Example 2.55 on p.42 in section 2C in "Measure, Integration & Real Analysis" by Sheldon Axler. Is my proof ok or not?

Question

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.

The following example is Example 2.55 on p.42 in section 2C in this book.

• Suppose $X$ is a set, $\mathcal{S}$ is a $\sigma$-algebra on $X$, and $w:X\to [0,\infty]$ is a function.
  Define a measure $\mu$ on $(X,\mathcal{S})$ by $$\mu(E)=\sum_{x\in E} w(x)$$ for $E\in\mathcal{S}.$ [Here the sum is defined as the supremum of all finite subsums $\sum_{x\in D} w(x)$ as $D$ ranges over all finite subsets of $E.$]

I tried to prove that $\mu$ is really a measure on $(X,\mathcal{S}).$
But I am not sure if my proof is ok or not.

My proof is here:

The only (finite) subset of $\emptyset$ is $\emptyset$ and $\sum_{x\in\emptyset} w(x)=0.$
So, $\mu(\emptyset)=0.$
Next we prove that $\mu(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)$ holds for every disjoint sequence $E_1,E_2,\dots$ of sets in $\mathcal{S}.$

1. We consider the case in which $w(x)=\infty$ for some $x\in\bigcup_{k=1}^\infty E_k.$
   In this case $\mu(\bigcup_{k=1}^\infty E_k)=\infty$ and $\sum_{k=1}^\infty \mu(E_k)=\infty.$
   So, $\mu(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)$ holds.
2. We consider the case in which $w(x)\in [0,\infty)$ for any $x\in\bigcup_{k=1}^\infty E_k$ and $\mu(E_{k_0})=\infty$ for some $k_0.$
   By the definition of $\mu$, if $E_1\subset E_2$ and $E_1,E_2\in\mathcal{S}$, then $\mu(E_1)\leq\mu(E_2).$
   So, $\infty=\mu(E_{k_0})\leq\mu(\bigcup_{k=1}^\infty E_k).$
   Therefore, $\mu(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)$ holds.
3. We consider the case in which $w(x)\in [0,\infty)$ for any $x\in\bigcup_{k=1}^\infty E_k$ and $\mu(E_{k})\in [0,\infty)$ for any $k.$
   Let $D$ be any finite subset of $\bigcup_{k=1}^\infty E_k.$
   Then, $D=(D\cap E_{k_1})\cup\dots\cup (D\cap E_{k_m})$ for some $\{k_1,\dots,k_m\}\subset\{1,2,\dots\}.$
   And $\sum_{x\in D} w(x) = \sum_{x\in D\cap E_{k_1}} w(x) +\dots+\sum_{x\in D\cap E_{k_m}} w(x)\leq\mu(E_{k_1})+\dots+\mu(E_{k_m})\leq\sum_{k=1}^\infty\mu(E_k).$
   Therefore, $\mu(\bigcup_{k=1}^\infty E_k)\leq\sum_{k=1}^\infty\mu (E_k).$
   Let $\epsilon$ be an arbitrary positive real number.
   Let $D_i$ be a finite subset of $E_i$ such that $\mu(E_i)-\frac{\epsilon}{2^i}<\sum_{x\in D_i} w(x)$ for each $i\in\{1,2,\dots\}.$
   Then, $\sum_{k=1}^m\left(\mu(E_k)-\frac{\epsilon}{2^k}\right)<\sum_{x\in D_1} w(x) + \dots + \sum_{x\in D_m} w(x)=\sum_{x\in D_1\cup\dots\cup D_m} w(x)\leq\mu(\bigcup_{k=1}^\infty E_k).$
   If $\sum_{k=1}^\infty\mu(E_k)=\infty$, then $\sum_{k=1}^\infty\left(\mu(E_k)-\frac{\epsilon}{2^k}\right)=\infty.$
   So, $\mu(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)=\infty$ holds.
   If $\sum_{k=1}^\infty\mu(E_k)\in [0,\infty)$, then $\sum_{k=1}^\infty\mu(E_k)-\epsilon\leq\mu(\bigcup_{k=1}^\infty E_k).$
   $\epsilon$ was an arbitrary positive real number.
   Therefore, $\sum_{k=1}^\infty\mu(E_k)\leq\mu(\bigcup_{k=1}^\infty E_k).$
   So, $\mu(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^\infty\mu(E_k)$ holds.