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My setup is the following.

Consider a sequence of random variables $(X_n)_n$ such that for every $\delta>0$ \begin{align*} \lim_{n,m\rightarrow\infty}\mathbb{P}[\sup_{m<k\leq n}|X_m-X_k|\geq \delta]=0 \end{align*}

And I have to prove that there is a random variable $X$ such that the sequence converges to $X$ almost surely.

What I have: The assumption implies the sequence is Cauchy in measure so there is a random variable $X$ such that the sequence converges to $X$ but in measure. Naturally, that is my aspirant to the a.s. limit. Furthermore, there exists a subsequence that converges a.s. to $X$ but I can't manage to prove the whole sequence converges.

Any suggestions? Thanks a lot!

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I believe we can do this without appealing to the existence of a random variable $X$ to which the sequence converges in measure. My argument is a little informal, so please let me know if you find something confusing or if you suspect something is incorrect.

Suppose that $(X_{n})_{n\in\mathbb{N}}$ does not have an a.s.-limit. Then $\{(X_{n})_{n\in\mathbb{N}}\text{ is not Cauchy}\}$ has positive measure. In particular, there exists some $j\in\mathbb{N}$ such that $\{\limsup_{n,m\to\infty}|X_{m}-X_{n}|>1/j\}$ has positive measure, for $$ \{(X_{n})_{n\in\mathbb{N}}\text{ is not Cauchy}\}=\bigcup_{j\in\mathbb{N}}\{\limsup_{n,m\to\infty}|X_{m}-X_{n}|>1/j\} $$ and if all the terms in the union have measure zero then countable subadditivity tells us that the lefthand side must have measure zero. $\{\limsup_{n,m\to\infty}|X_{m}-X_{n}|>1/j\}$ having positive measure contradicts your initial hypothesis.

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  • $\begingroup$ Thank you for your answer. I believe the idea is nice but I have problems comparing the sets $\{\limsup_{m,n}|X_n-X_m|\geq 1/j\}$ and $\{\sup_{m<k\leq n}|X_k-X_m|\geq1/j\}$ $\endgroup$
    – user152168
    Sep 16 '15 at 3:40
  • $\begingroup$ If $\omega\in\{\limsup_{n,m\to\infty}|X_{n}-X_{m}>1/j\}$, then for arbitrarily large $N,M\in\mathbb{N}$ there exist $n\geq N$ and $m\geq M$ such that $|X_{n}(\omega)-X_{m}|>1/j$. But we might not have $\omega\in\{\sup_{m<k\leq n}|X_{k}-X_{m}|>1/j\}$ if, for example, you take $n$ and $m$ to always differ by a fixed constant. So I think it depends on how you take $n,m\to\infty$. Sorry for the lack of clarity; I'm not sure exactly what's meant in this context by the limit as $n$ and $m$ approach $\infty$ at the same time! $\endgroup$ Sep 16 '15 at 14:10

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