It is stated in this Wikipedia article that convergence almost surely may be restated using the notion of the limit inferior of a sequence of sets, in that:

$$ \operatorname{Pr}\Big( \omega \in \Omega : \lim_{n \to \infty} X_n(\omega) = X(\omega) \Big) = 1. $$

can be restated as:

$$ \operatorname{Pr}\Big( \liminf_{n\to\infty} \big\{\omega \in \Omega : | X_n(\omega) - X(\omega) | < \varepsilon \big\} \Big) = 1 \quad\text{for all}\quad \varepsilon>0. $$

I am wondering why this is the case. I know that generally, it is NOT the case that:

$$ \omega\in\left\{\lim_{n\to\infty} X_n= X\right\}\iff \omega\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{|X_k-X|<\varepsilon \}, \text{ for all }\varepsilon > 0. $$

So why are the two statements above equal?

  • 3
    $\begingroup$ "I know that generally, it is NOT the case that: [...]" Why do you think that this is not the case...? $\endgroup$ – saz Jan 15 '17 at 20:05

Fix $\varepsilon>0$. Denote $$A=\{\omega \in \Omega:\lim_{n \to \infty}X_n(\omega)=X(\omega)\},$$ $$A_n^\varepsilon=\{\omega \in \Omega \, : \,|X_k(\omega)-X(\omega)|< \varepsilon\text{ for all } k \ge n\}.$$ Then the original statement can be written as $P(A)=1$, and the second statement can be written either as $\lim\limits_{n \to \infty} P(A_n^\varepsilon)=1$ or as $P(A^\varepsilon)=1$, where $A^\varepsilon \equiv \lim\limits_{n \to \infty}A_n^\varepsilon=\bigcup\limits_{n=1}^{\infty}A_n^\varepsilon$. (Notice that $\{A_n^\varepsilon\}$ is an increasing sequence of events. By the continuity theorem for monotone events, $P(A^\varepsilon)=\lim\limits_{n \to \infty} P(A_n^\varepsilon)$.)

  1. Suppose $P(A)=1$. Note that $A\subseteq A^\varepsilon$. Indeed, in detail, $A^\varepsilon$ is the following set of outcomes: \begin{align*} \quad \ A^\varepsilon & = \{\omega \in \Omega: \exists \,n_\varepsilon(\omega) \text{ such that } \omega \in A_k^\varepsilon, \ \forall k\, \ge n_\varepsilon(\omega)\}&\\ & =\{\omega \in \Omega: \exists \,n_\varepsilon(\omega) \text{ such that } |X_k(\omega)-X(\omega)| \le \varepsilon, \ \forall \, k \ge n_{\varepsilon}(\omega)\}& %& = \{\omega \in \Omega: \exists n_\varepsilon(\omega) \text{ such that } \omega \in A_k^\varepsilon, \ \forall k\ge n_\varepsilon(\omega)\}& \end{align*} Let $\omega \in A$. Then for a chosen $\varepsilon>0$, $\exists \, n_\varepsilon(\omega)$ such that $\forall \,(k \ge n_\varepsilon(\omega)) \ |X_k(\omega)-X(\omega)|\le\varepsilon$. Clearly then, $\omega \in A^\varepsilon$.
    The fact that $A\subseteq A^\varepsilon$ implies that $P(A^\varepsilon) \ge P(A)=1$ and, thus, because probability cannot be greater than $1$, $P(A^\varepsilon) = 1$.

  2. Suppose now that $P(A^\varepsilon)=1$ for any $\varepsilon>0$. Take $\varepsilon=1,\frac{1}{2},\frac{1}{3},\ldots$ Then $A^1\supset A^\frac{1}{2} \supset A^\frac{1}{3}\supset \ldots$ $-$ a decreasing sequence of events, for which $\lim\limits_{m \to \infty}A^\frac{1}{m}=\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}$. By the continuity theorem, $$P\left(\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}\right) = P\left(\lim\limits_{m \to \infty}A^\frac{1}{m}\right)=\lim\limits_{m \to \infty}P\left(A^\frac{1}{m}\right)=1.$$ Now notice that $A=\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}$ and, thus, $P(A)=1$.


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