# Interpreting almost sure convergence

I'm reading: https://en.wikipedia.org/wiki/Convergence_of_random_variables#Almost_sure_convergence and here it says that

Given a probability space $$(\Omega,\mathcal{F},P)$$ and a random variable $$X:\Omega \rightarrow \mathbb{R}$$ almost sure convergence stands for $$P\left(\omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega)=X\right)=1.$$ [...] almost sure convergence can also be defined as follows: $$P\left(\limsup_{n \rightarrow \infty} \left\{\omega \in \Omega: |X_n(\omega) - X(\omega)| > \varepsilon\right\}\right)=0, \quad \forall \; \varepsilon>0.$$

My question is, what is the intuition behind this equivalence? I understand the first definition, but why do we use $$\limsup$$ in the second one to make the equivalence work? Thanks

I don't really see intuition here, the equivalence just follows from using the definition of convergence. For a sequence of sets $$(A_n)$$ the set $$\lim \sup(A_n)=\{A_n\ \ i.o\}$$ is the set of elements which belong to infinitely many of the sets $$A_n$$. The formal definition of this set is $$\cap_{n=1}^\infty \cup_{k=n}^\infty A_k$$.

Assume $$X_n\to X$$ almost surely by the first definition and let any constant $$\epsilon>0$$. Define the sequence $$A_{n,\epsilon}:=\{\omega: |X_n(\omega)-X(\omega)|>\epsilon\}$$. Note that if $$\omega\in\lim\sup A_{n,\epsilon}$$ then it means that $$|X_n(\omega)-X(\omega)|>\epsilon$$ for infinitely many values of $$n$$, and hence $$X_n(\omega)$$ obviously does not converge to $$X(\omega)$$. So $$\lim\sup A_{n,\epsilon}\subseteq \{\omega: X_n(\omega)\nrightarrow X(\omega)\}$$, and by monotonicity of probability:

$$\mathbb{P}(\lim\sup A_{n,\epsilon})\leq \mathbb{P}(\{\omega: X_n(\omega)\nrightarrow X(\omega)\})=0$$

Second direction: Now assume $$X_n\to X$$ by the second definition. For each $$k\in\mathbb{N}$$ define $$B_k=\lim\sup A_{n,\frac{1}{k}}$$ where the sets $$A_{n,\epsilon}$$ are defined like before. Then by assumption $$\mathbb{P}(B_k)=0$$ for all $$k$$, and hence $$\mathbb{P}(\cup_{k=1}^\infty B_k)=0$$. Now suppose we have $$X_n(\omega)\nrightarrow X(\omega)$$ for some $$\omega$$. This implies that there must be some $$m\in\mathbb{N}$$ such that $$|X_n(\omega)-X(\omega)|>\frac{1}{m}$$ for infinitely many natural numbers $$n$$, and thus $$\omega\in B_m\subseteq\cup_{k=1}^\infty B_k$$.

In other words, we have the inclusion $$\{\omega: X_n(\omega)\nrightarrow X(\omega)\}\subseteq\cup_{k=1}^\infty B_k$$, and so $$\mathbb{P}(\{\omega: X_n(\omega)\nrightarrow X(\omega)\})=0$$.

• Thank you so much Aug 3 '20 at 1:54

Intuition

There is not much intuition to be gleaned here. The second definition comes from "massaging" the definition of the [non-random] limit of real numbers (since for a fixed $$\omega$$, the limit $$\lim_{n \to \infty} X_n(\omega)$$ is just a non-random limit).

The utility of the second definition is that it is easier to verify because it involves relatively simple sets $$\{|X_n(\omega) - X(\omega)| > \epsilon\}$$ (fixed $$\epsilon$$, fixed $$n$$). You only need to deal with one $$n$$ at a time to understand this set, and under certain circumstances, bounding the probability of this set for each $$n$$ can be enough to bound probability of the $$\limsup$$. By contrast, the set $$\{\lim_{n \to \infty} X_n(\omega) = X(\omega)\}$$ is difficult to deal with because of the limit inside the event.

Notation

Let $$A_{n, \epsilon} = \{|X_n(\omega) - X(\omega)| > \epsilon\}$$. Note that $$\limsup_{n \to \infty} A_{n, \epsilon} := \bigcap_n \bigcup_{k \ge n} A_{k,\epsilon}$$ by definition.

(1) $$\implies$$ (2)

Fix $$\epsilon > 0$$. If $$\omega \in \bigcap_n \bigcup_{k \ge n} A_{k, \epsilon}$$, then $$|X_n(\omega) - X(\omega)| > \epsilon$$ for infinitely many $$n$$, so $$\lim_n X_n(\omega) \ne X(\omega)$$. Thus $$P(\limsup_n A_{n, \epsilon}) \le P(\lim_n X_n(\omega) \ne X(\omega))$$ for each $$\epsilon$$. So if almost sure convergence holds in the sense of the first definition, then it holds in the sense of the second definition.

(2) $$\implies$$ (1)

Conversely, suppose $$\omega$$ is such that $$\lim_n X_n(\omega) \ne X(\omega)$$. If you write out the definition of a limit, this means there exists some $$\epsilon$$ such that $$|X_n(\omega) - X(\omega)| > \epsilon$$ for infinitely many $$n$$. That is, there exists $$\epsilon$$ such that $$\omega \in \bigcap_n \bigcup_{k \ge n} A_{k, \epsilon}$$. Then $$P(\limsup_n A_{n, \epsilon}) \ge P(\lim_n X_n(\omega) \ne X(\omega))$$ for this particular $$\epsilon$$. So if almost sure convergence holds in the sense of the second definition, it also holds in the sense of the first definition.