### Cauchy’s criterion on almost sure convergence

In this article, we will investigate interplays between various kinds of convergence of sequence of random variables. Specifically, we will discuss convergence a.s., convergence in probability, convergence in ${p}$th mean and convergence in distribution.

We can naturally formalize the notions of sequence of random variables being Cauchy in these contexts of convergence. We say: ${X_n}$ is Cauchy in probability if ${P(|X_m-X_n|>\epsilon)\rightarrow 0}$ as ${m, n\rightarrow\infty}$; ${X_n}$ is Cauchy a.s. if ${X_n}$ is Cauchy everywhere except a set of measure zero; ${X_n}$ is Cauchy in ${L^p}$ if ${\mathop{\mathbb E} |X_n-X_m|^p\rightarrow 0}$ as ${m,n\rightarrow\infty}$. Now we give some basic facts about the advantages of using Cauchy’s criterion.

Proposition 1 ${X_n\rightarrow X}$ a.s. if and only if ${P(\sup_{k\geq n}|X_k-X|\geq\epsilon)\rightarrow 0}$ for every ${\epsilon>0}$ as ${n\rightarrow\infty}$.

Proof: Let us consider the formulation of the set of points where ${X_n\not\rightarrow X}$; that is

$\displaystyle \{\omega: X_n(\omega)\not\rightarrow X(\omega)\}=\cup_{\epsilon>0}\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}$

In fact, we can use the denseness of real numbers to write

$\displaystyle \{\omega: X_n(\omega)\not\rightarrow X(\omega)\} =\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}$

So

$\displaystyle \begin{array}{rcl} X_n\xrightarrow{a.s.} X & \Leftrightarrow P(\{\omega: X_n(\omega)\not\rightarrow X(\omega)\}=0\\ & \Leftrightarrow P\left(\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}\right)=0\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}\right)=0\;\;\forall j\in{\mathbb N}\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)=0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+ \end{array}$

$\Box$

Proposition 2 ${X_n}$ is Cauchy a.s. if and only if ${P(\sup_{\substack{k\geq n\\l\geq n}}|X_k-X_l|\geq \epsilon)\rightarrow 0}$ as ${n\rightarrow\infty}$ for every ${\epsilon>0}$; or equivalently ${P(\sup_{k\geq 0}|X_{n+k}-X_n|\geq\epsilon)\rightarrow 0}$ as ${n\rightarrow\infty}$ for every ${\epsilon>0}$.

Proof: Using a similar argument, we obtain a chain of equivalences,

$\displaystyle \begin{array}{rcl} X_n \text{ is Cauchy a.s. }&\Leftrightarrow P(\{\omega: X_n \text{ is not Cauchy}\})=0\\ & \Leftrightarrow P\left(\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq 1/j\}\right)=0\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq 1/j\}\right)=0\;\;\forall j\in{\mathbb N}\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)=0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{\substack{k\geq 0}}\{\omega:|X_{n+k}-X_n|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+ \end{array}$

where the last equivalence is obtain from the following inequality

$\displaystyle \sup_{k\geq 0}|X_{n+k}-X_n|\leq\sup_{\substack{k\geq 0\\l\geq 0}}|X_{n+k}-X_{n+l}|\leq 2\sup_{k\geq 0}|X_{n+k}-X_n|$

This completes the proof. $\Box$

Remark: Similarly one can formulate another equivalent definition of Cauchy a.s., that is if ${P(\sup_{m\epsilon)\rightarrow 0}$ for all ${\epsilon>0}$ as ${n,m\rightarrow\infty}$, then ${X_n}$ is Cauchy a.s. Let ${\eta>0}$. Then there is an integer ${N, M}$ such that ${P(\sup_{m\epsilon)<\eta}$ whenever ${n>N, m>M}$. Sending ${N\rightarrow\infty}$ gives ${P(\sup_{k>m}|X_k-X_m|>\epsilon)<\eta}$ which is just the previous proposition.