Stein 1.16: The Boel Cantelli Lemma

$ \textit{The Borel-Cantelli Lemma} $

Let $(X,\mathcal{M},\mu)$ be a measure space. Given $\{E_k\}_{k=1}^\infty$ is a sequence of $\mu-$measurable subsets of $X$ and that $\sum_{k=1}^\infty \mu(E_k) < \infty$. Then, $\limsup E_k$ is also $\mu-$ measurable with measure 0.

$\textit{Preliminaries:}$ The $\limsup$ of a sequence of sets can be defined in several ways. The initial definition given in the book is $$\limsup E_k = \{x \in{X} : x\in{E_k} \quad \text{for infinitely many } k\}$$ Similarly, $$\liminf E_k = \{x \in{X} : x\in{E_k} \quad \text{for all but finitely many } k\}$$ The usual $\limsup$ of a sequence of real numbers $\{a_n\}$ gives us, in some sense, the number which is the largest accumulation point of $\{a_n\}$. Here, "largest" refers to comparison of real numbers with the ordering relation $" \leq "$. When looking at the generalized case when terms in our sequence are sets, the "largest accumulation set" is determined by the ordering relation on sets $" \subseteq "$. Equivalently, $$\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right) \quad \text{and} \quad \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$$ $\textit{Example 1}$: Let $X=\{0,1\}$ and let $E_k = \{\{-2\},\{4\},\{-1\}, \{0\},\{1\},\{0\},\{1\},...\}$.
So, $E_1 \cup E_2 \cup \cdots = \{0\} \cup \{1\} = \{0,1\}$. So $\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)= \bigcap_{k=1}^\infty \{0,1\} = \{0,1\}$...the set of $x\in{X}$ s.t. $x$ is in infinitely many $E_k$. On the other hand, $E_1 \cap E_2 \cap \cdots = \emptyset$ and so $\liminf E_k = \emptyset$...the only subset of $X$ that is in all but finitely many $E_k$. Note that in general, $\liminf E_k \subseteq \limsup E_k $: If $x$ is in all but finitely many $E_k$, then $x$ is in infinitely many $E_k$. However, if we consider our counting index to be $\mathbb{Z}$, $\{E_k\}_{k=-\infty}^{k+ \infty} $, then if, say, $x$ is in $E_k$ for all positive $k$, then $x$ is in infinitely many $E_k$, so $x\in{\limsup E_k }$. However, $x$ is still not in infinitely many $E_k$. So $x \not \in{\liminf E_k}$.

$\textit{Proof}$ :
First note that $\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)$ and $ \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$ are just combinations of countable unions and intersections of measurable $E_k$, so are measurable. $\sum_{k=1}^\infty \mu(E_k) < \infty \Longrightarrow \mu(E_k) \longrightarrow 0$ i.e. $\forall \epsilon > 0$, $\exists N\in{\mathbb{N}}$ s.t. as soon as $k \geq N$ then $\mu(E_k) \leq \dfrac{\epsilon}{2^k}$. Since $\bigcap_{N=1}^\infty(\bigcup_{k \geq N} E_k ) \subseteq \bigcup_{k \geq N} E_k $ then, since $\mu$ is a measure, it is monotonic and sub-additive: $$\mu(\limsup E_k) = \mu \left(\bigcap_{N=1}^\infty\left(\bigcup_{k \geq N} E_k \right) \right) \leq \mu \left( \bigcup_{k \geq N} E_k \right) \leq \sum_{k=N}^\infty \mu(E_k) $$ $$\leq \sum_{j=1}^\infty \mu(E_k) \leq \sum_{j=1}^\infty \frac{\epsilon}{2^k} = \epsilon \sum_{j=1}^\infty \frac{1}{2^k} = \epsilon$$ Since $\epsilon > 0 $ was chosen arbitrarily, we get $\mu(\limsup E_k) = 0$ $\blacksquare$