Does almost sure convergence implies that a sequence is cauchy with probability one? Well, if $P\left((X_{n}) \ \text{is cauchy}\right)=1$ then $X_n \longrightarrow X \ \text{a.s}$.
But is it true for converse? 
we have
$$(X_n\longrightarrow X) := \bigcap\limits_{ε\in\mathbb{Q^+}}\bigcup\limits_{N=1}^{\infty}\bigcap\limits_{n=N}(|X_n-X|\leq\epsilon) $$
And
$$(X_n \ \text{is cauchy}) := \bigcap\limits_{ε\in\mathbb{Q^+}}\bigcup\limits_{N=1}^\infty\bigcap\limits_{n=N}\bigcap\limits_{m=N}(|X_n-X_m|\leq\epsilon)$$.
we see that 
$(X_n \ \text{is cauchy}) \subset (X_n\longrightarrow X \ \text{a.s})$ 
so:  $ P(X_n \ \text{is cauchy}) \leq P(X_n\longrightarrow X)$ and since $P\left((X_{n}) \ \text{is cauchy}\right)=1$, hence $P(X_n\longrightarrow X)=1$. 

But what about the converse, that if $(X_n)$ converges a.s then $((X_n) \ \text{is cauchy})$ with probability one? 
intuitively, if $(X_n)$ converges almost surely to a limit X, then if we consider $N\in\mathbb{N}$ too large, then $\forall\epsilon>0\ \exists \ N\in\mathbb{N} \ \forall n,m\geq N :(|X_n-X_m|\leq \epsilon)$, which is cauchy, and it holds with probability one, right?
Hope someone can calrify this. Thanks in advance.
 A: No, both notions are not equivalent if we consider random variables with values in the extended real line, namely $\overline{\mathbb{R}}:=\mathbb{R}\cup \{-\infty ,+\infty \}$, as $\overline{\mathbb{R}}$ cannot be a metric space with the usual metric from $\mathbb{R}$. This means that sequences such that $\lim_{n\to \infty }X_n(\omega )\in\{-\infty ,+\infty \}$ cannot be defined as Cauchy as there is no metric in $\overline{\mathbb{R}}$ that generalizes the metric in $\mathbb{R}$.
However if $\Pr [\{\omega \in \Omega :\limsup_{n\to\infty}|X_n(\omega )|<+\infty \}]=1$ then both conditions, being almost sure convergent to some random variable $X$ and being almost sure Cauchy are equivalent because the set where this equivalence fails have zero probability. To be more precise: $\lim_{n\to \infty }X_n(\omega )\in \mathbb{R}$ if and only if $\{X_n(\omega )\}_{n\in\mathbb{N}}$ is Cauchy, therefore
$$
\left\{\omega \in \Omega : \lim_{n\to \infty }X_n(\omega )\in \mathbb{R}\right\}=\left\{\omega \in \Omega : \{X_n(\omega )\}_{n\in\mathbb{N}}\text{  is Cauchy}\right\}
$$
