Weak law vs strong law of large numbers - intuition I was wondering if my intuition behind the weak law (WLLN) and strong law of large numbers (SLLN) is correct.
The WLLN says that, if you consider a sequence $X_1,X_2,...$,of $i.i.d.$ random variables with finite mean, $E|X|<\infty$, then define $\bar{X}_n=\frac{X_1+X_2+...+X_n}{n}$.
As $n \rightarrow \infty$, the WLLN says that $\bar{X}_n\xrightarrow{p} E[X]=\mu$
The intution is that for $n$ larger and larger there is higher probability that $\bar{X}_n$ takes value closer and closer to $\mu$. However, it is still possible that for any $\epsilon>0$ , we have $|\bar{X}_n-\mu|>\epsilon$ occurs an infinite number of times although at infrequent intervals.
The SLLN claims that as $n \rightarrow \infty$, than $\bar{X}_n \xrightarrow{a.s.} E[X]=\mu$
However, we can still have $|\bar{X}_n-\mu|>\epsilon$, but this will happen a finite number of times.
Are my statements correct?

This is a supplement to Zhanxiong's comprehensive answer that addressed OP's concerns.
It would be apt to have a brief recollection of the concepts of almost everywhere convergence and convergence in probability.
$\bullet$ Consider a measure space $(X, \boldsymbol{\mathfrak A}, \mu).$ A sequence of $\boldsymbol{\mathfrak A}$-measurable extended real-valued functions $\langle f_n\rangle_{n\in\mathbb N}$ on $D\in \boldsymbol{\mathfrak A}$ converges almost everywhere to $f:=\lim_{n\to\infty} f_n(x)$ if there exists a null set $N\subset D$  such that $f$ exists for all $x\in D\setminus N $ and $f(x)\in \mathbb R$ for all $x\in  D\setminus N.$
$\bullet$ $f_n \overset{\textrm{a.e.}}{\to} f$ on $D$ iff $\mu\{D:f_n\nrightarrow f \} = 0.$ Now, by definition, $\{D:\lim_{n\to\infty} f_n = f\} = \bigcap_{m\in\mathbb N}\bigcup_{N\in\mathbb N}\bigcap_{p\in \mathbb N}\left\{D:|f_{N+p}- f|<\frac1m\right\}$; so using De Morgan's law
\begin{align}\mu\{D:f_n\nrightarrow f \} &= \mu\left\{D\setminus\bigcap_{m\in\mathbb N}\bigcup_{N\in\mathbb N}\bigcap_{p\in \mathbb N}\left\{D:|f_{N+p}- f|<\frac1m\right\}\right\}\\ &= \mu\left\{\bigcup_{m\in\mathbb N}\bigcap_{N\in\mathbb N}\bigcup_{p\in \mathbb N}\left\{D\setminus\left\{D:|f_{N+p}- f|<\frac1m\right\}\right\}\right\}\\&= \mu\left\{\bigcup_{m\in\mathbb N}\bigcap_{N\in\mathbb N}\bigcup_{p\in \mathbb N}\left\{D:|f_{N+p}- f|\geq\frac1m\right\}\right\} .\tag 1\label 1\end{align}
As countable union of null sets is null, from $\eqref 1$ dictates that for every $m,$ $$\mu\left\{\bigcap_{N\in\mathbb N}\bigcup_{n > N}\left\{D:|f_{n}- f|\geq\frac1m\right\}\right\} = 0.$$ That is, $$f_n \overset{\textrm{a.e.}}{\to} f~\textrm{on}~D\iff \mu\left\{\limsup_{n\to \infty}\left\{D:|f_{n}- f|\geq\frac1m\right\}\right\} = 0~~~\forall m\in \mathbb N.\tag 2\label 2 $$
$\bullet$ If $\mu(D)<\infty,$ by continuity property, $\eqref 2$ can be modified to $$f_n \overset{\textrm{a.e.}}{\to} f~\textrm{on}~D\iff \lim_{n\to \infty}\mu\left\{D:|f_{n}- f|\geq\frac1m\right\} = 0~\forall m\in \mathbb N.\tag 3\label 3$$
$\bullet$ $\langle f_n\rangle_{n\in\mathbb N}$ converges in measure to $f$ on $D$ if for every $\varepsilon >0, ~\eta >0,$ there exists $N_{\varepsilon,\eta}\in \mathbb N$ such that for $n\geq N_{\varepsilon,\eta} $ $$\mu\{D:|f_{n}- f|\geq \varepsilon\}<\eta,\tag 4$$ which is nothing but the re-statement of $\lim_{n\to\infty}\mu\{D:|f_{n}- f|\geq \varepsilon\} = 0$ for every $\varepsilon >0.$
$\bullet$ From $\eqref 3,$ it is clear that for $\mu(D)<\infty,$ almost everywhere convergence implies convergence in measure. This is always true for probability measures.
$\bullet$ What are these two types of convergences trying to convey?
It is clear almost everywhere/almost surely convergence is related with pointwise convergence. That is for a sequence of random variables $X_n$ on $\Omega,$ $$\mathbb P\left[\lim_{n\to \infty} X_n(\omega) = X(\omega)\right] = 1.$$



$\bullet$ Convergence in probability means $$\lim_{n\to\infty}\mathbb P[|X_n-X|<\varepsilon] = 1. $$  The sequence of probabilities $\langle\mathbb P[|X_n-X|<\varepsilon] \rangle_{n\in\mathbb N}$ is of concern here.

Convergence in probability requires that for any $\varepsilon > 0$ the sequence of functions must be within an $\varepsilon$-band of the function $X$ over a set of points from the sample space whose probability increases to one as $n \to\infty.$





References:
$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $2014, $ sec. $1\S6.$
$\rm[II]$ Introduction to Statistical Limit Theory, Alan M. Polansky, Taylor and Francis, $2011,$ sec. $3.2-3.3.$
