 Law
We can do this by defining set $A_k = \{|X_m - X_n| < \epsilon \text{ for all } m>n \ge k\}$ and then if $\sum P(A_k) < \infty$ then $P(A_k io)=0$. But, the trouble if $\sum P(A_k) = \infty$ then we can't claim that $P(A_k io)=1$ as $A_k$ need not be independent for different $k$.
Let $(E_n)_{n=1}^\infty$ be a sequence of independant events. If $\sum_{n=1}^\infty \mathbb{P}(E_n)<\infty$, you already showed that the probability of an infinite number of events happening is zero. Now, if instead $\sum_{n=1}^\infty \mathbb{P}(E_n)=\infty$, we try to show that the probability of an infinite number of events happening is one.
We have that $1-\mathbb{P}(\limsup_{n\to\infty}E_n)=\mathbb{P}(\liminf_{n\to\infty}E_n)=\lim_{n\to\infty}\mathbb{P}(\bigcap_{j=n}^\infty E_j^c)$. Because the events are independent, the probability of any of them happening at the same time is the product of their probabilities, i.e. $$\mathbb{P}(\bigcap_{j=n}^\infty E_j^c) = \prod_{j=n}^\infty (1-\mathbb{P}(E_j)).$$
Now here's a trick: because $1-x$ is always less than or equal to $e^{-x}$ for $x\in[0,1]$ (see Taylor expansion), we have that $\mathbb{P}(\bigcap_{j=n}^\infty E_j^c) \leq \prod_{j=n}^\infty e^{-\mathbb{P}(E_j)}=\exp{(-\sum_{j=n}^\infty}\mathbb{P}(E_j))=0$. Therefore $\mathbb{P}(\limsup_{n\to\infty}E_n)=1$.