Let $\{X_n\}$ be a collection of random variable with $X_{n+1} \geq X_n$ for all $n$ and $X_n \rightarrow X$ in probabilty. How to prove that $X_n \rightarrow X$ almost surely.

My partial answer:

Consider a probability space $(\Omega, B, P)$ Let $\varepsilon$ be a abritary positive constant. Define $E_n=\{\omega\in \Omega: |X_n(\omega) -X(\omega)|\geq \varepsilon\}$. Let $S=\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} E_k$. Since $X_n \rightarrow X$ in probabilty, we have \begin{align} P(S)=\lim_{n\rightarrow \infty} P(\cup_{k=n}^{\infty} E_k)= \lim_{n\rightarrow \infty}P(E_n)=0. \end{align} For $y \in \Omega \backslash S$, we have $|X_n(y)-X(y)|<\varepsilon$. Hence, $X_n \rightarrow X$ almost surely.

My poblem: I'm not sure about $\lim_{n\rightarrow \infty} P(\cup_{k=n}^{\infty} E_k)= \lim_{n\rightarrow \infty}P(E_n)$ and how to argue this by monotonicity of random variable.


2 Answers 2


If you proved $X \geq X_n$ ($n \in \mathbb{N}$), this would imply $$\mathbb{P} \left( \bigcup_{k=n}^{\infty} E_k \right) = \mathbb{P}(E_n)$$ since $E_k \subseteq E_n$ for $k \geq n$ (this follows directly from the monotonicity).

Here is an alternative proof: Since $X_{n+1} \geq X_n$ we know that $$Y := \sup_{n \in \mathbb{N}} X_n = \lim_{n \to \infty} X_n \in (-\infty,\infty]$$ exists. This implies in particular $X_n \to Y$ in probability and from the uniqueness of the limit, we conclude $X=Y$ a.s.. Thus $X_n \to X$ almost surely.

  • $\begingroup$ I have a question: why does the limit exist? Doesn't a monotone sequence have to be bounded in order to have a limit? $\endgroup$
    – VHarisop
    Apr 29, 2018 at 19:08
  • 2
    $\begingroup$ @VHarisop The limit exists in $(-\infty,\infty]$ (i.e. it might be infinite). $\endgroup$
    – saz
    Apr 29, 2018 at 19:10

I found this question in Resnick's book A probability path, exercise 6.7.1 (a). He suggests to think subsequences, so that's the approach I will follow. This result holds more generally for every monotone sequence $\{X_n\}$ of random variables. I will further assume the premises of your question.

Let $(\Omega,S,P)$ be a probability space and $\{X_n\}$ a sequence of random variables such that $X_{n+1}\ge X_n$ for all $n$, $X_n$ converging in probability to $X$. By Riesz-Weyl theorem $\exists$ a subsequence $\{X_{n_k}\}$ such that $X_{n_k}\rightarrow X$ almost surely i.e. there exists a negligible set $N\in S$ such that $X_{n_k}(\omega)\rightarrow X(\omega) \:\forall \omega\in \Omega\setminus N$. In other words, $\forall \omega\in \Omega\setminus N \:\:\:\exists \:\:\:s\in\mathbb{N}$ such that $\vert X_{n_k}(\omega)- X(\omega) \vert <\varepsilon \:\:\:\forall\:\:\:k\ge s$.

By monotonicity of the sequence $\{X_n\}$, then for all $n_s\le n$, $\:\:\exists m\ge s$ such that $n_m\le n \le n_{m+1}$. Hence

$$ X_{n_m}(\omega)\le X_{n}(\omega)\le X_{n_{m+1}}(\omega)\le X(\omega) $$

implying that

$$ \vert X_{n}(\omega)- X(\omega) \vert\le \vert X_{n_m}(\omega)- X(\omega) \vert $$

Apply the definition of limit, thus $\forall \omega\in \Omega\setminus N$, $\underset{n\rightarrow\infty}{\lim} X_{n}(\omega)=X(\omega)$. As a result, $X_n \rightarrow X$ almost surely.

  • 1
    $\begingroup$ I am curious about what theorem you are referring to as Riesz-Weyl, since as I see it, the existence of an a.s. convergent subsequence follows directly from the convergence in probability (equiv. being Cauchy in probability) and Borel-Cantelli lemma. I d take a subsequence, such that the probability of successive terms being separated more than negative powers of two is less than negative powers of two, check the series, that converges, note that by the lemma then they are closer than negative powers of two eventually a.s. therefore being Cauchy almost surely. $\endgroup$
    – xyz
    Apr 27, 2020 at 17:16

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