Why are the sets of rational and irrational numbers Borel sets (over the reals)?

I'm actually looking for more intuition behind what's going on than a simple answer. I understand that the Borel sets are the smallest sigma algebra that contain all of the open sets. I can see how open intervals and their unions are in this set, and how it makes sense to measure them. It's hard to see why you'd want to measure rational numbers (the probability that a normally distributed variable takes on a rational number?), and more importantly, why the sigma algebra should have to contain rational numbers (why can't we get away with a sigma algebra that doesn't contain the set of rational numbers?)

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if you care about countable unions and points, then you have to care about the rational numbers. What's so unreasonable about that? – Qiaochu Yuan Sep 21 '10 at 5:18
@Qiaochu, I see it now. – Neil G Sep 21 '10 at 5:20

If you want your sigma algebra to contain all of the one-point sets, so in particular if it's the Borel sigma-algebra generated by a $T_1$ topology, then taking countable unions yields all of the countable sets, including the set of rational numbers in your case. There are sigma-algebras of subsets of the real numbers that don't contain the rational numbers, but if you're OK with open intervals, then you have to be OK with points.
It isn't that we "Want them to be," but that they must be simply from definition: Single point sets are necessarily closed in the topology of $\mathbb{R}$, hence are in $\mathcal{B}$ - the Borel $\sigma$-algebra. Since $\mathbb{Q}$ is a countable union of single point sets, it must be in $\mathcal{B}$. Hence so is $\mathbb{Q}^c$.
If nothing else, this at least allows some nice examples. For instance, define $f(x)$ to be identically $1$ on $\mathbb{Q}$ and identically $0$ on $\mathbb{Q}^c$. $f$ is not Riemann integrable, however, $f$ is Lebesgue measurable, and integrates to zero in the Lebesgue sense (which is something one would expect).