Ex3-9

Chapter 3 Question 9
Let \{E_k\}_{k=1}^{\infty} be a sequence of sets so that \sum|E_k|_e is finite. Then for any positive number \epsilon>0, there exists an integer j>0 such that
\begin{align}
\sum_{k=j}^{\infty}\left|E_k\right|_e<\epsilon
\end{align}
From Theorem 3.4, we have
\begin{align}
\left|\bigcup_{k=j}^{\infty} E_k\right|_e\leq\sum_{k=j}^{\infty}\left|E_k\right|_e<\epsilon
\end{align}
From the definition of \lim \sup
\begin{align}
\lim \sup E_k=\bigcap_{i=1}^{\infty}\left(\bigcup_{k=i}^{\infty}E_k\right)
\end{align}
gives us that for each i, 
\begin{align}
\lim \sup E_k\subset \bigcup_{k=i}^{\infty} E_k
\end{align}
Then from the above part, we know for each \epsilon>0, there is a j such that
\begin{align}
\left|\bigcup_{k=j}^{\infty} E_k\right|_e<\epsilon
\end{align}
which implies
\begin{align}
\left|\lim \sup E_k\right|<\epsilon.
\end{align}
Since \epsilon can be arbitrary small, we have proven that \lim \sup E_k has measure zero.
Since 
\begin{align}
\lim \inf E_k \subset \lim\sup E_k
\end{align}
we can also get that \lim \inf E_k has measure zero.
Another proof(proof in other words)
Let E=\limsup E_{k} and F_{j}=\bigcup_{k=j}^{\infty} E_{k}. Then, F_{j} is a decreasing sequence of sets whose limit is \limsup E_{k}.
Then E \subset \bigcap_{j=n}^{\infty} F_{j} \subseteq F_{n} for all n, so |E|_{e} \leq\left|F_{n}\right|_{e} for all n by monotonicity of outer measure. 
Moreover, \left|F_{n}\right|_{e} \leq \sum_{k=n}^{\infty}\left|E_{k}\right|_{e} for all n. Since \left|F_{n}\right|_{e} \leq \sum_{k=n}^{\infty}\left|E_{k}\right|_{e} converges, for every \varepsilon >0, there exists N such that \sum_{k=n}^{\infty}\left|E_{k}\right|_{e}<\varepsilon for all n \ge N.
Thus, for every \varepsilon>0, we have |E|_{e} \leq \varepsilon, which implies that |E|_{e}=0.
Hence, E=\limsup E_{k} is measurable and has measure 0. Since \liminf E_{k} \subseteq \limsup E_{k}, \liminf E_{k} also has measure zero.