# Question about the set of irrationals.

1. Dec 4, 2012

### cragar

Is it possible to have a set that contain all the irrationals that has measure zero.
I dont know that much about measure theory. Or I guess we could just ask what is the measure of the irrationals. I know it is possible to have uncountable sets that have measure zero.

2. Dec 4, 2012

### ImaLooser

The measure of the set of irrational numbers is one.

So the measure of the rationals must be zero.

I seem to recall that the measure of any countable set is zero.

3. Dec 4, 2012

### pwsnafu

You didn't specify which measure you are talking about, so if we use the zero measure, then the answer is yes.

If you are talking about using the Lebesgue measure, the answer is no. Let A be all rationals between 0 and 1, and B be all irrationals between 0, and 1. Then
$m(A) + m(B) = m([0,1]) = 1$
But $m(A) = 0$, so B has non-zero measure.

Note that if C is any set that contains every irrational between 0 and 1 then $m(C) \geq m(B)$ by monotonicity.

4. Dec 4, 2012

### micromass

What? That's not true, is it?

5. Dec 4, 2012

### HallsofIvy

I suspect that Imalooser meant the set of all irrational numbers between 0 and 1.

6. Dec 4, 2012

### micromass

That would make sense.

7. Dec 5, 2012

### ImaLooser

Sorry, I'm so used to probability measures that I simply assumed that. Probability measures are normed so that the maximum measure is always 1 and the minimum is zero. Duh.

8. Dec 5, 2012

### cragar

do they have sets with different infinite measure?

9. Dec 5, 2012

### micromass

What do you mean with "they"??

Do you mean whether probability spaces have sets of infinite measure? The answer is no: the largest possible measure is 1.

10. Dec 5, 2012

### cragar

I guess I mean in ZFC are their sets that have infinite measure.
But I guess you said they dont

11. Dec 5, 2012

### micromass

First you need to specify what you mean with "measure". Which measure are you talking about? If you're talking about Lebesgue measure on $\mathbb{R}$ (which is basically the rigorous version of length), then there are sets of infinite measure. The set $\mathbb{R}$ itself has infinite measure.

I didn't say that. I said that in a probability space (that is when you work with a probability measure), then all sets have finite measure by definition. But when not working with a probability measure, then there might be sets of infinite measure.

12. Dec 5, 2012

### cragar

ok thanks for your response. Are their sets that have larger Lebesgue measure
than the set of reals.

13. Dec 5, 2012

### micromass

No, since they already have infinite measure.

In measure theory, there is only one kind of infinity. There is not an entire class of infinities like the infinites of Cantor.

14. Dec 5, 2012

### ImaLooser

Are you thinking of Cantor cardinalities? If so, the answer is yes. You can take the power set of any set, and the power set will have higher cardinality.

15. Dec 6, 2012

### pwsnafu

cragar explicitly asked about the Lebesgue measure, not cardinality.

To further what micromass said, the Lebesgue measure is a specific measure defined on certain subsets of ℝ. Therefore there can't be a Lebesgue measurable set that has measure larger than $m(\mathbb{R})$.