Two equivalent definitions of almost sure convergence of random variables. Let $\{X_n\}_{n=1,2,\cdots}$ is a sequence of random variables. There are equivalent definitions of almost sure convergence of random variables. How can one prove the equivalence?

$\mathbb{P}[\omega:\lim_{n\to\infty}X_n(\omega) = X(\omega)] = 1 \Leftrightarrow \lim_{n\to\infty}\mathbb{P}[\omega:\sup_{k>n}|X_k(\omega) - X(\omega)|>\epsilon] = 0$

 A: Well, let's denote by $Y_n$ the variable $\sup_{k>n} |X_k - X|$. Observe that


*

*$(Y_n)$ is a decreasing sequence of nonnegative random variables

*$Y_n \to 0$ iff $X_n \to X$

*"$\mathsf{P}\{Y_n > \epsilon\} \to 0$ for all $\epsilon$" means that $Y_n \to 0$ in probability


Now everything follows from the simple but useful fact: if a $(Y_n)$ is a monotone sequence, then its convergence in probability implies almost sure convergence (that the converse is true for any sequence is an extremely standard fact). This can be proved by different means, but the easiest proof that I know of is as follows:
Any monotone sequence converges (almost surely) to something (this is just standard calculus, and "almost surely" is actually irrelevant). Hence it also converges in probability to the same limit, hence if $Y_n \to Y$ in probability, then $Y$ must coincide with the almost sure limit.
A: The former implies
$$
\begin{eqnarray}
0 &=& P(\lim_{i\to\infty}X_i \neq X\,  \text{or} \lim_{i\to\infty}X_i \, \text{does not exists}) \\
&=& P(\omega:\exists n\in\mathbf{N}, \forall m\in\mathbf{N}, \exists i>m \,\,\,  \text{s.t.} \,\, |X_i(\omega) - X(\omega)| < 1/n)\\
&=& P(\bigcup_n \bigcap_m \bigcup_{i>m} \{\omega:|X_i(\omega) - X(\omega)| \ge 1/n\})\\
&\overset{\forall n}{\ge}& P(\bigcap_m \bigcup_{i>m} \{\omega:|X_i(\omega) - X(\omega)| \ge 1/n\})\\
&=&P(\limsup_{i\to\infty} \{\omega:|X_i(\omega) - X(\omega)| \ge 1/n\}) \\
&=&P(\lim_{i\to\infty} \sup_{j > i}\{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\})\\
&=&\lim_{i\to\infty} P(\sup_{j > i}\{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\}) \ge 0.\\
\end{eqnarray}
$$
The last line is justified by continuity of measures from above (see here). Now let's prove the converse. For some $n\in\mathbf{N}$,
$$
\begin{eqnarray}
0&=&\lim_{i\to\infty} P(\sup_{j > i}\{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\})\\
&=&P(\lim_{i\to\infty} \sup_{j > i}\{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\})\\
&=&P(\limsup_{i\to\infty} \{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\})\\
&=&P(\bigcap_i\bigcup_{j>i} \{\omega:|X_j(\omega) - X(\omega)| \ge 1/n\})\\
&=&P(\omega:\forall i, \exists j>i, \text{s.t.}\, |X_j(\omega) - X(\omega)| \ge 1/n)
\end{eqnarray}
$$
This implies, for each $n$,
$$
\begin{eqnarray}
1 &=& P(\omega: \exists i, \forall j>i, \text{s.t.}\, |X_j(\omega) - X(\omega)| < 1/n)\\
&=&P(\omega:\lim_{i\to\infty}|X_j(\omega) - X(\omega)| = 0)\\
&=&P(\omega:\lim_{i\to\infty}X_j(\omega) = X(\omega))
\end{eqnarray}
$$
Actually, I think this proof may need to be polished. But I hope it is not wrong in a bird's eye view at least.
A: Let us denote the probability space by the triple: $(\Omega, \mathbb{A}, \mathbb{P})$.
First we prove the former implies the latter:
Let us denote by $E$ the event that the sequence converges. That is: 
$$E \triangleq \{w \in \Omega | \lim_{n}Y_n(\omega) = Y(\omega)\} \in \mathbb{A} $$
Almost sure convergence implies that $\mathbb{P}(E) = 1$ (by definition of a.s. convergence).
We have:
\begin{align}
&\mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\}\right]= \\ 
&\mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \cap E \right] + \mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \cap E^C \right] \leq \\
&\mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \cap E \right] 
\end{align}
(In the last equality we applied monotonicity of measure and that $\mathbb{P}(E^C)=0$ ).
Continuity of measure from above (see:Wikipedia) implies we can take limits into probability measures:
\begin{align}
&\lim_{n}\left( \mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \cap E \right] \right)= \\
&\mathbb{P}\left[\lim_{n} \left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \cap E \right]=\mathbb{P}(\Phi) = 0 
\end{align}
Now we prove the latter implies the former. This is, again, easy by the  continuity of measure from above. Let $E$ be defined as above. We have:
\begin{align} 
0 &=\lim_{n}\left( \mathbb{P}\left[\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \right] \right) \\ 
&= \mathbb{P}\left[\lim_{n}\left\{w\in\Omega\middle| \sup_{k \geq n} |Y_k(\omega)-Y(\omega)|> \epsilon\right\} \right] = \mathbb{P}(E^c)
\end{align}
Where the first equality is continuity of measure and the last equality is just the definition of limits. Hence $P(E)=1$ as desired. 
