Proof.

1. (1)

   $\mu(F)=\mu(E)+\mu(F\del E)\geq \mu(E)$

2. (2)

   For each $n\in\N^{+}$ let $F_{n}\define E_{n}\del \cups{n-1}{i=1}E_{i}$ then $\curl{F_n}_{n\in\N}$ is pairwise disjoint and $\dcups{}{n\in\N}F_{n}=\cups{}{n\in\N}E_{n}$ hence

   \begin{align*}\mu \parens{\cups{}{n\in\N}E_n}=\mu \parens{\dcups{}{n\in\N}F_n }=\sums{}{n\in\N}\mu(F_{n})\leq \sums{}{n\in\N}\mu(E_{n})\end{align*}

   
   
3. (3)

   For each $n\in\N^{+}$ let $F_{n}\define E_{n}\del E_{n-1}$ then $\curl{F_n}_{n\in\N}$ is pairwise disjoint and $\dcups{}{n\in\N}F_{n}=\cups{}{n\in\N}E_{n}$ hence

   \begin{align*}\mu \parens{\cups{}{n\in\N}E_n}&=\mu \parens{\dcups{}{n\in\N}F_n }=\sums{}{n\in\N}\mu(F_{n})\\&=\limit{n\to\infty}\sums{}{m\leq n}\mu(F_{m})=\limit{n\to\infty}\mu(E_{n})=\SUP{n\in\N}\mu(E_{n})\end{align*}

   
   
4. (4)

   Since $\mu(E_{k})<\infty$

   \begin{align*}\limit{n\to\infty}\mu(E_{n})&=\mu(E_{k})-\limit{n\to\infty}\mu(E_{k}\del E_{k+n})\\&=\mu(E_{k})-\mu\parens{E_k\del\caps{}{n\in\N^+}E_{k+n}}=\mu\parens{\caps{}{n\in\N^+}E_{k+n}}=\mu\parens{\caps{}{n\in\N^+}E_n}\end{align*}

   
   
5. (5)

   By (1) and (3)

   \begin{align*}\mu\parens{\limitinf{n\to\infty}E_n}&=\mu\parens{\cups{}{n\in\N^+}\caps{}{m\geq n}E_m}=\limit{n\to\infty}\mu\parens{\caps{}{m\geq n}E_m}\\&\leq\limit{n\to\infty}\INF{m\geq n}\mu(E_{m})=\limitinf{n\to\infty}\mu(E_{n})\end{align*}

   
   
6. (6)

   By (1) and (4)

   \begin{align*}\mu\parens{\limitsup{n\to\infty}E_n}&=\mu\parens{\caps{}{n\in\N^+}\cups{}{m\geq n}E_m}=\limit{n\to\infty}\mu\parens{\cups{}{m\geq n}E_m}\\&\geq\limit{n\to\infty}\SUP{m\geq n}\mu(E_{m})=\limitsup{n\to\infty}\mu(E_{n})\end{align*}

   
   

$\square$