# Is the set of rationals a measurable set?

• ForMyThunder
In summary, Elias Stein showed that for every \epsilon>0, there exists an open \mathscr O with the property that m_*(\mathscr{O}-E) < \epsilon. However, 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. However, this set of rationals in the unit interval is the countable union of point sets so it MUST BE measurable.
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?

Hi ForMyThunder!

ForMyThunder said:
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!

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

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.

disregardthat said:
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!

micromass said:
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.

## 1. What is a measurable set?

A measurable set is a set that can be assigned a measure, which is a numerical representation of its size or extent. This measure is typically a real number that follows certain properties, such as being non-negative and additive.

## 2. How is the set of rational numbers defined?

The set of rational numbers, also known as the set of rationals, is defined as the set of all numbers that can be expressed as a ratio of two integers (a fraction). In other words, it is the set of all numbers that can be written in the form a/b, where a and b are integers and b is not equal to 0.

## 3. Is the set of rationals a finite or infinite set?

The set of rationals is an infinite set. This is because there are an infinite number of possible combinations of integers that can be used to express a rational number. Additionally, between any two rational numbers, there are infinitely many other rational numbers.

## 4. Why is it important to determine if the set of rationals is measurable?

Determining if a set is measurable is important in many areas of mathematics, such as measure theory, real analysis, and probability theory. It allows us to understand the properties and behavior of sets and their corresponding measures, which can be applied to various mathematical and scientific problems.

## 5. Can the set of rationals be assigned a measure?

Yes, the set of rationals can be assigned a measure. In fact, it is a countably infinite set, meaning that its elements can be placed in a one-to-one correspondence with the natural numbers. This allows us to assign a measure to the set of rationals by using a counting measure, which assigns a value of 1 to each element.

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