Let $(X_n)$ be a sequence of real random variables on $(\Omega,\mathcal A,\mathbb P)$. Then 1. and 2. are equivalent:

  1. There exists a random variable $X$, s.t. $X_n\to X$ $P$-almost sure for $n\to \infty$.
  2. $\sup_{m>n} |X_m-X_n|\to 0$ in probability for $n\to\infty$.

I tried showing $1.\Rightarrow 2.$:

I know that $X_n\to X$ $P$-almost sure means that $$P(\lim_{n\to\infty} X_n=X)=1$$ or equivalently $$(*)\quad \lim_{n\to\infty} P(\sup_{m\geq n} |X_m-X|\geq\varepsilon)=0\quad\forall\varepsilon>0$$

$\sup_{m>n} |X_m-X_n|\to 0$ in probability means \begin{equation}\lim_{n\to\infty} P(\sup_{m>n}|X_m-X_n|\geq\varepsilon)=0\quad\forall\varepsilon>0\end{equation}

This looks like a Cauchy-sequence in probability, but I don't know if I can deduce the convergence of this from $(*)$.

How can I go on proving this? Thanks for any input!

  • 1
    $\begingroup$ If $(X_n(\omega))$ is convergent for some $\omega \in \Omega$, then it is a Cauchy sequence hence $\sup_{m \geq n} |X_n(\omega)-X_m(\omega)|$ converges to $0$ as $n \to \infty$. Under the condition (1), this proves that $\sup_{m\geq n}|X_n-X_m|$ converges to $0$ almost surely. $\endgroup$
    – Siméon
    Jan 4, 2014 at 15:15

1 Answer 1


For the first part, use that almost sure convergence implies convergence in probability: we thus are reduced to show that $\sup_{m\geqslant n}|X_m-X|\to 0$ in probability, but there is in this case actually almost everywhere convergence.

Conversely, assume that $\mathbb P(\sup_{m\geqslant n}|X_m-X_n|\gt\varepsilon)\to 0$ for each positive $\varepsilon$. We have in particular that $(X_n)_n$ is Cauchy in probability, namely, $$\forall \varepsilon,\delta\gt 0,\exists n_0\mbox{ such that if }m,n\geqslant n_0, \mathbb P(|X_m-X_n|\gt\varepsilon)\lt\delta.$$ Taking $\delta$ of the form $2^{-k}$ we can extract an almost everywhere convergence subsequence to some random variable $X$. Then using Cauchy-ness, we can prove that $X_n\to X$ in probability. Now we go back to the assumption. This now reads $\mathbb P(\sup_{m\geqslant n}|X_m-X|\gt \varepsilon)\to 0$ for each positive $\varepsilon$. Define $Y_n:=\sup_{m\geqslant n}|X_m-X|$: the sequence $(Y_n)_n$ is non increasing and goes to $0$ in probability, hence almost everywhere.

  • $\begingroup$ Thanks for your answer. I see your argument, but I don't quite understand: Why do we know that there is an a.e. convergent subsequence? $\endgroup$
    – dinosaur
    Jan 4, 2014 at 18:03
  • 1
    $\begingroup$ This relies on Borel-Cantelli's lemma (construct $n_k$ inductively such that for each $k$, $\mathbb P(|X_{l}-X_{n_k}|\geqslant 2^{—k})\leqslant 2^{-k}$). $\endgroup$ Jan 4, 2014 at 18:19
  • $\begingroup$ @ Davide: I know the OP is 3years old, but I got the same problem recently. My question is, what is $X_l$ here? $\endgroup$
    – Extremal
    Oct 4, 2017 at 16:27

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