How do you prove that this semiring's set function is not σ-additive? My question concerns Exercise 1.2.1 from "Probability Theory, a Comprehensive Course" (click here) by Achim Klenke.
I have completed the question except for showing that µ is not σ-additive.
I have tried to create a counterexample by first creating an countably infinite series. Each member of the series is another countably infinite series of members of . All these members of  across all these series are pairwise disjoint but still have their collective union in .
Then I merged all the members into a single series but where they appear in a different order. Hoping that the sum of µ applied to each member would now converge to another value, I was disappointed (it still converged to the value of µ applied to the collective union). 
 A: You can easily extend $\mu$ to the field $F$ generated by $\mathcal A$, which is just the collection of finite unions of elements of $\mathcal A$ plus $\mathbb Q$ itself since $\mathcal A$ is a semiring. If $\mu$ were countable additive then it would also extend to a measure on the $\sigma$-field generated by $F$, which includes singletons and hence is simply the power set of $\mathbb Q$. We would then have $\mu(\{x\})=0$ and hence $\mu(B)=0$ for all sets $B$, leading to a contradiction.
(For future posts, it's best to write out your question in full rather than linking to it.)
EDIT: Since you have not yet seen the Caratheodory Extension Theorem, a more direct approach is needed. Let $B=(0,1]\cap\mathbb Q$. We know $B$ is countable so let $\{x_n\}_{n\ge1}$ be an enumeration with $x_1<\frac16$. For each $n$ define $B_n=(x_n-3^{-n},x_n]\cap\mathbb Q$. Then $\bigcup_{n=1}^\infty B_n=(x_1-\frac13,1]\cap\mathbb Q\in\mathcal A$, and so by monotonicity
$$\sum_{n=1}^\infty\mu(B_n)=\frac12<1=\mu(B)\le\mu\left(\bigcup_{n=1}^\infty B_n\right).$$
In particular, $\mu$ is not countably subadditive and so it cannot be countably additive.
