
# 05 The Borel-Cantelli Lemmas


Let $$(\Omega, {\cal F}, \prob)$$ be a probability space, and let $$A_1, A_2, A_3, \ldots \in {\cal F}$$ be a sequence of events. We define the following events derived from the sequence $$(A_n)_n$$:

\begin{eqnarray*}
\limsup A_n &=& \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty A_n, \\
\liminf A_n &=& \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty A_n.
\end{eqnarray*}

What is the meaning of these events? If we think of the sequence $$A_n$$ as representing a sequence of (not necessarily independent) probabilistic experiments, then we can translate the first event into words as

\begin{eqnarray*}
\limsup A_n &=& \textrm{the event that for all $$N\ge 1$$ there is an $$n\ge N$$ such}
\\ & & \textrm{that $$A_n$$ occurred''} \\ &=& \textrm{the event that infinitely many of the $$A_n's occurred''.} \end{eqnarray*} For this reason, the event \(\limsup A_n$$ is often denoted by $$\{ A_n\textrm{ infinitely often}\}$$ or $$\{A_n \textrm{ i.o.}\}$$.

The definition of the event $$\liminf A_n$$ can similarly be given meaning by writing
\begin{eqnarray*}
\liminf A_n &=& \textrm{the event that there exists an $$N\ge 1$$ such that $$A_n$$ }
\\ & & \textrm{occurred for all $$n\ge N''} \\ &=& \textrm{the event that all but finitely many of the \(A_n's occurred''.} \end{eqnarray*} \begin{exercise} Prove that for any \(\omega \in \Omega$$ we have
\begin{eqnarray*}
\ind_{\limsup A_n}(\omega) &=& \limsup_{n\to\infty} \ind_{A_n}(\omega), \\
\ind_{\liminf A_n}(\omega) &=& \liminf_{n\to\infty} \ind_{A_n}(\omega).
\end{eqnarray*}
\end{exercise}

\begin{thm}[Borel-Cantelli lemmas] \
\renewcommand{\labelenumi}{(\roman{enumi})}
\begin{enumerate}

1. \If\ \ $$\sum_{n=1}^\infty \prob(A_n) < \infty$$ then $$\prob(A_n\textrm{ i.o.}) = 0$$.
2. \If\ \ $$\sum_{n=1}^\infty \prob(A_n) = \infty$$ and $$(A_n)_{n=1}^\infty$$ are independent then $$\prob(A_n\textrm{ i.o.}) = 1$$.

\end{thm}

\begin{proof}

We essentially already proved part (i) in the first lecture, but here is a more general repetition of the same argument.

$\prob(A_n\textrm{ i.o.}) = \prob\left(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty A_n \right) %\\ %&=& \le \inf_{N\ge1} \prob\left( \bigcup_{n=N}^\infty A_n\right) \le \inf_{N\ge 1} \sum_{n=N}^\infty \prob(A_n).$
Since we assumed that $$\sum_{n=1}^\infty \prob(A_n)$$, converges, this last expression is equal to $$0$$.

Proof of (ii):
Consider the complementary event that the $$A_n's did \emph{not} occur for infinitely many values of \(n$$. Using De-Morgan's laws, we get

\begin{eqnarray*}
\prob(\{A_n\textrm{ i.o.}\}^c) &=& \prob\left(\left(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty A_n\right)^c \right) = \prob\left( \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty A_n^c \right)
\\ &\le &\sum_{N=1}^\infty \prob\left( \bigcap_{n=N}^\infty A_n^c \right).
\end{eqnarray*}

So, to show that this is 0 (under the assumptions that $$\sum_{n=1}^\infty \prob(A_n) = \infty$$ and that the events are independent), we show that $$\prob\left( \cap_{n=N}^\infty A_n^c \right) = 0$$ for all $$N\ge 1$$. Since the events are independent, the probability of the intersection is the product of the probabilities, so we need to show that

$\prod_{n=N}^\infty (1-\prob(A_n)) = 0,$

or equivalently (taking minus the logarithm) that

$-\sum_{n=N}^\infty \log(1-\prob(A_n)) = \infty.$

But \)-\log(1-x) \ge x\) for all $$x>0$$, so this follows from the assumption that the series of probabilities diverges.
\end{proof}