Measure on the set of rationals Consider the rational intervals defined as $$[a,b)_Q= \{r : a \leq r < b; r \in \mathbb{Q} \},a,b\in \mathbb{Q}, a<b. $$ Let $A$ be the class of all sets of rationals that can be produced as finite unions of rational intervals. Then $A$ is a ring not containing any sets of the form $\{a\}$. The $\sigma$-additive set function $\mu: A \to R$ is defined on the rational intervals the usual way:  $\mu([a,b)_Q)=b-a$. By this $\mu$ is defined on $A$.
For every rational number $a$ the set $\{a\}$ clearly belongs to the $\sigma$-ring generated by $A$. Indeed, $\{a\} = \bigcap_{n\in\mathbb{N}} [a,a+1/n)_Q$. As a result, one has on the one hand $\mu(\{r\}) = 0$ for all rationals and, on the other hand $\mu((a,b]_Q) = \sum_{r\in[a,b)_Q}\mu(\{r\}) = \sum_{r\in[a,b)_Q} 0 = 0$ and not $b-a$ as it ought to be by the definition.
Where is the mistake?

$\mu$ is not sigma additive on $A$: Let $\{r_n : n \geq 1\}$ list all rationals in $[0, 1)$. Let $I_n$ be an interval such that $r_n \in I_n$, $I_n \subseteq [0, 1)$ and $\mu(I_n) < 1/3^n$. Let $W_n = I_n \backslash (\bigcup_{m < n} I_m)$. Then $\{W_n : n \geq 1\} \subseteq A$ is a partition of $[0, 1)$ and $\sum_{n \geq 1} \mu(W_n) \leq 1/2$.
