Conditions where $\mu$ is semifinite and where $\mu$ is $\sigma$-finite This comes out of the book Real Analysis by Folland:

$\mu$ is semifinite if and only if $f(x) < \infty$ for every $x\in X$, and $\mu$ is $\sigma$-finite if and only if $\mu$ is semifinite and $\{x:f(x) > 0\}$ is countable.

Attempted proof for first statement - Let $X$ be any nonempty set, $M = P(X)$, and $f$ be any function from $X$ to $[0,\infty]$. Then $f$ determines a measure $\mu$ on $M$ by the formula $$\mu(E) = \sum_{x\in E}f(x)$$ (note this is how Folland sets up the above problem that he states the reader can verify) Now suppose $\mu$ is semifinite then $E\in M$ with $\mu(E) = \infty$. So $$\mu(E) = \sum_{x\in E}f(x) = \infty$$ then $f(x) < \infty$ for every $x\in X$ (Not sure if this is true but seems to make sense to me). 
OTOH, suppose $f(x) < \infty$ for every $x\in X$... Not really sure where to go from here... any suggestions is greatly appreciated.
Attempted proof for second statement - Let $\mu$ be $\sigma$-finite and let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$, and $\mu(E) = \infty$. Set $$F = \bigcup_{1}^{\infty} E_j \ \text{where} \ E_j\in M$$ Then there exists an $F\in M$ such that $F\subset E$ and $0 < \mu(F)\leq \mu(E) = \infty$ then $\mu$ is semifinite. Now define $$\sum_{x\in E'} f(x) := \sup\{\sum_{x\in E}f(x): E' \ \text{finite} \ , E' \subset E\}$$ Then observe that $\sum_{x\in E} f(x) < \infty$ which implies that $\{x\in E: f(x) > 0\}$ is countable.
OTOH, suppose $\mu$ is semifinite and $\{x: f(x) > 0\}$ is countable then.... not really sure where to go from here... any suggestions is greatly appreciated.
I would like to stick to how I set this proof up although if one would like to show me how they would prove it I would appreciate it.
 A: Let $X$ be any nonempty set, $M = P(X)$, and 
$f$ be any function from $X$ to $[0,+\infty]$. Then $f$ determines a measure $\mu$ on the $\sigma$-algebra $M$ by the formula $$\mu(E) = \sum_{x\in E}f(x)$$

Result 1: $\mu$ is semifinite if and only if $f(x) < +\infty$ for every $x\in X$.

Proof: 
($\Rightarrow$) Let us prove the counter-positive. Suppose there is $x_0\in X$ such $f(x_0)=+\infty$.  Then $\mu(\{x_0\})=+\infty$. So $\{x_0\}$ is a set of measure infinite and there is no subset $B\subseteq \{x_0\}$ such that $0<\mu(B)<+\infty$. So $\mu$ is not semifinite.  
($\Leftarrow$) Supose $f(x) < +\infty$ for every $x\in X$. Let $E\in M$ be any set such that $\mu(E)=+\infty$. Then 
$$ \sum_{x\in E}f(x)= \mu(E) = +\infty$$
So there is $x_0\in E$ such that $f(x_0)>0$.  So we have that $\{x_0\}\subseteq E$ and $$0<\mu(\{x_0\})=f(x_0)<+\infty$$ So, $\mu$ is semifinite.  

Result 2: $\mu$ is $\sigma$-finite if and only if $\mu$ is semifinite and $\{x:f(x) > 0\}$ is countable.

Proof: 
($\Rightarrow$) Let us prove the counter-positive. If $\mu$ is not semifinite then $\mu$ can not be $\sigma$-finite. This is a general result valide for any measure. In fact, if there is $E$ such that the measure of $E$ is infinite and any subset of $E$ has measure zero or infinite, then $E$ is not the countable union of subsets of finite measure. 
Suppose now $\mu$ is semifinite and $\{x:f(x) > 0\}$ is uncountable. 
Since $$\{x:f(x) > 0\}=\bigcup_{n\in \mathbb{N}}\left\{x:f(x) > \frac{1}{n+1}\right\}$$ there is $n$ such that $\left\{x:f(x) > \frac{1}{n+1}\right\}$ is uncountable. Let $E= \left\{x:f(x) > \frac{1}{n+1}\right\}$. It is clear that $\mu(E)=+\infty$ and if $B$ is a subset of $E$ such that $\mu(B)<+\infty$, then $B$ has a finite cardinality. So $E$ (which is uncountable) is not the countable union of subsets of finite measure (such subsets have finite cardinality). So $\mu$ is not  $\sigma$-finite. 
($\Leftarrow$) Suppose $\mu$ is semifinite and $\{x:f(x) > 0\}$ is countable. Using Result 1, we have that $f(x) < +\infty$ for every $x\in X$. Let $S=\{x:f(x) > 0\}$.  
Given any set $E \in M$, we have $\mu(E\setminus S)=0$, $E \cap S$ is countable and, for all $x\in E \cap S$, $\mu(\{x\})<+\infty$. We also have $$E=(E\setminus S) \cup \bigcup_{x\in E\cap S} \{x\}$$ So $E$ is a countable union of sets of finite measure. So $\mu$ is $\sigma$-finite. 
