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.
Best Answer
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
$$ \left\{\omega \in \Omega : \lim_{n\to \infty }X_n(\omega )\in \mathbb{R}\right\}=\left\{\omega \in \Omega : \{X_n(\omega )\}_{n\in\mathbb{N}}\text{ is Cauchy}\right\} $$