Consider a measure $\mu$: $ B_{\mathbb{R}}$ $\rightarrow$ $[0,\infty]$ satisfying $\mu((a,b))$=$b-a$ for any $b>a$. Show that $\mathbb{Q} \in B_\mathbb{R}$.
My proof: Enumerate the rationals $r_1$,$r_2$, $r_3$...
$\mathbb{Q}$ is the countable union of such singletons: $\mathbb{Q}=\bigcup\limits_{i=1}^{\infty} r_{i}$. Such singletons are closed sets and are hence Borel sets: $(r_1 \in B_\mathbb{R},r_2 \in B_\mathbb{R}...etc)$. (Another way of thinking of it is that $r_1 \in B_\mathbb{R}$, for example, because we can write $r_1$ as:$\bigcap\limits_{i=1}^{\infty} (r_1-1/i,r]$). Thus since $B_\mathbb{R}$ is a sigma algebra, it is closed under countable union, so $\mathbb{Q}\in B_\mathbb{R}$. Is my proof correct or is there anything I should fix? Thanks.
Yes, your argument is fine. It does not depend on $\mu$ though, it only depends on the definition of $\mathcal{B}_{\mathbb{R}}$. In a more succinct way, one can write
$$ \mathbb{Q} = \bigcup_{q \in \mathbb{Q}}\bigcap_{n \in \mathbb{N}}(q - \frac{1}{n}, q+ \frac{1}{n}) $$
from which we deduce that the rationals are a borel set: they are a countable union of countable intersections of borel sets.