Name for this condition: $\forall\epsilon > 0:\quad P(\exists m\ge n: |X_m - X|>\epsilon)\to 0,\quad n\to\infty$

Question

A sequence of random variables $X_n$ converges in probability to a random variable $X$ if $$\forall\epsilon > 0:\quad P(|X_n - X|>\epsilon)\to 0,\quad n\to\infty. \tag{1}$$

Note that the following condition is stronger: $$\forall\epsilon > 0:\quad P(\exists m\ge n: |X_m - X|>\epsilon)\to 0,\quad n\to\infty. \tag{2}$$

Question: Is there a name for condition $(2)$?

Answer 1

Condition (2) is equivalent to $$\tag{*}S_n:=\sup_{m\geqslant n}\left\lvert X_m-X\right\rvert \to 0\mbox{ in probability},$$ which is equivalent to $X_n\to X$ almost surely. Indeed, if $X_n\to X$ almost surely, then $\sup_{m\geqslant n}\left\lvert X_m-X\right\rvert \to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $\left(S_n\right)_{n\geqslant 1}$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.

Answer 2

I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.

If we denote by $A_n(\varepsilon)$ the event $\{|X_n-X|>\varepsilon\}$, then the event $\{\exists m\geq n: |X_n-X|>\varepsilon\}$ is just the union of events $$B_n(\varepsilon)=\bigcup_{m\geq n}A_m(\varepsilon).$$ Obviously the sequence $B_n$ decreases, in the sense that $B_{n+1}\subset B_n$ for all $n$. Elementary measure theory tells us that $$\lim_{n\to\infty}P(B_n(\varepsilon))=P\left(\bigcap_{n=1}^{\infty}B_n(\varepsilon)\right)$$ but $$\bigcap_{n=1}^{\infty}B_n(\varepsilon)=\bigcap_{n=1}^{\infty}\bigcup_{m\geq n}A_m(\varepsilon)=\overline{\lim_{n\to\infty}}A_n(\varepsilon)$$ where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation: $$\lim_{n\to\infty}P(B_n(\varepsilon))=P(\overline{\lim_{n\to\infty}}A_n(\varepsilon))$$ which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have $$P(\overline{\lim_{n\to\infty}}A_n(\varepsilon))\geq\overline{\lim_{n\to\infty}}P(A_n(\varepsilon)$$