# How is this $\mu_0$ not a premeasure?

I have a homework problem that I’ve been stuck on for some time. Define $\mathcal{A}$ to be an algebra on $\mathbb{Q}$ generated by the set $S = \{ (a, b] \cap \mathbb{Q} : a, b \in \mathbb{Q} \}$, and define the function $\mu_0 : \mathcal{A} \to [0, \infty]$ by
$$\mu_0 \left( \bigcup_{i=1}^{\infty} (a_i, b_i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (b_i – a_i)$$
for a disjoint collection of sets $\{ (a_i, b_i] \cap \mathbb{Q} \} \subseteq S$. The question asks to show that $\mu_0$ is not a premeasure on $\mathcal{A}$.

First I see that
$$\mu_0(\varnothing) = \mu_0 \left( \bigcup_{i=1}^{\infty} (i, i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (i – i) = 0.$$
Also,
$$\mu_0((a, b] \cap \mathbb{Q}) = b – a \quad (1)$$
by taking the union with infinitely many empty sets. So now for a disjoint collection of sets $\{ (a_i, b_i] \cap \mathbb{Q} \} \subseteq S$ whose union is in $\mathcal{A}$, why doesn’t $(1)$ imply we have $\sigma$-additivity:
$$\mu_0 \left( \bigcup_{i=1}^{\infty} (a_i, b_i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (b_i – a_i) = \sum_{i=1}^{\infty} \mu_0((a_i, b_i] \cap \mathbb{Q})?$$

#### Solutions Collecting From Web of "How is this $\mu_0$ not a premeasure?"

Let $(\varepsilon_n)_{n\in\mathbb{N}}$ be a sequence of rational numbers with $\sum_{n=1}^\infty\epsilon_n<1$, and let $(q_n)_{n\in\mathbb{N}}$ be an enumeration of $(0,1]\cap\mathbb{Q}$.

Define intervals $I_n$ for $n=1,2,\ldots$ as follows: If $q_n\in\bigcup_{k=1}^{n-1}I_n$, lelt $I_n=\emptyset$. Otherwise, let $I_n=(p_n,q_n]$ where $p_n$ is the largest of the numbers $0$, $q_n-\varepsilon_n$, $q_1$, … $q_{n-1}$ less than $q_n$.

Now $(0,1]\cap\mathbb{Q}$ is the disjoint union of the rational intervals $I_n\cap\mathbb{Q}$, and the sum of the lengths of these intervals is less than $1$.