Muhammad Haris Rao
wTheorem(Borel-Canetlli Lemma): Let $\left( \Omega, \mathcal{F}, \mathbf{P} \right)$ be a probability space and $\{ A_n \}_{n \ge 1}$ a sequence of events from $\mathcal{F}$. If $$ \sum_{n \ge 1} \mathbf{P} \left( A_n \right) < \infty $$ then $\mathbf{P} \left( A_n \text{ i.o} \right) = 0$ where $$ \{ A_n \text{ i.o} \} = \bigcap_{n \ge 1} \bigcup_{m \ge n} A_m $$ is the probability of infinitely many of the $\{ A_n \}_{n \ge 1}$ occuring.
Proof. Let $B_m = \bigcup_{n \ge m} A_m$. Clearly, $B_1 \supseteq B_2 \supseteq B_3 \supseteq \cdots$ so by continuity of probabilities, $$ \mathbf{P} \left( A_n \text{ i.o} \right) = \mathbf{P} \left( \bigcap_{n \ge 1} \bigcup_{m \ge n} A_m \right) = \mathbf{P} \left( \bigcap_{n \ge 1} B_n \right) = \lim_{n \to \infty} \mathbf{P}\left( B_n \right) = \lim_{n \to \infty} \mathbf{P} \left( \bigcup_{m \ge n} A_n \right) \le \lim_{n \to \infty} \sum_{m \ge n} \mathbf{P} \left( A_n \right) = 0 $$ since $\sum_{n \ge 1} \mathbf{P} \left( A_n \right) < \infty$.$\blacksquare$
Theorem(Second Borel-Canetlli Lemma): Let $\left( \Omega, \mathcal{F}, \mathbf{P} \right)$ be a probability space and $\{ A_n \}_{n \ge 1}$ a sequence of independent events from $\mathcal{F}$. If $$ \sum_{n \ge 1} \mathbf{P} \left( A_n \right) = \infty $$ then $\mathbf{P} \left( A_n \text{ i.o} \right) = 1$.
Proof. We have for every $n \ge 1$ that $$ \sum_{m \ge n} \mathbf{P} \left( A_m \right) = \infty $$ So then using the inequality $e^{-x} \ge 1 - x$ for all $x \in \mathbb{R}$ $$ 0 = \exp{(-\infty)} = \exp{ \left( - \sum_{m \ge n} \mathbf{P} \left( A_m \right) \right)} = \prod_{m \ge n} \exp{\left(-\mathbf{P}(A_n) \right)} \ge \prod_{m \ge n} \left( 1 - \mathbf{P} \left( A_n \right) \right) = \prod_{m \ge n} \mathbf{P} \left( A_m^c \right) = \mathbf{P} \left( \bigcap_{m \ge n} A_m^c \right) $$ Taking complements, we obtain then $$ 1 \le \mathbf{P} \left( \left( \bigcap_{m \ge n} A_m^c \right)^c \right) = \mathbf{P} \left( \bigcup_{m \ge n} A_m \right) $$ Then by continuity of probabilities, $$ \textbf{P} \left( A_n \text{ i.o.} \right) = \textbf{P} \left( \bigcap_{n \ge 1} \bigcup_{m \ge n} A_m \right) = \lim_{n \to \infty} \mathbf{P} \left( \bigcup_{m \ge n} A_m \right) \ge \lim_{n \to \infty} 1 = 1 $$ So it follows that $\textbf{P} \left( A_n \text{ i.o.} \right) = 1$ as was claimed. $\blacksquare$