
#1
Dec412, 03:10 AM

P: 2,453

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
Dec412, 03:51 AM

P: 571

So the measure of the rationals must be zero. I seem to recall that the measure of any countable set is zero. 



#3
Dec412, 04:03 AM

Sci Advisor
P: 778

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 nonzero 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
Dec412, 06:26 AM

Mentor
P: 16,581

question about the set of irrationals. 



#5
Dec412, 07:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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




#7
Dec512, 05:18 AM

P: 571





#8
Dec512, 06:13 AM

P: 2,453

do they have sets with different infinite measure?




#9
Dec512, 08:05 AM

Mentor
P: 16,581

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



#10
Dec512, 06:44 PM

P: 2,453

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



#11
Dec512, 06:50 PM

Mentor
P: 16,581





#12
Dec512, 06:57 PM

P: 2,453

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



#13
Dec512, 07:15 PM

Mentor
P: 16,581

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



#14
Dec512, 11:14 PM

P: 571





#15
Dec612, 12:23 AM

Sci Advisor
P: 778

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})##. 


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