1.16

This exercise is asking us to prove the Borel-Cantelli Lemma. In the measure theory settings, it states:

Suppose $\lbrace E_n \rbrace_{n=1}^{\infty}$ is a countable family of subsets of $\mathbb{R}^d$, and ($\dagger$) $$\sum_{k=1}^{\infty} m(E_k) < \infty$$ The set $E = \bigcap_{n=1}^{\infty} \bigcup_{k \geq n} E_k$ has measure zero.

Part a.) We first need to show $E$ is a measurable set. Recall that the measurable sets form a $\sigma$-algebra, $\mathcal{M}$, which is closed under countable unions and intersections.

Since each $E_k \in \mathcal{M}$, certainly $A_n = (\bigcup_{k \geq n} E_k) \in \mathcal{M} \hspace{0.25cm} \forall n$. It follows that since $E = \bigcap_{n=1}^{\infty} A_n$, that $E \in \mathcal{M}$.

Part b.) Assume to the contrary that $m(E) = \delta > 0$. Notice that if we define: $$S_N = \bigcap_{k=1}^N \bigcup_{n \geq k} E_n = \bigcup_{n \geq N} E_n $$ Then, certainly, $S_N \searrow E$, and $\forall N$, $$\delta \leq m(S_N) = m(\bigcup_{n \geq N} E_n) \leq \sum_{n=N}^{\infty} m(E_n)$$ Which certainly contradicts $\dagger$, completing the proof.