Let $X\neq \emptyset$ and $f:X \rightarrow [0, \infty]$ not identical infinity. Set $$ \sum_{x \in X} f(x)= \sup \left\{ \sum_{x \in F}f(x), F \subseteq X, F \mbox{ finite} \right\}.$$ $(i)$ Show that $\mu(E)= \sum_{x \in E} f(x)$ is a measure on $(X, P(X))$;
$(ii)$If $f(x) < \infty \> \forall x \in X$ and the set $\lbrace{ x \in X: f(x)>0 \rbrace}$ is at most countable show that $\mu$ is $\sigma$-finite.
My solution use this argument:
$(i)$ Let $E=\cup_n E_n $ with $E_n$ pairwise disjoint; If $x$ doesn't belong to $E$ then $x$ doesn't belong to any of the $E_n$ and this means $$ \mu(E)=0= \sum_{x \in \cup_n E_n} f(x)$$ Now let's consider that the set $E$ contains just one element $\overline{x}$. If $\overline{x} \in E$ then it exists a unique $\overline{n}$ such that $\overline{x} \in E_{\overline{n}}$. Computing the measure of $E$ leads to: $$ \mu(E)=f(\overline{x})=\mu(\cup_n E_n)=\mu(E_{\overline{n}})+ \sum_{n\neq \overline{n}} f(x)=\mu(E_{\overline{n}})+\sum_{n\neq \overline{n}} \mu(E_n)$$ that should hold because $\sum_{n\neq \overline{n}} \mu(E_n)=0$
$(ii)$ Being that $A=\lbrace{ x \in X : f(x)>0 \rbrace}$ is at most countable we can write $$ A=\cup_n \lbrace{x_n\rbrace}$$ and $\mu(\lbrace{x_n\rbrace})=f(x_n) <\infty$ from the hypothesis.
It's my solution acceptable? I didn't use the definition given with the $\sup$.
Here is $(i)$: It is clear that $\mu(E)\ge 0$. $\mu(\varnothing)=\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \varnothing, F \text{finite} \rbrace}$, then $F$ is necessarily empty thus $\sum_{x \in F}f(x)=0$.
Now let $(E_n)_{n\in \Bbb N}$ be disjoint sets we want to show $\mu(\bigcup_{n\in \Bbb N}E_n)=\sum_{n\in \Bbb N}\mu(E_n)$. We can assume $\mu(E_n)\neq \infty$ since if otherwise we have equality. If $F \subset \bigcup_nE_n$ then by the disjointness of the $E_n$ $F$ can be written as $F=F_1\cup F_2 \ldots F_n$ where $F_i$ is a finite subset of $E_i$ and are disjoint. Now $$\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \text{finite} \rbrace}=\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \text{finite} \rbrace}\ge \sup \lbrace{ \sum_{i=1}^n\sum_{x \in F_i}f(x), F \subseteq E_1 \cup \dots \cup E_n, F \text{finite} \rbrace}=\sum_{i=1}^n\sup \lbrace{ \sum_{x \in F_i}f(x), F_i \subseteq E_i, F_i \text{finite} \rbrace}$$ Now let $n \to \infty$ to get $\mu(\bigcup_{n\in \Bbb N}E_n)\ge\sum_{n\in \Bbb N}\mu(E_n)$.
For the reverse inequality, note that for any $\sum_{i=1}^n\sum_{x \in F_i}f(x) \in \left\{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \mbox{ finite} \right\}$ we have $\sum_{i=1}^n\sum_{x \in F_i}f(x)< \sum_{n\in \Bbb N}\sup \lbrace{ \sum_{x \in F_i}f(x), F_i \subseteq E_i, F_i \text{finite} \rbrace} $ thus $\mu(\bigcup_{n\in \Bbb N}E_n)=\sum_{n\in \Bbb N}\mu(E_n)$.
For $(ii)$ You need to write $X$ as a countable union of set with finite measure, I can't understand what you wrote. Write $X=A\cup A^c$, you have already found the measure of $A$ now find that of $A^c$.