$X_n \to X$ a.s. implies... 
Let $X_n$ be independent.
$ X_n \to X \, a.s.$ implies $ \sum_n P( |X_n-X|>\epsilon) < \infty$

I tried to prove as follows, which is wrong.
Note that $ X_n \to X \, a.s.$ is equivalent to $\forall \epsilon>0, \, P( |X_n-X|>\epsilon \quad \textrm{i.o})=0.$
Then, using contraposition of 2nd Borel-Cantelli lemma, I get to the conclusion.
Here, I made a mistake on application of BC lemma since independence of $\{X_n\}$ does not imply independence of $\{X_n-X\}$.
Anyone who could give me some hint about this?
Thanks in advance.

 A: But actually $X_{n} - X$ are also independent. This is because $X$ is a.s. constant by a simple application of Kolmogorov's 0-1 Law, but this is sort of overkill. So let us give a more direct proof. We first refer to the following lemma:

Lemma. $X_{1}, \cdots, X_{n}$ are independent if and only if
  $$ \Bbb{E}[ f_{1}(X_{1})\cdots f_{n}(X_{n}) ] = \Bbb{E} f_{1}(X_{1}) \cdots \Bbb{E} f_{n}(X_{n})$$
  for any bounded continuous functions $f_{1}, \cdots, f_{n}$.

Now let $f, g : \Bbb{R} \to \Bbb{R}$ be any bounded continuous functions. Then for any $m < n \ll k < l$, both $X_{m} - X_{k}$ and $X_{n} - X_{l}$ are independent. So
$$ \Bbb{E}[ f(X_{m} - X_{k}) g(X_{n} - X_{l})] = \Bbb{E} f(X_{m} - X_{k}) \Bbb{E} g(X_{n} - X_{l}). $$
Taking $l \to \infty$ followed by $k \to \infty$, the bounded convergence theorem says that
$$ \Bbb{E}[ f(X_{m} - X) g(X_{n} - X)] = \Bbb{E} f(X_{m} - X) \Bbb{E} g(X_{n} - X). $$
This implies that they are independent. (And this idea readily generalizes to show the mutual independence, not just the pairwise independence.)

Proof of Lemma. The 'only if' part is easy since $\sigma(X_{1}), \cdots, \sigma(X_{n})$ are independent. To prove the 'if' part, for any $a_{1}, \cdots, a_{n}$, let
$$ f_{i,k}(x)
= \int_{x}^{\infty} k 1_{[a_{i},a_{i}+1/k]}(t) \, dt
= \begin{cases}
1, \quad x \leq a_{i}, \\
0, \quad x \geq a_{i} + 1/k, \\
\text{linearly interpolated in-between}
\end{cases} $$
Then we easily check that $f_{i,k} \downarrow 1_{(-\infty, a_{i}]}$. So by the bounded convergence theorem,
  \begin{align*}
 \Bbb{P}(X_{1} \leq a_{1}, \cdots, X_{n} \leq a_{n})
&= \lim_{k\to\infty} \Bbb{E}[ f_{1,k}(X_{1}) \cdots, f_{n,k}(X_{n}) ] \\
&= \lim_{k\to\infty} \Bbb{E} f_{1,k}(X_{1}) \cdots \Bbb{E} f_{n,k}(X_{n}) \\
&= \Bbb{P}(X_{1} \leq a_{1}), \cdots,  \Bbb{P}(X_{n} \leq a_{n}).
\end{align*}
  This implies the independence of $X_{i}$'s (c.f. Theorem 2.1.4 in Durrett).

