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