Limit definition of Sets. Proposition 1.32 $X_{n}\xrightarrow{a.s.} X$ if and only if for any $\epsilon>0$ 
$P( | X_{n}- X |<\epsilon, \; \forall n\geq m )\rightarrow1$ $as$ $ m\rightarrow\infty$
Proof. Suppose first that $X_{n} \xrightarrow{a.s.} X$. Fix $\epsilon>0$, and $m\geq1$ define the event 
$A_m=\{| X_{n}-X |<\epsilon , \; \forall n\geq m\}$ 
Because $A_m,m\geq1$, is an increasing sequence of events, the continuity property of probabilities yields that 
$$\lim_{m\to \infty}P(A_m)=P(\lim_{m\to \infty}A_m)=P(| X_{n}- X |<\epsilon \text { for all n sufficiently large})\geq P(\lim_{n \to \infty}X_n=X)=1$$
To go the other way, assume that for any $\epsilon>0$
$P(| X_{n}-X |<\epsilon, \; \forall n\geq m)\rightarrow1$ as $m\rightarrow\infty$ 
let $\epsilon_{i}, i\geq1$ be a decreasing sequence of postive numbers that converge to $0 $ and let 
$A_{m,i}=\{| X_{n}-X |<\epsilon, \; \forall n\geq m\}$ 
Because $A_{m,i}\subset A_{m+1,i}$ and, by assumption, $\lim\limits_{m\to \infty}P(A_{m,i})=1$ ......


*

*The parts in bold that concern increasing and decreasing events confuse me thoroughly, I do not understand them :( 

*What does it actually mean to have $X_{n}\xrightarrow{a.s.} X$? 

 A: 1.1. Perhaps, you might better understand $A_m$'s as follows:
$$A_1 = \{\omega \in \Omega \mid |X_1(\omega) - X(\omega)| < \epsilon, |X_2(\omega) - X(\omega)| < \epsilon, |X_3(\omega) - X(\omega)| < \epsilon, \dots \}$$
$$A_2 = \{\omega \in \Omega \mid |X_2(\omega) - X(\omega)| < \epsilon, |X_3(\omega) - X(\omega)| < \epsilon, |X_4(\omega) - X(\omega)| < \epsilon, \dots \}$$
$$\vdots$$
$$A_m = \{\omega \in \Omega \mid |X_m(\omega) - X(\omega)| < \epsilon, |X_{m+1}(\omega) - X(\omega)| < \epsilon, |X_{m+2}(\omega) - X(\omega)| < \epsilon, \dots \}$$
$$\vdots$$
To be more elaborate, define:
$$B_1:= \{\omega \in \Omega \mid |X_1(\omega) - X(\omega)| < \epsilon\}$$
$$B_2:= \{\omega \in \Omega \mid |X_2(\omega) - X(\omega)| < \epsilon\}$$
$$B_3:= \{\omega \in \Omega \mid |X_3(\omega) - X(\omega)| < \epsilon\}$$
$$\vdots$$
$$B_m:= \{\omega \in \Omega \mid |X_m(\omega) - X(\omega)| < \epsilon\}$$
$$\vdots$$
Then $$A_1 = B_1 \cap B_2 \cap B_3 \cap \cdots = \bigcap_{k=1}^{\infty} B_k$$
$$A_2 = B_2 \cap B_3 \cap \cdots = \bigcap_{k=2}^{\infty} B_k$$
$$\vdots$$
$$A_m = B_m \cap B_{m+1} \cap \cdots = \bigcap_{k=m}^{\infty} B_k$$
$$\vdots$$
Hopefully, it is clear that $$A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m \subseteq \cdots$$

1.2 I have a feeling you made a typo. Did you mean
$$A_{m,i} = \{\omega \in \Omega \mid |X_m(\omega) - X(\omega)| < \epsilon_i, |X_{m+1}(\omega) - X(\omega)| < \epsilon_i, |X_{m+2}(\omega) - X(\omega)| < \epsilon_i, \dots \}$$
?
If so, then $$A_{m,i} \subseteq A_{m+1,i}$$ for the same reasons that $A_{m} \subseteq A_{m+1}$.



*This is almost sure convergence, the closest analogue to pointwise convergence from real analysis.


If we have random variables $X, X_1, X_2, \dots$ in $(\Omega, \mathscr F, \mathbb P)$, then $X_n$ converges almost surely to $X$ iff
$$P(\lim_{n \to \infty} X_n = X) := P(\omega \in \Omega \mid \lim_{n \to \infty} X_n(\omega) = X(\omega)) = 1$$
So if we collect all $(\omega_j)_{j \in J}$ s.t. $$\limsup X_n(\omega_j) = \liminf X_n(\omega_j) = X(\omega_j)$$, then we must have $$P(\bigcup_{j \in J} (\omega_j)) = 1$$
