Is the set of rationals a measurable set?

[ForMyThunder]

In Elias Stein's book Real Analysis, a measurable set $$E$$ is a set such that for every $$\epsilon>0$$, there exists an open $$\mathscr O$$ with the property that $$m_*(\mathscr{O}-E) < \epsilon$$. But for every open set that covers the rationals in, say, $$[0,1]$$ must cover the entire interval so that the set of rationals can't satisfy the conditions for a measurable set. But then, this set of rationals in the unit interval is the countable union of point sets so it MUST BE measurable. Where was the hole in my argument that the set isn't measurable?

 

[micromass]

Hi ForMyThunder!

But for every open set that covers the rationals in, say, $$[0,1]$$ must cover the entire interval

This is where your argument goes wrong. There are many open sets around the rationals that does not cover the entire set. For example, the set $\mathbb{R}\setminus \{\pi\}$ cover the rationals but not the entire real line.

In fact, since the rationals are countable, we can enumerate them as

$$\mathbb{Q}=\{q_1,q_2,q_3,...\}$$

then

$$G_\varepsilon=\bigcup_{n\geq 1}{B(q_n,\varepsilon/2^{n+1})}$$

is an open set containing the rationals and the open set has measure epsilon. Fine, I have no inuition about that open set and I can't imagine it, but that doesn't mean it doesn't exist!

 

[disregardthat]

$$G_\varepsilon=\bigcup_{n\geq 1}{B(q_n,\varepsilon/2^{n+1})}$$

is an open set containing the rationals and the open set has measure epsilon.

Hm, I can't see how that is correct. First of all, $\sum_{n \geq 1} \frac{1}{2^{n+1}} = 1/2$, and second of all, it is not a disjoint union. Also, I think the measure would depend on the how Q is ordered.

 

[micromass]

Hm, I can't see how that is correct. First of all, $\sum_{n \geq 1} \frac{1}{2^{n+1}} = 1/2$, and second of all, it is not a disjoint union. Also, I think the measure would depend on the how Q is ordered.

Sorry, that should be "measure at most epsilon", which is enough for our purposes: Thus

$$\lambda\left( \bigcup_{n\geq 1}{B(q_n,\epsilon/2^{n+1})} \right)\leq \sum_{n\geq 1}{\lambda(B(q_n,\epsilon/2^{n+1}))}=\sum_{n\geq 1}{\frac{\epsilon}{2^n}}=\epsilon$$

And of course the measure of the set, and the set itself will depend on the order of Q! For every different order, we get a different set!

 

[disregardthat]

And of course the measure of the set, and the set itself will depend on the order of Q! For every different order, we get a different set!

Not necessarily though, q_1,q_2,q_3,q_4,q_5,... and q_1,q_3,q_2,q_4,q_5,... would generate the same set if the balls around q_2 and q_3 are contained in the ball around q_1.

And furthermore, different sets can easily have the same measure! (not saying that you implied otherwise..) As an example, relabeling by reflection about 1/2 would yield a (potentially) different set with equal measure.

But this example is interesting regardless, it is quite difficult to conceive of a set containing small intervals around each point in a dense set while not containing its closure.