Monotone increasing sequence of random variable that converge in probability implies convergence almost surely

Question

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.

Answer 1

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.

Answer 2

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.