Partial Converse of Borel-Cantelli Lemma

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a Probability Space and suppose $({A}{n}){n=1}^{\infty} \subset \mathcal{B}$ is a Sequence of independent events. If $$\sum\limits_{n=1}^{\infty} \mathbb{P}(A_{n}) = \infty$$ then $$\mathbb{P}(A_{n} \text{ i.o. }) = 1$$

Observe that for any $M \in \mathbb{N}$ $$\begin{align*} \mathbb{P}(\bigcap\limits_{k \geq M} A_{k}^{C}) &= \prod\limits_{k=M}^{\infty} \mathbb{P}(A_{k}^{C})\\ &=\prod\limits_{k=M}^{\infty} (1 - \mathbb{P}(A_{k}))\\ &\leq \prod\limits_{k=M}^{\infty} \exp(- \mathbb{P}(A_{k})) \tag{3}\\ &=\exp\left(-\sum\limits_{n=M}^{\infty} \mathbb{P}(A_{n})\right)\\ &= 0 \tag{5} \end{align*}$$ where at (3) we used the fact that $1 - x \leq e^{-x}$ and at (5) we used the Continuity of Exponentiation. Therefore, $$\begin{align*} \mathbb{P}(\limsup\limits_{n \to \infty} A_{n}) &= \lim\limits_{n \to \infty} \mathbb{P}(\bigcup\limits_{k \geq n} A_{n}) & \text{(continuity from above)}\\ &=\lim\limits_{n \to \infty} (1 - \mathbb{P}(\bigcap\limits_{k \geq n} A_{n}^{C})) &\text{De Morgan’s Law}\\ & =\lim\limits_{n \to \infty} 1\\ &= 1 \end{align*}$$ $\blacksquare$

• Limsup of Set Sequence
• Linear and Exponential Inequality
• Continuity of Measures from Above
• De Morgan’s Law