Borel-Cantelli Lemma

Last updated Nov 1, 2022

Let $(X, \mathcal{M}, \mu)$ be a Measure Space and let $({A}{n}){n=1}^{\infty} \in \mathcal{M}$. If $$\sum\limits_{n=1}^{\infty} \mu(A_{n}) < \infty$$ then $$\mu(\limsup\limits_{n \to \infty} A_{n}) = 0$$

Since Convergence of Series implies tail goes to 0, we have that $$\begin{align*} 0 &= \lim\limits_{n \to \infty} \sum\limits_{k=n}^{\infty} \mu(A_{k})\\ &\geq \lim\limits_{n \to \infty} \mu(\bigcup\limits_{k \geq n}A_{k}) & \text{(subadditivity)}\\ &=\mu(\limsup\limits_{n \to \infty} A_{n}) &\text{(continuity from above)}. \end{align*}$$ where Continuity of Measures from Above holds because for $\epsilon = 1$, there exists $N \in \mathbb{N}$ so that $\forall n \geq N$, $\mu(\bigcup\limits_{k \geq n}A_{k}) \leq 1$.

TODO - from Durret - Probability Theory and Examples, uses Fubini’s Theorem and sum of Indicator Functions.

1. In the context of Probability Measures, we might write $\mathbb{P}(A_{n} \text{ i.o.}) = 0$.

• Limsup of Set Sequence