The counting measure over rational intervals

Question

This is an example from Ash's Probability and Measure Theory:

Let $\Omega$ be the rationals, $\mathcal{F}_0$ the field of finite disjoint unions of right-semiclosed intervals $(a,b]=\{\omega\in\Omega:a<\omega\leq b\}, a,b$ rational [counting $(a,\infty)$ and $\Omega$ itself as right-semiclosed]. Let $\mathcal{F}=\sigma(\mathcal{F}_0)$. Then, if $\mu(A)$ is the number of points in $A$ ($\mu$ is counting measure), then $\mu$ is $\sigma$-finite on $\mathcal(F)$ but not on $\mathcal{F}_0$

Actually, he gives the answer: since $\Omega$ is countable union of singletons, $\mu$ is $\sigma$-finite on $\mathcal{F}$. But every nonempty set in $\mathcal{F}_0$ has infinite measure, so $\mu$ is not $\sigma$-finite on $\mathcal{F}_0$.

It seems pretty clear after the 'But': the counting measure in each interval is infinite. But, I didn't understand the first sentence. How can this be possible? If I take any interval in $\mathcal{F}_0\subset\mathcal{F}=\sigma(\mathcal{F}_0)$, it is also in the $\mathcal{F}$. So, if it is infinite on the first one, shouldn't it be in the second one too? I'm confused about this concepts. Could anyone help me?

Answer 1

You are arguing that there are lots of sets on infinite measure in $\mathcal F,$ which is certainly true, but the author's point is that there are sets of finite measure (in particular, the singleton sets) in $\mathcal F.$ Thus, the rationals can be written as the countable union of sets of finite measure in $\mathcal F,$ but as you have noted, this is impossible with $\mathcal {F_0}$

Answer 2

I'm guessing the definition of sigma-finite on $\mathcal F$ you are given is something like: $\Omega$ can be covered by a countable union of finite-measure elements of $\mathcal F.$

The key here is that $\mathcal F$ has more sets than $\mathcal F_0,$ and thus more finite-measure sets with which you can put together a countable covering. In fact, $\mathcal F$ contains all singletons (as you can straightforwardly write a singleton as the countable intersection of half-open intervals). And you can write $\Omega$ as the countable union of singletons, so you're done.

On the other hand $\mathcal F_0$ contains no singletons, nor any finite non-empty set, as your solution says.