Second Borel–Cantelli lemma

Statement

Let $A_1, A_2, \dots$ be mutually independent events in a probability space $(\Omega, \mathcal{F}, P)$ with $\sum_{n=1}^{\infty} P(A_n) = \infty$. Then $$P\left(\limsup_{n \to \infty} A_n\right) = 1.$$ So almost surely, infinitely many of the events $A_n$ occur.

Remarks

Proof: the complement of $\limsup_n A_n$ is $\bigcup_n \bigcap_{m \ge n} A_m^c$. By independence and $1 - x \le e^{-x}$, $P\left(\bigcap_{m = n}^{N} A_m^c\right) = \prod_{m=n}^{N} (1 - P(A_m)) \le \exp\left(-\sum_{m=n}^{N} P(A_m)\right)$. The right side tends to $0$ as $N \to \infty$. So each $\bigcap_{m \ge n} A_m^c$ has probability $0$, and so does their union. Independence is necessary. For $A_n = A$ with $0 < P(A) < 1$ for all $n$, the sum diverges but $P(\limsup_n A_n) = P(A) < 1$. With the first lemma, it follows that for independent events, $P(\limsup_n A_n)$ is $0$ or $1$.

Explanations

L1 human

If independent events have chances that add up to infinity, then infinitely many of them happen, with probability $1$. For example, in an infinite sequence of tosses of a fair coin, the pattern "ten heads in a row" appears infinitely often.