Prove $X_n\rightarrow X$ a.s iff $\forall \epsilon > 0$, $P({|X_n-X|>\epsilon, i.o})=0$ I am trying to prove,
if $X_n\rightarrow X$ a.s iff $\forall \epsilon > 0$, $P({|X_n-X|>\epsilon, i.o})=0$
Starting with the right side, If $X_n\rightarrow X$ a.s then we get
$P(\lim_{n\to\infty} X_n=X)=1$, $\forall m\geq 0$ then $P(|X_i-X|>1/m, i.o)=0$\
Now how do we conclude like that? I mean why did we use $1/m$? From now, what will be the next step?
 A: Suppose $X_n\to X$ almost surely. Then $\mathbb P\left(\lim_{n\to\infty}X_n=X\right)=1$, or to be more precise,
$$
\mathbb P\left(\lim_{n\to\infty} \{\omega\in\Omega : X_n(\omega) = X(\omega)\} \right) = 1.
$$
We can rewrite this as:
$$
\mathbb P\left(\lim_{n\to\infty} \{\omega\in\Omega : |X_n(\omega)-X(\omega)|=0\} \right) = 1.
$$
So for any $\varepsilon>0$, we may choose $N$ such that $\mathbb P\left(\{\omega\in\Omega: |X_n(\omega)-X(\omega)|<\varepsilon\}\right) = 0$. It follows that
$$
\mathbb P\left(\limsup_{n\to\infty} \{|X_n-X|>\varepsilon\}\right) = \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k-X|>\varepsilon\}\right) = 0.
$$
Conversely, suppose that $\mathbb P(\limsup_{n\to\infty}\{|X_n-X|>\varepsilon\})=0$ for all $\varepsilon>0$. Then for any $\varepsilon>0$, we have
$$
\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k-X|>\varepsilon\}\right) = 0.
$$
This implies that there exists $N$ such that $\mathbb P\left(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|=0\}\right)=n$ for $n\geqslant N$, and hence $X_n\to X$ almost surely.
A: Why the probability of ${ω∈Ω:|Xn(ω)−X(ω)|<ε})$ is 0? I think that it has to be 1. And can you argue why you know that It follows that
P(lim supn→∞{|Xn−X|>ε})=P(⋂n=1∞⋃k=n∞{|Xk−X|>ε})=0.
