How do i prove that "countably monotone + Finitely Additive" implies "Premeasure" on a semiring? 
Let $S$ be a semi-ring of subsets of $X$
Let $\mu:S \rightarrow [0,\infty]$ be a set function on $X$
If $\mu$ is countably monotone and finitely additive, then $\mu$ is a premeasure.

I know that this must be true, since $\mu$ can be extended to $\mu^*$ on a ring generated by $S$ which is countably monotone and finitely additive.
However, i don't know how to prove this directly.
Help..
 A: I'm the OP and since i have a low bounty, i cannot comment on Karene's question, so i'm writing this as an answer. The set $S$ Karene constructed is not a semi-ring on $X$. Since $\mathbb{N} \setminus \{a\}$ cannot be a finite union of elements of $S$.
I have proved the following:

Let $X$ be a set.
Let $S$ be a semi-ring on $X$.
Let $\mu:S\rightarrow [0,\infty]$ be a function such that $\mu(\emptyset)=0$.
Then, $\mu$ is countably additive if and only if it is countably monotone and finitely additive.

The proof for if part does not require any choice, but the proof for only if part requires choice.
The key idea for the proof is that $\{A\subset X: A \text{ is a finite disjoint union of elements in } S\}$ is indeed a ring on $X$.
Finite additivity makes it possible to construct a function $\mu^*$ on this ring such that $\mu^* (A) = \sum_{i=1}^n \mu(A_i)$ whenever $A$ is a disjoint union of a finite sequence $A_i$ in $S$.
A: According to the definition I was given:
Definitions:
Semi-ring: From Wikipedia.
Countably monotone: From this question.
Pre-measure: From the comment of user Number9 below.
Example: Then consider the set $\mathbb{N}$, the collection $S$ of sets $\emptyset$ together with all the singleton subsets of $\mathbb{N}$, and $\mathbb{N}$. 
Define $\mu(\emptyset)=0$, $\mu(\{a\})=1/2^a$, and $\mu(\mathbb{N})=1$.
We have that $S$ is a semi-ring: It contains the emptyset, intersections of any two sets in $S$ is the emptyset, and differences of any two sets are in there.
We  have that $\mu(\mathbb{N})=1<\sum_{n=0}^{\infty}\mu(\{n\})=1+1/2+1/2^2+...=2$
Finite additivity is trivially satisfied because the only finite disjoint unions of elements of $S$ that are also in $S$ are finite unions like $\emptyset\cup\{a\}$ or $\emptyset\cup\mathbb{N}$.
The issue is being that since finite additivity is being satisfied trivially, it is not forcing $\mu(\mathbb{N})$ up from the monotonicity that follows from finite additivity.
Can we consider then $\mu$ defined on a ring of sets?
A: Here's the direct proof the OP sought. 
Let $A_1,A_2,\ldots$ be disjoint sets in the $S$. Let $A = \bigcup_{i=1}^{\infty} A_i \in S$. By countable monotonicity (or even countable subadditivity), we have 
$$
\mu(A) \leq \sum_{i=1}^{\infty} \mu(A_i). 
$$
Now we prove the other direction.
By induction, we can write 
$$
A \setminus \bigcup_{i=1}^{n} A_i = \left( A \setminus \bigcup_{i=1}^{n-1} A_i \right) \setminus A_n = \bigcup_{i=1}^{m_n} D_{i,n}
$$
where $D_{1,n},\ldots,D_{m_n,n}$ are disjoint sets in $S$. Moreover, each $D_{1,n},\ldots,D_{m_n,n}$ is disjoint from each $A_1,\ldots,A_n$. Thus the sets $D_{1,n},\ldots,D_{m_n,n},A_1,\ldots,A_n$ are disjoint sets in $S$. Also
$$
A = \left( \bigcup_{i=1}^{m_n} D_{i,n} \right) \cup \left( \bigcup_{i=1}^{n}  A_i \right).  
$$
Therefore finite additivity gives 
$$
\mu(A) = \sum_{i=1}^{m_n} \mu(D_{i,n}) + \sum_{i=1}^{n} \mu(A_i) \geq \sum_{i=1}^{n} \mu(A_i).  
$$
Letting $n \to \infty$ gives the desired result.  
