Prove that if $\mu$ is $\sigma$-subadditive then $\mu$ is a measure. 
Let $S$ be a semiring and let $\mu: S\to [0, \infty]$ be a finite additive measure. Prove that if $\mu$ is $\sigma$-subadditive then $\mu$ is a measure.

I know the following:
Since $\mu$ is finite additive, we know that $\mu(\emptyset) =0$ and $\displaystyle \mu\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n \mu(A_i)$ for every disjoint $A_1, A_2 , \dots, A_n \in S$ with $\displaystyle \bigcup_{i=1}^n A_i \in S$.
This is where I am stuck. Can anybody help by pointing me in the right direction as to how I can go about continuing this proof? I know I have not provided much to go by, I am completely new to measure theory.
 A: Updated answer: As pointed out in detail in the comments by Ramiro – as $S$ is a semiring – $\bigcup\limits_{k=1}^{\infty}A_k \in S$ does not ensure that $\bigcup\limits_{k=1}^{n}A_k \in S$ for any $n \in \mathbb{N}$ and the finite additivity of $\mu$ cannot be directly used. One rather has to extend the $S$ to the ring $R(S)$ generated by $S$ and extend $\mu$ on $S$ to the unique extension $\mu '$ on $R(S)$.
Note that finite additivity also holds for $\mu'$ and $\mu'$ coincides with $\mu$ on $S$.
Finally this gives us:
\begin{align*}  \mu\left (\bigcup_{i=1}^\infty A_i \right) &\leqslant \sum_{i=1}^\infty \mu(A_i)  \tag{ by $\sigma$-subadditivity}  \\  &= \lim_{n \to \infty} \sum_{i=1}^n  \mu(A_i) \\&=\lim_{n \to \infty} \sum_{i=1}^n  \mu'(A_i) \tag { $\mu'$ and $\mu$ coincides on $S$} \\ &= \lim_{n \to \infty} \mu'\left(\bigcup_{i=1}^n A_i\right)  \tag { note: $\bigcup_{i=1}^n A_i$ is in $R(S)$} \\ &\leqslant \mu'\left (\bigcup_{i=1}^\infty A_i \right) \tag{ by monotonicity of $\mu'$} \\ &=\mu\left (\bigcup_{i=1}^\infty A_i \right) \tag { $\mu'$ and $\mu$ coincides on $S$} \end{align*}
Original answer: If we can show that the content $\mu$ is $\sigma$-additive, then $\mu$ is a measure on the semiring.
So let $\{A_k\}_{k \geq 1}$ a sequence of pairwise disjoint sets in $S$, so that $A := \bigcup\limits_{k=1}^{\infty}A_k \in S$. Then we have
$$\mu(A) \underbrace{\leq}_{\sigma-\text{subadditive}} \sum_{k=1}^{\infty}{\mu(A_k)} = \lim_{n \to \infty}{\sum_{k=1}^{n}{\mu(A_k)}} \underbrace{=}_{\text{finite additivity}} \lim_{n \to \infty}{\underbrace{\mu(\bigcup\limits_{k=1}^{n}A_k)}_{\leq \mu(A) \text{ because of monotonicity of }\mu}} \leq \mu(A)$$
As we have the same expression on the left and right hand side, all inequalities must already be equalities, which provides us with
$$\mu(A) = \sum_{k=1}^{\infty}{\mu(A_k)}.$$
