> [!theorem] First Borel-Cantelli Lemma > > Let $(\Omega, \cf, \mu)$ be a [[Measure Space|measure space]] and $\seq{E_n} \subset \cf$. If > $ > \sum_{n \in \nat}\mu(E_n) < \infty > $ > then > $ > \mu\paren{\limsup_{n \to \infty}E_n} = 0 > $ > *Proof*. Recall that > $ > \limsup_{n \to \infty}E_n = \bigcap_{n \in \nat}\bigcup_{k \ge n}E_k \subset \bigcap_{n \ge N}\bigcup_{k \ge n}E_k \quad \forall N \in \nat > $ > Since the sum is finite, for any $\varepsilon > 0$, there exists $N \in \nat$ such that $\sum_{n \ge N}\mu(E_n) < \varepsilon$. By monotonicity and subadditivity > $ > \begin{align*} > \mu\paren{\limsup_{n \to \infty}E_n} &\le \mu\paren{\bigcap_{n \ge N}\bigcup_{k \ge n}E_k} \\ > &\le \mu\paren{\bigcup_{k \ge N}E_k} \\ > &\le \sum_{k \ge N}\mu(E_k) < \varepsilon > \end{align*} > $ > As this applies for all $\varepsilon > 0$, $\mu\paren{\limsup_{n \to \infty}E_n} = 0$. > [!theoremb] Second Borel-Cantelli Lemma > > Let $(\Omega, \cf, \mathbf{P})$ be a [[Probability|probability space]], and $\seq{E_n} \subset \cf$ be [[Probabilistic Independence|mutually independent]] events. If > $ > \sum_{n \in \nat}\bp\bracs{E_n} = \infty > $ > then $\bp\bracs{\limsup_{n \to \infty}E_n} = 1$. > > *Proof*. By definition of the limits, > $ > \paren{\limsup_{n \to \infty}E_n}^c = \liminf_{n \to \infty}(E_n^c) = \bigcup_{n \in \nat}\bigcap_{k \ge n}E_k^c > $ > By subadditivity > $ > \bp\bracs{\bracs{\limsup_{n \to \infty}E_n}^c} \le \sum_{n \in \nat}\bp\bracs{\bigcap_{k \ge n}E_k^c} > $ > and it's sufficient to show that all summands on the right are zero. > > Denote $p_n = \bp\bracs{E_n}$, then by monotonicity and mutual independence, > $ > \bp\bracs{\bigcap_{k \ge n}E_k^c} \le \bp\bracs{\bigcap_{k = n}^NE_k^c} = \prod_{k = n}^N(1 - p_k) \quad \forall N \in \nat > $ > Since $1 - p_n \le e^{-p_n}$, > $ > \bp\bracs{\bigcap_{k \ge n}E_m^c} \le \prod_{k \ge n}(1 - p_k) \le e^{-\sum_{k \ge n}p_n} = 0 > $