Let $\Omega$ be the sample space, and, $X$ and $\{X_n\}_{n\geq1}$ be random variables. We say $X_n$ converges almost surely to $X$ through the following definition.
$$\begin{array}{ccl}X_n\overset{a.s.}{\longrightarrow}X &\Leftrightarrow&\mbox{(1) }\exists\mbox{ null set }N\mbox{ such that }\forall\,\omega\in\Omega\setminus N,\,\underset{n\rightarrow\infty}{\lim}X_n(\omega)=X(\omega)\mbox{ finite} \\
& & \\
&\Leftrightarrow&\mbox{(2) }\forall\,\varepsilon>0,\,\underset{m\rightarrow\infty}{\lim}\mathscr{P}\left\{|X_n-X|\leq\varepsilon,\,\forall\,n\geq m\right\}=1 \\
& &\qquad\mbox{or, }\underset{m\rightarrow\infty}{\lim}\mathscr{P}\left\{|X_n-X|>\varepsilon,\,\mbox{for some}\,n\geq m\right\}=0 \\
& & \\
&\Leftrightarrow&\mbox{(3) }\forall\,\varepsilon>0,\,\mathscr{P}\left\{|X_n-X|>\varepsilon,\,\mbox{i.o.}\right\}=0. \\ \end{array}$$
The first one is the basic definition of almost surely convergence. But, it might be hard to be used since we must check the convergence for every element in $\Omega$. Thus, here we introduce some usual tools for checking almost surely convergence derived by the definition. Finally, we know that the Borel-Cantelli Lemma plays an important role as the most useful tool here.
(1) $\iff$ (2)
$$\begin{array}{ccl}X_n\overset{a.s.}{\longrightarrow}X
& \iff & \mathscr{P}\{\omega\mid\underset{n\rightarrow\infty}{\lim}X_n(\omega)=X(\omega)\}=1 \\
& \iff & \mathscr{P}\{\omega\mid\forall\, \varepsilon > 0,\, \exists\, m(\varepsilon)\mbox{s.t.}\forall\,n\geq m, \, |X_n(\omega) - X(\omega)| < \varepsilon\} = 1 \\
& \iff & \mathscr{P}\{\bigcap_{\varepsilon > 0}\bigcup_{m > 0}\bigcap_{n\geq m}(\omega : \, |X_n(\omega)-X(\omega)| < \varepsilon)\}=1 \\
\mbox{Let }\varepsilon=\frac{1}{k} & \iff &\forall\,k\geq1,\mathscr{P}\{\bigcup_{m=1}^\infty\bigcap_{n\geq m}(\omega : \, |X_n(\omega)-X(\omega)| < \frac{1}{k})\}=1 \\
& \iff & \forall\,k\geq1,\mathscr{P}\{\liminf_n A_n\}=1,\,A_n=\{\omega:\, |X_n(\omega)-X(\omega)| < \frac{1}{k}\}\\
& \iff & \forall\,k\geq1,\mathscr{P}\{\limsup_n A^c_n\}=0 \\
& \iff & \forall\,k\geq1,\mathscr{P}\{\bigcap_{m\geq1}\bigcup_{n\geq m} A^c_n\}=0 \\
& \iff & \forall\,k\geq1,\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{\bigcup_{n\geq m} A^c_n\}=0\;\because\bigcup_{n\geq m} A^c\downarrow\bigcap_{m\geq1}\bigcup_{n\geq m} A^c. \\
\end{array}$$ Thus, we have $$\forall\,\varepsilon>0,\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{|X_n(\omega)-X(\omega)|>\varepsilon\,\mbox{for some}\,n\geq m\}=0.$$
& \iff & \mathscr{P}\{\omega\mid\underset{n\rightarrow\infty}{\lim}X_n(\omega)=X(\omega)\}=1 \\
& \iff & \mathscr{P}\{\omega\mid\forall\, \varepsilon > 0,\, \exists\, m(\varepsilon)\mbox{s.t.}\forall\,n\geq m, \, |X_n(\omega) - X(\omega)| < \varepsilon\} = 1 \\
& \iff & \mathscr{P}\{\bigcap_{\varepsilon > 0}\bigcup_{m > 0}\bigcap_{n\geq m}(\omega : \, |X_n(\omega)-X(\omega)| < \varepsilon)\}=1 \\
\mbox{Let }\varepsilon=\frac{1}{k} & \iff &\forall\,k\geq1,\mathscr{P}\{\bigcup_{m=1}^\infty\bigcap_{n\geq m}(\omega : \, |X_n(\omega)-X(\omega)| < \frac{1}{k})\}=1 \\
& \iff & \forall\,k\geq1,\mathscr{P}\{\liminf_n A_n\}=1,\,A_n=\{\omega:\, |X_n(\omega)-X(\omega)| < \frac{1}{k}\}\\
& \iff & \forall\,k\geq1,\mathscr{P}\{\limsup_n A^c_n\}=0 \\
& \iff & \forall\,k\geq1,\mathscr{P}\{\bigcap_{m\geq1}\bigcup_{n\geq m} A^c_n\}=0 \\
& \iff & \forall\,k\geq1,\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{\bigcup_{n\geq m} A^c_n\}=0\;\because\bigcup_{n\geq m} A^c\downarrow\bigcap_{m\geq1}\bigcup_{n\geq m} A^c. \\
\end{array}$$ Thus, we have $$\forall\,\varepsilon>0,\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{|X_n(\omega)-X(\omega)|>\varepsilon\,\mbox{for some}\,n\geq m\}=0.$$
(2) $\iff$ (3)
Without loss of th generality, let $X=0$. Define $A_m=\bigcap_{n\geq m}\{|X_n|\leq\varepsilon\}$, hence, $A_m^c=\bigcup_{n\geq m}\{|X_n|>\varepsilon\}$ and $$\{|X_n|>\varepsilon\mbox{ i.o.}\}=\bigcap_{m\geq1}\bigcup_{n\geq m}\{|X_n|>\varepsilon\}=\bigcap_{m\geq1}A_m^c.$$ By (2), we have $X_n\overset{a.s.}{\longrightarrow}0\iff\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{A_m^c\}=0$. Thus by the monotone property of sets,
Without loss of th generality, let $X=0$. Define $A_m=\bigcap_{n\geq m}\{|X_n|\leq\varepsilon\}$, hence, $A_m^c=\bigcup_{n\geq m}\{|X_n|>\varepsilon\}$ and $$\{|X_n|>\varepsilon\mbox{ i.o.}\}=\bigcap_{m\geq1}\bigcup_{n\geq m}\{|X_n|>\varepsilon\}=\bigcap_{m\geq1}A_m^c.$$ By (2), we have $X_n\overset{a.s.}{\longrightarrow}0\iff\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{A_m^c\}=0$. Thus by the monotone property of sets,
$$A_m^c\downarrow\mbox{ w.r.t. }m\Rightarrow \mathscr{P}\{|X_n|>\varepsilon\mbox{ i.o.}\}=\mathscr{P}\{\bigcap_{m\geq1}A_m^c\}=\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{A_m^c\}=0.$$
$\Box$
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